PSEB 10th Class Science Important Questions Chapter 6 Life Processes

Punjab State Board PSEB 10th Class Science Important Questions Chapter 6 Life Processes Important Questions and Answers.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes

Long Answer Type Questions

Question 1.
What are the kinds of heterotrophic organisms on the basis of nutrition?
Answer:
Heterotrophic nutrition is a mode of nutrition in which organisms obtain readymade organic food from outside source. The organisms that depend upon outside sources for obtaining organic nutrients are called heterotrophs. It is a characteristic feature of all animals and non-green plants, that are unable to utilize carbon and synthesi s organic compounds necessary for life, but depends upon organic sources of carbon. They are thus dependent upon autotrophic organisms (Plants) and are called as heterotrophs.

It is of the following types :

  • Saprophytic nutrition: In this type of nutrition, an organism lives upon dead organic sources such as dead plants and dead animals. These usually secrete dissolving and digesting enzymes and absorb the liquefied molecules so formed e.g. yeast, bread moulds and dung moulds etc.
  • Parasitic nutrition: In this type of nutrition, an organism lives totally at the expense of others and derives its food material and shelter from the other .These organisms which derive food material are called parasites and the organism from which food is derived is called as host. This type of nutrition is termed as parasitic or holozoic nutrition: It is also known as parasite-host relationship e.g. Cuscuta, Ascaris etc.
  • Holozoic nutrition. It is a mode of heterotrophic nutrition which involves intake of solid pieces of food. Since solid food is taken in, Holozoic nutrition is also called ingestive nutrition. Holozoic nutrition (GK. “Holo”-Whole; “Zoon”-Animal) is found in animals and protists. The food may consist of another animal, plant or its parts.

Depending upon the source of food, holozoic organisms are of three types :

  1. Herbivores: These organisms obtain their food from plants e.g. cow, rat, deer and goat etc.
  2. Carnivores: These organisms take the flesh of other organisms as their food e.g. tiger, cheetah, snake, eagle etc.
  3. Omnivores: These can feed on plants and flesh of other organisms e.g. human, cockroach, crow etc.

Question 2.
What is photosynthesis? Describe the significance of photo-synthesis.
Answer:
Photosynthesis (Photos-Light, Synthesis-putting together) may be defined as an anabolic process in which green plants manufacture complex organic food substances (carbohydrate) from simple inorganic compounds like carbon dioxide and wrater in presence of sunlight with the aid of chlorophyll and evolve out oxygen as a byproduct of the process. Thus photosynthesis is a process in which radiant energy is converted into chemical energy
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 1
In other words photosynthesis is a series of oxidation- reduction reaction in which CO2 is reduced and H2O is oxidized to produce carbohydrates and oxygen.

  • Chloroplasts are the actual sites for photosynthesis.
  • All green parts of a plant are involved in photosynthesis.
  • Leaves are the most important organs of photosynthesis.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes 2
Section of Leaf showing site of photosynthesis

Significance of photosynthesis:

  • Photosynthesis is a source of all our food and fuel. It is the only biological process that acts as the driving vital force for the whole animal kingdom and for the non-photosynthetic organisms.
  • It drives all other processes of biological and abio- logical world. It is responsible for the growth and sustenance of our biosphere.
  • It provides organic substances, which are used in the production of fats, proteins, nucleoproteins, pigments, enzymes, vitamins, cellulose, organic acids, etc. Some of them become structural parts of the organisms.
  • It makes use of simple raw materials such as CO2, H2O and inexhaustible light energy for the synthesis of energetic organic compounds.
  • It is significant because it provides energy in terms of fossil fuels like coal and petrol obtained from plants, which lived millions and millions of years ago.
  • Plants, from great trees to microscopic algae, are engaged in converting light energy into chemical energy, while man with all his knowledge in chemistry and physics cannot imitate them.

Question 3.
Which organelle is responsible for photosynthesis? Describe the role of chlorophyll.
Answer:
Chloroplast is the organelle responsible for photosynthesis. Chloroplasts contain green pigment called as chlorophyll. Photosynthetic pigments occur in the granum. They constitute the pigment system called photosystem. About 250 to 400 pigment molecules are present in a photosystem.

The primary function of photosystems is to trap light energy and converts it to chemical energy.

  • Chloroplast was discovered by Schimper.
  • Number of chloroplasts is variable in different species of plants.
  • In lower plants like algae they are 1 or 2 number.
  • In higher plants their number varies from 40 -100 per palisade cell or more.
  • Chloroplasts also have variable shapes, for example cup-shaped, ribbon-shaped etc. in algae while it is discoidal in higher plants.

A typical structure of chloroplast is a double membranous structure having two parts.

  • Grana: It is a lamellar system consisting of stacks of granum lamella each bounded by a membranous box called as thylakoid. They are 40 – 60 per cell. Number of thylakoids per grana is 50 or more. Chlorophyll molecules are found inside the thylakoid membrane where they trap solar energy in the form of small energy packets called ‘photon’ or ‘quanta’. Grana are interconnected to each other by a channel called as stroma lamellae or Fret’s channel.
  • Stroma: It is a non-pigmented proteinaceous matrix in which grana remain embedded. It contains enzymes for dark reaction.
  • Mechanism of Photosynthesis: Photosynthesis is formation of organic food from carbon dioxide and water with the help of sunlight inside chlorophyll containing cells. Oxygen is produced as by-products.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes 3
Oxygen comes from water. Hydrogen of water is used to reduce carbon dioxide to form carbohydrate.

Photosynthesis occurs in two main steps:

  • The first step is dependent upon light and the second step is not dependent upon light. Hence, the former is called light reaction or photochemical phase while the latter, the dark reaction or biosynthetic phase of photosynthesis.
  • Water is split up during photosynthesis by the process called photolysis. It provides reductant for carbon dioxide. Oxygen is liberated. All the liberated oxygen, therefore, comes from water.
  • The photolysis of water in photosynthesis was discovered by Hill and hence it is also known as Hill reaction.
  • The light reaction of photosynthesis is followed by the dark reaction. In this, CO2 is first fixed by ribulose diphosphate and from this fixed CO2 phosphoglyceric acid is formed. The phosphoglyceric acid thus formed ultimately forms carbohydrates.
  • The basic organic compound formed in photosynthesis is often considered to be glucose. The storage product of plants is commonly starch.
  • A chemical equation of photosynthesis is

PSEB 10th Class Science Important Questions Chapter 6 Life Processes 4
In plants and most algae it occurs in the chloroplasts i.e., light reaction and dark reaction.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes

Question 4.
Discuss the steps of light reaction and dark reaction.
Answer:
Steps of light reaction.
Two main steps of light reaction are :
(a) Photolysis of water
(b) Conversion of fight energy into chemical energy.

  • Light reaction takes places in the thylakoid membranes and the intergranal lamellae of the chloroplast in the presence of fight.
  • Two photosystems (PSI and PSII) work in a coordinated manner.
  • H2O splits into H+ and OH.
  • H+ is used to reduce NADP to NADPH2 which is used in dark reaction.
  • Photophosphorylation takes place in two ways—cyclic and non-cyclic.
  • Light energy is converted into chemical energy.
  • The end-products are ATP and NADPH2.
  • As a result of photolysis of water, oxygen is evolved as a by-product.
    4H2O → 4 OH + 4 H+
    4 OH + 4e → 4 OH
    OH → 2H2O + O2

Summary of dark reaction :

  • Dark reaction takes place in the stroma of the living chloroplast.
  • Atmospheric CO2 is absorbed.
  • The end products of fight reaction (ATP and NADPH2) are also used.
  • All green plants operate C3 photosynthetic pathway. Some monocot plants like maize, sugarcane operate both C3 and C4 photosynthetic pathways.
  • First end product of photosynthesis is sugar.

Question 5.
Explain process of nutrition in Amoeba.
Answer:
Nutrition in Amoeba. Amoeba is omnivorous, i.e. it feeds on smaller animals, plants, micro-organisms and fragments of larger organisms. Nutrition is holozoic. Ingestion can occur at any place on the surface since a regular mouth is absent. Ingestion occurs through phagocytosis or engulfing the food particle in an invagination of the body. The engulfed food particle comes to fie inside a food vacuole. The latter is surrounded by a membrane.
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 5
Digestion of food within food vacuole

Digestion: The digestion is intracellular and the food vacuoles act as temporary stomach for digestion.
Absorption: It occurs by diffusion and distribution takes place by cyclosis.
Assimilation: Assimilation of digested material occurs in a single cell.
Egestion: The undigested food is eliminated through the surface of the cell, where the food vacuole containing the undigested food bursts and discharges its contents to the outside.

Question 6.
Difference between the followings :
(i) Autotrophic and Heterotrophic nutrition
(ii) Herbivore and Carnivore.
Answer:
(i) Autotrophic and Heterotrophic nutrition :

Autotrophic Nutrition Heterotrophic Nutrition
1. Food is self-manufactured. 1. Food is obtained readymade from outside.
2. An external source of energy is required for synthesis of food. 2. An external source of energy is not required. The required energy is present in the food obtained from outside.
3. Inorganic substances constitute the raw materials for manufacturing food. 3. Inorganic substances are not much required.
4. Chlorophyll is present for trapping light energy. 4. Chlorophyll is absent
5. Digestion is absent. 5. An external or internal digestion is required for conversion of complex organic materials into simpler and soluble ones.
6. Organisms performing autotrophic nutrition function as producers. Examples: Green plants, some bacteria, some protists. 6. Organisms performing heterotrophic nutrition function as consumers. Animals, many protists and monerans.

(ii) Herbivore and Carnivore

Herbivore Carnivore
Animals winch cat only plants. e.g. Cow, goat etc. They feed on flesh of other animals, e.g. Lion, vulture etc.

Question 7.
Describe human alimentary canal. Draw a labelled diagram of human alimentary canal.
Or
Describe the human alimentary canal with the help of a suitable diagram.
Answer:

  • Mammalian (human) alimentary canal comprises mouth, buccal cavity, pharynx, oesophagus, stomach, small intestine (duodenum, jejunum and ileum), large intestine (caecum, colon and rectum) and anus.
  • The mouth, bounded by two lips open into oral cavity.
  • Buccal cavity contains teeth and tongue, and receives saliva from 3 pairs of salivary glands. Teeth are meant for cutting and mastication of food.
  • Pharynx: It is a vertical tube. It is a cross passage for food and air. It has uvula and epiglottis which closes the internal nares and glottis respectively during swallowing of food to ensure the passage of food into oesophagus (food pipe).
  • Oesophagus: It is a 25 cm long narrow muscular straight tube. It opens into stomach. Oesophagus propels the swallowed food into stomach.
  • Stomach: It is a sac-like structure situated in the upper part of abdominal cavity below the diaphragm. Large part of this sac is situated left of the median line.
  • Small Intestine: It is the longest part of alimentary canal. It is thin-walled and highly coiled tubular structure. It is about 3-3.5 metres long and occupies most part of abdominal cavity. It is coiled upon itself. Its inner lining is thrown into numerous villi.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes 6
Human digestive system.

Large intestine: The large intestine is about 1.5 metres long. It is divided into following parts, i.e. the vermiform appendix, the colon and the rectum. Caecum is a blind tube and represented by vermiform appendix (5-8 cm) and is present below the junction of small and large intestine. Rectum is the last part and opens to the outside by anus guarded by anal sphincter.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes

Question 8.
Describe the process of digestion of food in man.
Answer:
Digestion: The process of conversion of non-diffusible form of food into the simple and diffusible form by chemical and mechanical processes in the alimentary canal is called digestion.

  • The process of digestion starts in the mouth cavity and is completed in the intestine.
  • In the mouth, food gets mixed up with saliva secreted by salivary glands.
  • Saliva contains an enzyme ptyalin (salivary amylase) which breaks polysaccharide starch into disaccharide maltose.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes 7

  • The food from the mouth cavity called bolus passes into the stomach through the oesophagus.
  • The gastric glands of the stomach secrete gastric juice which contains hydrochloric acid, protein digesting
  • enzyme-pepsin Rennin, mucus and small amount of gastric lipase are also components of gastric juice.
  • Pepsin breaks down proteins into peptones and proteoses in acidic medium of stomach.
  • Rennin undies the milk.
  • Muscles present on the wall of stomach churn and propel the food called chyme forward into duodenum.
  • The digested food moves from stomach to duodenum of the small intestine.
  • Duodenum receives bile from juices from liver and pancreatic juice from pancreas.
  • The pancreatic juice contains trypsin, amylase and lipase.
  • The proteins, fats and carbohydrates are further digested into diffusible form amino acids, glycerol and fatty acids, glucose and fructose.
  • The intetinal juice consists of amylolytic, proteolytic and lipolytic enzymes.
  • Finally, the digestion is completed in the ileum with the secretion of the intestinal juice by intestinal glands.
  • Emulsion form of food called chyle is ready for absorption.

Question 9.
Name the constituents of blood and state the functions of each.
Answer:

Constituents Functions
1. Plasma, (i) Serum It contains proteins as well as organic and inorganic substances in solution.
(ii) Fibrinogen It serves to carry the nutritive and waste materials, antibodies, enzymes and hormones.
2. Red blood corpuscles (R.B.Cs.) Clotting of blood.
(Erythrocytes) They help to transport oxygen.
3. White blood corpuscles (W.B.Cs) They help to defend our body against bacteria, as well as the toxins which these organisms may produce. They also help to remove useless dead tissues from the blood.
4. Blood Platelets or Thrombocytes They play a vey important role in bringing about the coagulation of blood.

Question 10.
State the functions of blood.
Answer:
Blood performs a number of functions in the body, the most important of which are as follows :

  • Blood supplies nutrients and oxygen to various organs and cells of the body.
  • It carries the waste matter formed in the cells to the excretory organs.
  • It regulates the temperature of the body.
  • It supplies hormones to different parts of the body.
  • It prevents the body from various diseases by destroying the pathogenic germs.
  • It prevents excessive loss of nutrients from cuts and wound by forming a clot.

Question 11.
Describe the structure of human heart.
Answer:
Heart.

  • It is a highly efficient, pumping organ of body. Human heart consists of 4 chambers: upper smaller right and left atria (auricles) with thinner wall and lower larger right and left ventricles with thicker walls.
  • Atria open into the respective ventricles by atrioventricular apertures guarded by atrioventricular valves.
  • The two atria are separated from each other by interatrial septum, and the two ventricles are separated from each other by interventricular septum.
  • The sinoatrial node (SAN) or the pacemaker is located in the upper wall of right atrium.

Valves of the heart.
Valves are muscular flaps which prevent the blood to flow back through it. Two types of heart valves are distinguished :

  1. The Atrioventicular valves: These valves separate the atria from the ventricles. The right side of the heart possesses the tricuspid valve or right atrioventricular valve and the left side of the heart possesses the bicuspid or mitral valve.
  2. Semilunar valves: These are located in the arteries leaving the heart. The pocket-shaped pulmonary semilunar valves lies in the opening where the pulmonary trunk leaves the right ventricle and aortic semilunar valve lies at the opening between the left ventricle and aorta.

Question 12.
Draw Human heart and label its parts.
Or
Draw a sectional view of the human heart and label on it, Aorta, Right Ventricle and Pulmonary veins.
Answer:
Human Heart
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 8
V.S. of Human heart.

Question 13.
Explain respiratory system of human.
Answer:
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 9
Human respiratory system

PSEB 10th Class Science Important Questions Chapter 6 Life Processes

Question 14.
Describe the mechanism of breathing in human beings.
Answer:
Mechanism of breathing
Breathing. It is simply inhaling fresh air rich in oxygen and exhaling foul air rich in carbon dioxide. Respiratory system of man consists of :

  1. Respiratory tract.
  2. Respiratory organs (lungs).

Respiratory tract. It is the tract or the path through which fresh air enters the body reaches the lungs and foul air (rich in CO2) leaves the lungs to come out. It consists of Nasal cavity → Pharynx → Larynx → Trachea → Bronchi → Bronchioles → Alveoli (of Lungs).
Breathing is accomplished through changes in size and air pressure of chest cavity. It involves Inspiration and Expiration.

Inspiration:
Intake of fresh air is called inspiration. It occurs when the chest cavity is increased in size and therefore decrease in pressure.

Expiration:
The expulsion of foul air (rich in CO2) from lungs is called expiration. It results when the chest cavity is reduced in size.

Question 15.
Draw a well labelled diagram of human excretory system.
Answer:
Human Excretory System.
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 10
Human excretory system.

Question 16.
Describe the structure of human urinary system.
Answer:
The urinary system consists of the following :

  1. Kidneys
  2. Ureters
  3. Urinary bladder
  4. Urethra.

1. Kidneys: The kidneys are a pair of bean-shaped delicate organs. They are situated one on each side of the mid-dorsal line of the abdominal cavity, just below the level of the stomach.

2. Ureters: They are two tubes about 30 cm long, emerging from each kidney with the pelvis of which they are continuous. The ureters run downwards and inwards and open into the urinary bladder.

3. Urinary bladder: It serves as a reservoir for the urine. It is a hollow muscular organ lined by stratified epithelium. Its average capacity for storage is about 500 mm. It is situated in the cavity of the pelvis just behind the pubic symphysis.

4. Urethra: The urethra in two sexes differ. The male urethra is about 20 cm in length. The female urethra is a short duct of about 4 cm long and it extends from the urinary bladder to the external urethra orifice which is in the vestibule just above and anterior to the vaginal orifice.

Short Answer Type Questions

Question 1.
What are life processes?
Answer:
Life processes: Every living organisms takes food, derives energy, removes wagte materials from their bodies and responds to changes in their environment. These activities are called life processes. In all living organisms there occur the basic life processes such as nutrition, respiration, transportation, excretion and reproduction, which are necessary for survival.

Question 2.
What is nutrition? Briefly explain the two major kinds of nutrition.
Or
Define autotrophic nutrition.
Answer:
Nutrition: All living organisms need matter to build up the body and energy to operate the metabolic reaction that sustains life. The materials which provide these two primary requirements of life are called nutrients or foods. The sum total of processes by which organisms obtain matter and energy is termed nutrition.

Modes of Nutrition:
1. Autotrophic or Holophytic nutrition: All green plants and certain protozoans (Euglena) have evolved a mechanism to directly use the energy7 of sunlight for preparing organic food in their own body from simple raw materials i.e. C02 and H20. These simple inorganic materials are transformed into glucose and oxygen is evolved.
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 11
2. Heterotrophic Nutrition: Animal, fungi, (Amoeba) and many bacteria cannot utilize solar energy. They use chemical bond-energy of organic molecules synthesized by other organisms in building their own organic molecules. Such a mode of feeding is termed as heterotrophic nutrition, and the organisms having it are called heterotrophs.

It is of three types :

  1. Saprophytic
  2. Parasitic
  3. Holozoic.

Question 3.
Write a note on saprophytic nutrition.
Answer:
Saprophytic nutrition (Sapros = rotten; phyton = plant) It this organism releases some juices to soften or digest the food and then absorbs the nutrient. Thus they decompose the dead organic matter into simpler substance. Fungi (yeast, moulds, mushrooms) and many bacteria are saprophytic in nutrition.

The saprophytic mode of nutrition can best be exemplified by the common bread mould, Rhizopus. It converts the complex organic food materials of bread i.e. starch into soluble sugars with the help of starch digesting enzymes. These soluble sugars are then absorbed by the fungus.

Question 4.
Explain the two main steps of photosynthesis.
Answer:
Photosynthesis occurs in two main steps.

  • The first step is dependent upon light and the second step is not dependent upon light. Hence, the former is called light reaction or photochemical phase while the latter, the dark reaction or biosynthetic phase of photosynthesis.
  • Water is split up during photosynthesis by the process called photolysis. It provides reductant for carbon dioxide. Oxygen is liberated. All the liberated oxygen, therefore, comes from water.
  • The photolysis of water in photosynthesis was discovered by Hill and hence it is also known as Hill reaction.
  • The light reaction of photosynthesis is followed by the dark reaction. In this, CO2 is first fixed by ribulose diphosphate and from this fixed CO2 phosphoglyceric acid is formed. The phosphoglyceric acid thus formed ultimately forms carbohydrates.
  • The basic organic compound formed in photosynthesis is often considered to be glucose. The storage product of plants is commonly starch.
  • A chemical equation of photosynthesis is

PSEB 10th Class Science Important Questions Chapter 6 Life Processes 12
In plants and most algae it occurs in the chloroplasts i.e., light reaction and dark reaction.

Question 5.
Give a brief account of factors affecting the process of photosynthesis.
Answer:
Factors affecting the process of photosynthesis.

  • Temperature: The rate of photosynthesis increases with increase in temperature upto a maximum of 35°C. However, the rate starts decreasing if the temperature rises beyond 30°C.
  • Water: The rate of photosynthesis is slow in water-deficient conditions.
  • Carbon dioxide: The rate of photosynthesis increases with an increase in carbon dioxide concentration upto a certain level, beyond which there is no effect on the rate of photosynthesis.
  • Anatomy of leaf
  • Chlorophyll contents.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes

Question 6.
What are stomata? Explain role in respiration.
Or
How respiration takes place in roots and stems in plants?
Answer:
Respiration in plants. Plants, during the process of photosynthesis, give off oxygen which is utilized during respiration.

Plant respiration occurs at slower rate.

  • In the leaves Stomata are the openings located on the surface of the leaves which are guarded by two kidney-shaped guard cells. Through stomatal opening, air can pass into or out of leaves.
  • In woody stems. Lenticelo are the breathing pores located on the surface of the woody stems through which air can pass. Plants do not have any specialised ventilation mechanism.
  • In roots. Roots take up oxygen present in between particles through root hairs by diffusion. Root hairs are simple extensions of epidermal cells of root in contact with soil.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes 13
In the older parts of roots: Older portions of root hairs lacks root hairs. They are covered with protective layer of dead cells having very small openings called lenticels through which gaseous exchange take place between soil and inner living cell.

Question 7.
Demonstrate with experiment that O2 is evolved during photosynthesis.
Answer:
Take a beaker filled with water. Add a pinch of baking soda (NaHCO3) to it and put a Hydrilla plant (Aquatic plant) in it. Cover the plant with a funnel. Invert a test tube containing water over the stem of the funnel. Keep this apparatus in the bright sunlight. After some time bubbles start emerging out from the plant, which gets collected in the upper part of the test tube. Remove the test tube and test the gas with a lighted splinter, it keeps on glowing showing that the gas is a supporter of combustion. Thus, the experiment clearly shows that O2 is evolved during photosynthesis.
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 14

Question 8.
To demonstrate that photosynthesis take place in the presence of chlorophyll
Procedure :

  • Take a potted plant with variegated leaves-for example, money plant or crotons.
  • Keep the plant in a dark room for three days so that all the starch gets used up.
  • Now keep the plant in the sunlight for about six hours.
  • Pluck a leaf from the plant. Mark the green areas in it and trace them on a sheet of paper.
  • Dip the leaf in boiling water for a few minutes.
  • After this immerse it in a beaker containing alcohol.
  • Carefully place the above beaker in a water-bath and heat till the alcohol begins to boil.
  • What happens to the colour of the leaf? What is the colour of the solution?
  • Now dip the leaf in a dilute solution of iodine for a few minutes.
  • Take out the leaf and rinse off the iodine solution.
  • Observe the colour of the leaf and compare this with the tracing of the leaf done in the beginning (Fig.).

PSEB 10th Class Science Important Questions Chapter 6 Life Processes 15
What can you conclude about the presence of starch in various areas of the leaf?

  • Observation and Conclusion: Only green patches of variegated leaf take on blue colour. Other parts remain unchanged.
  • Photosynthesis take place only in chlorophyll-containing patches of leaf.

Question 9.
To demonstrate that CO2 necessary for photosynthesis.
Answer:
Procedure:

  • Take two healthy potted plants which are nearly of the same size.
  • Keep them in a dark room for three days.
  • Now place each plant on separate glass plates. Place a watch-glass containing potassium hydroxide by the side of one of the plants.

The potassium hydroxide is used to absorb carbon dioxide.
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 16
Experimetal set-up (a) with potassium hydroxide (b) without potassium hydroxide

  • Cover both the plants with separate bell-jars as shown in Figure.
  • Use vaseline to seal the bottom of the jars to the glass plates so that the set-up is air-tight.
  • Keep the plants in sunlight for about two hours.
  • Pluck a leaf from each plant and check for the presence of starch as in the above activity.
  • Do both the leaves show the presence of the same amount of starch?

What can you conclude from this activity?
Observation and Conclusion. The plant containing CO2 in the surrounding carry out photosynthesis. Other plant does not. It proves that CO2 is essential for photosynthesis.

Question 10.
Demonstrate with experiment that light is essential for photosynthesis.
Answer:
Take a de-starched potted plant, which has been kept in dark for 3 to 4 days. Cover one of its leaves completely with a carbon paper so that no light falls on it. Keep the plant in light for 4 to 6 hours. Test the covered leaf and uncovered leaf for starch with iodine test. The covered leaf will show negligible amount of starch, while the uncovered leaf will give positive test for starch. The process clearly shows that light is necessary for photosynthesis.

Question 11.
How is respiration different from breathing?
Answer:
Differences between Breathing and Respiration

Breathing Respiration
1. It is ventilation or bringing in of oxygenated air and giving out deoxygenated air. 1. Respiration of animals includes breathing, gaseous exchange and catabolic breakdown of food.
2. It is a physical process. 2. Respiration is both a physical and physiological process.
3. Breathing does not liberate energy. 3. It liberates energy.
4. It is restricted to organs where gaseous exchange occurs between blood and atmospheric air. 4. Respiration involves every living ceil of the body.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes

Question 12.
List three differences between respiration in plants and respiration in animals.
Answer:
Differences in respiration in plants and animals

Respiration in plants Respiration in animals
1. All the cells of plant parts (root, stem, leaves) perform the respiration individually. 1. It is performed by specific respiratory organs for all the cells of body.
2. There is little transport of gases from one part to the other. 2. Transport of gases is maximum.
3. Rate of respiration is low. 3. Rate of respiration is high.

Question 13.
What is the function of epiglottis?
Answer:
Epiglottis is a flap like structure present at the top of glottis. It closes glottis during swallowing of food thus checks the entry of food into respiratory passage.

Question 14.
Why is food necessary for living organisms?
Answer:
Food provide energy to raw materials for growth and maintenance.

Utility of components of food.

  • Carbohydrates are mainly used for producing energy.
  • Fats serve as stored concentrated fuel for energy production.
  • Proteins are mainly used to build up tissues.
  • Mineral salts and vitamins regulate metabolic processes and growth.
  • Water is essential for all biological activities.

Question 15.
Explain the mechanism of breathing in human.
Answer:
Inhalation or Inspiration

  • The entry of air from outside into alveoli of lungs through respiratory tract is called inhalation.
  • The air enters when thoracic cavity expands due to contraction of intercostal muscles attached to ribs and peripheral muscles of the diaphragm.
  • Thus the thorax moves upward, outward and forward.
  • It increases the volume of thoracic cavity and the pressure decreases.
  • Thus air from outside rushes into alveoli of lungs through nostrils, nasal chambers, trachea, bronchi and bronchioles.
  • The alveolar sac gets filled with oxygen rich air.

Exhalation or Expiration is concerned with the expelling of carbon dioxide from lungs.

  • It takes place when the volume of the thoracic cavity decreases and the pressure of the contained air in the thoracic cavity increases.
  • Air passes out through the respiratory tract from the lungs.

Question 16.
What is the effect of sternuos exercise on rate of breathing and why?
Answer:
Normally man breathes about 15-18 times per minute but during hard exercise the breathing rate increases to 20 to 25 times per minute. It is due to the fact that body needs more of energy thus requires more of oxygen.

Question 17.
Give an outline of Calvin-Benson Cycle.
Answer:
Melvin Calvin and Andy Benson discovered this cycle, hence it is called Calvin cycle.
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 17
Calvin-Benson Cycle

Question 18.
Label the parts in the figure.
Answer:
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 18
Answer:

  1. Pseudopodium
  2. Food particle
  3. Ingestion
  4. Food Vacuole
  5. Waste
  6. Digestion of food.

Question 19.
Describe the role of intestinal juice.
Answer:
Intestinal juice too is alkaline (pH 8.3).

It has many enzymes :

  • Intestinal amylase hydrolyses the remaining starch and glycogen to maltose.
  • Maltase changes maltose to glucose.
  • Sucrase converts sucrose into glucose and fructose.
  • Lactase hydrolyses lactose to glucose and fructose.
  • Dipeptidases hydrolyse dipeptides to amino acids.
  • Intestinal lipase splits emulsified fats into fatty acids and glycerol.
  • Alkaline emulsion of digestion products formed in the small intestine is called chyle.

Question 20.
Explain absorption of food in the small intestine.
Answer:
Absorption. The process of diffusion of digested food into blood present in the blood capillaries of smail intestine is called absorption. Inner lining is thrown in fold called villi. They increase surface area for absorption of food.
Glucose, amino acid, vitamins, mineral salts and water diffuses into blood present in blood capillaries of numerous villi of the small intestine. The fatty acids and glycerol diffuses into lymph present in lymph vessels called lacteals. The digested food is carried to the liver by the hepatic portal vein. The fatty acid and glycerol unite in lymph to form fat. Most of the fat passes as a milky emulsion. After absorption, the undigested- food passes into large intestine.

Question 21.
State the functions of stomach and large intestine.
Functions of Stomach
Answer:

  • Storage of food.
  • Mechanical breakdown of food.
  • Partial digestion of food.

Functions of large intestine
Colon:

  • Its wall absorbs the water from undigested food.
  • Absorption of digested food also takes place in this region which has not been absorbed by ileum.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes

Question 22.
Differentiate the following :
Respiration and Photosynthesis
Answer:
Differences between Respiration and Photosynthesis

Respiration Photosynthesis
1. It is a catabolic process in which food substrates are broken down. 1. It is an anabolic process in which food substrates are synthesized.
2. It takes place in all living cells. 2. It is carried out only by the chlorophyll containing cells of plants.
3. CO2 and H2O are produced. 3. CO2 and H2O are used.
4. CO2 is given out. 4. O2 is released as a byproduct.
5. Chemical energy is converted into ATP and some energy is lost as heat. 5. Radiant energy’of light is converted into 1 chemical energy.

Question 23.
What is fermentation? How is it important?
Answer:
The slow decomposition of organic matter into simpler substances in the presence of enzymes is known as fermentation. It is a type of anaerobic respiration. Fermentation literally means a chemical change accompanied by effervescence. The anaerobic breakdown of glucose to carbon dioxide and ethanol is a form of respiration referred to fermentation. It is normally carried by yeast cells and accounts for the production of alcohol in alcoholic bevefages. In fermentation process, if glucose is converted into ethanol then it is called ethanolic fermentation. When glucose is converted into organic acids such as lactic acid, then this type of fermentation is known as lactic acid fermentation. It is carried out by the bacterium Bacillus acidilacti.

Question 24.
Briefly explain the human respiratory system and also label the parts 1, 2, 3, 4, 3 and 6.
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 19
Answer:
Human Respiratory System. Respiratory system of human beings and other mammals consists of air passage or respiratory tract, a pair of lungs.

Respiratory tract is made up of nostrils nasal cavity, pharynx, larynx, trachea and bronchi.

  1. Left Lung
  2. Heart
  3. Diaphragm
  4. Trachea
  5. Larynx
  6. Nasal Cavity

The lungs are a pair of brownish grey coloured spongy structures situated in the thoracic cavity. The left lung consists of two lobes while the right lung consists of three lobes.

Each lobule of a lung consists of bronchioles which terminate into a bunch of spherical thin walled air sacs, called alveoli.

Each alveolus or air sac has a diameter of 75 to 300 microns and has a very thin wall. The walls of the alveoli are elastic and are supplied with capillaries. Gases are exchanged between the capillaries and the air sacs through these thin walls.

Question 25.
What are the different modes of respiration in animals?
Answer:
In animals such as earthworm, respiration is by skin.

  • The insects have an elaborate tracheal system of respiration.
  • Fishes respire through gills.
  • Respiratory system of human beings and other mammals consists of air passage or Respiratory tract, and a pair of lungs.
  • Respiratory tract is made up of nasal cavity, pharynx, larynx, trachea and bronchi. The lungs are a pair of brownish grey coloured spongy structures situated in the thoracic cavity.
  • The left lung consists of two lobes while the right lung consists of three lobes.
  • Each lobule of a lung consists of bronchioles which terminate into a bunch of spherical thin-walled air sacs, called alveoli.

Question 26.
Differentiate between respiration and combustion.
Answer:
Difference between respiration and combustion.

Respiration Combustion
1. It occurs in living cells. 1. It occurs in non living and dead cells.
2. It is a complex biochemical process controlled by several enzymes. 2. It is a chemical process.
3. Heat produced is much less. 3. A large amount of heat is produced.
4. There is no flame or light produced. 4. This process is usually accompanied with flame and light.

Question 27.
What are parasitic nutrition?
Answer:
Parasitic nutrition (Para = besides ; sitos = food). In this an organism (parasite) depends upon the organism (host) for its nutritional requirements. Many bacteria, viruses, fungi, some non-green plants and many animals have this mode of nutrition.

For example, a fungus Puccinia is a parasite on wheat and barberry plants; Cuscuta or dodder plant grows as a parasite on many plants; tapeworms and round worms are parasites in the body of man etc.

Parasites are of two kinds :

  1. Ectoparasites and
  2. Endoparasites

Question 28.
Differentiate between Light reaction and Dark reaction in photosynthesis.
Answer:
Difference between light reaction and Dark reaction

Light reaction Dark reaction
1. Light reaction is light induced chemical reaction. 1. Dark reaction requires no light and is purely enzyme controlled reaction.
2. Energy rich compounds like ATP and NADPH2 are synthesized. 2. The energy rich compounds are used to produce the organic compounds.
3. Oxygen is liberated. 3. No liberation of oxygen.
4. It takes place in the grana of the chloroplasts. 4. It takes place in the stroma of chloroplasts.

Question 29.
What are the functions of liver?
Answer:
Functions of liver

  1. Role in digestion. Bile produced by liver helps in the digestion of food as follows.
    • It emulsifies the fats with its salts.
    • It prevents decomposition of food by checking the growth of bacteria.
    • It neutralizes the acid coming from the stomach along with food and provide alkaline medium in the intestine required for action of enzymes of pancreas and intestinal glands.
  2. Regulation of Blood Sugar: The liver separates the excess of sugar from the blood and stores it in its cells as glycogen (animals starch).
  3. Formation of Glycogen from non-carbohydrates Sources.
  4. Deamination: In the liver, the amino acids coming from the alimentary canal are sorted out, ammonia is formed. Ammonia is converted to less toxic urea.
  5. Excretion: Liver collects haemoglobin of the worn-out red blood corpuscles and changes it into bile pigments.

Question 30.
Differentiate between Saprophytic and Parasitic nutrition.
Answer:

Saprotrophic Nutrition Parasitic nutrition
Many organisms absorb fluid food through the body surface. This is called saprotrophic nutrition. Bacteria and fungi flourish on dead, decaying organic matter of both plant and animal origin. They secrete digestive enzymes onto this matter. The enzymes hydrolyze the organic matter into simple soluble products that are then absorbed. This method of taking up organic food is known as saprophytic nutrition. The organisms obtain nutrients from a living host without helping it any way.
Examples. Liver fluke lives in the bile duct of sheep and absorbs nutrient. Other examples include several fungi, bacteria and a few higher non-green plants such a cuscuta.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes

Question 31.
Discuss the fate of food in the oral cavity of man.
Answer:
Fate of food in mouth cavity. The partial digestion of food occurs due to the action of Ptyalin enzyme. It acts on starches and forms maltose.
The sufficiently masticated, partially digested food forms bolus. It is swallowed into oesophagus through gullet by raising the throat aided by the muscles of pharynx.

Question 32.
What is a digestive gland? Name the various digestive glands of man and their secretions.
Answer:
Digestive gland. A gland that secretes digestive juice which is helpful in the digestion of food is called a digestive gland.

Digestive glands of Man

Name of digestive gland Name of digestive juice/Secretion
1. Salivary gland Saliva
2. Gastric glands Gastric juice
3. Pancreas Pancreatic juice
4. Liver Bile
5. Intestinal glands Intestinal juice

Question 33.
Write the enzymes of the pancreatic juice, the substrates they digest and the products of their digestive action.
Answer:
Enzymes of the pancreatic juice

Name of Enzyme Substrate Name of end products
1. Amylase Starch, glycogen Maltose and Isomaltose
2. Trypsin Proteoses, Peptones and Proteins Peptides and amino acid
3. Lipase Emulsified lipids Glycerol and fatty acids

Question 34.
What is meant by photosynthesis? What are the basic requirements for the process of photosynthesis?
Answer:
The manufacture of organic compounds from carbon dioxide and water in the presence of sunlight inside the chlorophyll containing cells of the plant is called photosynthesis. The overall reaction of photosynthesis is
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 20
Basic requirements for photosynthesis. CO2, water, chlorophyll and solar energy. Photosynthesis is a photo-biochemical process in which energy rich compounds such as carbohydrates (glucose) are synthesized from simple inorganic compounds like CO2 and water in the presence of sunlight and chlorophyll. Oxygen is a by-product.

Question 35.
What are the different ways in which glucose is oxidized to provide energy in various organisms?
Answer:
Breaking down of glucose involves two step process. In the first step, it is broken into three carbon molecule called pyruvate. The pyruvate is further broken down into energy in following different ways in various organisms :

  • Aerobic Respiration: In this case, pyruvate is broken down into water and carbon dioxide along with release of energy. It commonly occurs in mitochondria of cells.
  • Anaerobic Respiration in Yeast: In yeast cells during fermentation pyruvate is converted into ethanol and carbon dioxide in the absence of oxygen.
  • Anaerobic Respiration in Muscles: Due to lack of oxygen, e.g. during vigorous running or exercise, in human muscles, pyruvate is converted into lactic acid.

Question 36.
Explain how exchange of materials takes place between Blood and Tissues.
Answer:
Arteries supply fresh blood with 02 and food materials to different body organs. Inside the body organ the artery divides into smaller branches called arterioles. The arterioles further divide into extremely thin walled blood capillaries. The blood capillaries form an extensive network inside the body organ. They make their way through the tissue cells. Blood plasma along with the dissolved materials comes out of the thin walls of the blood capillaries and collects into the tissue. It is then called tissue fluid, which acts as an intermediate medium between blood and tissue cells.

The tissue fluid contains different materials such as oxygen, amino acids, glucose, mineral ions and proteins etc. which are needed by the body cells. The body cells take up the required materials from the tissue fluid and release their wastes such as C02 and nitrogenous wastes. These enter through the blood capillary wall and dissolve into the blood plasma or enter into the red blood cells and are carried away.

Question 37.
Outline inhalation-exhalation cycle.
Answer:

  • Inhalation: Lowering of diaphragm → Rising of rib cage → Gas (O2) passes to Alveoli
  • Exhalation: Air is forced out → Rising of diaphragm → Lowering of ribcage

Question 38.
Leaves of a healthy potted plant are coated with vaseline to block the stomata. Will this plant remain healthy for long? State three reasons to support your answer.
Answer:
No, the plant will not remain healthy because no exchange of gases will take place. It will lead to :

  • low respiration
  • no photosynthesis occurs
  • no transpiration.

Hence plant will not remain healthy and may die eventually.

Question 39.
What are
(i) stomata and
Answer:
Stomata are tiny apertures found on the surface of the leaf, which regulate the exchange of respiratory gases and transpiration.

(ii) lenticels?
Answer:
Lenticels are the raised pores in the woody plants that allow the exchange of gases between the atmosphere and the internal tissues.

Question 40.
Briefly explain breathing, external resperation exchange of gases and tissue resperation.
Answer:

  • Breathing involves inhaling of oxygen rich fresh air and exhaling of carbon dioxide rich foul air. The respiratory surface is richly supplied with blood for this purpose. Oxygen of the inhaled air is taken up by blood while carbon dioxide of the blood passes into the air for exhalation.
  • The exchange of gases between the blood and the air at the respiratory surface is. known as external respiration.
  • Oxygen absorbed by the blood at the respiratory surface is taken to various parts of the body through arteries. Blood loses the oxygen contained in it to tissue fluid, from where it picks up carbon dioxide. The latter is brought to the respiratory surface by blood.
  • Tissue respiration, also called internal respiration, is the exchange of gases between the tissue cells and the blood involving uptake of oxygen by tissue cells, oxidation of respiratory substrate and elimination of carbon dioxide by the cells.

Question 41.
Why is mitochondria termed ‘power house’ of the cell?
Answer:

  • Most of the aerobic respiration occurs inside the mitochondria and therefore, the latter are also called power houses of the cells.
  • Mitochondria are site of synthesis, storage and transport of ATP.
  • ATP (Adenosine triphosphate) is the energy currency of the living organisms.
  • Adenosine triphosphate. The energy released during cellular respiration is immediately used to synthesise a molecule called ATP from ADP and inorganic phosphate

PSEB 10th Class Science Important Questions Chapter 6 Life Processes 21
ATP is used to fuel all other activities in the cell. Therefore, it is said to be the energy currency for . most cellular processes.

Question 42.
The figure shows different ways in which glucose is oxidised to provide energy in various organisms?
Fill up the blanks : A _________ B _________ C _________, D. _________ E _________ F _________
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 22
Answer:
Different pathways to provide energy from glucose
A. glucose
B. Pyruvate
C. Absence of Oxygen
D. Presence of oxygen
E. Lactic Acid F. Carbondioxide

Question 43.
Fill in the blanks 1, 2, 3, 4, 5, 6 in the figure.
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 23
Double circulation of blood in birds and mammals.
Answer:

  1. Right vetricle
  2. Lungs
  3. Left Auricle
  4. Left Ventricle
  5. Body parts (except Lungs)
  6. Right Auricle

PSEB 10th Class Science Important Questions Chapter 6 Life Processes

Question 44.
State differences between artery and vein.
Answer:
Differences between Artery and Vein

Artery Vein
1. An artery carries blood from the heart to different organs of the body. 1. A vein collects blood from different organs of the body and brings it back to the heart.
2. Blood flows under great pressure. 2. Blood flows under less pressure.
3. It has a thick muscular wall. 3. Wall is thin
4. It is non collapsible. 4. It is collapsible.
5. It contains oxygenated blood (Exception pulmonary artery). 5. It contains deoxygenated blood (Exception pulmonary vein)
6. Valves are absent. 6. Valves are present.
7. Mostly deep seated. 7. Mostly superficial.

Question 45.
Why are WBCs called ‘soldiers of the body’?
Answer:
WBC (White Blood Corpuscles) or leucocytes engulf and destroy the foreign particles in the body. Hence they are called ‘soldiers of the body.’

Question 46.
What are hypertension and hypotension?
Answer:
Hypertension. It is the high blood pressure which is caused due to emotions such as worry, excitement, fear etc.
Hypotension. It is the low blood pressure when it falls below the normal level.

Question 47.
Name the three major types of blood vessels. Explain briefly.
Answer:
The three main types of blood vessels are :

  • Arteries have thick elastic walls and their diameter may be even 1 cm. These blood vessels carry the blood from the heart to the various parts of the body.
  • Capillaries. Arteries divide into thin arterioles and arterioles further ramify into capillaries (1 micron diameter). The wall of a capillary is made up of a single layer of cells. The muscles and elastic fibres are absent in the capillaries. The walls of these capillaries are so thin that the exchange of food materials, waste materials and gases takes place between the blood and protoplasm of cells (liver, lung etc.) through them.
  • Veins. The capillaries again reunite to form venules and venules unite to form veins. These venules and veins return the blood to the heart.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes 24
Diagram showing the relationship among blood vessels

Question 48.
Write a note on lymphatic system in human beings.
Answer:
Lymphatic system: The lymphatic system comprises colourless fluid, the lymph; a network of fine channels, the lymphatic capillaries; tubes of varied sizes, the lymphatic vessels; and the lymph nodes. Tissue or interstitial fluid present in the spaces between tissue cells is formed by filtration of protein-free fluid from the blood. Tissue fluid passes into lymphatic capillaries to form lymph. The latter is carried by lymphatic vessels to the veins. Lymphatic vessels have lymph nodes which filter lymph, removing microorganisms and cellular debris and adding lymphocytes.

Question 49.
Write functions of lymph.
Answer:
Functions of lymph

  • It drains excess tissue fluid from the extracellular spaces back into the blood.
  • Some of the fluid from the digestive tract is absorbed in the lymph. The lymphatic vessels store this fluid temporarily, and release it gradually so that the kidneys do not face a sudden pressure of urine excretion.
  • It carries carbon dioxide and nitrogenous waste materials that diffuse into the tissue fluid to the blood.
  • It takes lymphocytes and antibodies from the lymph nodes to the blood.

Question 50.
Give differences between blood and lymph.
Answer:
Differences between Blood and Lymph

Blood Lymph
1. It consists of plasma, erythrocytes, leucocytes and platelets. 1. It consists of plasma and leucocytes (lymphocytes most abundant).
2. It is red in colour due to the presence of haemoglobin in erythrocytes. 2. It is colourless as haemoglobin is absent,
3. Its plasma has more proteins, calcium and phosphorus. 3. Its plasma has fewer proteins and less calcium and phosphorus.
4. It transports materials in the body. 4. It acts as middle man between blood and body tissue.

Question 51.
How are xylem and phloem well suited for transport of materials in plants? Explain.
Or
How are water and mineral transported in plants?
Answer:
Xylem and phloem are well suited to carry water, minerals and food in plants. Vessels in the xylem are cylindrical in shape with their ends open and are placed one above the other so as to form a continuous column stretching from roots to leaves. So, the water and minerals absorbed by the roots are carried upwards to the leaves. This is known as transportation.

Similarly, phloem has sieve tubes that are also cylindrical but the ends are not open instead covered with sieve (perforated) plate. These tubes are also placed one above the other, forming a continuous column from leaves to other parts of the plant body. The food synthesized in the leaves is carried to other paxts of the plant body through phloem. Sucrose is the main form in which carbohydrates are translocated in plants.
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 25
Xylem vessels that transport water and mineral sahs

PSEB 10th Class Science Important Questions Chapter 6 Life Processes 26
Phioem tubes that conduct prepared food.

Question 52.
Mention any three methods adopted by plants to minimise the transpiration rate.
Answer:
Three methods adopted by plants to minimise the rate of transpiration are :

  1. In some cases leaves are rolled to cover stomata (e.g. some grasses)
  2. The stomata may be sunken (e.g. Nerium)
  3. In some cases, leaves may be dropped or absent as in most cacti.

Question 53.
Write a short note on root pressure.
Answer:
Root pressure: During absorption, water is forced into the xylem vessels by the surrounding cortical cells with a certain force. This induces a pressure which is responsible for ascent of sap to many feet in xylem. This pressure which is developed due to the activity of root is called as root pressure.
Root pressure is a vital phenomenon and depends upon the activity of living root cells. The magnitude of root pressure varies from 2-8 atm.

Question 54.
How is transpiration useful to plants?
Answer:
Advantages of transpiration

  • It has cooling effect on the plants as excess of sun’s energy is dissipated.
  • It helps in the removal of excess of water from the plant.
  • It causes ascent of sap.
  • It helps to maintain water cycle.
  • It increases the amount of sugar and mineral content in the fruit.
  • It is needed to permit photosynthesis to take place.

Question 55.
What are the disadvantages of transpiration?
Answer:

  • More plants die from excessive water loss by transpiration.
  • Due to high rate of transpiration plants suffer from loss of turgidity.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes

Question 56.
What is translocation?
Answer:
Translocation: The long distance transport of the organic food from a source to a sink is known as translocation.

Question 57.
(a) What is blood pressure?
Answer:
Blood pressure is the force that blood exerts against the wall of a vessel. This pressure is much greater in arteries than in veins. The pressure of blood inside the artery during ventricular systole is called systolic pressure and pressure in artery during ventricular diastole is called diastolic pressure.

(b) What are normal value of systolic and diastolic blood pressure
Answer:
The normal systolic pressure is about 120 mm of Hg and diastolic pressure is 80 mm of Hg.

(c) How is it measured?
Answer:
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 27
Measurement of blood pressure

(d) Name the instrument used to measure blood pressure.
Answer:
Blood pressure is measured with an instrument called sphygmomanometer.
High blood pressure is also called hypertension and is caused by the constriction of arterioles, which results in increased resistance to blood flow. It can lead to the rupture of an artery and internal bleeding.

Question 58.
Write a note on mechanism of blood clotting.
Answer:
Blood Clotting: At the site of injury, blood platelets disintegrate and release enzyme thromboplastin. Thromboplastin in the presence of calcium ions, transforms the inactive prothrombin into active thrombin. Thrombin converts soluble fibrinogen into insoluble fibrin. This forms a meshwork of fibrils which entangle blood corpuscle and transforms the liquid blood into a gel or clot.
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 28
Mechanism of Blood Clotting
Unless the blood clots at the site of the injury, there will be loss of blood (blood haemorrhage).
Such a loss of blood will lead to death of the person.

Question 59.
Briefly describe excretory system.
Answer:

  • Kidneys are a pair of bean-shaped reddish brown organs which lie in the lumbar part of abdomen along the dorsal wall, one on either side of the vertebral column.
  • Each kidney receives a renal artery from dorsal aorta and sends a renal vein to inferior vena cava.
  • The excretory waste products ,are filtered out in the kidney.
  • Each kidney contains about 1.2 million excretory units called uriniferous tubules or nephrons.
  • Uriniferous tubule is a long, twisted, narrow, tubular structure which consists of ‘ Bowman’s capsule, neck, proximal convoluted tubule, loop of Henle and distal convoluted tubule.
  • Bowman’s capsule is a blind cup-shaped end of uriniferous tubule with a tuft of blood capillaries called glomerulus.
  • Ureters are two distensible tubes which connect the kidneys with the urinary bladder.
  • Urinary bladder is a median pear-shaped bag-like structure that occurs in she pelvic region of abdominal cavity.
  • Urinary bladder can hold 300-800 ml of urine.
  • Urethra is a tubular connection between the urinary bladder and the external Opening of urinary tract.

Question 60.
Label the parts of human excretory system.
Answer:
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 29

  1. Kidney
  2. Ureter
  3. Urinary Bladder.

Question 61.
How does the excretion take place in Amoeba?
Answer:
The excretory product of Amoeba is ammonia. Special excretory organelle is lacking in Amoeba. C02 and ammonia are excreted by diffusing in solution through plasma membrane. The concentration of ammonia is always higher in Amoeba than in the surrounding water. The water enters through plasma membrane by “endosmosis”. Ammonia is formed in cytoplasm by metabolism.

Surplus water enters contractile vacuole. This surplus water can rupture the animal’s body. Thus size of contractile vacuole increases, when the contractile vacuole is fully expanded with water, it moves towards the periphery. As the contractile vacuole comes in close contact with the plasma membrane, it bursts. Thus excess of water (surplus water) is discharged in the surrounding water. This phenomenon of controlling the amount of water in the body is called as “osmoregulation”.
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 30
Excretion in Amoeba

Question 62.
State one main function of the following :
(i) Glomerulus
Answer:
The filtration of blood in the nephron takes place in the glomerulus.

(ii) Malpighian capsule
Answer:
Malpighian capsule is concerned with ultrafiltration.

(iii) Sweat gland
Answer:
Sweat glands produce sweat containing urea, uric acid and salts. Sweat evaporates to bring down the body temperature to normal.

(iv) Nephron of kidney tubule
Answer:
Through the kidney tubule or nephrons, filtration of urea, uric acid, water and some salts occur from the blood.

(v) Loop of Henle.
Answer:
Loop of Henle is useful in the absorption of water and secretion of urea.

Question 63.
Name the chief organs of excretion in man. Mention the waste products that they excrete.
Answer:
The chief excretory organs and the waste products removed by them are :

  1. Kidneys – Urea in the form of urine.
  2. Lungs – Carbon dioxide.
  3. Skin – Water and salts as sweat.

Question 64.
Name the following :
(i) A process by which the unwanted nitrogenous wastes are eliminated from the body.
Answer:
Excretion

(ii) Major excretory organs of man.
Answer:
Kidneys

(iii) The structural and functional units of kidney.
Answer:
Nephrons

(iv) A tuft of blood capillaries found in the Bowman’s capsule of nephron.
Answer:
Glomerulus

Question 65.
Name the following :
(i) The structure that brings urine from the kidney to the urinary bladder.
Answer:
Ureter

(ii) Thin membranous sac serving as the reservoir of urine.
Answer:
Urinary bladder

(iii) Any two organic constituents of normal human urine.
Answer:
Urea, creatinine

(iv) The chief nitrogenous waste product in the human urine and the organ which produces it.
Answer:
Urea, liver

(v) Name two excretory products formed by the liver.
Answer:
Bile pigments (Bilirubin, Biliverdin), urea.

Question 66.
What do you understand by artificial kidney? Name the principle on which it works.
Or
Write a note on haemodialysis.
Answer:
Artificial Kidney or Haemodialysis. In case of acute kidney failure, the poisonous materials may accumulate in the body fluids which will cause oedema and finally lead to death of the patient. In such cases, artificial kidney is used. It works on the principle of dialysis and separates wastes from the blood. The process is called haemodialysis.

Artificial kidney contains a number of tubes with a semipermeable lining.
Its functioning is similar to kidney, but it is different since there is no reabsorption involved.

In this, blood from an artery is diverted through a cellophane tube, having pores equal to those of glomerular capillaries, placed in a circulating bath. Concentration of bath-fluid is kept equal to that of normal plasma. The pores in the cellophane tube allow small sized wastes like urea, ammonium salts etc. to pass through but do not allow the passage of blood cells, proteins, fats etc. Diffusion of small and useful substances like glucose, amino acids etc is prevented by keeping their concentration in the dialysis fluid equal to the normal plasma. Blood from dialyser is returned to the body through a vein.
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 31
Flow of blood through an artificial kidney for haemodialysis.

Question 67.
What are the functions of tongue?
Answer:
Functions of tongue.

  • It helps in mastication of food.
  • It bears taste buds and helps in the sensation of taste of food.
  • It takes part in the modification of sound production.
  • It acts as a brush and cleans the teeth.
  • It aids in deglutition of food.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes

Question 68.
(i) Label the part A-E
Answer:
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 32
A. Interventricular System B. Right Ventricle C. Right Atrium D. Left Ventricle E. Aorta

(ii) Write Function of E.
Answer:
Function of E-Aorta carries blood to all parts of body.

Very Short Answer Type Questions

Question 1.
Why are maintenance processes required?
Answer:
They are required to prevent breakdown.

Question 2.
List three characteristics of living organisms.
Answer:

  1. Growth
  2. Movements
  3. Repair and maintenance of their structures.

Question 3.
What are life processes?
Answer:
The processes which together perform the maintenance jobs are collectively termed as life processes.

Question 4.
What is the basic requirement of maintenance?
Answer:
Energy is needed by living organisms for maintenance.

Question 5.
What are sources of energy for living organisms?
Answer:
Carbon based molecules i.e. food obtained from environment.

Question 6.
List the common reactions required to obtain energy from carbon based molecules.
Answer:
Oxidising-reducing reactions.

Question 7.
Name any four life processes required for maintenance.
Answer:

  1. Nutrition
  2. Respiration
  3. Transportation
  4. Excretion.

Question 8.
What are nutrients?
Answer:
The substances which provide materials for growth, energy and maintenance are called nutrients.

Question 9.
What is nutrition? Why is it necessary?
Answer:
Nutrition. The sum total of processes by which living organisms obtain food materials and prepare them for use in the growth, repair and providing energy is termed nutrition.

Question 10.
What is food?
Answer:
Food provides energy. It Provides raw materials for growth and maintenance.

Question 11.
What is holozoic nutrition?
Answer:
Holozoic nutrition. When the nutrients are ingested as solid organic food matter, it is called holozoic nutrition.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes

Question 12.
What is the source of food for heterotrophs?
Answer:
All heterotrophs obtain food from autotrophs.

Question 13.
Name the process which prepares food is autotrophs.
Answer:
Photosynthesis.

Question 14.
Why are green plants called producers?
Answer:
Green plants prepare their food from CO2 and H2O in the presence of sunlight and chlorophyll. All other organisms obtain food from green plants, thus called as producers.

Question 16.
In which spectrum of light maximum photosynthesis occurs?
Answer:
Red light.

Question 17.
Where is chlorophyll present in cells of leaves?
Answer:
Chloroplast.

Question 18.
How does oxygen produced during photosynthesis enter the atmosphere?
Answer:
Oxygen passes out of green leaves through stomata and diffuses into the atmosphere.

Question 19.
Where does photolysis occur in plant?
Answer:
Photolysis occurs in chloroplasts present in the cell.

Question 20.
Write chemical reaction of photosynthesis.
Answer:
PSEB 10th Class Science Important Questions Chapter 6 Life Processes 33

Question 21.
What are the advantages of cooked food?
Answer:
Human beings consume cooked food. Cooking makes it soft, palatable, tasty and easier to digest.

Question 22.
What is the role of CO2 during photosynthesis?
Answer:
CO2 provides carbon for synthesis of glucose (C6H12O8) during photosynthesis.

Question 23.
What is the role of stomata in green leaves?
Answer:
Stomata are minute pores through which exchange of gas occurs.

Question 24.
Name the minerals obtained from soil by plants.
Answer:
Nitrogen, phosphorus, potassium, iron, magnesium and other minerals.

Question 25.
What is the role of nitrogen plants?
Answer:
Nitrogen is constituent of proteins and nitrogen bases of nucleic acids.

Question 26.
What is the food of Amoeba?
Answer:
Food of Amoeba consists of small protozoans, algae, rotifers, bacteria, and diatoms. It also feeds upon bits of organic matter.

Question 27.
Is food vacuole of Amoeba, temporary structure or a permanent one?
Answer:
Food vacuole is a temporary structure.

Question 28.
What is saliva?
Answer:
Saliva is a digestive juice secreted by salivary glands present in oral cavity.

Question 29.
Name the enzyme present in saliva.
Answer:
Salivary amylase.

Question 30.
Name the organ through which blood passes into stomach from oral cavity.
Answer:
Pharynx and oesophagus.

Question 31.
Name the largest part of alimentary canal.
Answer:
Small intestine.

Question 32.
Which of the animals need long large intestine?
Answer:
Animals eating grasses need large intestine for digestion of cellulose.

Question 33.
What is the nature of food entering small intestine from stomach?
Answer:
Acidic in nature.

Question 34.
Name the juices which convert acidic food into alkaline in small intestine.
Answer:
Bile and pancreatic juice.

Question 35.
Name the digestive juice secreted by liver.
Answer:
Bile juice.

Question 36.
Name the digestive juice secreted by pancreas.
Answer:
Pancreatic juice.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes

Question 37.
List the enzymes which help in the digestion of proteins.
Answer:

  • Pepsin
  • Trypsin
  • Chymotryposin
  • Peptidases.

Question 38.
What is approximate length of human alimentary canal?
Answer:
9 to 10 metres.

Question 39.
Name the largest gland present in human body.
Answer:
Liver.

Question 40.
Name the gland which is exocrine as well as endocrine in nature.
Answer:
Pancreas.

Question 41.
What are three parts of small intestine?
Answer:

  1. Duodenum
  2. Jejunum
  3. Ileum.

Question 42.
Differentiate between chyme and chyle.
Answer:

  • Chyme is a semisolid, semidigested acidic food which passes from stomach into duodenum.
  • Chyle is emulsion form, completely digested, alkaline food present is small intestine ready for absorption.

Question 43.
Name the four enzymes present in pancreatic juice.
Answer:

  1. Pancreatic amylase
  2. Pancreatic lipase
  3. Trypsin
  4. Chymotrypsin.

Question 44.
Maximum water is absorbed in which part of alimentary canal.
Answer:
Large intestine.

Question 45.
What is digestion?
Answer:
Digestion. Chemical and mechanical break down of complex, non-diffusible form of food into simple diffusible form of food by action of enzyme.

Question 46.
What is assimilation?
Answer:
The absorption and digestion of food or nutrients by the body or any biological system.

Question 47.
What is egestion?
Answer:
The act or process of discharging undigested food as faeces from a cell in case of unicellular organisms.

Question 48.
What are villi present is small intestine and not in stomach?
Answer:
Small intestine is site of absorption of digested food, villi increase the surface area for absorption of food. No absorption of food occurs in stomach thus villi are absent.

Question 49.
What happens to pyrnvate produced during anaerobic respiration?
Answer:
Pyruvate produced at the end of glycolysis is converted to C02 and ethanol.

Question 50.
What is aerobic respiration?
Answer:
Aerobic respiration is the process of producing cellular energy involving oxygen cells break down food in the mitochondria in a long, multistep process that produces roughly 36 ATP.

Question 51.
What is anaerobic res-piration?
Answer:
A form of incomplete intracellular breakdown of sugar or other organic compounds in the absence of oxygen that releases energy.

Question 52.
What is ATP?
Answer:
ATP is Adenosine triphosphate. It is the energy currency of life. ATP is a high- energy molecule found in every cell. Its job is to store and supply the cell with needed energy.

Question 53.
Expand NADP.
Answer:
NADP. Nicotinamide Adenine Dinucleotide.

Question 54.
Name three animals which respire through skin.
Answer:

  1. Earthworm
  2. Leech
  3. Frog.

Question 55.
Where does Krebs cycle occur in the body?
Answer:
Krebs cycle is completed in mitochondria of cells.

Question 56.
What is hypoxia?
Answer:
Hypoxia is a condition of shortage of oxygen in the body due to strangulation on cyanide poisoning.

Question 57.
Name the respiratory sub-strate.
Answer:
Glucose.

Question 58.
What is the end product of aerobic respiration?
Answer:
Carbon dioxide and water.

Question 59.
What is fermentation?
Answer:
Incomplete breakdown of glucose in the absence of oxygen in microbes such as bacteria and yeast. It forms CO22, ethanol and energy is released in small amount.

Question 60.
Man breaths how many times per minute?
Answer:
12-15 times.

Question 61.
What are the functions of ATP?
Answer:
ATP provide energy for all metabolic reactions in the body such as movements, synthesis, cell division.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes

Question 62.
What is bark?
Answer:
The tissue ontside the outer most covering of the stem or root is called bark.

Question 63.
Name the tissue which transport food in plants.
Answer:
Phloem.

Question 64.
What is xylem?
Answer:
Xylem is a complex tissue which transport water and minerals from roots to other parts of the plant.

Question 65.
Name two substances which enter the root through the root hairs.
Answer:

  1. Water
  2. Soluble minerals from soil.

Question 66.
What is ascent of sap?
Answer:
The process by which water absorbed by the root is carried to aerial parts of plant is called ascent of sap.

Question 67.
Name the transport tissue of body.
Answer:
Blood and lymph.

Question 68.
Name the three types of blood vessels.
Answer:

  1. Arteries
  2. Veins
  3. Capillaries.

Question 69.
List types of circulation in human body.
Answer:
Double circulation involving pulmonary circulation and systemic circulation.

Question 70.
Name the fluid medium of blood.
Answer:
Plasma.

Question 71.
Why is the S-A node called pace-maker of the heart?
Answer:
S-A node, being self-excitatory, initiates a wave of contraction in the heart.

Question 72.
Name the organs which play role in circulation of blood.
Answer:
Heart. It is a muscular, pumping organ. It pumps the blood in the body.

Question 73.
Name the types of cells which destroy harmful bacteria in the body.
Answer:
White blood corpuscles (W.B.C.)

Question 74.
Name the instrument used to measure blood pressure.
Answer:
Sphigmomanometer.

Question 75.
Name the artery which carry impure (deoxygenated) blood from heart to lungs.
Answer:
Pulmonary artery.

Question 76.
Expand ECG.
Answer:
Electrocardiogram.

Question 77.
Write normal blood pressure in human body.
Answer:
120/80 → Systolic = 120 and Diastolic = 80

Question 78.
Name the largest artery.
Answer:
Aorta.

Question 79.
What is excretion?
Answer:
Excretion. Elimination of nitrogenous waste materials from body is called excretion.

Question 80.
How many nephrons are present in each kidney?
Answer:
About 10 lakhs.

Question 81.
What are the excretory structures of amoeba?
Answer:
Contractile vacuole.

Question 82.
What is a malpighian body (renal corpuscle)?
Answer:
Bowman’s capsule and glomerulus.

Question 83.
Name the reservoir of urine in the body.
Answer:
Urinary bladder.

Question 84.
What is micturition?
Answer:
Act of passing out of urine from urinary bladder is called micturition.

Question 85.
Name the structures which store wastes in plants.
Answer:
Central vacuoles.

Question 86.
What are resins and gums?
Answer:
These are storage wastes of plants.

Question 87.
Define autotrophic nutrition.
Answer:
The organisms prepare their own food from raw materials like C02 and H20 in the presence of sunlight. It takes place in green plants containing chlorophyll.

Question 88.
Define heterotrophic nutrition.
Answer:
The mode of taking readymade organic food material is called heterotrophic nutrition. It may be holozoic (ingestive) or saprophytic and parasitic (absorptive).

Multiple Choice Questions

Question 1.
Among the following which is a parasitic plant?
(A) Plasmodium
(B) Cuscuta
(C) Amoeba
(D) Rhizobium.
Answer:
(B) Cuscuta

Question 2.
Dark reaction and light reaction of photosynthesis takes place is:
(A) stroma and grana of chioroplast respectively
(B) grana and stroma of chioroplast respectively
(C) grana only
(D) stroma only.
Answer:
(A)stroma and grana of chioroplast respectively

PSEB 10th Class Science Important Questions Chapter 6 Life Processes

Question 3.
Chemical reaction takes place during dark reaction of photosynthesis is:
(A) photo1ysi
(B) hydrolysis
(C) carbon dioxide is bonded with RUBP
(D) nitrogen fixation.
Answer:
(C) carbon dioxide is bonded with RUBP

Question 4.
Plants are green in colour because:
(A) they absorb green light only
(B) they reflect green light
(C) they absorb green light but reflect all other lights
(D) none of the above are correct.
Answer:
(B) they reflect green light.

Question 5.
The nutrition in Mucor is:
(A) parasitic
(B) autotrophic
(C) saprophytic
(D) holozoic.
Answer:
(C) saprophytic

Question 6.
In amoeba the digestion is intracellular because :
(A) amoeba is unicellular
(B) amoeba is multicellular ‘
(C) amoeba is found in pond
(D) amoeba is microscopic animal.
Answer:
(A) amoeba is unicellular

Question 7.
Which of the following has no digestive enzyme?
(A) Saliva
(B) Bile
(C) Gastric juice
(D) Intestinal juice.
Answer:
(B) Bile

Question 8.
C02 acceptor during dark reaction of photosynthesis is :
(A) RUBP
(B) PEP
(C) NADPH
(D) ATP.
Answer:
(A) RUBP

Fill in the blanks:

Question 1.
Viruses show _________ movements.
Answer:
Molecular.

Question 2.
Growth, _________ and repair and _________ are characteristics of life.
Answer:
movements, maintenance.

Question 3.
_________ is needed by living organisms for movements and maintenance.
Answer:
Energy.

PSEB 10th Class Science Important Questions Chapter 6 Life Processes

Question 4.
_________ is the largest gland of body.
Answer:
Liver

Question 5.
_________ and are the raw materials for photosynthesis.
Answer:
C02 and H20.

Question 6.
RBC transport _________ in the body.
Answer:
Oxygen.

Question 7.
Xylem and _________ are the main conducting tissues in plants.
Answer:
Phloem.

Question 8.
Translocation of food takes place through _________ of phloem.
Answer:
Sieve tubes.

PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements

Punjab State Board PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements Important Questions and Answers.

PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements

Long Answer Type Questions

Question 1.
What do you understand by Dobereiner’s triads? Give some examples to support it.
Answer:
In 1817, a German Chemist, Dobereiner, gave a classification in which the similar elements were arranged in groups of three called Triads. The arrangement was such that the atomic mass of the middle element was almost the average of the atomic masses of the first and third elements. For example, if A, B and C are the elements present in the triad, then

Atomic mass of B =\(\frac{\text { Atomic mass of } \mathrm{A}+\text { Atomic mass of } \mathrm{C}}{2}\)

Examples of triads.
A few examples of triads are listed as given ahead :
1. Lithium, Sodium. Potassium
Atomic mass of Lithium (Li) = 7
Atomic mass of Potassium (K) = 39
Atomic mass of Sodium (Na) = \(\frac{39+7}{2}\) = 23
Actual atomic mass of sodium = 23

This group of triads is called Alkali Metal Group. All the elements present are metals, have valency equal to one (1) and dissolve in water to form soluble hydroxides called alkalis.

2. Calcium, Strontium, Barium
Atomic mass of Calcium (Ca) = 40
Atomic mass of Barium (Ba) = 137
Atomic mass of Strontium (Sr) = \(\frac{137+40}{2}=\frac{177}{2}\) = 88.5
Actual atomic mass of strontium = 88

This group of triads is called Alkaline Earth Metals Group. All the elements present are metals, have valency equal to two (2) and their oxides dissolve in water to form hydroxides which are alkaline in nature.

Question 2.
(a) What was Dobereiner’s basis of classifying elements?
Answer:
Dobereiner gave a classification in which the elements were arranged in a group of three elements called triads. The arrangement was such that the atomic masses of the middle elements were almost the average of the atomic masses of the first and third elements.

(b) What is the basis of classification of elements according to Mendeleev?
Answer:
The major contribution in the classification of the elements is by Mendeleev, a Russian chemist. He studied in detail the properties of the elements and made a very important observation. He stated that when elements are arranged in order of increasing atomic masses, the elements with similar properties recur after a definite gap. He based his classification of law called Mendeleev’s Periodic Law. The law may be stated as:

Physical and chemical properties of the elements are periodic function of their atomic weights or atomic masses.

PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements

Question 3.
Give a brief discussion of the Mendeleev’s classification of the elements.
Answer:
Mendeleev’s classification of the elements is based upon the Mendeleev’s periodic law. The law helped him to develop a table called Mendeleev’s Periodic Table. The table has been divided into vertical columns which are called Groups and horizontal rows which are known as Periods.

These are briefly discussed as follows :
1. Groups: These are the vertical rows. There are in all eight groups. The elements present in first seven groups are called Normal Elements. The elements present in group VIII are called the Transition Elements. Each group (I to VII) has been further divided into sub-groups which are called A and B. The inert gas or noble gas elements (He, Ne, Ar, Kr, Xe) were not known at that time. Therefore, they were not shown in the table. All the elements placed in a group have the same valency. Ail the elements present in a sub-group have the similar properties. For example, group I-B includes element Cu (Copper), Ag (Silver) and Au (Gold). They have similar properties.

2. Periods: In the periodic table horizontal rows are called periods. There were in all six periods in the original periodic table. The seventh period was added later on and this is not shown in the periodic table. The properties of the elements present in a period change systematically. For example, in every period, the first element is a typical metal. As we move from left to right, the metallic character gradually decreases and non- metallic character increases. For example, in period 2, the first element Li (Lithium) is a metal while the last element F (Fluorine) is a non-metal. The Mendeleev’s Periodic Table is shown below :

Mendeleev’S Periodic Table
PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements 1

Question 4.
(a) Why do we classify elements?
Answer:
To simplify and systematize the study of known elements.

(b) What were the two criteria used by Mendeleev in creating his Periodic Table?
Answer:

  1. Mendeleev arranged the elements in order of increasing atomic masses.
  2. Mendeleev considered the compounds formed by the elements oxygen and hydrogen.

(c) Why did Mendeleev leave some gaps in his Periodic Table?
Answer:
For the discovery of new elements.

(d) In Mendeleev’s Periodic Table, why was there no mention of Noble gases like7 Helium, Neon and Argon?
Answer:
Because noble gases were not known at that time.

(e) Would you place the two isotopes of chlorine, Cl-35 and Cl-37 in different slots because of their different atomic masses or in the same slot because their chemical properties are the same? Justify your answer.
Answer:
In the same slot because Cl-35 and Cl-37 have same chemical properties.

Question 5.
How did Mendeleev’s Periodic Table help in the discovery of new elements?
Answer:
When Mendeleev gave the periodic table, only 63 elements were known. The classification was based on two major properties :

  • Elements are arranged in order of increasing atomic masses.
  • Elements present in a group have similar properties.

Many elements were not known at the time the periodic table was given. Therefore, Mendeleev left gaps for these elements in the periodic table. But the properties of these elements could be predicted. For example, let us suppose that Rb (Rubidium) a member of group IA was not known when Mendeleev framed the periodic table. But its properties could be predicted. It was expected to be a metal with valency equal to 1. It was expected to be soluble in water to form a soluble hydroxide RbOH which is an alkali. This helped in the discovery of the element. In this manner, a number of elements could be discovered.

PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements

Question 6.
Discuss some major merits of the Mendeleev’s Periodic Table.
Answer:
Merits of Mendeleev’s Periodic Table
Mendeleev’s periodic table was the first proper systematic classification of the elements. The important merits of the table are listed as follows :
1. Systematic study of elements. With the classification of elements into groups and periods, their study became quite systematic. For example, if the properties of one particular element in a group are known, those of the other elements could be predicted. Actually, elements placed in a group are expected to show similar characteristics.

2. Correction of wrong atomic masses. The periodic table helped in correcting the atomic masses of some of the elements because the elements were arranged in order of their increasing atomic masses.

3. Prediction of new elements. At the time Mendeleev gave the periodic table, only 63 elements were known. While arranging these elements in groups and periods, certain gaps were left. These gaps represented some undiscovered elements. But the properties of these unknown elements could be predicted from their positions in the respective groups. This helped, later on, to discover these elements.

Question 7.
On the basis of Mendeleev’s Periodic Table given following, answer the questions that follow the table :
PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements 2
(a) Name the element which is in
(i) 1st group and 3rd period
Answer:
Sodium

(ii) VII group and 2nd period.
Answer:
Fluorine

(b) Suggest the formula for the following :
(i) oxide of nitrogen
Answer:
N2O5

(ii) hydride of oxygen.
Answer:
H2O.

(c) In group VIII of the Periodic Table, why does cobalt with atomic mass 58.93 appear before nickel having atomic mass 58.71?
Answer:
Because the elements with similar properties could be grouped together.

(d) Besides gallium, which two other elements have since been discovered for which Mendeleev had left gaps in his Periodic Table?
Answer:
Scandium and Germanium.

(e) Using atomic masses of Li, Na and K, find the average atomic mass of Li and K and compare it with the atomic mass of Na. State the conclusion drawn from this activity.
Answer:
Average atomic mass of Li and K = \(\frac{6.939+39.102}{2}=\frac{46.04}{2}\) = 23.02
Atomic mass of Na = 22.99
Hence Atomic mass of Na = Average of atomic masses of Li and K = \(\frac{46.04}{2}\) = 23.02

Question 8.
Point out the major defects in Mendeleev’s Periodic Table.
Answer:
Defects in Mendeleev’s Periodic Table:
Mendeleev’s periodic table was quite helpful in the classification of the elements.

But it had certain defects also. These are discussed as follows :
1. Position of hydrogen. Hydrogen was placed at the top of group LA. It is a non-metal where all other elements included in the group are metals.

2. Position of isotopes. The periodic table is based on the basis of the atomic masses of the elements and the elements with different atomic masses must be given separate places in the table. If this is correct, all the isotopes of an element must be allotted separate positions. For example, there are three isotopes for hydrogen and they must be given three separate places in the table. But only one position for hydrogen has been given.

3. Wrong order of atomic masses of some elements. In the table, the elements are arranged in order of increasing atomic masses. This means that the element with higher atomic mass must be placed after the element with the lower atomic mass. But in the table, there are some anomalies. For example, Co (Cobalt) with atomic mass 58-9 should be placed after Ni (Nickel) with atomic mass 58-7. But it has been placed before nickel.

4. Elements with similar properties placed in different groups. In the periodic table, it has been found that the elements with similar properties are placed in different groups. For example, copper and mercury have many common properties. But copper has been placed in group I B and mercury in group II B.

5. No similarity in the elements placed in sub-groups. The elements present in different sub-groups of the same group are expected to have common properties. But these are quite different. For example, elements in group I A are very soft and reactive metals but elements in group IB are hard and less reactive in nature.

6. No explanation for the cause of periodicity. Mendeleev was not in a position to explain why the elements included in a group show similar properties.

Question 9.
Give a brief description of Long Form of Periodic Table.
Answer:
The Long form of periodic table has been formed by arranging the elements in order of increasing atomic numbers. It is based upon Modern periodic law which states that the properties of the elements are the periodic function of their atomic numbers. Just as in case of the Mendeleev’s table, this periodic table has also been divided into Periods and Groups.
(A) Groups
These are the vertical columns. In all, there are eighteen groups in the table. The details of the groups are as follows :

Long Form of Periodic Table of Modern Periodic Table
PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements 3

  • Group 1: The elements present in group I or 1 are called Alkali Metals.
  • Group 2: The elements which are present in group IIA or 2 are called Alkaline Earth Metals.
  • Groups 13 to Group 18: There are in all six groups. The Groups 13 to 16 are named after the first element present in the family. For example, Group IIIA or 13 is called Boron Family because first member is boron.
  • Group 17 consists of a family called Halogen Family.
  • The group 18 is also called zero group because the elements have zero valency. These elements are ail gases. They have very little tendency to take part in chemical combination. These are also called Noble Gases.

In the latest Long Form of Periodic Table, groups of Alkali metals and Alkaline erttH metals are given numbers 1 and 2. The transition of elements are numbered from its,2 The non-metals are included in groups 13 to 18.

Group No. Name of Family
Group 13 Boron Family
Group 14 Carbon Family
Group 15 Nitrogen Family
Group 16 Oxygen Family
Group 17 Halogens
Group 18 or zero group  Noble Gases
  • Group 3 to Group 12: There are in all ten groups. These are all metals and are called Transition elements. When we go down each group, the metallic character further increases.
  • Group 3: Group 3 also includes fourteen elements belonging to Lanthanide family. These are called Lanthanides because they start after Lanthanum (La) with Z = 57. These are present in the 6th period as shown in the table.
  • It also includes another fourteen elements called Actinides. These are present in 7th period. These are so called as they come after Actinium (Ac) with Z = 89. These are placed at the bottom of the table for convenience.

(B) Periods
Periods are the horizontal rows which are present in the Long Form of Periodic Table. Different periods have different number of elements and their atomic numbers are continuous. There are in all seven periods. The seventh period is still incomplete. The number of elements which are included in each period are given below :

Period No. of elements Name of the Period
1 2 Shortest Period
2 8 Short Period
3 8 Short Period
4 18 Long Period
5 18 Long Period
6 32 Longest Period
7 20 Incomplete Period

PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements

Question 10.
What is periodicity? What is the cause of periodicity?
Answer:
Periodicity may be defined as the repetition of the similar properties of the elements placed in a group and separated by definite gaps of atomic numbers (8, 8, 18, 18, 32).

Cause of Periodicity. The properties of the elements, particularly the chemical properties, are linked with number of electrons present in the outermost shell of their atoms which is also called Valence shell. Elements with similar valence shell electronic configurations are expected to have similar properties.

It may be noted that all the elements which are present in a group have the same number of electrons in the valence shells of their atoms. In other words, the same valence shell electronic arrangement gets repeated after definite gaps of atomic numbers (8, 8, 18, 18, 32). Therefore, the elements placed in a group show similar properties.

Example, Let us write the electronic distribution of the first four members of the alkali metals present in group I.
PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements 4
All the four elements have one electron each in the valence shell of their atoms. They have, therefore, similar properties.

Question 11.
(a) What were the two major shortcomings of Mendeleev’s periodic table? How have these been removed in the modern periodic table?
Answer:
The shortcomings of Mendeleev’s periodic table are :

  • Isotopes of an element find different positions in periodic table.
  • Some chemically similar elements have been separated and some dissimilar elements are placed together.

In these Modern Periodic Table
Modern Table

  • Isotopes of an element occupy the same position because they have same atomic number.
  • The similar elements are grouped together and dissimilar elements are separated.

(b) Two elements X and Y have atomic numbers 12 and 16 respectively. Write the electronic configuration for these elements. To which period of the modern periodic table do these two elements belong? What type of bond will be formed between them and why?
Answer:

Element Electronic Configuration Period
X12 2, 8, 2 3rd
Y16 2, 8, 6 3rd

They will form ionic bonds because two electrons are transferred from X to Y so that they get their octets complete :
PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements 5

Question 12.
Why is Long Form of Periodic Table regarded better than Mendeleev’s Periodic Table?
Or
How could Modern Periodic Table remove various anomalies of Mendeleev’s Periodic Table?
Answer:
Long Form of Periodic Table is regarded better than the Mendeleev’s periodic table due to the following reasons:

  • It is based upon atomic number which is considered better than the atomic mass because the properties of the elements are related to the atomic number.
  • It explains why the elements placed in a group show similar properties but Mendeleev’s Periodic Table gives no explanation for the same.
  • All groups in the Periodic Table are independent groups and there are no sub¬groups as in Mendeleev’s Periodic Table.
  • Many defects in the Mendeleev’s Periodic Table have been removed.
  • There is no confusion regarding the position of isotopes because all the isotopes of an element have the same atomic number.
  • The periodic table is more systematic than the Mendeleev’s table and is easy to remember.

Question 13.
The following tables shows the position of six elements A, B, C, D, E and F in the periodic table.
PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements 6
Using the above table answer the following questions :
(a) Which element will form only covalent compounds?
Answer:
E

(b) Which element is a metal with valency 2?
Answer:
D

(c) Which element is non-metal with valency of 3?
Answer:
B

(d) Out of D and E, which one has a bigger atomic radius and why?
Answer:
D

(e) Write a common name for the family of elements C and F,
Answer:
The noble gases.

Question 14.
The question refers to the elements of the periodic table with atomic number from 3 to 18.
PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements 7
(a) Which of these :
(i) is/are noble gas?
Answer:
H, P

(ii) is it a halogen?
Answer:
G, O

(iii) is an alkali metal?
Answer:
A, I

(iv) is it a metal with valency 2?
Answer:
A, I

(b) Write the electronic arrangement of G.
Answer:
G has the electronic configuration = 2, 7

(c) If A combines with F, what would be the formula of resulting compound?
Answer:

Short Answer Type Questions

Question 1.
Atomic number is considered to be a more appropriate parameter than atomic mass classification of elements in a periodic table. Why?
How does atomic size of elements vary on moving from
(i) Left to right in a period.
(ii) from top to bottom in a group.
Give reasons for your answers.
Answer:

  • This is because atomic number is a more fundamental property of an atom.
  • As we move from left to right along a period atomic radius decreases due to increase in effective nuclear charge.
  • As we move from top to bottom in a group, atomic radius increases. This is due to addition of new electronic shells.

PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements

Question 2.
Define periodic law. Why was it necessary to change the basis of classification from atomic masses to atomic numbers?
Answer:
Periodic Law: The properties of elements are a periodic function of their atomic numbers. It was necessary to change the basis of classification from atomic masses to atomic numbers because atomic number and not atomic mass is the fundamental property of an element.

Question 3.
What do you understand by the term periodicity? Do the properties of two elements placed in a group the same? Illustrate.
Answer:
The repetition of the similar properties of the elements placed in a group and separated by definite gaps of atomic numbers (8, 8, 18, 18, 32) is called periodicity. The elements placed in a group show similar properties, e.g. consider group 1 elements.

Element Symbol Electronic configuration
Lithium (3) Li – 2, 1
Sodium (11) Na – 2, 8, 1
Potassium (19) K – 2, 8, 8, 1
Rubidium (37) Rb – 2, 8, 18, 8, 1
Cesium (55) Cs – 2, 8, 18, 18, 1
Francium (87) Fr – 2, 8, 18, 32, 18, 8, 1

These elements show similar properties because they have similar outer electronic configurations.

Question 4.
What was wrong with Dobereiner’s classification of elements?
Answer:
Dobereiner classified the elements in group of three in such a way that the atomic mass of the middle element was the mean of the first and the third elements. But he could not find many triads of elements. Therefore, the classification was rejected.

Question 5.
What properties do ail elements in the same column of the periodic table as boron have in common?
Answer:
The elements of Boron family in the periodic table show

  • Tricovalency
  • Form trihalides
  • Form trioxides
  • React with halogens to form halides.

Question 6.
Indicate the atomic number of elements of period 3 of Modern periodic table :
(a) non-metals
Answer:
14, 15, 16, 17

(b) elements forming negative ions.
Answer:
15, 16, 17

(c) elements with high melting points.
Answer:
11, 12, 13, 14

(d) elements forming positive ions.
Mention the atomic number only.
Answer:
11, 12, 13

Question 7.
Define atomic radius. Give its units.
Answer:
Atomic radius. It may be defined as the distance between the centre of nucleus and the outermost shell of an isolated atom.

Also the atomic radius of a non-metallic element is defined as half the distance between the nuclei of two atoms bound by a single covalent bond.
Units = Å or pm (picometre)
e.g. atomic radius of hydrogen atom = 37 pm.

Question 8.
How does atomic radius vary down a group and along a period?
Answer:
Variation in a group. The atomic radius generally increases from top to bottom in a group due to the addition of a new shell.
Variation along a period. The atomic radius decreases on moving from left to right due to the increase in nuclear charge.

Question 9.
Write down the electronic configuration of elements with atomic numbers 2, 14, 17, 19. Indicate the group of the periodic table to which they belong.
Answer:
The information is being given in a tabular form.

Atomic number Electronic configuration K L M N Group
2 2 18
14 2, 8, 4 14
17 2, 8, 7 17
19 2, 8, 8, 1 1

Question 10.
Locate the following group in the periodic table :
(a) Alkali metals
Answer:
Alkali metals: Group 1 or IA

(b) Halogens
Answer:
Halogens: Group 17 or VIIA

(c) Alkaline earth metals
Answer:
Alkaline earth metals: Group 2 or IIA

(d) Noble gases.
Answer:
Noble gases: Group 18 or zero

PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements

Question 11.
What properties do the elements in the same vertical column of the periodic table as fluorine have in common?
Answer:
These are :

  • They form diatomic molecules F2, Cl2, Br2 I2
  • They are non-metals.
  • They show a valency of one.

Question 12.
Write the chemical electronic configuration of nitrogen (N = 7) and phosphorous (P = 15).
Answer:
PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements 8

Question 13.
(i) Name the members of the alkaline earth family.
Answer:
The members of the alkaline earth family are :
Be, Mg, Ca, Sr, Ba, Ra.

(ii) To which group do they belong?
Answer:
They belong to group 2.

(iii) Which member is radioactive in nature?
Answer:
The radioactive member in them is Radium (Ra).

(iv) Which member is the least reactive?
Answer:
The least reactive member of the family is Beryllium (Be).

Question 14.
Why are the members of group 1 called alkali metals?
Answer:
The members of group 1 are called alkali metals because all of them are water soluble. They react with water to form soluble hydroxides. The soluble hydroxides of the metals are called alkalies.
PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements 9

Question 15.
An atom has the electronic configuration 2, 8, 7.
(a) What is the atomic number of this element?
Answer:
The atomic number of the element = Total number of electrons
= 2 + 8 + 7 = 17

(b) To which of the following elements would it be chemically similar? (atomic numbers are given) N (7), F (9), P (15), Ar (18).
Answer:
The electronic configurations of given atom and N, F, P and Ar are.

Atom Electronic configuration
K L M
2 8 7
N 2 5
F 2 7
P 2 8 5
Ar 2 8 8

Since F has same number of electrons in the outermost shell as the given atom. Hence, the given atom is chemically similar to F.

Question 16.
What physical and chemical properties of elements were used by Mendeleev in creating his periodic table? List two observations which posed a challenge to Mendeleev’s Periodic Law.
Answer:
He selected the compounds of the elements with oxygen and hydrogen. He gave a table based upon atomic weights of the elements.

Limitations of Mendeleev’s Classification

  • The position of hydrogen was uncertain.
  • The isotopes of elements were not given proper positions in the periodic table.

Question 17.
(a) What are amphoteric oxides? Choose the amphoteric oxides from amongst the following oxides :
Na2O, ZnO, Al2O3, CO2, HaO
Answer:
The oxides which can react both with acids as well as bases to produce salts and water.

(b) Why is it that non-metals do not displace hydrogen from dilute acids?
Answer:
This is because non-metals cannot lose electrons and cannot reduce H+ ions from acid to H2.

PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements

Question 18.
What are noble gas elements? Why are they so called?
Answer:
Noble gas elements are the. elemehts present in group 18 of the periodic table which is also called zero group, It means that the valency of the elements is zero. Actually, whereas the first member helium has two electrons in its only shell, the atoms of the remaining elements (Neon, Argon, Krypton, Xenon and Radon) have eight electrons in their outermost shells. They do not have any tendency to combine with atoms of other elements. Hence, they show zero valency. These are also called noble gases because they do not take part in chemical combination.

Question 19.
How is metallic character of an element defined? How does the metallic character of the elements change in a group?
Answer:
The metallic character of an element may be expressed in terms of its tendency to lose electrons and to form positive ion.
M (Element) → M+ + e

In a group the metallic character increases’ downwards. For example, among the elements of group 2, Beryllium (Be) is the least metallic. At the same time, radium (Ra) which is the last element is maximum metallic in nature.

Question 20.
Why do the elements present in a group show similar chemical properties?
Answer:
The properties of the elements, particularly the chemical properties are related to valence shell electronic distribution. The elements with the same valence shell electronic distribution have the similar chemical properties. For example, the members, of alkaline earth metal family (Group 2) have two electrons in the valence shell of their atoms. They therefore, show similar chemical properties.

Question 21.
How does the reactivity of the metals vary in a group?
Answer:
In a group, containing metals, the reactivity increases down the group. For example, in the metals of group 1 (Alkali metals), Lithium reacts with water very slowly. Sodium is more reactive and potassium is. still more reactive than sodium.

Question 22.
Name the elements present in the second period. Give their electronic configuration.
Answer:
The second period of the Long Form of Periodic Table has eight elements. The first element is Lithium and the last element is Neon. The electronic configuration of the elements are given below:
PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements 10

Question 23.
Why do not the elements present in a period show same valency?
Answer:
The valency of the element is related to the number of electrons in the outermost energy shell of its atom. Since the elements present in a period have different number of valence electrons or outermost electrons, they show different valencies.

Question 24.
The metallic character of the elements in a period decreases from left to the right. Justify.
Answer:
In every period, as we move from left to right, the metallic character of the elements decreases gradually. This is shown with the elements present in the third period.
PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements 11

Question 25.
Give symbols for :
(a) a metal belonging to second group of the periodic table.
Answer:
The metal belonging to second group is calcium (Ca).

(b) a metal belonging to the third group of the periodic table.
Answer:
The metal belonging to third group is aluminium (Al).

(c) two non-metals belonging to the halogen family.
Answer:
The two non-metals of halogen family are fluorine (F) and chlorine (Cl).

Question 26.
Write electronic structures of:
(i) Potassium
(ii) Lithium
(iii) Fluorine.
Answer:
The electronic configurations of the atoms are given below :
PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements 12

Question 27.
Name two other elements which are in the same family as
(i) carbon
Answer:
Carbon belongs to the group 16. Two other elements are silicon (Si) and germanium (Ge).

(ii) fluorine
Answer:
Fluorine belongs to group 17. Two other elements are chlorine (Cl) and bromine (Br).

(iii) sodium.
Answer:
Sodium belongs to group 1. Two other elements are lithium (Li) and potassium (K).

PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements

Question 28.
Carbon (atomic number 6) and silicon (atomic number 14) are elements in the same group of the periodic table. Give the electronic arrangements of the carbon and silicon atoms and state the groups in which these elements occur.
Answer:
The required information may be given in a tabular form as follows :

Element Atomic no. Electronic arrangement Group
Carbon (C) 6 2, 4 14
Silicon (Si) 14 2, 8, 4 14

Question 29.
Sodium and aluminium have atomic numbers of 11 and 13 respectively. They are separated by one element in the periodic table and have valencies of 1 and 3 respectively. Chlorine and potassium are also separated by one element in the periodic table (their atomic numbers are 17 and 19 respectively) and yet both have valency of one. Explain your answer.
Answer:
Sodium and aluminium: The electronic configurations of the elements are given below :
Sodium (Na) = 2, 8, 1
Aluminium (Al) = 2, 8, 3

The valency in this case is given by the number of valence electrons. Therefore, the valency of sodium is 1 and that of aluminium is 3.

Chlorine and potassium: The electronic configurations of the elements are given below :
Chlorine (Cl) = 2, 8, 7
Potassium (K) = 2, 8, 1

Question 30.
Give the atomic number and electronic distribution of:
(i) The third alkali metal
Answer:
Potassium (19) 2, 8, 8, 1

(ii) The second alkaline earth metal
Answer:
Magnesium (12) 2, 8, 2

(iii) The first halogen
Answer:
Fluorine (9) 2, 7

(iv) The second noble gas.
Answer:
Neon (10) 2, 8.

Question 31.
Observe the following elements in the Modern Periodic Table.
PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements 13
Name the elements A, B, C and D. Also indicate noble gas
Answer:
(A) Lithium
(B) Chlorine
(C) Neon
(D) Potassium C is a noble gas

Question 32.
Match the following :

(a) Fluorine (i) Metalloid
(b) Neon (ii) Halogen
(c) Sodium (iii) Noble gas
(d) Arsenic (iv) Alkali metal

Answer:

(a) Fluorine (ii) Halogen
(b) Neon (iii) Noble gas
(c) Sodium (iv) Alkali metal
(d) Arsenic (i) Metalloid

PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements

Question 33.
How many electrons can be present in the valence shells of metal atoms and non-metal atoms?
Answer:
Metal atoms have 1, 2 or 3 electrons in their valence shells whereas non-metal atoms have 4 to 7 electrons in their valence shells.

Question 34.
How are the various groups of the Modern Periodic Table designated according to the IUPAC system and old system?
Answer:
The designations of various groups of the Modern periodic table are :
PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements 14

Question 35.
What are the uses of Modern Periodic Table?
Answer:

  • Systematic study of the elements. In the periodic table, the elements with similar properties are placed together in the same group. If we know the properties of one element of the group, the properties of other elements belonging to the same group can be predicted. Thus, there is no need to study the properties of all the elements.
  • Properties of an element can be predicted from the position of the element in the periodic table. For example, if the element belongs to group IA or IIA it is likely to be a reactive metal, and if it belongs to group VII A it is likely to be a reactive non-metal.
  • It has led to the discovery of many new elements.

Very Short Answer Type Questions

Question 1.
How many elements have been discovered so far?
Answer:
114.

Question 2.
How are elements classified?
Answer:
The elements have been classified on the basis of their properties.

Question 3.
Why are group IA elements called alkali metals?
Answer:
This is because all these elements are metals and their oxides and hydroxides give alkaline solutions in water.

Question 4.
What is the basis of Modern Periodic table?
Answer:
It is based upon Modern Periodic law.

PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements

Question 5.
Name the family to which halogens belongs?
Answer:
Halogen family.

Question 6.
Name the second elements of group 14.
Answer:
Silicon.

Question 7.
How many valence electrons and present in halogen elements?
Answer:
Seven.

Question 8.
How many elements are present in 4th period?
Answer:
18.

Question 9.
How many electrons are present in Mg2+ ion?
Answer:
Ten.

Question 10.
Out of Na and Mg which was larger size?
Answer:
Na

Question 11.
What is the valency of nitrogen?
Answer:
Three.

Question 12.
Out of Na and K which is more reactive?
Answer:
Sodium (Na).

Question 13.
Name the group number of halogen family.
Answer:
Group 17 (or VII-A).

Question 14.
Name the last element of third period?
Answer:
Argon.

PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements

Question 15.
What is Dobereiner’s Triad?
Answer:
A group of three elements having similar properties is called Dobereiner’s Triad.

Question 16.
A, B and constitute the Dobereiner s Traid. Atomic mass of A and C are 7 and 23 respectively. Calculate atomic mass of B.
Answer:
\(\frac{7+23}{2}=\frac{30}{2}\) = 15

Question 17.
Name the elements discovered after Mendeleev’s Periodic Table?
Answer:
Scandium (Se), Gallium (Ga), Germanium (Ge) etc.

Question 18.
Name the groups and periods in Mendeleev’s Periodic Table?
Answer:
8 groups and 7 periods.

Question 19.
How does atomic radii as we move from left along a period in the Periodic Table?
Answer:
It decreases.

Question 20.
An element has the electronic configuration 2, 8, 3. What is its group number in Modern Periodic Table?
Answer:
Group 13.

Question 21.
Give the basis of Dobereiner’s classification.
Answer:
Doberenier Triads.

Question 22.
Give the characteristics of Dobereiner’s Triads.
Answer:
The atomic mass of the central element is the average of masses of the other two elements.

Question 23.
What is the drawback of Doberenier’s Triads?
Answer:
All the known elements could be grouped into Triads.

Question 24.
There are three alements A, B and C. The atomic masses of A and C are 7 and 39. What is the atomic mass of B on the bais of Dobereiner’s Traids?
Answer:
Atomic mass of B = \(\frac{7+39}{2}\) = 23.

Question 25.
There are three elements X, Y, Z. Atomic masses of X and Z are 35.5 and 127. What will be atomic mass of Y on the basis of Dobereiner’s Traids?
Answer:
Atomic mass of Y
\(\frac{35.5+127}{2}=\frac{162.5}{2}\) = 81.25

Question 26.
Write Newland’s Law of Octaves for classification of elements.
Answer:
When the elements are arranged in order of increasing atomic masses, eighth elements has properties similar to the first element.

Question 27.
How many element were classified by Newland.
Answer:
Upto mass number 40.

Question 28.
Indicate the group number and period number of P in the modern Periodic Table.
Answer:
Group-15, Period-3.

Question 29.
An element has the electronic configuration 2, 8, 8, 2. Indicate its group and Period in the Modern Periodic Table.
Answer:
Group-12, Period-4.

Question 30.
An element M is in the group 13 of Modern Periodic Table write the formula of its oxide.
Answer:
M2O3.

Question 31.
Give the groups and periods in the Modern Periodic Table.
Answer:
Group-18, Periods-7.

Question 32.
Give the electronic configuration of 17Cl35. Also indicate its position in the Periodic Table.
Answer:
Electronic configuation of 17Cl35 = 2, 8, 7
Period number = 3
Group number =17. (VII A)

Question 33.
Give the name and electronic configuation of element with atomic number 9.
Answer:
Fluorine and it has the electronic configuation = 2, 7.

Question 34.
What is Modern Periodic Law?
Answer:
It states that the properties of the elements are the periodic functions of their atomic numbers.

Question 35.
Who gave Newland’s Law of Octaves.
Answer:
Newland.

Question 36.
Define Mendeleev’s periodic law.
Answer:
It states that the properties of the elements are the periodic functions of their atomic masses.

Question 37.
What is the basis of Mendeleev’s Modern Table?
Answer:
It is based upon Mendeleev periodic law and in this table the elements are arranged in order of increasing atomic masses.

Question 38.
How many groups are present in Mendeleev’s Period Table.
Answer:
Eight.

Question 39.
Name the next elements after P in Modern Periodic Table.
Answer:
Sulphur (S).

Question 40.
Give the group number of nitrogen and phosphorus.
Answer:
They belong to group number 15.

Question 41.
Out of Mg and Al which is more metallic?
Answer:
Mg.

Question 42.
Out of Be, Mg, Ca, Al which does not belong to same group?
Answer:
Al.

Question 43.
In which group noble gases are present?
Answer:
Group 18 or group zero.

PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements

Question 44.
Na and S are present in the third period of Modern Periodic Table. Which is more metallic and why?
Answer:
Na is more metallic due to larger size than S.

Question 45.
What is metallic character?
Answer:
It is the tendency of an atom of the element to form positive ions by losing electrons.

Question 46.
What is the trend in metallic character on moving from left to right along a period?
Answer:
It decreases.

Multiple Choice Questions:

Question 1.
Who gave Law of octaves?
(A) Newland
(B) Dobereiner
(C) Mendeleef
(D) Lother Mayer.
Answer:
(A) Newland

Question 2.
In Mendeleef’s periodic table which element was discovered in the gap between Boron and Aluminium ?
(A) Na
(B) Ca
(C) Ga
(D) Ba.
Answer:
(C) Ga

Question 3.
According to Mendeleef’s periodic law, the elements are arranged in order of A increasing:
(A) Atomic numbers
(B) Decreasing atomic number
(C) Increasing atomic masses
(D) Decreasing atomic masses.
Answer:
(C) Increasing atomic masses

Question 4.
Which element occupied gap left in Mendeleers periodic table?
(A) Germanium
(B) chlorine
(C) Oxygen
(D) Silicon.
Answer:
(A) Germanium

Question 5.
An element has the electronic configuration 2, 8, 2. It is present in group:
(A) 2
(B) is
(C) 8
(D) 10.
Answer:
(A) 2

Question 6.
Which element shows metallic character?
(A) 2, 8, 2
(B) 2, 8, 4
(C) 2, 8, 8
(D) 2, 7.
Answer:
(A) 2, 8, 2

Question 7.
Which shell is largest shell?
(A) K
(B)L
(C) M
(D) N.
Answer:
(D) N

Fill in the blanks :

Question 1.
Out of Na and Mg, ______ has bigger size.
Answer:
Na.

Question 2.
Number of elements known in Mendeleefs periodic table were ______
Answer:
63.

Question 3.
Oxygen and sulphur belong to same ______
Answer:
group.

Question 4.
The elements of group 17 are called ______
Answer:
halogens.

PSEB 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements

Question 5.
The valency of the members of noble gas family is ______
Answer:
zero.

Question 6.
The halogens belong to group ______
Answer:
17.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Punjab State Board PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life Important Questions and Answers.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Long Answer Type Questions

Question 1.
Describe the structure of the plant cell.
Answer:
A generalized plant cell consists of three distinct parts viz, cell wall, cell membrane or plasma membrane and protoplasm.
1. Cell wall. It is the outermost covering of a cell made up of a non-living substance called cellulose. Is is permeable. It provides shape, strength and protection to cell.

2. Cell membrane or Plasma membrane. It is thin, delicate and elastic protoplasmic covering lying internal to the cell wall. The plasma membrane controls the entrance and exit of molecules and ions.

3. Cytoplasm. It consists of a number of living and non-living structures. The living structures are called cytoplasmic organelles or cell organelles and include structures like endoplasmic reticulum, ribosomes, golgi apparatus, mitochondria, lysosomes, plastids, vacuoles etc.
(a) Plastids. Plastids are spherical or discoid bodies containing different types of pigments in them. They are of three types viz. chloroplasts, chromoplasts and leucoplasts. The chloroplasts contain a green pigment in them, called chlorophyll. They play a vital role in the process of photosynthesis. The chromoplasts contain pigments other than chlorophyll. They impart various colours to flowers and fruits. Leucoplasts are colourless plastids. They are concerned with the storage of starch.
(b) Mitochondria are power house of cell.
(c) Ribosomes are of 70 S type. They are site for protein synthesis.
(d) Endoplasmic reticulum is network of membranes.
(e) Vacuoles. These are fluid-filled bubble-like structures bounded by a membrane, called tonoplast. They are mostly found in plant cell. Vacuoles provide turgidity and rigidity to cell.

4. Nucleus. It is a dense spherical body bounded by a membrane, called nuclear membrane. Embedded within the nuclear sap are found a number of thread-like structures called chromosomes with genes on them. The nucleus controls the various metabolic activities of the cell.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 2.
Draw a well labelled diagram of ultrastructure of plant cell.
Answer:
Ultrastructure of plant cell
PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life 1

Question 3.
Describe the structure of nucleus. Write its functions.
Answer:
A true nucleus is present in all the eukaryotic cells except mammalian RBCs, sieve tube cells, tracheids and vessels. It directs life processes of cells. It is formed of four components:
(a) Nuclear membrane. It is a two-layered envelope around the nuclear sap. It is porous (with nuclear pores) and semipermeable membrane. Outer membrane is studded with ribosomes and may be continuous with RER. It regulates exchange of materials between nucleoplasm and cytoplasm.
PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life 2
(b) Nucleoplasm. It is a semifluid colloidal substance in which nucleoli and chromatin fibres are present.
(c) Nucleolus. It is dense, spherical, naked and darkly stained structure and is the site of formation and storehouse of rRNAs.
(d) Ovomatin fibres. These are long, fine and darkly stained threads which collectively form nuclear reticulum. During prophase of mitosis and meiosis, these condense to form a species-specific number of rods, called chromosomes. These are with genes which are chemically formed of DNA and act as units of heredity and variations.
Functions of the nucleus:
The nucleus performs the following important functions:

  • The nucleus controls all the cellular activities of the cell.
  • Tire nucleus plays an important role in transmission of hereditary characters from parents to offsprings.
  • The nucleus plays a vital role in cell division.

Question 4.
Write short notes on:
(a) Chloroplast
(b) Endoplasmic reticulum
(c) Golgi bodies
Answer:
Structure and functions of a chloroplast:
A chloroplast is formed of two membranes and two chambers, so resembles a mitochon-drion. Outer membrane is freely permeable while inner membrane is semi permeable. Inner chamber is filled with a denser and granular proteinaceous (about 50%) ground substance called stroma or matrix. It is site of dark reaction or CO2 fixation of photosynthesis and has a number of grana, 70 S ribosomes and DNA molecules (circular and naked).
1. Grana are site for light reactions of photosynthesis.
2. Function of Chloroplasts: These are sites of photosynthesis, so are called “kitchens of cell.”
PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life 3
(b) Endoplasmic reticulum
There are two forms of endoplasmic reticulum. They are the following:

  • Rough type (RER): Ribosomes are associated with rough endoplasmic reticulum.
  • Smooth type (SER): Golgi apparatuses are associated with smooth endoplasmic reticulum.

Components of ER
PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life 4
Functions of Rough and Smooth Endoplasmic reticulum

Functions of Endoplasmic Reticulum:

  1. The endoplasmic reticulum acts as secretory, storage, circulatory and nervous system of the cell. Its functions depend upon its location, nature and cell type.
  2. It provides an ultrastructural skeletal framework of the cell.
  3. ER provide channels for quick transport of materials.
  4. It provides surface for attachment of ribosomes.
  5. It provides materials, lipids and proteins for biogenesis of membranes.

Functions of Smooth Endoplasmic Reticulum:

  1. Detoxification of drugs and poison.
  2. Formation of visual pigment from Vitamin A in retinal cells.
  3. Synthesis of fats inside the cells of adipose tissue.
  4. Synthesis of steroids and hormones.

(c) Golgi complex or Golgi Apparatus:
The golgi complex is usually located near the cell nucleus and in animal cells it is frequently disposed around the centriole pair. It normally consists of numerous sets of membrane-bound, smooth surface cisternae. Each cisterna is flat, disc-shaped and curved like a shallow bowl.
PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life 5
It has lumen which is bounded by a single membrane. Each set of flattened disc-shaped cistemae forms a structure that resembles stack of plates, called a Golgi stack or dictyosome (as referred in the plants).

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 5.
Discuss the structure and function of mitochondrion.
Answer:
Structure. Mitochondria are ‘power house’ of cells. It has two mitochondrial membranes and two chambers. Outer mitochondrial membrane is smooth and porous while inner mitochondrial membrane is produced into finger-like processes called cristae to increase the surface area for the distribution of respiratory enzymes of electron transport chain.

Cristae are studded with regularly placed knob-like oxysomes which take part in energy releasing reactions. Inner chamber is filled with a granular matrix having respiratory enzymes. Matrix present in inner chamber also contains DNA, RNA and ribosomes.
PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life 6
Functions of mitochondria:

  1. These are called power houses or power plants or ATP mills as these are site of aerobic oxidation (cellular respiration) of glucose to produce energy-rich ATP molecules.
  2. They provide biological intermediates for the synthesis of various biomolecules such as fatty acids, amino acids, steroids etc.

Question 6.
Write a note on prokaryotic cell.
Answer:

  1. The prokaryotes have a single membrane.
  2. Cells which have respiratory enzymes associated with plasma membrane.
  3. Circular DNA without proteins, forming a single prochromosome called nucleoid and lying in direct contact with cytoplasm.
  4. They have 70 S ribosomes.
  5. They lack membrane-bound organelles and sap vacuoles.
  6. They do not show cyclosis, phagocytosis, pinocytosis and exocytosis.
  7. Cell division by fission or budding.
  8. Examples. Bacteria, Blue green algae and Mycoplasma.

Question 7.
Write a short note on eukaryotic cell.
Answer:

  1. The eukaryotic cells have double membrane and plasma membrane devoid of respiratory enzymes.
  2. These cells have linear DNA associated with proteins forming two to many chromosomes enclosed by a nuclear envelope. Thus proper well defined nucleus is present.
  3. There are present 80 S ribosomes.
  4. Many membrane-bound organelles such as mitochondria, chloroplasts, lysosomes etc., and sap vacuoles present.
  5. Form mitotic apparatus in cell division and undergo meiosis.
  6. They may show cyclosis, phagocytosis, pinocytosis and exocytosis.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 8.
Draw the ultrastructure of a typical animal cell.
Answer:
An animal cell has three functional regions. Plasma membrane, cytoplasm and nucleus.
PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life 7

Short Answer Type Questions

Question 1.
What is cell and cell theory?
Answer:
Cell. 1. Structural and functional unit of life is called cell.
2. A cell is mass of protoplasm formed of different types of inorganic and organic biomolecules and externally bounded by a thin and semipermeable plasma membrane.
Cell Theory. M.J. Schleiden and Theodore Schwann (1938-39) proposed cell theory. The cell theory can be stated as follows:
(a) The body of all living organisms is composed of cells. Cell is a unit of structure and function of life.
(b) New cells are formed from the pre-existing cells only (R. Virchow).
(c) The life is passed from one generation to the next generation through a single cell.

Question 2.
What is a prokaryotic cell?
Answer:
Prokaryotic cell, (pro = primitive, karyon = nucleus) It is characterized by an incipient nucleus called nucleoid formed of single and circular chromosome. It is formed of double-stranded DNA but not associated with histone proteins. It is not enclosed by nuclear membrane.

  • Membrane-bound cell organelles such as mitochondria, chloroplasts etc. are absent.
  • Examples: Bacteria, Blue-green algae, Mycoplasma etc.

Question 3.
What is a eukaryotic cell?
Answer:
A eukaryotic (eu = good; karyon = nucleus) cell has true nucleus which has two or more linear chromosomes lying in the nucleoplasm and is surrounded by a double and porous nuclear membrane. Membrane bound organelles present. 80 S Ribosomes present (S = Svedberg unit).

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 4.
Explain structural and functional organisation of unicellular and multicellular organism.
Answer:

  • Unicellular organism, in such organisms, a single cell carries out all life processes with the help of organelles. They are also called acellular organisms.
  • Multicellular organism. A multicellular organism is made up of large number of cells.
  • The cells are differentiated to share varied functions performed by unicellular organisms. Whatever the case may be, all multicellular organisms start their life from a single cell.
  • The multicellular organism represents itself in two forms – as an individual and as a member of assembly of cells.
  • Advantages of Multicellularity:
    • The division of labour in a multicellular organism increases the efficiency of the organism.
    • It promotes the chance of survival.

Question 5.
What do you mean by:
1. Organelle
2. Inclusions
Answer:
1. Organelle, it is a living structure of cell with a specific function.
2. Inclusions. The lifeless structures in a cell, which may or may not have a definite function are called inclusions. These are vacuoles, excretory products, starch grains etc.

Question 6.
Distinguish between cell wall and cell membrane.
Answer:
Differences between cell wall and cell membrane

Cell wall Cell membrane
1. It is present only in plant cells where it lies outside the cell membrane.

2. Dead in nature and permeable.

3. Composed of cellulose.

4. It is thick in nature.

1. It ocurs in animal cells and plant cells.

2. Living membrane and is semi-permeable.

3. Composed of lipids and proteins.

4. Comparatively very thin.

Question 7.
Name the two nucleic acids present in cell.
Answer:

  • DNA (Deoxyribose nucleic acid): It is the genetic material. It is passed on from parents to next generation, It regulates the functioning of cells.
  • RNA (Ribose nucleic acid): It is structural component of ribosome. It plays role in protein systhesis.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 8.
Why are mitochondria called “power house of cell”?
Answer:
Mitochondria are site of generation, transport and storage of ATP. Adenosine triphosphate (ATP) is the energy currency of the cell.

Question 9.
Explain lysosomes.
Answer:

  • Lysosomes. They are single membrane bound vescicular structures. They are present in cytoplasm of eukaryotic cells. They contain digestive enzymes.
  • They exist in four different forms thus exhibit polymorphism.
  • They are involved in autolysis and hence called as ‘suicidal bags.’

Question 10.
List the functions of vacuoles.
Answer:
Functions of vacuoles

  1. They act as storage sacs for solid or liquid contents of cell.
  2. They help the cell to remain turgid and rigid.
  3. Vacuoles play an important role in growth.
  4. In some freshwater unicellular organisms, specialised vacuoles play an important role in expelling excess water and some wastes from, cells. It is called osmoregulation.

Question 11.
What is centrosome? List two functions of centrosome.
Answer:
Centrosome: It consists of granule like two centrioles; surrounded by clear area of cytoplasm called centrosphere. It is present in animal cells only.
Functions of centrosome:

  • They help in spindle formation during cell divison in animal cell.
  • They act as basal bodies and give rise to motile structures, i.e. cilia and flagella.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 12.
Differentiate smooth endoplasmic reticulum (SLR) and rough endoplasmic reticulum (RER).
Answer:
Differences between SER and RER:

SER RER
1. It does not bear ribosomes over the surface of its membranes which are smooth.

2. It is engaged in the synthesis of glycogen, lipids and steroids.

3. It gives rise to sphaerosomes.

1. It possesses ribosomes attached to its membranes which are rough.

2. The reticulum takes part in the synthesis of proteins and enzymes.

3. It helps in the formation of lysosomes through golgi apparatus.

Question 13.
What are ribosomes? Write their kinds and chemical composition.
Answer:
Structure of Ribosome. The ribosomes are ribonucleoproteinic, obovate, spheroid without membrane structures. They are having two sub-units, one is larger sub-unit having dome-shaped structure and the other smaller sub-unit forming a cap-like structure. They are composed of ribosomal RNA (rRNA) and proteins.
Type of ribosomes:
1. 70 S ribosomes. They are present in both prokaryotic and eukaryotic cell.
2. 80 S ribosomes. They are present in eukaryotic cell.
PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life 8
Functions of ribosomes:
They are the site of protein synthesis as they contain enzymes involved in protein synthesis.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 14.
Write functions of cell wall.
Answer:
Functions of cell wall:

  1. It provides shape and rigidity to the cell
  2. It protects the protoplast.
  3. It permits cells of bacteria, fungi and plants to withstand very dilute external environment without bursting.
  4. It increases the size of cell by continuous absorption.
  5. It is involved in the movement of materials in and out of the cell.

Question 15.
Give the functions of plasma membrane.
Answer:
Functions of plasma membrane:

  1. It maintains individuality of cells.
  2. It provides shape to the cell.
  3. It keeps the cell contents in place and distinct from the environmental materials.
  4. It protects the cell from injury.
  5. It regulates the flow of material into and out of the cell. It allows only the selected substances to move across it.
  6. It forms organelles of cell.

Question 16.
Define the following:

  1. Cytoplasm
  2. Hyaloplasm
  3. Nucleoplasm.

Answer:

  1. Cytoplasm: The clear, translucent, colourless, viscous, fluid in which are present various cell inclusions and organelles is called cytoplasm.
  2. Hyaloplasm: The clear homogeneous ground substance of cytoplasm is termed hyaloplasm (cytoplasmic matrix).
  3. Nucleoplasm: Clear transparent homogeneous fluid inside the nucleus is called ucleoplasm.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 17.
State the differences between cilia and flagella.
Answer:
Differences between cilia and flagella:

Cilia Flagella
1. They are short hair-like structures with an average length of 5-10 mm.

2. They are present all over the surface of cell of the organism.

3. They are more in number usually 100 to a few thousand per cell.

4. They beat perpendicularly and simultaneously.

5. The cilia produce a sweeping or pendular stroke.

1. They are long, whip-like structures with an average length of 150 mm.

2. They are usually present on one end of cell.

3. They are less in number usually 1 or 2 per cell.

4. They beat freely.

5. The flagella produce undulatory motion.

Question 18.
State the differences between grana and stroma.
Answer:
Differences between grana and stroma:

Grana Stroma
1. It consists of flatfened sac-like structures which are piled up one above the other like stacks of coins.

2. They contain photosynthetic pigments on them.

3. They are made of lipoproteins.

1. It is matrix or ground substance of chloroplast in which different structures like grana and lamellae are embedded.

2. Pigments are absent.

3. It is composed of watery proteinaceous substance.

Question 19.
Write functions of cell wall.
Answer:
Function of cell wall:

  1. It protects the cell from mechanical damage and infection.
  2. It helps in cell-to-cell interaction.
  3. It acts as a barrier to unwanted molecules.
  4. It provides rigidity and mechanical support to the cells.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 20.
How will you examine cells of onion peel.
Answer:
Examination of Onion Feel

  1. Take out the onion peel from the inner side of the scale of the bulb.
  2. Put this transparent peel into water and cut a small piece.
  3. Stain it with safranin solution. It will make the nucleus more distinct.
  4. Put it on the slide and place it in glycerine.
  5. Observe under low power microscope, then change it to high power and study the following:

Observation:

  • Regular, rectangular-shaped cells are placed close to each other.
  • Cell has a prominent cell wall, cytoplasm and nucleus. The vacuole occupies the central position.

Question 21.
What are possible shapes of cells? Explain with simple sketches.
Answer:
Shapes of cells

  1. Shapes of the cells differ not only in different organisms but also in different parts of same organism. They may be oval, spherical, discoid, cuboidal, columnar, spindle-shaped etc.
  2. The shape of the cell is usually related with its function.
  3. Some other factors controlling the shape of cell are function, age, pressure, cell wall and internal or external skeleton.
  4. Amoeba and leucocytes go on changing their shapes to enable them to engulf the materials.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life 9

Question 22.
Differentiate between diffusion and osmosis
Answer:
Differences between diffusion and osmosis:

Diffusion Osmosis
1. Diffusion is the net downward movement of a given substance (ions, atoms or molecules, etc.) along the free energy gradient from the place of its higher concentration to an area of its lesser concentration.

2. The diffusion may occur in any medium and the diffusing particles may be solid, liquid or gas.

3. Presence of semipermeable membrane is not required.

Osmosis is a special type of diffusion of solvent molecules from low concentration of solution to higher concentration of solution when the two are separated by a semipermeable membrane.

2. The osmosis occurs in liquid medium and only the solvent molecules move from one place to another.

3. Presence of semi-permeable membrane in between the two solutions is required.

Question 23.
Which organelle is called as protein factory of the cell?
Answer:
Ribosomes provide space for the synthesis of proteins in the cell. Hence, they are called protein factories. It synthesises various proteins and enzymes for the cell. Haemoglobin is an example of a protein made by free ribosomes in the young red blood corpuscles.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 24.
Differentiate Ribosome and Lysosome.
Answer:
Differences between ribosome and lysosome:

Ribosome Lysosome
1. Ribosome is an oval to rounded granular structure which is not covered ever by any membrane.

2. It consists of two unequal nucieoprotein subunits.

3. It takes part in protein synthesis.

1. Lysosome is a rounded membrane lined vesicle containing number of hydrolytic enzymes.

2. It does not consists of nucieoprotein.

3. It takes part in digestion, scavenging and defence against pathogens.

Question 25.
What are plastids? Name its three types.
Answer:
Plastids. These are double membrane bound organelles found in plant cell. These are of three types:

  • Chromoplasts
  • Chloroplasts
  • Leucoplasts.

Question 26.
Write functions of plastids.
Answer:
Functions of plastids

  1. Chloroplasts are the site of photosynthesis so called as kitchen of the cell.
  2. Chromoplasts provide colour to flowers and fruits.
  3. Leucoplasts are site of storage of food.
  4. Granum is site of light reaction, while stroma is the site of dark reaction.

Question 27.
What are the functions of lysosomes?
Answer:
Functions of lysosomes:

  1. Take part in intracellular digestion of foreign particle.
  2. They provide energy during starvation.
  3. They help in defence against bacterial and viral infection.
  4. Lysosome removes cellular debris.

Question 28.
Name the various organelles of cell and mention the most important function of each.
Answer:
Names and functions of cellular organelles:

Name of cell organelle Function
1. Plasma membrane Protects the cell organelles and separates the contents of cell from external environment.
2. Cell wall (plant cell only) Provides strength and rigidity to the cell. It permits the bacterial and plant cells to withstand very dilute external medium without bursting.
3. Cytoplasm Contains machinery and metabolites for carrying out the instructions sent from nucleus.
4. Cytoplasmic matrix Contains raw materials and enzymes for metabolism.
5. Endoplasmic reticulum It is responsible for transportation of extracellular and intracellular chemical molecules. It is also the site for synthesis of various molecules.
6. Ribosomes These are sites for protein synthesis.
7. Golgi complex Complexing, packaging and dispatching various materials to various targets inside and outside the cells.
8. Mitochondria

 

Site for respiratory reactions and energy liberation in the form of ATP.

 

9. Lysosomes (animal cells only) Centrosome (animal cells only) Plastids (plants only).
Cilia and flagella.
Vacuoles.
Nucleus.
Helps in waste disposal system and autolysis. Forms spindle during cell division.
Responsible for photosynthesis.
Provide movement to cell.
Storage sacs for liquid and solid contents. Regulates growth and reproduction of cell.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 29.
Differentiate cytoplasm and nucleoplasm.
Answer:
Differences between cytoplasm and nucleoplasm

Cytoplasm Nucleoplasm
1. Part of protoplasm that lies outside 1. Part of protoplasm that lies inside the
2. It appears to be homogeneous semi-fluid jelly-like substance 2. It is transparent.
3. It contains number of inorganic substances forming clear true solution and organic substances lipids, protein and carbohydrates forming a polyphasic colloidal system. 3. Nucleoplasm is colloidal substance having similar composition to cytoplasm, but contains more of nucleo- tides.
4. Suspended in it are various organelles and inclusions. 4. Suspended in it is chromatin.

Question 30.
What will happen to a cell if its nucleus is removed?
Answer:
Cell will die with passage of time as nucleus is the controlling centre of all functions of cell.

Very Short Answer Type Questions

Question 1.
What is the structural and functional unit of life?
Answer:
Cell.

Question 2.
Name three functional regions of a cell.
Answer:

  • Plasma membrane
  • Nucleus
  • Cytoplasm.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 3.
Name the orgnelles present in the cells.
Answer:
Mitochondria, Ribosomes, Endoplasmic Reticulum, Golgi bodies, Centrosome, Lysosomes, Plastids.

Question 4.
Name the organelle present in plant cell only.
Answer:
Plastids.

Question 5.
What is the main component of nucleus?
Answer:
Chromatin network which takes up the shape of chromosomes at the time of cell division.

Question 6.
Give six examples of single celled organisms.
Answer:
Paramecium, Amoeba, Chlamydomonas, Euglena, Trypanosoma and Malarial parasite.

Question 7.
Who coined the term ‘Protoplasm’ for the fluid substance of cell?
Answer:
Purkinje (1839).

Question 8.
How it was made possible to observe the complex structures of cell?
Answer:
Discovery of electron microscope.

Question 9.
Name the organelle which takes part in protein synthesis.
Answer:
Ribosomes.

Question 10.
Which is the longest cell in plants.
Answer:
Sclerenchyma.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 11.
Name the smallest and largest cells.
Answer:

  • Smallest cell – Mycoplasma.
  • Largest cell-Ostrich egg.

Question 12.
Name the smallest and largest cell of human body.
Answer:

  • Smallest cell – of kidney.
  • Largest cell-neuron (nerve cell).

Question 13.
List the factors which denote size of cell.
Answer:

  1. Structure
  2. Location
  3. Function.

Question 14.
What are the two main components of protoplasm?
Answer:
Cytoplasm and nucleoplasm.

Question 15.
Which cell organelle is called suicidal bag?
Answer:
Lysosome.

Question 16.
What is the primary function of lysosome?
Answer:
A centre of intracellular digestion, so is called digestive bag.

Question 17.
Name the cell organelle responsible for protein synthesis.
Answer:
Ribosomes.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 18.
Expand DNA.
Answer:
Deoxyribose nucleic acid.

Question 19.
Expand RNA.
Answer:
Ribose nucleic acid.

Question 20.
Who examined thin slice of cork under microscope?
Answer:
Robert Hooke.

Question 21.
What is cork?
Answer:
Cork is obtained from the bark of a tree.

Question 22.
Who coined the term cell?
Answer:
Robert Hooke.

Question 23.
Who examined the living cells for first time?
Answer:
A.V. Leeuwenhoek.

Question 24.
What are unicelluar organisms?
Answer:
The organisms formed of single cell e.g. Amoeba.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 25.
What are multicellular organisms?
Answer:
The organisms formed of a large number of cells are termed as multicellular organisms e.g. human beings.

Question 26.
Write examples of multicellular organisms.
Answer:
Fungi, plants and animals.

Question 27.
Write two features of cells of onion peel.
Answer:
Presence of cell wall and large central vacuole.

Question 28.
What is the contribution of Robert Brown in cell biology?
Answer:
Discovered nucleus in the cells of orchid.

Question 29.
What are the two types of cells on the basis of nature of nucleus? *
Answer:
Prokaryotic and Eukaryotic cells.

Question 30.
Name any two prokaryotic cells.
Answer:

  • Bacterial cell
  • Blue-green algae

Question 31.
Who proposed the cell theory?
Answer:
M.J. Schleiden and Theodore Schwann.

Question 32.
What is cell theory?
Answer:
All plants and animals are made up of cells and cell is the structural and functional unit of life.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 33.
Give the contribution of Rudolf Virchow in the field of biology.
Answer:
He stated “omnis cellula e cellula” which means new cells are formed by the growth and division of pre-existing cells.

Question 34.
What is the number of cells present in an adult person?
Answer:
100 trillion (1014).

Question 35.
Which is the smallest measuring unit in the field of cell biology?
Answer:
Angstrom (A).

Question 36.
In which year electron microscope was invented?
Answer:
1940.

Question 37.
What is the nature of plasma membrane?
Answer:
It is formed of lipids and proteins.

Question 38.
Give the primary function of plasma membrane.
Answer:
It regulates exchange of materials between the cytoplasm and extracellular fluid.

Question 39.
Give one term for a semifluid ground substance present between plasma membrane and nucleus.
Answer:
Cytoplasm.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 40.
What is diffusion?
Answer:
Diffusion. The movement of substance from a region of higher concentration to a region of lower concentration is called diffusion.

Question 41.
What is the term for movement of water into and out of cell across a semipermeable membrane?
Answer:
Osmosis.

Question 42.
Define plasmolysis.
Answer:
Loss of water from a plant cell resulting in shrinkage or contraction of cell away from cell wall.

Question 43.
Give the common name of the mitochondria.
Answer:
Powerhouses or ATP mills.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 44.
Expand ATP.
Answer:
ATP. Adenosine triphosphate

Question 45.
Name two semi-autonomous cell organelles.
Answer:
Mitochondria and plastids.

Question 46.
Which type of enzymes are located inside the mitochondria?
Answer:
Respiratory enzymes.

Question 47.
Name the largest-sized cell organelle.
Answer:
Plastid (chloroplast).

Question 48.
Name three types of plastids.
Answer:
Leucoplasts, chloroplasts and chromoplasts.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 49.
Which cell organelle is called kitchen of cell and why?
Answer:
Chloroplast. It is site of photosynthesis in green plants. During this process plant prepare their own food by trapping solar energy and use COz and water as raw materials.

Question 50.
What are chromoplasts?
Answer:
These are coloured plastids with yellow, orange or red coloured plastids.

Question 51.
Write the chemical composition of cell wall.
Answer:
Cellulose.

Question 52.
What is the role of cell wall?
Answer:
Cell wall provide structural support to plant cell.

Question 53.
Which chemical is used to strain plant cell?
Answer:
Iodine, Safranin, methylene blue.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 54.
Give one major difference between SER and RER.
Answer:
SER is without ribosomes and is involved in lipid synthesis, while RER is with ribosome-studded cistemae and is involved in protein synthesis.

Question 55.
List two general functions of ER.
Answer:

  • ER acts as channels for transport of materials within the cell.
  • It also act as cytoskeleton.

Question 56.
Write one function of SER.
Answer:
Detoxification of many poisons and drugs.

Question 57.
Give the primary function of Golgi body.
Answer:
Cell secretion and acts as condensation membrane.

Question 58.
Which plastids are involved in storage of food?
Answer:
Leucoplasts.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 59.
What is the main difference between cell organelles and cell inclusions?
Answer:
Cell organelles are living structures of cytoplasm, while cell inclusions are non-living structures of cytoplasm.

Question 60.
Name the vacuolar membrane.
Answer:
Tonoplast.

Question 61.
Why is nucleus called director of cell?
Answer:
It controls and directs all the cellular activities.

Question 62.
Which organelle plays central role in cellular reproduction?
Answer:
Nucleus.

Question 63.
What is nature of nuclear membrane?
Answer:
It is two-layered, lipoproteinaceous, porous and semipermeable.

Question 64.
What is primary function of nuclear membrane?
Answer:
Regulates exchange of materials between nucleoplasm and cytoplasm.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 65.
What are the functional units of chromosomes?
Answer:
Genes.

Question 66.
What is the chemical nature of a chromosome?
Answer:
DNA and basic protein histones, so is called Deoxyribonucleoprotein (DNP).

Question 67.
Give the primary function of DNA of chromatin fibre.
Answer:
DNA acts as genetic material and helps in transmission of characters from par¬ents to offsprings.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 68.
What is the function of vacuole in plant cell?
Answer:
Vacuoles are full of cell sap and provide rigidity and turgidity to cell vacuoles, temporarily stores wastes.

Question 69.
Write functions of contractile vacuole in plant cell.
Answer:
Osmoregulation and expelling wastes from the cell.

Question 70.
Name the organelle which contains chlorophyll.
Answer:
Chloroplast.

Question 71.
What are two functions of plasma membrane?
Answer:

  • Provides shape to cell
  • Keeps intact the contents of cell.

PSEB 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Question 72.
What is osmosis?
Answer:
It is a special type of diffusion of solvent molecules from low concentration of solution to higher concentration solution separated by semipermeable membrane.

PSEB 10th Class Science Solutions Chapter 5 Periodic Classification of Elements

Punjab State Board PSEB 10th Class Science Book Solutions Chapter 5 Periodic Classification of Elements Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements

PSEB 10th Class Science Guide Periodic Classification of Elements Textbook Questions and Answers

Question 1.
Which of the following statements is not a correct statement about the trends when going from left to right across the periods of periodic table.
(а) The elements become less metallic in nature.
(b) The number of valence electrons increases.
(c) The atoms lose their electrons more easily.
(d) The oxides become more acidic.
Answer:
(c) The atoms lose their electrons more easily.

Question 2.
Element X forms a chloride with the formula XCl2, which is a solid with a high melting point. X would most likely be in the same group of the periodic table as :
(a) Na
(b) Mg
(c) Al
(d) Si
Answer:
(b) Mg

Question 3.
Which element has :
(а) two shells, both of which are completely filled with electrons?
Answer:
Neon

(b) the electronic configuration 2, 8, 2?
Answer:
Magnesium

(c) a total of three shells, with four electrons in its valence shell?
Answer:
Silicon

(d) a total of two shells, with three electrons in its valence shell?
Answer:
Boron

(e) twice as many electrons in its second shell as in its first shell?
Answer:
Carbon.

Question 4.
(a) What property do all elements in the same column of the periodic table as Boron have in common?
Answer:
All elements of this column have 3 electrons in their valence shell like Boron.

(b) What property do all elements in the same column of the periodic table as Fluorine have in common?
Answer:
All elements of this column have 7 electrons in their valence shell like fluorine.

Question 5.
An atom has electronic configuration 2, 8, 7.
(а) What is the atomic number of this element?
Answer:
17

(b) To which of the following element would it be chemically similar? (atomic numbers are given in parenthesis).
N (7), F (9), P (15), Ar (18).
Answer:
F (9) (2, 7)

Question 6.
The position of three elements A, B and C in the periodic table are as shown below :
PSEB 10th Class Science Solutions Chapter 5 Periodic Classification of Elements 1
(а) State whether A is a metal or non-metal.
Answer:
A is non-metal.

(b) State whether C is more reactive or less reactive than A.
Answer:
C is less reactive than A.

(c) Will C be larger or smaller in size than B.
Answer:
C has smaller size than B.

(d) Which type of ion, cation or anion, will be formed by element A?
Answer:
Anion, A-.

PSEB 10th Class Science Solutions Chapter 5 Periodic Classification of Elements

Question 7.
Nitrogen (atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the periodic table. Write the electronic configurations of these two elements. Which of these will be more electronegative? Why?
Answer:

K L M
N7 has electronic configuration 2 5
P15 has electronic configuration 2 8 7

Nitrogen is more electronegative than Phosphorus due to smaller size.

Question 8.
How does the electronic configuration of an atom relate to its position in the Modern Periodic Table?
Answer:
By knowing the electronic configuration of an element, we can know its period number from the number of shells present in its atom and from number of electrons in the valence shell of its atoms we can know its group number, e.g. let us consider the case of sodium atom.

Atomic number of sodium = 11
Its electronic configuration = 2, 8, 1 (K) (L) (M)
∴ Number of shells = 3
∴ Sodium belongs to 3rd period.

Also sodium atom has one electron in its valence shell.
∴ It is present in first group.
∴ Sodium lies in the first group and third period of Modern Periodic Table.

Question 9.
In the Modern Periodic Table, calcium (atomic number 20) is surrounded by elements with atomic numbers 12,19, 21 and 38. Which of these have physical and chemical properties resembling calcium?
Answer:
Element with atomic numbers 12 and 38 because they have two electrons in their valence shells like calcium (2, 8, 8, 2).

Question 10.
Compare and contrast the arrangement of elements in Mendeleev’s Periodic table and the Modern Periodic table.
Answer:
Similarities :

  • In both the elements are arranged in groups and periods.
  • In both similar elements are placed in same group.
  • Both the classification make the study of elements simple and systematic.

Differences :

Mendeleev’s Periodic Table Modern Periodic Table
1. The elements are arranged in order of increasing mass numbers. 1. The elements are arranged in order of increasing atomic numbers.
2. It has 8 vertical columns called groups. 2. It contains eighteen vertical columns called groups.
3. Groups like group VIII ‘have been divided into sub groups A and B. 3. Each group is an independent group.
4. Inert gases are not included in this table. 4. Inert gases are included in this periodic table.

Science Guide for Class 10 PSEB Periodic Classification of Elements InText Questions and Answers

Question 1.
Did Dobereiner’s triads also exist in the columns of Newlands’ octaves? Compare and find out.
Answer:
Yes, Dobereiner’s triads also existed in columns of Newlands’ octaves. These are:

H Li Be
F Na Mg
Cl K Ca

Question 2.
What were the limitations of Dobereiner’s classification?
Answer:

  • Dobereiner could find only three triads from the elements known at that time.
  • It is applicable only to a few elements.

PSEB 10th Class Science Solutions Chapter 5 Periodic Classification of Elements

Question 3.
What were the limitations of Newlands’ law of octaves?
Answer:

  • It is applicable upto calcium only.
  • Sometimes two elements were put in the same slot.
  • After the discovery of noble gas, law of octave is not valid.

Question 4.
Use Mendeleev’s periodic table to predict the formulae for the oxides of following elements :
K, C, Al, Si, Ba.
Answer:
K2O, CO2, Al2O3, SiO2, BaO.

Question 5.
Besides Gallium, which other elements have since been discovered to fill the gaps left by Mendeleev in his periodic table? (any two)
Answer:
Scandium and Germanium.

Question 6.
What were the criteria used by Mendeleev in creating his periodic table?
Answer:

  • The formulae of hydrides and oxides formed by an element wrere treated as one of the basic properties of an element for its classification.
  • It is based upon Mendeleev’s periodic law which predicts that the properties of the elements are the periodic function of their atomic masses.

Question 7.
Why do you think the noble gases are placed in a separate group?
Answer:
These gases were discovered very late because they are very inert and placing them in a separate group, does not disturb the existing order put forward by Mendeleev.

Question 8.
How could Modern Periodic Table remove various anomalies of Mendeleev Periodic Table?
Answer:

  • Isotopes of an elements occupy same position in the periodic table due to same atomic number.
  • There is a logical separation of elements into subgroups.
  • It is based upon the fundamental property of an element i.e. atomic number.

Question 9.
Name two elements you would expect to show same kind of chemical reactivity as magnesium. What is the basis for your choice?
Answer:
Calcium and Strontium because they have same number of valence electrons.

Question 10.
Name :
(а) three elements that have only a single electron in their outermost shells.
Answer:
Lithium, Sodium and Potassium.

(b) two elements that have two electrons in their outermost shells.
Answer:
Magnesium, Calcium.

(c) three elements with filled outermost shells.
Answer:
Neon, Argon, Krypton.

Question 11.
(a) Lithium, Sodium, Potassium are all metals that react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements?
Answer:
One electron in their valence shells and are metals.

(b) Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common?
Answer:
Eight electrons in their valence shells.

Question 12.
In the modern periodic table, of the first ten elements, which are metals?
Answer:
Lithium and Beryllium.

PSEB 10th Class Science Solutions Chapter 5 Periodic Classification of Elements

Question 13.
By considering their position in the periodic table, which one of the following elements would you expect to have the most metallic characteristics?
Ga Ge As Se Be.
Answer:
Gallium.

PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds

Punjab State Board PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds Important Questions and Answers.

PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds

Long Answer Type Questions

Question 1.
What is allotropy? Name the allotropes of carbon. Are they chemically same? Compare their physical properties.
Answer:
Allotropy: The phenomenon of existence of two or more different physical forms of the same chemical element, with same chemical properties is called allotropy.

Allotropes of carbon

  • Diamond
  • Graphite

Similarity in chemical composition:
If both the allotropes in equal amount are heated in air then both will release same amount of carbon dioxide and will not leave any residue. Therefore this proves that both are same in chemical composition.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 1

Comparison of Physical Properties

Property Diamond Graphite
1. Appearance Transparent Black, shiny
2. Hardness Very hard Soft, slippery/ smooth in touch
3. Heat conductivity Very less Medium conductor
4. Electric conductivity Bad conductor Good conductor
5. Density (kg/m3) 3 – 510 2 – 250
6. Purity Purest Less pure than diamond
7. Melting point 3500°C 3000°C
8. Uses/Applications Jewellery, drilling purpose Lubricants. Drycells, Electric Arcs, Pencils, Lead,

Question 2.
What is covalent bond? Write its properties.
Answer:
The chemical bond formed by sharing of electrons between atoms of same or different non-metal elements is known as covalent bond.

Properties of Covalent bond/bonds.

  • They hax e strong bonds within molecules but their inter molecular bond is weak.
  • They have low boiling points.
  • They have low melting points.
  • Their compounds are bad conductors of electricity.

Question 3.
(a) Why does carbon form largest number of compounds?
Answer:
This is due to its tetracovalency and catenation.

(b) Why are some of these called saturated and other unsaturated compounds?
Answer:
The compounds containing C – C single bond are called saturated compounds and the compound containing
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 2
bonds are called unsaturated compounds.

(c) Which of these two is more reactive?
Answer:
Unsaturated compounds are more reactive than saturated compounds.

(d) Write the names of the compounds
(i) CH3 – CH2 – Br
Answer:
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 3

PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 4
Answer:
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 5

Question 4.
Write about nomenclature of carbon compounds.
Answer:
The names of compounds in a homologous series are based on the name of basic carbon chain modified by a prefix’ ‘phrase before’ or ‘suffix’ ‘phrase after’ indicating the nature of the functional group. For example, the names of various alcohols are methanol, ethanol, propanol and butanol.

Naming of carbon compound can be done by the following method :

  • Identify the number of carbon atoms in the compound. A compound having three carbon atoms will have the name propane.
  • In case a functional group is present, it is indicated in the name of the compound with either a ‘prefix’ or a ‘suffix’. Like chloro propane, bromo propane, propanal, propanone etc.
  • If the name of functional group is to be given on the basis of suffix, the name of the carbon chain is modified by deleting the final ‘e’ and adding the appropriate suffix. For example, a chain having three carbons with a Ketone group will be given name in the following manner.
    Propane-‘e’ = propan + ‘one’ = propanone.
  • If the carbon chain is unsaturated then the final ‘ane’ in the name of carbon chain is substituted by ‘ene’ or ‘yne’. For example, a three-carbon chain with a double bond would be called propene and if it has triple bond, it would be called propyne.

PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds

Question 5.
Write down the chemical properties of carbon compounds.
Answer:
The main chemical properties of carbon compounds are :
1. Burning: Carbons in all its allotropic forms burns in the presence of oxygen and produces heat, light and C02, For example :
C + O2 → CO2 + heat and light
CH4 + 2O2 → CO2 + 2H2O + Heat and light
CH3CH2OH + 3O2 → 2CO2 + 3H2O + Heat and light.
Saturated hydrocarbons burn with a clear flame whereas unsaturated hydrocarbons burn with sooty yellow flame.

2. Oxidation: Carbon compounds can easily be oxidised by burning. Alkaline potassium permanganate or acidified potassium permanganate converts alcohols into acids.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 6

3. Addition reaction: Unsaturated hydrocarbons make saturated hydrocarbons by adding hydrogen to itself in the presence of catalysts like nickel and palladium. The reaction is commonly used in hydrogenation of vegetable oils using nickel catalyst.

Vegetable oils generally have long unsaturated carbon chain while animal fats have saturated carbon chains.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 7

4. Substitution reaction: Saturated hydrocarbons are fairly unreactive and are inert in the presence of most reagents. However, in the presence of sunlight, chlorine gets combined with hydrocarbons in a quick reaction. Chlorine replaces very quickly the hydrogen atoms in the chemical reaction one by one due to which number of products are usually formed with the higher homologous of alkanes.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 8

Question 6.
(a) What is an alcohol? Give two examples.
Answer:
Alcohol is a simple compound of carbon, hydrogen and oxygen. If one hydrogen atom is replaced by hydroxyl (-OH) group in alkene, alcohol is produced. General formula of alcohol is CnH2n+2

Examples :

  • In Methane (CH4) when hydrogen atom is replaced by hydroxyl (- OH) group, methanol (CH3OH) is formed.
  • Ethanol (C2H5OH) is produced with the help of ethane

(b) How is synthetic ethanol produced? Also write.
Answer:
Synthetic ethanol is prepared by reacting ethane with water in the presence of phosphoric acid (H3P04).
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 9
Properties of Ethanol :

  • It is a colourless liquid with specific colour.
  • Its boiling point is 351 K and melting point is 156 K.
  • It is soluble in water in any proportion.
  • It doesn’t show any reaction with litmus because it is neutral.
  • It burns with blue flame in air to produce carbon dioxide and water.
    C2H5OH + 3O2 → 2CO2 + 3H2O
  • It reacts with oxygen or potassium dichromate (K2Cr2O7) to produce ethanoic acid.
    C2H5OH + O2 → CH3COOH + H2O
  • It reacts with sodium metal.
    2C2H5OH + 2Na → 2C2H5ONa + H2
  • It reacts with acetic acid in the presence of concentrated H2SO4. When this mixture is heated and poured in ice cold solution of sodium carbonate, a sweet smell is produced.

PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 10

Question 7.
Give important properties of monocarboxylic acids. Also give its uses.
Or
Give important properties of ethanoic acid (Acetic acid). Also give its uses.
Answer:
Important properties of monocarboxylic acid :
1. Ethanoic acid or acetic acid is colourless liquid and is soluble in water, 5-8% solution of acetic acid in water is called vinegar.

2. Pure ethanoic acid melts at 290 K and freezes during winter in cold climates and looks like ice. Hence it is called glacial acetic acid.

3. Esterification: When an acid is heated with an alcohol in the presence of cone. H2S04, an ester is produced. This is a slow reversible reaction and is called esterification.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 11
Esters are sweet-smelling substances.

4. Action with a base
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 12

5. Action with carbonates and bicarbonates
Ethanoic acid decomposes carbonates and bicarbonates producing salt, carbon dioxide and water.
2CH3COOH + Na2CO3 → 2CH3COONa + H2O + CO2
CH3COOH + NaHCO3 → CH3COONa + H2O + CO2

Uses of ethanoic acid :

  • It is used in the manufacture of dyes, perfumes and rayon.
  • It is used in the manufacture of plastics, rubber and silk industries.
  • It is used as a solvent.
  • It is used as a vinegar in cooking, as food dressing and for preparing pickles.
  • It is used for the manufacture of chemicals like acetone, acetic anhydride, etc.
  • It is used for making white lead [2PbCO3. Pb(OH)2]

Question 8.
Explain the following terms :
(i) Esterification
Answer:
Alcohols are treated with carboxylic acids in the presence of concentrated sulphuric acid to form ester and this process in known as esterification.

Method: Mix ethyl alcohol with acetic acid in a test tube. Few drops of concentrated sulphuric acid is added and test tube is heated mildly in hot tub of water. Instantaneously sweet smell of ester is diffused in whole room.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 13
It is an example of esterification. Ester is used in ice-cream, cold drinks, medicines, make up etc.

(ii) Saponification
Answer:
Saponification: Breaking down of fats is called saponification.

It is performed by heating vegetable or animal oils with 40% solution of caustic soda. Fats and base react to produce soap and glycerol.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 14
Concentrated common salt solution is added to crystallize soap from water. Soap floats on the surface of water, after cooling. Soap is extracted from water and desired colour and smell are added and is given desired shape.

(iii) Decarboxylation
Answer:
Decarboxylation: Methane is produced when sodium or potassium salt of ethanoic acid are heated with sodium hydroxide and calcium oxide in 3 : 1 ratio.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 15
It is a useful method to prepare methane. It is known as decarboxylation because one molecule of CO2 is removed.

(iv) Polymerization.
Answer:
Polymerization: When a large number of small molecules join together at a specific temperature and pressure to form a big molecule, this process is termed as polymerization. These small molecules are known as ‘monomer’ whereas big combined molecule is known as ‘polymer.’
For example. Molecules of ethene join together at 2000 atm pressure and 200°C to form polythene.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 16

Question 9.
Describe a method for the preparation of soap.
Answer:
Soaps are sodium or potassium salts of higher fatty acids. When the naturally occurring esters called fats or oils are heated with NaOH solution, they undergo hydrolysis to form sodium salt of higher fatty acid (called soap) and glycerol.

Manufacture of Soap. Soap can be made easily in the laboratory. Heat fat or oil with sodium hydroxide solution. After a few minutes, and constant stirring, the oil and water layers get mixed.

Add 5-10 g of common salt to it, stir the mixture and allow’ it to cool. On cooling, pale yellow a solid forms as a cake called soap.

The same principle is used for making soap in soap industry. Some other substances like perfumes, disinfectants and medicines are added to soap to give it desired characteristics.

Example :
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 17

PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds

Question 10.
What do you understand by isomers? Explain it with an example.
Or
Write down the isomers of butane.
Answer:
Isomers. The compounds with same molecular formula but different structural formula are called isomers and the phenomenon is called isomerism. There is no change in the structure of methane, ethane, propane if the atoms of carbon and hydrogen are rearranged but in case of alkane molecule if number of carbon is more than three then more than one arrangement is possible.

In this one long carbon chain is formed and the others are branched. In case of butane the chain is linked with atleast three carbon atoms with other carbon atoms. This type of alkanes are termed as iso-alkane. In a chain having no branch, carbon atom cannot be linked with two carbon atoms. Alkanes of this type are called normal alkane (n-alkane).

Two Isomers of Butane:
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 18

Isomers of Pentane
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 19

Short Answer Type Questions

Question 1.
Explain why carbon forms mostly covalent compounds?
Answer:
Carbon has four electrons in its outermost shell. In order to get stable electronic configuration, it has to lose or gain four electrons, but it is difficult to lose or gain four electrons from energy considerations. Hence, Carbon complete its octet by sharing its valence electrons with other atoms forming covatent bonds resulting in the formation of covalent compounds.

Question 2.
In order to form large amount of carbonic compounds which major elements are used other than hydrogen and oxygen?
Answer:
The number of carbonic compounds is very large which was not possible only with the help of hydrogen and oxygen. Along with these two elements some other elements like nitrogen, sulphur, oxygen, halogens (Cl, Br, I, F) etc. also combine and because of it the caybonic compounds have crossed the number of many million. These combine together on the basis of functional groups and make new compounds.

Question 3.
‘Carbon is a unique element’, why?
Answer:
Carbon is the only known element out of all other known elements, that has got capability to make long chains. Each long chain of carbon atoms provides such a simple base with which other atoms can join by various methods resulting in the composition of various compounds.

Carbon atoms can make long chains or rings in three ways because of having catenation property.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 20

Question 4.
Explain the structure of diamond and write why is it so hard?
Answer:
In diamond each carbon atom remains at the centre of regular tetrahedral and it remains bound with covalent bond to four carbon atoms which are situated at four corners of tetrahedral. Thus, all electrons of carbon are bound atoms and none is found in free state. So it forms a strong three-dimensional tetrahedral structure and because of it diamond is the hardest substance with very high density.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 21

Question 5.
What are the uses of diamond?
Answer:
Uses of Diamond

  • Diamond being the hardest substance is used for cutting other substances.
  • Diamond because of its shine and lustre is used for making ornaments.
  • It is used for drilling hard rocks.
  • Sharp-edged diamonds are used by eye surgeon in catract surgery.
  • Being extra ordinarly sensitive for heat radiations and having property to remove immediately harmful radiations. These are used in making absolute thermometers.

Question 6.
Why do diamonds glitter?
Answer:
Diamond is a transparent substance whose refractive index is very high. The rays of light passing through it get diverted much. According to its cut base edges light rays undergo total internal reflection. When these edges are polished, diamond glitters very beautifully.

PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds

Question 7.
How are diamonds made naturally and artifically?
Answer:
Naturally, diamonds are made by carbon present at the depth of about 150 km, where pressure is 70,000 atm and the temperature is about 1500°C. These diamonds come up near the surface with special kimberlite rocks. Artifically, diamonds are made by heating graphite under high pressure in the presence of special catalysts.

Question 8.
Write the structure of graphite. Why is it so soft?
Answer:
Each carbon atom is bonded with three neighbouring carbon atoms with covalent bonds in graphite. They are always in the same plane giving hexagonal array. As compared to diamond, the distance between carbon atoms is more. The distance between upper and lower layers does not permit carbon atoms to make covalent bonds and because of this fourth electrons remains free. These layers can easily slide one upon the other and graphite attains the property of a lubricant. It is very soft and slippery.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 22

Question 9.
Why are diamonds used in jewellery? Give reasons.
Answer:
Refractive index of diamond is 2.45 which is the highest of all known substances, because of this, it shines brightly. Its critical angle is 24°, which is very less. This is why, the light rays passing through it deviates much. A small deviation in it causes beautiful spectrum of colours. When the bases of diamond are polished, it produces an astonishing brightness. That is why diamonds are used in making jewellery.

Question 10.
Why is graphite a good conductor of electricity?
Answer:
Each carbon atom of graphite is surrounded by three carbon atoms joined with covalent bond which makes hexagonal layers over each other. In these layers distance between the carbon atoms is more, that is why the possibility of fourth covalent bond between the carbon atoms of opposite layers is reduced. The fourth covalent electron is set free. The graphite has easy flow of electrons and it acts as a good conductor of electricity.

Question 11.
What are the uses of graphite?
Answer:
Uses of Graphite

  • It is good conductor of electricity so- it is used in making dry cells and electric arcs as electrods.
  • It is used in making pencils, black colour and black paint.
  • It has lubricant property so it is used in machines to keep them lubricating at high temperature.
  • It is used in making crucibles to boil some metals because of its high melting point.

Question 12.
Why are allotropes of carbon in form a molecule known as fullerenes?
Answer:
The allotropes of carbon in which 60 carbon atoms join to form a molecule are termed as fullerene. US architect Buckminster Fuller designed three dimensional domes which were supported by pentagonal and hexagonal structures.

Since fullerene molecules looked like these domes, that is why they are known as fullerene.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 23

Question 13.
Draw the electron dot structures of methane, ethane, ethene, butane.
Or
(i) Write molecular formula of butane.
(ii) Draw the structure of Propanal.
Answer:
Methane
Formula -CH4
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 24

Ethane
Formula -C2H6
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 25

Ethene
Formula -C2H4
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 26

Or
(i) Butane
Formula – C4H10
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 27

(ii) Propanal
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 28

PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds

Question 14.
Write two tests of alcohol.
Answer:

  1. Ester test: When alcohol is heated with acetic acid, in the presence of sulphuric acid, sweet smelling ester is produced.
  2. Sodium test: Alcohol reacts with sodium metal and always produces hydrogen gas.

Question 15.
(a) What do you understand by homologous series? Explain with an example.
Answer:
Homologous series. A series of compounds in which the same functional group substitutes for hydrogen in a carbon chain is called a homologous series. Each successive member differ by (- CH2) group. All the members of a homologous series can be obtained by similar process.

General Formula : CnH2n+2
For example : Homologous series of alkane is as follows
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 29

(b) Write the first three homologous compounds of methane.
Answer:
Homologous Compounds of Methane
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 30

Question 16.
Define homologous series. What are its main properties?
Answer:
Homologous series. A series of compounds in which the same functional group substitutes for hydrogen in a carbon chain is called homologous series. In this series the adjacent members differ by (-CH2) group. Each member of a homologous series is called homolog.

Properties of Homologous Series :

  • All the members of a homologous series can be represented by a common formula.
  • Two successive members differ by one carbon atom and 2 hydrogen atoms.
  • All the members of a homologous series possess similar chemical properties.
  • There is a small difference between all the members of a homologous series.
  • There is always a difference of 14 a.m.u. in atomic mass of two adjacent homologous.

Question 17.
What are alkanes? Write the properties of its main members. What do we conclude from its properties?
Answer:
Alkanes. The hydrocarbons with structural/general formula CnH2n+2 are called alkanes. The members of this group form an homologous series. Hydrocarbons like methane, ethane, propane and butane form many carbonic compounds which are collectively known as alkane. In these compounds, the number of carbon and hydrogen atoms differ by (- CH2) for the next member of family.

Properties of Alkanes
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 31

Conclusions:

  • Physical properties of alkane are based on their mass.
  • First four members of alkane are in gasous state and alkanes having more than six carbon atoms are in liquid state.
  • Alkanes having high molecular mass are found in solid state.
  • Melting and boiling point of alkanes increase with increase in molecular mass.

Question 18.
What happens when hydrogen gas is passed through vegetable oils in the presence of nickel?
Answer:
Vegetables oils have double bond and polymerisation is possible. When nickel is used as catalyst, vegetable oil converts to vegetable ghee when hydrogen is passed through it because of hydrogenation. Vegetable ghee is solid like butter at room temperature.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 32

Question 19.
Write the chemical properties of ethanoic acid.
Answer:
Ethanoic acid reacts with some metals, metal carbonates, hydroxides and bicarbonates.
1. Action with sodium:
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 33

2. Action with sodium hydroxide and potassium hydroxide:
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 34

3. Action with carbonate and hydrogen carbonate :
It reacts with carbonates and hydrogen carbonate to form water, salt and carbon dioxide.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 35

PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds

Question 20.
Which type of oils should be used for cooking? Why?
Answer:
Animal fats have saturated fatty acids, which are harmful for health, whereas oils have long unsaturated carbon chains which are not harmful. Hence unsaturated fatty acids oils are used for cooking.

Question 21.
Write the ill-effects of using alcohol.
Answer:

  • It is an addictive substance.
  • It affects the sensitivity of nervous system.
  • When a drunken person drives the vehicle he/she is not able to take the right decisions which leads to accident.

Question 22.
Why is consumption of methanol (CH3OH) very fatal?
Answer:
Even a very small amount of methanol can prove fatal. In liver, it oxides to form methanale, which vigorously react with cells of liver, by which protoplasm gets emanating and swells in the same manner as boiled egg. Consumption of methanol also lead to blindness.

Question 23.
Write the IUPAC names of ethyl alcohol and acetic acid. Write the name of products formed when ethyl alcohol and acetic acid react in the presence of concentrated sulphuric acid. Write its chemical reaction.
Answer:

Compounds IUPAC Names
1. Ethyl alcohol ethanol
2. Acetic acid ethanoic acid

Chemical reaction :
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 36

Question 24.
How will you prepare :
(a) ethyl alcohol from ethene
(b) acetic acid from ethyl alcohol?
Give reactions.
Answer:
(a) Ethene is heated with water in the presence of phosphoric acid to prepare ethyl alcohol
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 37
(b) Acetic acid is prepared by oxidation of ethyl alcohol
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 38

Question 25.
Write in brief the various industrial alcohols.
Answer:

  • Absolute alcohol: It is 100% pure ethanol. It is produced by fractional distillation of ethanol which is prepared by fermentation of carbohydrates.
  • Denatured alcohol: It is 95% ethanol and is used in industries. It is made poisonous by adding methanol, copper sulphate and pyridine, so as to make it unfit for drinking.
  • Power alcohol: It is mixture of Benzene and Ether in 20% Ethanol and 50% gasolene. It is used to save petroleum.
  • Alcoholic beverages: Alcohol is used as addictive drink. It is sold under various names like, rum, whisky, brandy, Jinn, etc.

Question 26.
Why is cleansing effect of soap not effective in hard water?
Answer:
Soaps are not effective in hard water for washing purposes because hard water has salts of calaium and magnesium. When soaps are dissolved in hard water, calcium and magnesium ions l’eact with water to form calcium and magnesium salts of fatty acids.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 39
That is why a large amount of soap gets wasted. Calcium and magnesium salts being insoluble in water make precipitates and they get attached with the clothes and cause hinderance to the cleaning action and the clothes are not cleaned properly.

Question 27.
Distinguish between soaps and detergents.
Answer:

Soaps Synthetic detergents
1. Soaps are sodium or potassium salts of higher fatty acids e.g. Sodium Stearate. 1. Synthetic detergents are Sodium alkyl Sulphates or Sodium alkyl benzene sulphates.
2. Soaps are prepared from vegetable oils or animal fats. 2. Synthetic detergents are prepared from the hydrocarbons obtained from petroleum.
3. Soaps have relatively weak cleansing action. 3. They have strong cleansing action.
4. Soaps form curdy white precipitates with calcium and magnesium salts present in hard water and hence, are not used in hard water. 4. Calcium and magnesium salts of detergents are soluble in water. Therefore no curdy white precipitates are obtained in hard water and hence synthetic detergents can be used even in hard water.
5. Soaps cannot be used in acidic medium as they are decomposed into carboxylic acids in acidic medium. 5. They can be used in acidic medium as they are the salts of strong acid and are not decomposed in acidic medium.
6. Soaps do not cause water pollution. 6. Synthetic detergents cause water pollution.
7. Soaps are biodegradable. 7. Some of the synthetic detergents are non biodegradable.

Question 28.
Write the formation of ethanol by fermentation process.
Answer:
Fermentation occurs in the presence of Bio-chemical catalysts at normal temperature by which sugar molecules convert into alcohol and carbon dioxide. These catalysts are known as enzymes, which means ‘yeast’ or ‘in the fermentation’.

Alcohol is formed by fermentation of sugar or starch. In a beaker, grape juice or glucose mixture is heated at 20° – 30°C in the presence of yeast. Sugar or starch molecules break into smaller molecules which produce carbon dioxide due to process of fermentation. This carbon dioxide is expelled out but air is not given inlet. A dilute mixture of ethanol is formed in water during fermentation. Ethanol is purified by distillation.

Chemical reaction :
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 40

Question 29.
Write the name of the reaction in which ethanoic acid and ethanol react together to form one product. Write the common name of the product.
Answer:
When ethanoic acid reacts with ethanol in the presence of concentrated sulphuric acid, ester is produced and this process is known as esterification.
CH3COOH + C2H5OH → CH3COOC2H5 + H2O

Question 30.
Observe the figure given below :
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 41
(i) Write the reaction taking place
Answer:
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 42

(ii) What is this reaction called?
Answer:
Esterification

Question 31.
What is indicated in the figure given below and give use of it.
Answer:
The figure indicates formation of micelle. Micelles are used to remove oily dirt from clothes.
1 – Hydrophilic end
2 – Oil droplet.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 43

Question 32.
Show the bonds of Hydrogen, Oxygen and Nitrogen atoms respectively.
Answer:
1. Hydrogen: Two atoms of hydrogen by sharing of their electrons make hydrogen molecule (H2).
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 44

2. Oxygen: A double bond is made between two atoms of oxygen. Each atom of oxygen shares two electrons with other atom to form .an octet.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 45

3. Nitrogen: Each atom of nitrogen shares three electrons to form an octet.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 46

Question 33.
(i) Write molecular formula of Propane.
Answer:
C3H8

(ii) Write IUPAC name of
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 47
Answer:
Bromoethane.

Question 34.
What is detergent? Write the formation of synthetic detergents. Write its advantage.
Answer:
Materials used for cleaning purposes are known as detergents. Since long, soaps have been used as detergents but now-a-days synthetic detergents are more popular. The detergent molecule has two ends, one which is made of sulphate (-SO4) or sulphonate (SO3Na) group and it is hydrophilic by nature whereas other which is made of hydrocarbon is hydrophobic.

Synthetic detergent produces ample amount of lather even in hard water. These do not form insoluble salts of calcium or magnesium.

Question 35.
Which substances are used to produce synthetic detergents?
Answer:
Following substances are needed for production of synthetic detergents :

  • Long chained hydrocarbons obtained from petroleum and coal.
  • Concentrated sulphuric acid.
  • Sodium hydroxide.

Procedure: Hydrocarbons are treated with cone, sulphuric acid to get hydrocarbonic sulphuric acid. It is further reacted with NaOH which forms synthetic detergents. About 15 to 30% of the total weight of washing powder is detergent. Rest of the others are various chemicals which provide other qualities.

PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds

Question 36.
What are the components of washing powder?
Answer:

  • Synthetic detergents (15 to 30%).
  • Sodium sulphate and sodium nitrate.
  • Sodium tri-poly phosphate or sodium carbonate.
  • Sodium perborate.

Very Short Answer Type Questions

Question 1.
Which element is present necessarily in the following items?
Food, clothes, medicines and hooks.
Answer:
Carbon.

Question 2.
On which element do all living processes depend?
Answer:
Carbon.

Question 3.
What percentage of carbon is present in minerals found under the crust of earth?
Answer:
0.02 %.

Question 4.
What percentage of CO2 is present in atmosphere?
Answer:
0.003%.

Question 5.
How many electrons are present in the outer shell of carbon?
Answer:
4.

Question 6.
Why does carbon need four electrons to gain or lose?
Answer:
To attain noble gas configuration.

Question 7.
How does carbon gain four electrons?
Answer:
By making C4 anion.

Question 8.
How does carbon lose four electrons?
Answer:
By making C4+ cation.

Question 9.
Which type of bond is shared by a pair of two hydrogen atoms?
Answer:
Single bond.

Question 10.
How is single bond between two atoms expressed?
Answer:
By single straight line.

Question 11.
What is the atomic number of nitrogen?
Answer:
7.

Question 12.
How many electrons are given by each atom of Nitrogen to complete the octet?
Answer:
3.

PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds

Question 13.
Which gas is used as a fuel the most?
Answer:
Methane gas.

Question 14.
Methane is the main constituent gas of which two fuel gases?
Answer:
Biogas, C.N.G.

Question 15.
Write the full form of C.N.G.
Answer:
Compressed Natural Gas.

Question 16.
How is the structure of diamond?
Answer:
Strong three-dimensional structure.

Question 17.
How is the structure of graphite?
Answer:
Hexagonal.

Question 18.
Which of these two is a good conductor of electricity diamond or graphite?
Answer:
Graphite.

Question 19.
Which is the hardest substance?
Answer:
Diamond.

Question 20.
How does it feel by touching graphite?
Answer:
Soft and slippery.

Question 21.
Which element has the ability to form large number of compounds?
Answer:
Carbon.

Question 22.
Give names of two elements which are placed in the group of carbon and have same valency.
Answer:
Valency of carbon is 4. This group has silicon and germanium.

Question 23.
What is the simplest form of hydrocarbon?
Answer:
Methane (CH4).

Question 24.
In which types of bonds can carbon atoms join?
Answer:
Single, double, triple bond and ring-shaped.

Question 25.
Write the names of five elements which can make bond with carbon to form new compounds.
Answer:
Oxygen, Hydrogen, Nitrogen, Sulphur, Chlorine.

PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds

Question 26.
Write the formula of Ethane.
Answer:
C2H6.

Question 27.
Write the structural formula of Ethane.
Answer:
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 48

Question 28.
Give the next higher homo-logues of:
(i) C3H6
Answer:
C4H8

(ii) C6H8.
Answer:
C7H10.

Question 29.
Give the structure of ethanol.
Answer:
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 49

Question 30.
Name the functional group in propanone.
Answer:
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 50

Question 31.
Give the structure of simplest ketone.
Answer:
Simplest keton is acetone. It has the structure :
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 51

Question 32.
Give the name and structure of four carbon atoms in aldehyde.
Answer:
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 52

Question 33.
Name the products obtained when ethanol undergoes complete combustion.
Answer:
Carbon dioxide (CO2) and Water (H2O).

Question 34.
Give the structural formula of ethanol.
Answer:
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 53

Question 35.
Give the name and structure of functional group present in acetic acid (CH3COOH).
Answer:
Carboxyl group
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 54

Question 36.
Which of the following compounds belong to same homo-Iogous series?
C2H6O2, C2H6O2, C2H6, CH4O
Answer:
C2H6 O, (C2H5OH) And CH4O or (CH2OH).

Question 37.
Give the structural formula of an ester.
Answer:
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 55

Question 38.
Give the names and structural formulas of acid and alcohol from which ethyl acetate is obtained.
Answer:
Acetic acid
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 56
and ethyl alcohol, CH3CH2OH

Question 39.
Give the electron-dot stucture of methane.
Answer:
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 57

Question 40.
Define covalent bond.
Answer:
It is formed by mutual sharing of electron pair between two atoms.

Question 41.
Explain what covalent compounds do not conduct electric current?
Answer:
This is because there are neither free ions nor free electrons.

Question 42.
Name the bond formed between metal atom and non-metal atoms.
Answer:
Ionic or electro valent bond.

Question 43.
What are allotropes?
Answer:
When an element exists in two or more form having different physical pruperties but, some chemical properties, there forms are called allotropes.

Question 44.
Define catenation.
Answer:
It is the property due to which a large number of atoms of the same element get linked together through covalent bonds forming long straight chains, branched chains and rings of different sizes.

PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds

Question 45.
Give two oxidising agents.
Answer:

  1. Acidified potassium permanganate.
  2. Acidified potassium dichromate.

Question 46.
Give the electron dot structure of ethane.
Answer:
Dot structure of ethane :
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 58

Question 47.
Give the structural formula of propane.
Answer:
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 59

Question 48.
Give the electron dot structure of ethene.
Answer:
Dot structure of ethene.
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 60

Question 49.
Define functional group.
Answer:
It is an atom or group of atoms which when present in a molecule gives special properties to it. e.g. C2H5OH has the functional group, -OH (hydroxyl group).

Question 50.
If two organic compounds have some functional group, predict they have similar physical or chemical properties.
Answer:
Similar chemical properties.

Question 51.
Give the general formulae of alkanes, alkenes and alkynes.
Answer:
Alkanes: Cn H2n+2
Alkaenes: CnH2n
Alkynes: CnH2n-2, where n = No. if carbon atoms).

Question 52.
Give the structural formula of isomers of C4H10 (butane).
Answer:
Butane has two structural isomers :
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 61

Question 53.
Give the first four members of homologous series of alcohols.
Answer:
CH3OH, C2H5OH. C3H7OH, C4H5OH.

Question 54.
Define a Catalyst.
Answer:
It is a substance which can increase the rate of a reaction but remains unchanged in mass and composition at the end of reaction.

Question 55.
What happens when a piece of sodium metal is added to ethanol? Give chemical equation also.
Answer:
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 62

Question 56.
What happens when ethanol is heated with conc. H2SO4 at 443K?
Answer:
Ethanol undergoes acidic dehydration to give ethene
PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds 63

Question 57.
What is denatured spirit or alcohol?
Answer:
It is prepared by adding poisonous substance like methanol, acetone, pyridine or copper sulphate to ethanol. It is unfit for drinking purposes.

Question 58.
Give three uses of ethanoic acid.
Answer:

  1. In the manufacture of vineger
  2. In the manufacture white lead
  3. As a reagent in laboratory.

Question 59.
What is the atomic number of hydrogen?
Answer:
One.

Question 60.
Give the electronic configuration of carbon atom.
Answer:
(2, 4).

Question 61.
Give name of four carbon compounds.
Answer:
Methane, Chloroform, Ethanol and ethanoic acid.

Question 62.
Give two allotropic forms of carbon.
Answer:
Diamond and graphite.

Question 63.
Give the name of simplest ketone.
Answer:
Propanone (CH3 COCH3).

PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds

Question 64.
Give the name of first member of homologous series of alkynes.
Answer:
Ethyne (C2H2).

Question 65.
Give the name and formula of acid present in vinegar.
Answer:
Ethanoic acid (CH3 COOH).

Multiple Choice Questions:

Question 1.
Ethanoic acid has the formula:
(A) C2H5OH
(B) CH3COCH3
(C) CH3COOH
(D) C2H5COOH.
Answer:
(C) CH3COOH

Question 2.
Carboxylic acids contain functional group:
(A) -CHO
(B) -CH2OH
(C) -COOH
(D) -OH.
Answer:
(C) -COOH

Question 3.
The general formula of the alkyne is:
(A) CnH2n-2
(B) CnH2n+2
(C) CnH2n
(D) CnH2n+2.
Answer:
(A) CnH2n-2

Question 4.
Propanone has the functional group :
(A) -OH
(B) -CHO
(C) C=O
(D) -COOH.
Answer:
(C) C=O

Question 5.
Vinegar contains acetic acid:
(A) 5-8%
(B) 15-20%
(C) 21-29%
(D) 30-40%.
Answer:
(A) 5-8%

Question 6.
The reaction of acetic acid with ethyl alcohol is called :
(A) Polymerisation
(B) Saponification
(C) Hydrogenation
(D) Esterification.
Answer:
(D) Esterification.

Fill in the Blanks :

Question 1.
The functional group present in ethyl alcohol is ___________
Answer:
The functional group present in ethyl alcohol is alcoholic group (-OH)

Question 2.
CH3—CH2—OH represents ___________
Answer:
Ethyl alcohol.

Question 3.
Two atoms of the same element combine to form a molecule. The bond between them is known as ___________ bond.
Answer:
Covalent.

PSEB 10th Class Science Important Questions Chapter 4 Carbon and its Compounds

Question 4.
In the formation of oxygen molecule, the oxygen atoms share ___________ electrons each.
Answer:
Two.

Question 5.
The number of single covalent bonds in the molecule of ammonia is ___________
Answer:
Three.

PSEB 5th Class Hindi Solutions Chapter 4 हम सुमन एक उपवन के

Punjab State Board PSEB 5th Class Hindi Book Solutions Chapter 4 हम सुमन एक उपवन के Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Hindi Chapter 4 हम सुमन एक उपवन के

Hindi Guide for Class 5 PSEB हम सुमन एक उपवन के Textbook Questions and Answers

I. बताओ

प्रश्न 1.
उपवन के फूलों को क्या-क्या समान चीजें प्राप्त होती हैं ?
उत्तर:
उपवन के फूलों को एक-सी धरती, मिट्टी, धूप, जल, सूर्य की किरणें, चाँद की चाँदनी, भँवरों का गुंजन आदि सब कुछ एक समान रूप में प्राप्त होती हैं।

PSEB 5th Class Hindi Solutions Chapter 4 हम सुमन एक उपवन के

प्रश्न 2.
उपवन की शोभा किनसे बनती है ?
उत्तर:
उपवन की शोभा फूलों से बनती है।

प्रश्न 3.
फूलों ने हँसकर जीना किससे सीखा है ?
उत्तर:
फलों ने हँसकर जीना काँटों से सीखा है।।

प्रश्न 4.
‘एक सूत्र में बँधकर हमने, हार गले का बनना सीखा’, इस काव्य पंक्ति का क्या अर्थ है ?
उत्तर:
इस पंक्ति का अर्थ है कि अलग-अलग होते हुए भी हम सब एक ही धागे से बँधकर हार के रूप में गले की शोभा बने हुए हैं।

प्रश्न 5.
फूल अपनी सुगन्ध किसे देता है ?
उत्तर:
फूल अपनी सुगन्ध सबको देता है।

II. इन प्रश्नों के उत्तर चार-पाँच पंक्तियों में लिखो

प्रश्न 1.
कवि ने फूल के द्वारा क्या सन्देश दिया है? पाँच वाक्यों में लिखो।
उत्तर:
कवि ने फूल के द्वारा सन्देश दिया है कि जिस प्रकार फूल अलग-अलग रंगों के होते हैं, अलग-अलग क्यारियों में पलते हैं, लेकिन सभी मिलकर उपवन की शोभा बढ़ाते हैं, इसी प्रकार अलग-अलग जातियों, प्रान्तों में रहते हुए सभी लोग भारत देश की शोभा बढ़ाते हैं।

फूलों के समान ही सभी मनुष्यों को परस्पर मिलजुल कर रहना चाहिए। फूलों के समान ही मनुष्यों को अपने अच्छे कर्मों, सद्व्यवहार और सदाचरण की महक चारों ओर बिखरानी चाहिए।

प्रश्न 2.
प्रकृति जैसे मिट्टी, वायु, सूरज, चाँद, भँवरे अपने गुण देने में कोई भेदभाव नहीं करते। वे निस्वार्थ भाव से देते हैं। इसी प्रकार पाँच उन प्राकृतिक वस्तुओं के नाम लिखो जो सबको समान भाव से देते रहते हैं।
उत्तर:
सर्दी, गर्मी, वर्षा, फल, फूल, नदियों का कल-कल करता बहता जल आदि ऐसे प्राकृतिक उपहार हैं जो सभी प्राणियों को समान भाव से मिलते हैं।

III. सरलार्थ करो

काँटों से खिलकर हम सबने,
हँस-हँस कर है जीना सीखा,
एक सूत्र में बँधकर हमने,
हार गले का बनना सीखा।
सबके लिए सुगन्ध हमारी
हम शृंगार धनी-निर्धन के।
उत्तर::
यह पद्यांश ‘हम सुमन एक उपवन के’ नामक कविता से लिया गया है। इसमें कवि कहता है कि हम सभी देशवासियों ने मुसीबतों तथा कठिनाइयों का डटकर सामना करना सीखा है। हमें रास्ते के काँटों की कोई परवाह नहीं है। हम सभी एक ही धागे में बँधे हुए हैं। हमने एक हार बनकर भारत माँ की शोभा बढ़ायी है। हम सब एक हैं। हमारी खुशब सभी के लिए समान है। हम अमीरग़रीब सभी की शोभा हैं। हम किसी से कोई भेदभाव नहीं रखते।

PSEB 5th Class Hindi Solutions Chapter 4 हम सुमन एक उपवन के

IV. वाक्यों में प्रयोग करो

(i) उर की कली खिलना = खुश होना
(ii) गले का हार बनना = अतिप्रिय, सदा साथ लगे रहना।
उत्तर:
(i) उर की कली खिलना = खुश होनाअपने जन्म दिन पर चाचा जी से सुन्दर घड़ी पाकर मेरे तो उर की कली खिल गई।
(ii) गले का हार बनना = अतिप्रिय-अपने अच्छे कार्यों से मन्त्री तो राजा के गले का हार बन गया।।

V. समानार्थक शब्द लिखो

(i) सुमन = प्रसून
(ii) उपवन = ……………
(iii) धरती = …………
(iv) जल = …………
(v) पवन = ………..
(vi) सूरज = …………
(vii) चाँद = ……………
(viii) भ्रमर = ………………..
(ix) गगन = ………..
उत्तर:
समानार्थक शब्द
(i) सुमन = प्रसून, पुष्प।
(ii) उपवन = बाग़, बगीचा।
(ii) धरती = धरा, पृथ्वी।
(iv) जल = पानी, नीर।
(v) पवन = हवा, वायु।
(vi) सूरज = सूर्य, रवि।
(vii) चाँद = राकेश, चन्द्र।
(vii) भ्रमर = भंवरा, अलि।
(ix) गगन = आकाश, आसमान।

VI. विपरीत शब्द मिलाओ

(i) धरती – दुर्गन्ध।
(ii) धूप – अन्धेरी
(iii) चाँदनी – निर्धन
(iv) सुगन्ध – छाया
(v) धनी – आकाश।
उत्तर:
विपरीतार्थक शब्द:
(i) धरती – आकाश।
(ii) धूप – छाया।
(iii) चाँदनी – अन्धेरी।
(iv) सुगन्ध – दुर्गन्ध।
(v) धनी – निर्धन।

VII. नए शब्द बनाओ

(i) सूत्र = त् + रत्र ………… , ……………
(ii) श्रृंगार = श् + ऋ = शृ, श्रृंग ……………., ……………
उत्तर:
(i) त्र = पुत्र, पत्र।
(ii) शृ = शृंगाल।

अध्यापन संकेत : अध्यापक इस कविता को पढ़ाते हुए बच्चों को बताए ‘कि देखो कवि की कल्पना का चमत्कार। एक फूल के माध्यम से एक ओर तो प्रकृति की नि:स्वार्थ देन की ओर संकेत किया है तो दूसरी ओर देश की एकता का सन्देश है। अलग-अलग प्रांतों, भाषा, वेशभूषा, त्योहार, संस्कृति के होते भी हम सब एक हैं। हमारा एक माली, ईश्वर है जो हमें सुख-दुःख में जीना सिखाता है।

बहुवैकल्पिक प्रश्न

पूछे गए प्रश्नों के सही विकल्प पर (✓) निशान लगाएं

प्रश्न 1.
कवि ने मनुष्यों को किसके समान माना है ?
(क) फूल
(ख) धूल
(ग) भूल
(घ) मूल।
उत्तर:
(क) फूल

PSEB 5th Class Hindi Solutions Chapter 4 हम सुमन एक उपवन के

प्रश्न 2.
सारा संसार किससे सींचा जाता है ?
(क) फूल
(ख) धूल
(ग) जल
(घ) बादल।
उत्तर:
(ग) जल

प्रश्न 3.
सबसे मिलकर किसकी शोभा बढ़ती है ?
(क) घर की
(ख) उपवन की
(ग) बाहर की
(घ) धरती की।
उत्तर:
(ख) उपवन की

प्रश्न 4.
फूलों ने हंसकर जीना किससे सीखा है ?
(क) कांटों से
(ख) बातों से
(ग) उपवन से
(घ) लोगों से।
उत्तर:
(क) कांटों से

प्रश्न 5.
फूल अपनी सुगंध किसे देता है ?
(क) राजा को
(ख) रानी को
(ग) भंवरे को
(घ) सबको।
उत्तर:
(घ) सबको।

हम सुमन एक उपवन के Summary

हम सुमन एक उपवन के पाठ का सार

कवि हम सब मनुष्यों को फूल के समान मान कर कहता है कि यह धरती हम सबकी है जिस पर हमने जन्म लिया है। इस पर रहते हुए समान रूप से धूप-पानी हमने प्राप्त किया है। हवा के झूलों में हम झूले हैं। सूर्य और चाँद ने हमारे प्रति एक-सा अच्छा व्यवहार किया है। हमें भंवरों-सी मीठी आवाज़ प्राप्त हुई है। चाहे हमारे रूप-रंग अलग-अलग हैं पर हम सब धरती रूपी इस उपवन की शोभा हैं। इस आसमान के नीचे रहने वाले हम सब का ईश्वर रूपी माली एक ही है। कष्टों में रहकर भी हमने हंस-हंस कर जीना सीखा है। हम चाहे अमीर हों या गरीब हमने एक साथ मिल-जुल कर रहना सीखा है।

पद्यांशों के सरलार्थ

1. हम सब सुमन एक उपवन के।
एक हमारी धरती सबकी।
जिसकी मिट्टी में जनमे हम,
मिली एक ही धूप हमें है।
सींचे गए एक जल से हम।
पले हुए हैं झूल-झूलकर ।
पलनों में हम एक पवन के।

शब्दार्थ:
सुमन = फूल। उपवन = बगीचा, बाग़। पलना = बच्चों का झूला। पवन = हवा।

सरलार्थ:
यह पद्यांश ‘हम सुमन एक उपवन के’ कविता से लिया गया है। इसमें कवि कहता है कि हम सभी देशवासी एक ही बाग़ में खिले फूलों के समान हैं। एक ही धरती पर रहते हैं। इस धरती की मिट्टी में हमारा जन्म हुआ है। हम सब को एकजैसी धूप मिली है। हम सब ने एक ही जैसा जल पिया है। हम सभी एक ही वायु के पलनों में झूल झूल कर बड़े हुए हैं। हम सब को एक ही जैसी वायु मिली है। कहने का भाव है कि हम सब एक हैं।

भावार्थ:
कवि का मानना है कि हम सब एक समान हैं।

2. सूरज एक हमारा,
जिसकी किरणें उर की कली खिलाती,
एक हमारा चाँद,
चाँदनी जिसकी हम सबको नहलाती।
मिले एक से स्वर हमको हैं
भ्रमरों के मीठे गुंजन के।

शब्दार्थ:
सूरज = सूर्य। उर = हृदय। स्वर = आवाज़। भ्रमर = भौरे। गुंजन = गूंज, भँवरे की आवाज़।

सरलार्थ:
यह पद्यांश ‘हम सुमन एक उपवन के’ नामक कविता से लिया गया है। इसमें कवि कहता है कि हम सब का एक ही सूरज है। उसकी किरणों से हमारे मन की कली खिल उठती है। हम सब का चाँद भी एक ही है। हम सभी उसकी शीतल चाँदनी का आनन्द लेते हैं। भंवरों के मीठे-मीठे गीतों की तरह हम सब को एक जैसी आवाजें मिली हुई हैं।

भावार्थ:
ईश्वर ने हमें एक-सी सुख-सुविधाएं – और बोलने की शक्ति दी है।

PSEB 5th Class Hindi Solutions Chapter 4 हम सुमन एक उपवन के

3. रंग-रंग के रूप हमारे
अलग-अलग है क्यारी-क्यारी,
लेकिन हम सबसे मिलकर ही है
उपवन की शोभा सारी।
एक हमारा माली,
हम सब रहते नीचे एक गगन के।

शब्दार्थ;
रूप = आकृति। शोभा = सुन्दरता।। गगन = आकाश।

सरलार्थ:
यह पद्यांश ‘हम सुमन एक उपवन के’ नामक कविता से लिया गया है। इसमें कवि कहता है कि भारतवर्ष एक बाग़ की तरह है। हम सब भारतीय हैं, इस बाग़ के भिन्न-भिन्न फूल हैं। हमारे रंग-रूप भले ही अलग-अलग हैं फिर भी हम सब एक हैं। हमारे प्रदेश बाग़ की अलग-अलग क्यारियों की तरह हैं। परन्तु हम सब मिलकर ही इस भारत रूपी बाग़ की सारी शोभा बढ़ाते हैं। हमारा माली एक ही है और हम एक ही आकाश के नीचे मिलकर रहते हैं। भाव यह है कि अलग-अलग होते हुए भी हम सब एक हैं।

भावार्थ:
कवि का मानना है कि चाहे हम सबके रूप-रंग अलग-अलग हैं पर हमारा ईश्वर रूपी माली एक ही है।

4. कांटों में खिलकर हम सबने
हँस-हँस कर है जीना सीखा,
एक सूत्र में बंधकर हमने
हार गले का बनना सीखा।
सबके लिए सुगंध हमारी
हम श्रृंगार धनी-निर्धन के।

शब्दार्थ:
सूत्र = धागा, डोरी। सुगन्ध = खुशबू। श्रृंगार = सजावट। निर्धन = ग़रीब धनी = अमीर।

सरलार्थ:
यह पद्यांश ‘हम सुमन एक उपवन के’ नामक कविता से लिया गया है। इसमें कवि कहता है कि हम सभी देशवासियों ने मुसीबतों तथा कठिनाइयों का डटकर सामना करना सीखा है। हमें रास्ते के काँटों की कोई परवाह नहीं है। हम सभी एक ही धागे में बँधे हुए हैं। हमने एक हार बनकर भारत माँ की शोभा बढ़ायी है। हम सब एक हैं। हमारी खुशब सभी के लिए समान है। हम अमीरग़रीब सभी की शोभा हैं। हम किसी से कोई भेदभाव नहीं रखते।

भावार्थ;
हम सबने कष्टों को झेलते हुए एक साथ मिल-जुल कर जीना सीखा है।

PSEB 5th Class Maths Solutions Chapter 9 Volume Intext Questions

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 9 Volume Intext Questions and Answers.

PSEB 5th Class Maths Solutions Chapter 9 Volume Intext Questions

Page No. 195

(a) Tick (✓) the object which takes more space:

Question 1.
PSEB 5th Class Maths Solutions Chapter 9 Volume Intext Questions 1
Solution:
PSEB 5th Class Maths Solutions Chapter 9 Volume Intext Questions 2

PSEB 5th Class Maths Solutions Chapter 9 Volume Intext Questions

Question 2.
PSEB 5th Class Maths Solutions Chapter 9 Volume Intext Questions 3
Solution:
PSEB 5th Class Maths Solutions Chapter 9 Volume Intext Questions 4

Question 3.
PSEB 5th Class Maths Solutions Chapter 9 Volume Intext Questions 5
Solution:
PSEB 5th Class Maths Solutions Chapter 9 Volume Intext Questions 6

PSEB 5th Class Maths Solutions Chapter 9 Volume Intext Questions

Question 4.
PSEB 5th Class Maths Solutions Chapter 9 Volume Intext Questions 7
Solution:
PSEB 5th Class Maths Solutions Chapter 9 Volume Intext Questions 8

Question 5.
PSEB 5th Class Maths Solutions Chapter 9 Volume Intext Questions 9
Solution:
PSEB 5th Class Maths Solutions Chapter 9 Volume Intext Questions 10

(b) Write in ascending order according to the space each object covers.
PSEB 5th Class Maths Solutions Chapter 9 Volume Intext Questions 11

PSEB 5th Class Maths Solutions Chapter 9 Volume Intext Questions

Solution:
PSEB 5th Class Maths Solutions Chapter 9 Volume Intext Questions 12

PSEB 5th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.4

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.4

Question 1.
Solve the following :
(a) 450 × 6
(b) 963 × 9
(c) 529 × 23
(d) 988 × 38
(e) 912 × 56
(f) 806 × 56
(g) 252 × 54
(h) 1888 × 19
(i) 2005 × 34
(j) 1560 × 64
(k) 10569 × 8
(l) 10210 × 9
(m) 230 × 150
(n) 400 × 225
Solution:
PSEB 5th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.4 2
PSEB 5th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.4 3

Question 2.
Find the product of the following:
(a) 4045 × 23
(b) 1609 × 30
(c) 363 × 134
(d) 455 × 208
(e) 105 × 120
(f) 1440 × 25
(g) 1530 × 61
(h) 3817 × 12
(i) 1908 × 35
(j) 1000 × 29
Solution:
PSEB 5th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.4 4
PSEB 5th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.4 5

PSEB 5th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.4

Question 3.
Find the digits in place of * :
PSEB 5th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.4 6
Solution:
PSEB 5th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.4 7

PSEB 5th Class Maths MCQ Chapter 9 Volume

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 9 Volume MCQ Questions and Answers.

PSEB 5th Class Maths Chapter 9 Volume MCQ Questions

Multiple Choice Questions

Tick (✓) the correct answer :

Question 1.
Volume of cube with side 9 cm is :
(a) 81 cubic cm
(b) 90 cubic cm
(c) 729 cubic cm
(d) 8 cubic cm.
Answer:
(c) 729 cubic cm

PSEB 5th Class Maths Solutions MCQ Chapter 9 Volume

Question 2.
Find volume of cuboid with length 6 cm, breadth 4 cm and height 2 cm :
(a) 24 cubic cm
(b) 28 cubic cm
(c) 64 cubic cm
(d) 48 cubic cm.
Answer:
(d) 48 cubic cm.

Question 3.
Which is not the standard unit of volume?
(a) cubic cm
(b) sq. m
(c) cubic mm
(d) cubic metre.
Answer:
(b) sq. m

Question 4.
A cuboid with all sides equal is called ……………
(a) square
(b) cube
(c) cuboid
(d) rectangle.
Answer:
(b) cube

PSEB 5th Class Maths Solutions MCQ Chapter 9 Volume

Question 5.
Count the number of cubes and find the volume of cuboid.
PSEB 5th Class Maths Solutions MCQ Chapter 9 Volume 1
Solution:
Volume of cuboid = 8 cubic cm.

Question 6.
A trolly is full of bricks. Length of the trolly is 400 cm, breadth is 200 cm and height is 100 cm. Length, breadth and height of a brick are 20 cm, 10 cm and 6 cm respectively. How many bricks are there in the trolly?
Solution:
Length of trolly, L = 400 cm
Breadth of trolly B = 200 cm
Height of trolly, H = 100 cm
Volume of trolly = L × B × H
= 400 cm × 200 cm × 100 cm.
Length of a brick, l = 20 cm
Breadth of a brick; b = 10 cm
Height of a brick, h = 6 cm
Volume of a brick = 20 cm × 10 cm × 6 cm
Number of bricks in the trolly
= \(\frac{\text { Volume of trolly }}{\text { Volume of each brick }}\)
= \(\frac{400 \times 200 \times 100}{20 \times 10 \times 6}\)
= \(\frac{20 \times 20 \times 50}{3}\) = \(\frac{20000}{3}\)
= 6666 (Approx)

PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds

Punjab State Board PSEB 10th Class Science Book Solutions Chapter 4 Carbon and its Compounds Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Science Chapter 4 Carbon and its Compounds

PSEB 10th Class Science Guide Carbon and its Compounds Textbook Questions and Answers

Question 1.
Ethane, with the molecular formula C2H6 has :
(a) 6 covalent bonds
(b) 7 covalent bonds
(c) 8 covalent bonds
(d) 9 covalent bonds.
Answer:
(b) 7 covalent bonds

Question 2.
Butanone is a four-carbon compound with the functional group :
(a) carboxylic acid
(b) aldehyde
(c) ketone
(d) alcohol.
Answer:
(c) ketone

Question 3.
While cooking, if the bottom of the vessel is getting blackened on the outside, it means that :
(а) the food is not cooked completely
(b) the fuel is not burning completely
(c) the fuel is wet
(d) the fuel is burning completely.
Answer:
(b) the fuel is not burning completely

Question 4.
Explain the nature of the covalent bond using the bond formation in CH3Cl.
Answer:
The formation of CH3Cl can be represented as :
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 1
Carbon forms single covalent bonds with three H- atoms and one Cl – atom by sharing one electron pair with each C-H bonds are non-polar. But C – Cl bond is polar because C and H leave almost same electronegativity whereas Cl has more electronegativity than carbon.

PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds

Question 5.
Draw the electron dot structures for :
(a) ethanoic acid
Answer:
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 2

(b) H2S
Answer:
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 3

(c) propanone
Answer:
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 4

(d) F2.
Answer:
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 5

Question 6.
What is an homologous series? Explain with an example.
Answer:
A series of compounds having similar structural formulae, same functional group and hence similar chemical properties is called a homologous series. In the homologous series any two adjacent members differ by CH2 unit in their molecular formulae.

For example homologous series of aldehydes (or alkanals) can be represented as :

H – CHO Methanal
CH3 – CHO Ethanal
CH3 – CH2 – CHO Propanal
CH3 – CH2 – CH2 – CHO Butanal
CH3 – CH2 – CH2 – CH2 – CHO Pentanal and so on.

Question 7.
How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties?
Answer:
Differences between ethanol and ethanoic acid

Ethanol Ethanoic acid
1. It is a colourless liquid having a pleasant smell. 1. It is colourless liquid having vineger like smell.
2. It has no action with a litmus solution. 2. It turns blue litmus solution red.
3. It has no action with sodium hydrogen carbonate solution. 3. It decomposes sodium hydrogen carbonate solution giving brisk effervescence of carbon dioxide gas.

Question 8.
Why does micelle formation take place when soap is added to water? Will a micelle be formed in other solvents such as ethanol also?
Answer:
Soap molecule has two ends, one is hydrophilic, and it dissolves in water, while the other end is hydrophobic, and it dissolves in hydrocarbons. When soap is at the surface of water , the hydrophobic ‘tail’ of soap will not be soluble in water and the soap will align along the surface of water with the ionic end in water and the hydrocarbon ‘tail’ pointing out of water.
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 6
Inside water, these molecules have a unique orientation which keeps the hydrocarbon portion out of the water. This is achieved due to the formation clusters of molecules in which the hydrophobic tails are in the interior of the cluster and the ionic ends are on the surface of the cluster. This formation is called a micelle.

Such micelles can be formed in other polar solvents like ethanol.

Question 9.
Why are carbon and its compounds used as fuels for most applications?
Answer:
Carbon and its compounds are used as fuels for most applications because they bum producing a large amount of heat and light.

PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds

Question 10.
Explain the formation of scum when hard water is treated with soap.
Answer:
When soap is added to hard water, the soluble calcium and magnesium salts present in it react with soap to give insoluble calcium salt of soap which produces scum.
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 7

Question 11.
What change will you observe if you test soap with litmus paper (red and blue)?
Answer:
Soap solution will turn red litmus paper blue because soap is alkaline in nature.

Question 12.
What is hydrogenation? What is its industrial application?
Answer:
The addition of hydrogen to unsaturated hydrocarbons in the presence of catalysts like palladium, platinum, nickel etc. to give saturated hydrocarbons is called hydrogenation.
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 8
This reaction is used for hydrogenation of liquid vegetable oils using a nickel catalyst to get artificial or vanaspati ghee.

Question 13.
Which of the following hydrocarbons undergo addition reaction : C2H6, C3H8, C3H6, C2H2 and CH4.
Answer:
Out of C2H6, C3H8, C3H6, C2H2 and CH4; C3H6 and C2H2 undergo addition reactions because they contain multiple bonds.

Question 14.
Give a test that can be used to differentiate chemically between butter and cooking oil.
Answer:
Distinction between Butter and Cooking oil:

Butter Cooking Oil
1. It is solid at room temperature. 1. It is liquid at room temperature.
2. Mix equal volumes of HgCl2 solution in 50% alcohol and 5% iodine solution in alcohol. To this add lg of butter. Violet colour does not fade away. 2. Mix equal volumes of HgCl2 solution in 50% alcohol and 5% solution of iodine in alcohol. To this add 1 ml of cooking oil. Violet colour fades away.
3. Take 2 g of butter in a test tube. To this add 1ml of cone. HCl and a few drops of 2% furfural solution in alcohol. Shake and allow to stand for 5 – 10 minutes. No rose red coloration appears. 3. Take 2 g of cooking oil in a test tube. To this add 1 ml of cone. HCl and a few drops of 2% .furfural solution in alcohol. Shake it and allow to stand for 5 – 10 minutes. Rose red coloration is obtained.

Question 15.
Explain the mechanism of the cleansing action of soaps.
Answer:
Mechanism of cleansing action of Soap:
Soaps are sodium or potassium salts of higher fatty acids e.g. sodium palmitate, C15H31COONa+, sodium stearate, C17H35COONa+ etc. A molecule of soap consists of two parts :

  1. a long chain hydrocarbon part (C15H31, C17H35 …. etc.) which is soluble in oil and
  2. ionic part on polar group, – COONa+ which is soluble in water. Thus a molecule of soap can be represented as :

PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 9
The long hydrocarbon chain is insoluble in water but soluble in oil and greases whereas the ionic or polar part is soluble in water. Soap has a capacity to clean a dirty piece of cloth whereas ordinary water cannot. The dirty clothes contain greasy and oily substance (dirt). Soap molecules dissociate in water to give carboxylate ion (RCOO) and cation (Na+). When soap added to dirty clothes dipped in water, the hydrocarbon part of carboxylate group dissolving in greasy or oily dirt particles where the polar (COO) group remain attached to water. In this way each oil droplet acquires negative charge.
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 10
The cleansing action of Soap

These negative charged oil droplets called micelles cannot coalesce and hence form a stable emulsion water. These small droplets along with dirt can be easily washed away with water. Thus the soap helps in removing greasy dirt by producing a stable oil in wrater type emulsion. Also the soap reduces surface tension of water. Hence cloth is wetted more effectively and is cleaned.

Science Guide for Class 10 PSEB Carbon and its Compounds InText Questions and Answers

Question 1.
What would be the electron-dot structure of carbon dioxide which has the formula CO2?
Answer:
In carbon dioxide, carbon atoms are bonded with two oxygen atoms. The atomic number of carbon is 6, and it has four electrons in the outer shell.

To make an octet it requires four electrons. Oxygen requires only two electrons in the outer shell. Therefore electron-dot structure will be :
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 11
Every oxygen atom is joined to carbon atom by double bond.

Question 2.
What would be the electron-dot structure of a molecule of sulphur which is made up of eight atoms of sulphur? (HINT. The eight atoms of sulphur are joined together in the form of ring.)
Answer:
The atomic number of sulphur is 16
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 12
Sulphur has 6 electrons in the outermost shell and to complete an octet it requires 2 electrons.
∴ Sulphur atom will share 2 electrons. It’s chemical formula is S8.
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 13

PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds

Question 3.
How many structural isomers can you draw for pentane?
Answer:
Three ; n-Pentane, iso-pentane, neo-pentane.
Structural isomers can be drawn for pentane.
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 14

Question 4.
What are the two properties of carbon which lead to the huge number of carbon compounds we see around us?
Answer:

  1. Catenation: The carbon atoms have an astonishing property to combine and form bond with other carbon atoms to form long chain compounds. This property is known as catenation. In this, either long chain of carbon are in ring form or the carbon atoms join in single, double or triple bond.
  2. Tetravalency: Carbon has four electrons in the outermost shell. That is why its valency is four and it has got capacity to make bonds with other elements. Oxygen, Hydrogen, Nitrogen, Sulphur, Chlorine and many other elements can make new compounds with the help of carbon.

Question 5.
What will be the formula and electron dot structure of cyclopentane?
Answer:
Molecular formula of cyclopentane = C5 H2 × 5 = C5H10
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 15

Question 6.
Draw the structures for following compounds :
Are structural isomers possible for bromopentane?
(i) Ethanoic acid
Answer:
Ethanoic acid (CH3COOH)
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 16

(ii) Bromopentane
Answer:
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 17
Due to exchange of position of carbon with bromine, many isomers of bromopentane are possible.
For example :
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 18

(iii) Butanone
Answer:
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 19

(iv) Hexanal.
Answer:
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 20

Question 7.
How would you name the following compounds?
(i) CH3 – CH2 – Br
Answer:
Bromoethane

PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 21
Answer:
Methanal

PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 22
Answer:
Hex-1-yne.

PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds

Question 8.
Why is the conversion of ethanol to ethanoic acid an oxidation reaction?
Answer:
This is because in this reaction oxygen gets added to ethanol.
PSEB 10th Class Science Solutions Chapter 4 Carbon and its Compounds 23

Question 9.
A mixture of oxygen and ethyne is burnt for welding. Can you tell why a mixture of ethyne and air is not used?
Answer:
A mixture of ethyne and air is not burnt for welding. This is because air also contains nitrogen along with oxygen. Nitrogen will also burn in oxygen producing oxides of nitrogen such as nitre oxide (NO) and nitrogen dioxide (NO2) which cause pollution.

Question 10.
How would you distinguish experimentally between an alcohol and a carboxylic acid?
Answer:
The following two tests are used :

  • Litmus test: Treat the given compound with blue litmus solutions. If the blue litmus solution turns red, it is a carboxylic acid and if does not turn red, it is an alcohol.
  • Sodium bicarbonate test: Add some sodium bicarbonate solution to the given compound. If their is a brisk evolution of a colourless and odourless gas (CO2) which turns freshly prepared lime water milk, it is carboxylic acid and if their is no effervescence, it is an alcohol.

Question 11.
What are oxidising agents?
Answer:

  • The substances which can oxidise other substances by giving oxygen are called oxidising agents.
  • Examples: Alkaline potassium permanganate solution, acidified potassium dichromate solution, etc.

Question 12.
Would you be able to check if water is hard using a detergent?
Answer:
No, we can’t check whether the water is hand or soft using a detergent.

Question 13.
People use a variety of methods to wash clothes. Usually after adding the soap, they ‘beat’ the clothes on a stone, or beat it with a paddle, scrub with a brush, or the mixture is agitated necessary to get clean clothes?
Answer:
This is because when soap molecules dissolve in the dirt, the dirt is somewhat loosened from the clothes, and in order to remove it from clothes, the clothes have to be beaten on a stone or beaten with a paddle or scrubbed with a brush or mixture has to be agitated in washing machines.