202110207-APEX-STUDENT-WORKBOOK-PHYSICAL_SCIENCE-G09-PART1

∴ F W.B = m.v/t According Newton’s third law of motion, ∴ F B.W = –F W.B m.v/t= –m.v/t Total force = m.v/t + m.v/t = 2 m.v/t Q5. If a car is travelling westwards with a constant speed of 20m/s, what is the resultant force acting on it? [Refer to TB page 35 Q11] A. If the car is traveling westward (constant direction) at constant speed, then the velocity is constant. That means the acceleration is zero. And that means (Newton’s Second Law) that the net (resultant) force on the car is also zero. SESSION 2. SECOND LAW OF MOTION 49

SESSION 3 THIRD LAW OF MOTION 3.1 Mind Map 3.2 Terminology i. Action and reaction pair: Equal and opposite forces that act on different objects. 3.3 Key Concepts i. Two opposing forces are called the action and reaction pair. ii. Newton’s third law of motion: If one object exerts a force on another object, the second object exerts a force on the first one with equal magnitude but in opposite direction. SESSION 3. THIRD LAW OF MOTION 50

3.4 Conceptual Understanding Q1. Illustrate an example of the third Law of motion. [Refer to TB page 35 Q7] A. The third law of motion is observed when a passenger jumps out of a rowing boat. As the passenger jumps forward, the force on the boat moves it backwards. Q2. A horse continues to apply a force in order to move a cart with a constant speed. Explain. [Refer to TB page 36 Q4, Try These] A. A horse continues to apply a force in order to move a cart with a constant speed. This enables the cart move forward. Otherwise, if the action is always equal to the reaction, the cart would not have moved. 3.5 Asking Questions and Making Hypothesis Q1. Divya observed a horse pulling a cart. She thought that cart also pulls the horse with same force in opposite direction. As per third law of motion the cart should not move forward. But her observation of moving cart raised some questions in her mind. Can you guess what questions are raised in her mind? [Refer to TB page 35 Q9] A. The questions that arose in the mind of Divya might be: i. Why did the cart move when the horse began running? ii. Is the third law of motion false? iii. What physical condition enabled the cart to move in the direction in which the horse was running? iv. Does the cart continue to move even if the horse stops pulling it? v. What is the action and reaction in this system? Q2. If a fly collides with the windshield of a fast moving bus: [Refer to TB page 35 Q8] (a) Is the impact force experienced same for the fly and the bus? Why? (b) Is the same acceleration experienced by the fly and the bus? Why? A. (a) The impact force experienced by the by and the bus is the same. According to Newton’s third law of motion, the total momentum before impact and after impact remains unchanged and as the masses of the fly and car are both constant, the forces are equal but opposite in direction. (b) The accelerations experienced by the fly and the bus are different since the masses of the two bodies are different but the forces are the same. SESSION 3. THIRD LAW OF MOTION 51

SESSION 4 CONSERVATION OF MOMENTUM AND IMPULSE 4.1 Terminology i. Momentum – Product of mass and velocity. ii. Conservation of momentum –In an isolated system in absence of a net external force, the total momentum is conserved. iii. Impulse – Product of force and time. iv. Impulsive forces – Forces that act for a very short time. 4.2 Solved Examples Q1. A cannon of mass m1 = 12000 kg located on a smooth horizontal platform fires a shell of mass m2 = 300 kg in horizontal direction with a velocity v2 = 400m/s. Find the velocity of the cannon after it is shot. [Refer to TB page 33] A. Since the pressure of the powder gases in the bore of the cannon is an internal force so, the net external force acting on cannon during the firing is zero. Let v1 be the velocity of the cannon after shot. The initial momentum of system is zero. The final momentum of the system = m1 v1 + m2 v2 From the conservation of linear momentum, we get m1 v1 + m2 v2 = 0 m1 v1 = – m2 v2 or v1 = –m2 v2 / m1 Substituting the given values in the above equation, we get SESSION 4. CONSERVATION OF MOMENTUM AND IMPULSE 52

v =– (300kg)x (400m/s) = –10 m/s. 12000kg Here ’–’ sign indicates that the canon moves in a direction opposite to the direction of bullet. 4.3 Key Concepts i. The product of mass and velocity is called as momentum. ii. When two bodies collide, the total momentum is unchanged before and after the colli- sion.The total quantity of momentum remains constant in a system until it is subjected to external influence. This is known as law of conservation of momentum. iii. An indefinitely large force acting for a very short time but producing a finite change of momentum is called impulse. iv. Product of net force and interaction time is called impulse. v. Impulse is equivalent to rate of change of momentum. vi. Force exerted over a limited time is called impulsive force. 4.4 Conceptual Understanding Q1. What is the momentum of a 6.0 kg bowling ball with a velocity of 2.2 m/s? [Refer to TB page 35 Q3] A. Momentum of a body p = mv Mass of bowling ball, m = 6.0 kg Velocity of the ball, v =2.2 m/s Momentum of bowling ball = 6.0 x 2.2 kg –m/s = 13.2 kg –m/s Q2. Two ice skaters initially at rest, push each other. If one skater whose mass is 60 kg has a velocity of 2 m/s, what is the velocity of other skater whose mass is 40 kg? [Refer to TB page 37 Q13, Try These] SESSION 4. CONSERVATION OF MOMENTUM AND IMPULSE 53

A. From the conservation of linear momentum, we have m1v1 + m2v2 = 0 or m1v1 = −m2v2 (1) m1 = mass of first skater = 60 kg v1 = velocity of first skater = 2m/s m2 = mass of second skater = 40 kg v2 = velocity of second skater = ? Substituting the above values in equation (1) 60 × 2 = –40 × v ⇒ v2 = − 60X2 = −3 m/s 40 ∴ Velocity of second skater = –3 m/s 4.5 Communication Through Drawing and Model Making Q1. Take some identical marbles. Make a path or a track keeping your note books on either side so as to make a path in which marbles can move. Now use one marble to hit the other marbles. Take two, three marbles and make them to hit the other marbles. What can you explain from your observation? [Refer to TB page 37 Q10, Try These] A. Students’ Activity SESSION 4. CONSERVATION OF MOMENTUM AND IMPULSE 54

—— CCE Based Practice Questions —— AS1-Conceptual Understanding [] Very Short Answer Type Questions [] 1. State true or false. [] [Refer to Session 2.2 ] (i) Force = mass of the body x acceleration due to gravity. (ii) Quantity of matter contained in a body is called weight. (iii) One Newton = 1kg x 1 m/s2. 2. Fill in the blanks. [Refer to Session 2.2 ] (i) Larger is the mass is the acceleration. (ii) Momentum depends on and . (iii) SI unit of momentum is or . (iv) Larger the net force is the acceleration. 3. Answer the following questions in one sentence. [Refer to Session 2.3 ] (i) A vehicle has a mass of 1500 kg. What must be the force between the vehicles and the road if the vehicle is to be stopped with a negative acceleration of 1.7 m/s2? 4. Fill in the blanks. [Refer to Session 2.3 ] (i) For every action there is and reaction. CHAPTER 2. LAWS OF MOTION 55

(ii) The two opposing force is known as and . (iii) Newton’s third law is applicable to interaction between bodies. (iv) The reaction force is indicated by sign. (v) The force exerted between two bodies is in magnitude and opposite in . 5. Answer the following questions in one sentence. [Refer to Session 2.1 ] (i) Who formulated laws of motion? (ii) Do all bodies have same inertia? (iii) Define inertia. (iv) Is it possible to have a world with no friction as imagined by Galileo? (v) How many laws of motion were proposed by Newton? 6. Fill in the blanks. [Refer to Session 2.1 ] (i) Mass is considered as the of inertia. (ii) If the mass of the object is greater, the more it resists its . (iii) Inertia depends on the of the object. 7. State true or false. [Refer to Session 2.4 ] (i) Law of conservation of momentum states in the absence of a net external force on the system, the momentum of the system remains changed. [] CHAPTER 2. LAWS OF MOTION 56

(ii) A system is said to be isolated when net external force acting on it is zero. ] [ 8. Fill in the blanks. . [Refer to Session 2.4 ] (i) A system is said to be isolated when net external force acting on it is (ii) When the resultant force acting on a system is zero, the total momentum of the system . This is called . (iii) The velocity with which the gun moves in backward direction is called as velocity. (iv) According to second law of motion Fnet = . (v) When the lift is stationary or moves with uniform velocity the acceleration is zero and the net force is _________________ . (vi) The change in momentum over a longer time requires force. (vii) Forces exerted over a limited time are called ______________________________. Short Answer Type Questions 9. Answer the following questions in 3-4 sentences. (i) [(Session 2.2)] State Newton’s second law of motion. (ii) [(Session 2.2)] What is the momentum of a 6 kg bowling ball with velocity 2.2 m/s? (iii) [(Session 2.2)] Write the equation to show how force and acceleration are related. 10. Answer the following questions in 3-4 sentences. (i) [(Session 2.3)] State Newton’s third law of motion. (ii) [(Session 2.3)] Explain the motion of a rocket. CHAPTER 2. LAWS OF MOTION 57

11. Answer the following questions in 3-4 sentences. (i) [(Session 2.1)] State first law of motion. (ii) [(Session 2.1)] State the relationship between inertia and mass. (iii) [(Session 2.1)] What happens if you kick a ball high? (iv) [(Session 2.1)] Give an example to show the relationship between inertia and mass. 12. Answer the following questions in 3-4 sentences. (i) [(Session 2.4)] State the law of conservation of momentum. (ii) [(Session 2.4)] Define impulse and impulsive force. (iii) [(Session 2.4)] Why are air bags used in cars? (iv) [(Session 2.4)] Derive an equation to say impulse is equal to change in momentum. Long Answer Type Questions 13. Answer the following questions in 6-8 sentences. (i) [(Session 2.2)] A force of 500 dynes acts on a mass of 0.05 kg over a distance of 200 m. Assuming that the mass is initially at rest, find the final velocity and time for which the force acts. 14. Answer the following questions in 6-8 sentences. (i) [(Session 2.3)] Identify the action and reaction process involved in the act of walking. 15. Answer the following questions in 6-8 sentences. (i) [(Session 2.1)] Explain the reason – a. A rider falls forward, when a galloping horse suddenly stops. b. Before taking a long jump, a boy runs a certain distance. c. While jumping out of a bus, it’s important to run along the same direction of the moving bus. CHAPTER 2. LAWS OF MOTION 58

(ii) [(Session 2.1)] Two objects have masses 8 kg and 2500 gm. Which one has more inertia? Why? 16. Answer the following questions in 6-8 sentences. (i) [(Session 2.4)] An electron of mass of 9 × 10-31 kg is moving with a linear velocity of 6 × 107 ms-1 . Calculate the linear momentum of the electron. (ii) [(Session 2.4)] A body has a linear momentum of 5 Ns. If the velocity of the body is 200 ms-1, find the mass of the body. AS2-Asking questions and making hypothesis Short Answer Type Questions 17. Answer the following questions in 3-4 sentences. (i) [(Session 2.1)] When driver applies the brake of the bus, we move forward. Why do you think this happens? (ii) [(Session 2.1)] Sanjay says the car possesses more inertia than a bicycle. Is that true? If so, why and if not, why? AS3-Experimentation and field investigation Long Answer Type Questions 18. Answer the following questions in 6-8 sentences. (i) [(Session 2.3)] Write the experiment to show the action and reaction forces acting on two different objects. AS4-Information skills and projects Very Short Answer Type Questions 19. Answer the following questions in one sentence. [Refer to Session 2.2 ] (i) On what factors does momentum depend? AS5-Communication through drawing and model making Short Answer Type Questions CHAPTER 2. LAWS OF MOTION 59

20. Answer the following question. (i) [(Session 2.3)] Observe the given diagram carefully. What are the momenta of mar- bles before and after collision? AS6-Appreciation and aesthetic sense, Values Long Answer Type Questions 21. Answer the following questions in 6-8 sentences. (i) [(Session 2.1)] Why do you appreciate Galileo Galilei? AS7-Application to daily life, concern to bio diversity Short Answer Type Questions 22. Answer the following questions in 3-4 sentences. (i) [(Session 2.2)] Solve the following problems – Two ice skaters initially at rest push each other. If one skater whose mass is 60 kg has a velocity of 2 m/s. What is the velocity of other skater whose mass is 40 kg? (ii) [(Session 2.2)] Calculate the momentum of a bullet having mass of 25 g is thrown using hand with a velocity of 0.1 m/s. Long Answer Type Questions 23. Answer the following questions in 6-8 sentences. (i) [(Session 2.2)] A car of mass 1000 kg develops a force of 500 N over a distance of 49 m. If initially the car is at rest find – a. Final velocity b. Time for which it accelerates 24. Answer the following questions in 6-8 sentences. (i) [(Session 2.4)] Which has more momentum between a metal sphere of mass 10 g moving with a velocity of 400 m/s and a cricket ball of mass 400 g thrown with a speed of 90 km/h? CHAPTER 2. LAWS OF MOTION 60

Objective Questions AS1-Conceptual Understanding 25. Choose the correct answer. (i) An object will remain in same motion as long as (A) it is pulled. (B) no external force is applied to it. (C)it’s acceleration is constant. (D)external force is applied to it. (ii) The property of matter that resists change in its state of motion or rest is called (A) acceleration (B) force (C) inertia (D)centripetal force (iii) Greater the mass of the body (A) more it will resist the change in state. (B) lesser it will resist the change in state. (C)mass does not affect the motion. (D)none of the above (iv) Larger the force the acceleration. (A) lesser (B) greater (C) same (D) zero CHAPTER 2. LAWS OF MOTION 61

(v) The following represents Newton’s second law of motion best: (A) p = mv (B) m = fa (C)m = a/f (D)a = fm (vi) A force acts on body of mass 1 kg and producing an acceleration of 1m/s2. This force will be (A) 1 N (B) 2 N (C)3 N (D)4 N (vii) The product of net force and interaction time is called as (A) impulse (B) momentum (C) acceleration (D) inertia AS2-Asking questions and making hypothesis 26. Choose the correct answer. (i) Which of the following object displays Newton’s third law of motion on operation? (A) Helicopter (B) Bicycle (C) Rocket (D) Car (ii) A body is moving with mass 5 kg and velocity 10 m/s. Another body with mass 25 kg and velocity 2 m/s is moving along the same line. Which of the following statements holds true? (A) Both bodies experience different forces. (B) Both bodies experience same forces. (C)Both bodies have different momentum. (D)Both bodies have same momentum. (B) f x m AS4-Information skills and projects (D)m x v 27. Choose the correct answer. (i) Linear momentum is given by (A) m x a (C)f x a CHAPTER 2. LAWS OF MOTION 62

3. IS MATTER PURE? SESSION 1 TYPES OF SUBSTANCES - MIXTURES, SOLUTIONS, COLLOIDS AND SUSPENSIONS 1.1 Mind Map 1.2 Terminology i. Pure substances – Pure substances are defined as substances that are made of only one type of atom or only one type of molecule (a group of atoms bonded together). ii. Mixture – A mixture is a combination of two or more pure substances in which each pure substance retains its individual chemical properties. iii. Heterogeneous mixture – A mixture in which the components are not uniformly dis- tributed is called a heterogeneous mixture. iv. Homogeneous mixture – A mixture in which the components are uniformly distributed is called a homogeneous mixture. v. Suspension – A suspension is a heterogeneous mixture in which solute–like particles settle out of a solvent–like phase some time after their introduction. SESSION 1. TYPES OF SUBSTANCES - MIXTURES, SOLUTIONS, COLL... 63

vi. Solution – A solution is a homogeneous mixture of two or more substances. The major component of a solution is called the solvent, and the minor, the solute. vii. Emulsions – Emulsions are mixtures which consist of two liquids that do not mix and they settle into layers when they are left undisturbed. viii. Colloidal dispersions – Colloids are heterogeneous mixtures in which the particle size is too small to be seen with the naked eye, but is big enough to scatter light. The colloid has the dispersed phase and the medium in which they are distributed is called the dispersion medium. ix. Saturated solution – At any particular temperature, a solution that has dissolved as much solute as it is capable of dissolving is said to be a saturated solution. x. Concentration of solution – Concentration of solution is defined as the amount of solute dissolved in a specific (fixed) amount of solvent. Units of concentration are Molarity, Molality, Normality. xi. Tyndall effect – The Tyndall effect, also known as Tyndall scattering, is light scattering by particles in a colloid or else particles in a very fine suspension. 1.3 Solved Examples Q1. A solution contains 50 g of common salt in 200g of water. Calculate the concentration in terms of mass by mass percentage of the solution. [Refer to TB page 42] A. Mass of solute (salt) = 50 g Mass of solvent (water) = 200 g Mass of solution = Mass of solute + Mass of solvent = 50 g + 200 g = 250 g Mass percentage of a solution = Mass o f solute × 100 = 50 × 100 = 20% Mass o f solution 250 1.4 Key Concepts i. All matter around us can be classified into two groups – pure substances and mix- tures. ii. A mixture is composed of different substances mixed in any proportion and not chem- ically combined. iii. A mixture in which the components are uniformly distributed is called a homogeneous mixture. iv. A mixture in which the components are not uniformly distributed is called a heteroge- neous mixture. SESSION 1. TYPES OF SUBSTANCES - MIXTURES, SOLUTIONS, COLL... 64

v. A homogeneous mixture of two or more substances is called a solution. Solution = solute + solvent. vi. The substance which is present in large quantity is called solvent and the substance which is in less quantity is called solute. vii. The amount of solute present in unit volume or per unit mass of a solution or solvent is called the concentration of the solution. viii. Materials that are insoluble in a solvent and have particles that are visible to naked eyes form a suspension. A suspension is a heterogeneous mixture. ix. Emulsions are mixtures which consist of two liquids that do not mix and they settle into layers when they are left undisturbed. x. A heterogeneous mixture in which the size of constituent particles is too small to be individually seen by naked eyes is called a colloid. But these particles are big enough to scatter light. xi. The scattering of a beam of light by the particles of a colloidal solution is called Tyndall effect. 1.5 Conceptual Understanding Q1. Explain the following giving examples: (a) Saturated solution (b) Pure substance (c) Colloid (d) Suspensions. [Refer to TB page 54 Q2] A. (a) Saturated solution: If we take 50 ml of water and add a spoonful of salt to it and stir well, the salt dissolves in water and no trace of it is left behind. If we continue adding more and more of salt keeping the solution at room temperature, at a certain point no more of the salt goes into the solution but settles at the bottom. It is called a saturated solution. Definition: At any particular temperature, a solution that has dissolved as much solute as it is capable of dissolving is said to be a saturated solution. SESSION 1. TYPES OF SUBSTANCES - MIXTURES, SOLUTIONS, COLL... 65

(b) Pure substance : The general term “Pure substance” means a substance free from adulteration. But in chemistry it means that the substance is homogeneous and the composition doesn’t change no matter which part of the substance has been taken as a sample for examination. E.g. Gold bar (c) Colloid : A mixture which contains very small particles of a substance but appears to be homogeneous and the particles of which scatter a beam of visible light is called a colloid. E.g . Milk (d) Suspension : If we add a little chalk powder to 20 ml. of water, we find that the particles of chalk do not dissolve, but remain suspended throughout the volume of wa- ter. It is a heterogeneous mixture and the particles are visible to the naked eye. Such heterogeneous mixtures are called suspensions. E.g. Syrups, Chalk powder added to water. Q2. Classify each of the following as a homogeneous or heterogeneous mixture. Give rea- sons. [Refer to TB page 54 Q6] Soda water, wood, air, soil, vinegar, filtered tea. A. S.No. Substance Nature of mixture 1 Soda water Homogeneous 2 Wood Heterogeneous 3 Air Homogeneous 4 Soil Heterogeneous 5 Vinegar Homogeneous 6 Filtered tea Homogeneous In the homogeneous solutions the particles are uniform throughout and in heteroge- neous they are not uniform. SESSION 1. TYPES OF SUBSTANCES - MIXTURES, SOLUTIONS, COLL... 66

Q3. How would you confirm that a colourless liquid given to you is pure water? [Refer to TB page 54 Q3] A. i. A pure substance has a characteristic melting point or boiling point at a given pressure. For example, pure water is always colourless, odourless and tasteless and boils at 373 K at normal atmospheric pressure. Fractional Distillation can be used to find whether a given solution is pure or not. On fractional distillation, the pure liquid will completely evaporate as it will have its characteristic boiling point and condense thus separate. Nothing will be left in the distillation flask. While if it is impure, the liquid will distill off leaving behind the impure part. ii. Observe the smell. We should not find any smell. iii. Observe with a naked eye, we should not find any suspended particles or fumes or air bubbles. iv. Pass a beam of light. It should not scatter. If the liquid satisfy all the above conditions then the given colourless liquid is pure. Q4. Which of the following materials fall in the category of a pure substance? Give reasons. [Refer to TB page 54 Q4] (a) Ice (b) Milk (c) Iron (d) Hydrochloric acid (e) Calcium oxide (f) Mercury (g) Brick (h) Wood (i) Air. A. (a) Ice: Ice is a pure substance. It is formed by two pure substances, namely hydrogen and oxygen which have chemically reacted to form the substance water. (b) Milk: Milk contains particles of cream and a little water too. So it is not a pure substance. (c) Iron: Iron is a pure substance. It contains molecules of only one element, namely iron. (d) Hydrochloric acid: It is formed when a gas called hydrogen chloride is dissolved in water. So it contains pure substances namely hydrogen, chlorine and oxygen. So it is a pure substance by itself. (e) Calcium oxide: Calcium oxide is a compound formed by calcium and oxygen. So it is a pure substance. (f) Mercury: Mercury is a shining metal in liquid state. It contains only one type of particles, namely mercury. So it is a pure substance. (g) Brick: Brick is prepared by using clay. As the clay contains humus minerals and other organic substances it is not a pure substance. (h) Wood: Wood is an organic material whose components are carbon and other pure substances. So wood is not a pure substance. SESSION 1. TYPES OF SUBSTANCES - MIXTURES, SOLUTIONS, COLL... 67

(i) Air: Air is a mixture of several gases namely nitrogen, oxygen, carbon dioxide, inert gases in addition to dust particles and several impurities. So, air is not a pure substance. Except Milk, Brick, Air and Wood remaining materials can be treated as pure substance. If we take any small part of Ice, Iron, Hydrochloric acid, Calcium oxide, Mercury and test for their components, we find that the composition is same throughout them. Q5. Identify the solutions among the following mixtures: (a) Soil (b) Sea water (c) Air (d) Coal (e) Soda water. [Refer to TB page 54 Q5] A. (a)Soil: Soil is a heterogeneous mixture of humus, fine particles of rocks, minerals and other organic substances. So it is not a solution. (b) Sea water: Sea water contains common salt dissolved in water. So it is a solution of solid in liquid. (c) Air: Air is a mixture of several gases. It contains nitrogen, oxygen, carbon dioxide, water vapour and inert gases. These gases are mixed up such that we cannot see their particles separately. So it is a solution of several gases. (d) Coal: Coal contains mostly carbon besides several organic substances, which is an element. So it is not a solution. (e)Soda water: Soda water is obtained by passing carbon dioxide into water at high pressure. We cannot recognise the carbon dioxide present in soda water by performing the identifying test for carbon dioxide. So soda water is a solution of carbon dioxide in water. Q6. Classify the following substances in the below given table: Ink, soda water, brass, fog, blood, aerosol sprays, fruit salad, black coffee, oil and water, boot polish, air, nail polish, starch solution, milk. [Refer to TB page 55 Q8] SESSION 1. TYPES OF SUBSTANCES - MIXTURES, SOLUTIONS, COLL... 68

A. Solution Suspension Emulsion Colloidal dispersion Black Fruit salad Oil and Fog coffee water Soda Nail polish Ink water Brass Milk Liquid Starch solution Boot polish Air Aerosol Sprays Blood Q7. Determine the mass by mass percentage concentration of a 100 g salt solution which contains 20 g salt. [Refer to TB page 55 Q9] A. Percentage of solution = Mass of solute × 100 ÷ Mass of solution Mass of solute = 20 g Mass of solution = 100 g So, percentage of solution = 20 × 100 ÷ 100 = 20 % 20 % NaCl solution. Q8. Calculate the concentration in terms of mass by volume percentage of the solution con- taining 2.5 g of potassium chloride in 50 mL of potassium chloride solution? [Refer to TB page 55 Q10] A. Mass of solute = 2.5 g Volume of KCl solution = 50 mL Mass/volume % = Mass of solute × 100 ÷ Volume of solution = 2.5 × 100 ÷ 50 = 5% SESSION 1. TYPES OF SUBSTANCES - MIXTURES, SOLUTIONS, COLL... 69

1.6 Experimentation and Field Investigation Q1. Take a solution, a suspension, a colloidal dispersion in different beakers. Test whether each of these mixtures shows the Tyndall effect by focusing a light at the side of the container. [Refer to TB page 55 Q12] A. The scattering of light by particles is called Tyndall effect. It is shown by suspensions and colloidal solutions. A true solution does not show Tyndall effect as the particle size is too small to scatter light. Q2. Which of the following will show Tyndall effect? How can you demonstrate Tyndall effect in them? [Refer to TB page 55 Q11] i. Salt solution ii. Milk iii. Copper sulphate solution iv. Starch solution A. Milk and starch solution will show Tyndall effect as they are colloidal solutions. SESSION 1. TYPES OF SUBSTANCES - MIXTURES, SOLUTIONS, COLL... 70

SESSION 2 SEPARATING THE COMPONENTS OF A MIXTURE 2.1 Mind Map 2.2 Terminology i. Evaporation – The phenomena of change of a liquid into vapours at any temperature below its boiling point is called evaporation. ii. Centrifuge – Centrifuge is a machine that uses a force pulling dense objects away from the center to separate particles or to draw off moisture. An example of a centrifuge is a machine that separates cream and milk. iii. Immiscible liquids – An immiscible liquid is one which doesn’t dissolve but forms a layer over another liquid and can be separated easily. iv. Miscible liquids – A liquid is said to be miscible if it dissolves completely in another liquid and is difficult to separate like alcohol is miscible in water. v. Chromatography – A technique for the separation of a mixture by passing it in solution or suspension through a medium in which the components move at different rates. vi. Distillation – Process in which the components of a substance or liquid mixture are sep- arated by heating it to a certain temperature and condensing the resulting vapours. vii. Fractional distillation – The process of separating the constituents of a liquid mixture by heating it and condensing separately the components according to their different boiling points. SESSION 2. SEPARATING THE COMPONENTS OF A MIXTURE 71

2.3 Key Concepts i. Heterogeneous mixtures can be separated into their constituents by using simple physical methods like filtration, handpicking, sieving etc. ii. Homogeneous mixtures and solutions can also be separated into their constituents, but needs advance methods and more complex machinery. 2.4 Conceptual Understanding Q1. Which separation techniques will you apply for the separation of the following? [Refer to TB page 54 Q1] a. Sodium chloride from its solution in water. b. Ammonium chloride from a mixture containing sodium chloride and ammonium chlo- ride. c. Small pieces of metal in the engine oil of a car. d. Different pigments from an extract of flower petals. e. Butter from curd. f. Oil from water. g. Tea leaves from tea. h. Iron pins from sand. i. Wheat grains from husk. j. Fine mud particles suspended in water. A. a. Sodium chloride from its solution can be separated by evaporation. SESSION 2. SEPARATING THE COMPONENTS OF A MIXTURE 72

b. Ammonium chloride is a substance which undergoes sublimation . So, on heating the mixture and cooling the vapours of sublimated ammonium chloride we can separate the constituents of this mixture. c. Small pieces of metal in the engine oil of a car can be separated by filtration through a filter paper. d. Different pigments from an extract of flower petals can be separated by using chro- matography. e. Butter can be separated from curd by churning or centrifuging the mixture using a centrifuge. f. Oil from water can be separated by using a separating funnel. As oil and water are immiscible, the denser liquid settles at the bottom while the lighter liquid (oil) floats on it. We can make the heavier water run down into a beaker from the separating funnel and later we can collect oil in another beaker. g. Tea leaves from tea can be separated by using a strainer and by filtration of the mixture. h. Iron pins are attracted by a magnet. So by passing a magnet through the mixture the iron pins stick to the magnet since they are attracted by the magnet. We can remove them from the magnet and separate them. i. Wheat grains from husk can be separated by winnowing . Standing at a higher level, if we drop the mixture down when the wind is blowing, the lighter particles of husk are blown by the wind to a little distance and fall as separate heap. The heavy wheat grains fall nearer as another heap. j. Fine mud particles suspended in water can be separated by decantation. if we keep the water containing mud particles in a vessel and allow it to settle for some time, the mud particles collect at the bottom of container and the pure water settles above these mud particles. SESSION 2. SEPARATING THE COMPONENTS OF A MIXTURE 73

2.5 Communication Through Drawing and Model Making Q1. Draw the figures of arrangement of apparatus for distillation and fractional distillation. What do you find the major difference in these apparatus? [Refer to TB page 55 Q13] A. Apparatus for distillation – Apparatus for fractional distillation – SESSION 2. SEPARATING THE COMPONENTS OF A MIXTURE 74

Observation: The major difference in the arrangement of apparatus for fractional dis- tillation over distillation is that, in fractional distillation we use a fractionating column between the distillation flask and the condenser. The beads taken in the fractionating column provide maximum possible surface area for the vapours to cool and condense repeatedly. 2.6 Application to Daily Life, Concern to Bio Diversity Q1. Write the steps you would use for making tea. Use the words given below and write the steps for making tea: Solution, Solvent, Solute, Dissolve, Soluble, Insoluble, filtrate and residue. [Refer to TB page 55 Q14] A. (1) About 25 ml. of water is taken in a vessel and heated on the flame of a gas stove. (2) Just when the water begins to boil, two teaspoonful of tea powder is added to the water and is continued heating for one to two minutes. The water is the solvent and the tea powder is the solute here. (3) The tea powder is insoluble in water. It is brewed in water and the product is tea decoction. (4) Then a strainer is used to separate/ filter out the insoluble tea powder from the solution and the liquid is collected in a separate glass. This is called filtrate, which is the black tea. (6) The residue left in the strainer is undissolved tea powder and is thrown away. (7) Appropriate amount of sugar is added to the black tea. Sugar is soluble in black tea. (8) Add appropriate amount of boiled milk.The tea is ready. SESSION 2. SEPARATING THE COMPONENTS OF A MIXTURE 75

SESSION 3 TYPES OF PURE SUBSTANCES 3.1 Mind Map 3.2 Terminology i. Elements – An element is a form of matter that cannot be broken down by chemical reactions into simpler substances. ii. Compounds – A compound is a substance composed of two or more different types of elements, chemically combined in a fixed proportion. 3.3 Key Concepts i. Pure substances are the substances which cannot be separated by any further methods of separation. ii. Pure substances can be either elements or compounds. An element is a form of matter that cannot be broken down by chemical reactions into simpler substances. SESSION 3. TYPES OF PURE SUBSTANCES 76

A compound is a substance composed of two or more different types of elements, chemically combined in a fixed proportion. iii. The properties of compounds differ from their constituent elements. But in a mixture the properties of its constituent elements or compounds are exhibited. 3.4 Conceptual Understanding Q1. Classify the following into elements, compounds and mixtures. [Refer to TB page 55 Q7] (a) Sodium (b) Soil (c) Sugar solution (d) Silver (e) Calcium carbonate (f) Tin (g) Silicon (h) Coal (i) Air (j) Soap (k) Methane (l) Carbon dioxide (m) Blood. A. Elements Compounds Mixtures Sodium Calcium carbonate Soil Silver Methane Sugar Solution Tin Carbon di oxide Coal Silicon Soap Air, Blood SESSION 3. TYPES OF PURE SUBSTANCES 77

—— CCE Based Practice Questions —— AS1-Conceptual Understanding Very Short Answer Type Questions 1. State true or false. [Refer to Session 3.1 ] (i) If the amount of solute present in a solution is less than saturation level, it is called saturated solution. [] (ii) If the amount of solute present is more, it is called concentrated solution. ] [ (iii) When a solid is completely dissolved in a solution, it is said to be a suspension. ] [ (iv) Mixtures are of two types. [] (v) A solution is a homogeneous mixture of two or more substances. [ ] 2. Match the following. Column B [(Session 3.1)] [ ] a. Diluted Column A i. Components of mixture are uniformly distributed ii. Component of solution that dissolves [ ] b. Saturated other component in it iii. Amount of solute present in saturated [ ] c. Homogenous solution at certain temperature iv. No more solute can be dissolved in so- [ ] d. Solvent lution v. In a solution, amount of solute present [ ] e. Solubility is little. CHAPTER 3. IS MATTER PURE? 78

3. State true or false. [ ] [Refer to Session 3.3 ] [ ] [ ] (i) Pure substance is classified into elements and compounds. [ ] [ ] (ii) Soil is a compound. (iii) Tin is an element. (iv) Calcium carbonate is a mixture. (v) Blood is a compound. 4. Answer the following questions in one sentence. [Refer to Session 3.3 ] (i) Classify the following into elements and compounds. a) Sodium – b) Methane – c) Silicon – d) Calcium carbonate– e) Tin – CHAPTER 3. IS MATTER PURE? 79

5. Match the following. Column B [(Session 3.2)] [ ] a. Separating funnel Column A i. Separation of salt from water ii. Separation of salt by ammonium chlo- [ ] b. Chromatography ride iii. Separation of dyes from ink [ ] c. Evaporation iv. Separation of two miscible liquids [ ] d. Sublimation v. Separation of immiscible liquids [ ] e. Distillation 6. Answer the following questions in one sentence. [Refer to Session 3.2 ] (i) How do we use the following methods in our daily life? a) Centrifugation – b) Distillation – c) Chromatography – d) Separating funnel – e) Evaporation – (ii) Which of the following methods would you use to separate: a) Sodium chloride from solution of water– b) Butter from curd– c) Oil from water– d) Iron pins from sand– e) Fine mud particles suspended in water– CHAPTER 3. IS MATTER PURE? 80

Short Answer Type Questions 7. Answer the following questions in 3-4 sentences. (i) [(Session 3.1)] Define homogeneous and heterogeneous mixture. (ii) [(Session 3.1)] Distinguish between dispersed phase and dispersion medium. (iii) [(Session 3.1)] Write the dispersed phase and dispersion medium for: a) Clouds b) Milk c) Cheese 8. Answer the following questions in 3-4 sentences. (i) [(Session 3.3)] Define pure substance. (ii) [(Session 3.3)] Define element and compound. (iii) [(Session 3.3)] Classify the following as element or compound. a) Oxygen b) Copper c) Water d) Iron e) Salt f) Methane Long Answer Type Questions 9. Answer the following questions in 6-8 sentences. (i) [(Session 3.1)] Tyndall effect can be observed when sunlight passes through the canopy of a dense forest. Why do you think this phenomenon happens? 10. Answer the following questions in 6-8 sentences. (i) [(Session 3.2)] Describe the following separation techniques. Give one example for each. a. Sublimation b. Evaporation c. Chromatography d. Distillation e. Fractional distillation CHAPTER 3. IS MATTER PURE? 81

AS2-Asking questions and making hypothesis Short Answer Type Questions 11. Answer the following questions in 3-4 sentences. (i) [(Session 3.2)] What kind of mixtures can be separated using sublimation? AS3-Experimentation and field investigation Short Answer Type Questions 12. Answer the following questions in 3-4 sentences. (i) [(Session 3.1)] Iron fillings mixed with sulphur powder can be separated using a mag- net. But the same Iron fillings when heated with sulphur does not show magnetic property. What is the reason behind these two different observations? 13. Answer the following questions in 3-4 sentences. (i) [(Session 3.2)] How would you separate different colours in a pigment? Long Answer Type Questions 14. Answer the following questions in 6-8 sentences. (i) [(Session 3.2)] You have been given a mixture that contains: water, ethanol, coconut oil, and salt. You are asked to separate out each of these substances. Write the processes that would need to be performed in order to separate all of these, and also write the order in which these have to be performed. AS4-Information skills and projects Very Short Answer Type Questions 15. Answer the following questions in one sentence. [Refer to Session 3.1 ] (i) Complete the table. CHAPTER 3. IS MATTER PURE? 82

Mixtures Homogeneous/Heterogeneous Soda water Homogeneous Wood Air Homogeneous Soil Vinegar Filtered tea 16. Answer the following questions in one sentence. [Refer to Session 3.3 ] (i) Complete the table. Pure substance/ not a pure Substance substance Mercury Brick Milk Not a pure substance Calcium oxide Pure substance Iron Air CHAPTER 3. IS MATTER PURE? 83

Long Answer Type Questions 17. Answer the following questions in 6-8 sentences. (i) [(Session 3.3)] Write the Dispersion medium, Dispersed phase and the Colloid type for each of the examples given in the table. Examples Dispersion Dispersed Colloid type medium phase Foam Shaving cream Pumice stone Jelly Face cream Liquid AS5-Communication through drawing and model making Long Answer Type Questions 18. Answer the following question. (i) [(Session 3.3)] Draw a flowchart to show the classification of different states of matter. Give examples. 19. Answer the following question. (i) [(Session 3.2)] Draw a neat diagram to show the separation of the different compo- nents of air. CHAPTER 3. IS MATTER PURE? 84

AS6-Appreciation and aesthetic sense, Values Long Answer Type Questions 20. Answer the following questions in 6-8 sentences. (i) [(Session 3.2)] What is the role of these scientists in understanding about compounds and elements? i. Isaac Newton ii. Hennig Brand iii. Sir Humphry Davy iv. Robert Boyle v. Antoine Lavoisier AS7-Application to daily life, concern to bio diversity Short Answer Type Questions 21. Answer the following questions in 3-4 sentences. (i) [(Session 3.2)] What are the uses of chromatography? Long Answer Type Questions 22. Answer the following questions in 6-8 sentences. (i) [(Session 3.2)] Some solid substances if left in air, directly change into gaseous state. What is the name of this phenomenon? Give three examples of these type of sub- stances observed in your daily life. CHAPTER 3. IS MATTER PURE? 85

Objective Questions AS1-Conceptual Understanding 23. Choose the correct answer. (i) A mixture of oil in water is an example of (A) colloidal solution (B) suspension (C) emulsion (D)homogeneous solution (ii) Aerosol is an example of (B) suspension (A) solution (D)colloidal dispersion (C) emulsion (iii) The substances which cannot be separated any further by any methods of separa- tion are called as (A) pure substances (B) impure substances (C) mixtures (D) solutions AS2-Asking questions and making hypothesis 24. Choose the correct answer. (i) Varun mistakenly mixed water into a can containing cooking oil. Now he wants to separate both of them. What technique will you suggest him? (A) Separating funnel (B) Filtration (C) Evaporation (D)Paper chromatography AS3-Experimentation and field investigation 25. Choose the correct answer. (i) Different pigments from an extract of flower petals can be separated by using CHAPTER 3. IS MATTER PURE? 86

(A) chromatography (B) sublimation (C) decantation (D) centrifugation (ii) A solution prepared by the addition and dissolution of maximum amount of solute in a solvent is called (A) saturated solution (B) unsaturated solution (C)aqueous solution (D)heterogeneous solution (iii) The components of ink can be efficiently separated by the technique of (A) distillation (B) filtration (C)paper chromatography (D) centrifugation (iv) can be broken down by chemical or electrochemical processes. (A) Elements (B) Atoms (C) Molecules (D) Compounds AS4-Information skills and projects (B) Suspensions (D)Heterogeneous solutions 26. Choose the correct answer. (i) show Tyndall effect. (A) Emulsions (C)Colloidal solutions AS7-Application to daily life, concern to bio diversity 27. Choose the correct answer. (i) The human blood is an example of (B) element (A) compound (C)homogeneous Mixture (D) molecule CHAPTER 3. IS MATTER PURE? 87

4. ATOMS AND MOLECULES SESSION 1 HOW DALTON’S ATOMIC THEORY CAME INTO BEING 1.1 Mind Map 1.2 Terminology i. Laws of conservation of mass – \"Matter is neither created nor destroyed during a chemical reaction. More simply, the mass of products is equal to the mass of the reactants in a chemical reaction.” ii. Law of constant proportions –“A given chemical substance always contains the same elements combined in a fixed proportions by mass.” iii. Postulate – A statement accepted as true for the purposes of argument or scientific investigation; also, a basic principle. SESSION 1. HOW DALTON’S ATOMIC THEORY CAME INTO BEING 88

1.3 Key Concepts i. By the end of 18th century, scientists have recognized the difference between elements and compounds. ii. Afterwards the scientists proved into how and why elements combine and what happens when they combine. iii. “The total mass of the reactants is equal to the total mass of the products in a chemical reaction”. This is known as law of conservation of mass. 1.4 Conceptual Understanding Q1. Explain the process and precautions in verifying law of conservation of mass. [Refer to TB page 71 Q1] A. Law of conservation of mass states that mass of a substance can neither be created nor be destroyed in a chemical reaction. Through this experiment, we can prove the mass of the reactants is equal to the mass of the products. Procedure and precautions to be taken in verifying law of conservation of mass: Procedure: i. Prepare a solution by dissolving approximately 2 g of lead nitrate in 100 ml of distilled water. ii. Prepare another solution by dissolving approximately 2 g of potassium iodide in 100 ml water. iii. Take 100 ml solution of lead nitrate in 250 ml conical flask. iv. Also take 4 ml solution of potassium iodide in the test tube. v. Hang the test tube in the flask carefully, without mixing the solutions. Put a cork on the flask. (see figure 1) vi. Weigh the flask with its contents carefully by the spring balance.(see figure 2) vii. Now tilt and swirl the flask, so that the two solutions mix. viii. Weigh the flask again by the same spring balance as shown in figure 3. SESSION 1. HOW DALTON’S ATOMIC THEORY CAME INTO BEING 89

Conclusion: The mass of substances before and after a chemical reaction is same. The mass of reactants = Mass of products. Precautions: 1. Care should be taken while handling chemicals. 2. Glass apparatus should be used carefully. 3. The content of the conical flask should not mix before weighing first time. Q2. 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.144 g of oxygen and 0.096 g of boron. Calculate the percentage composition of the compound by weight. [Refer to TB page 71 Q2] A. Boron and oxygen compound ⇒ Boron + Oxygen 0.24 g ⇒ 0.096 g + 0.144 g For boron 0.24 g of compound contains ⇒ 0.096 g of Boron 100 g contains = ? ⇒ 100 × 0.096 ÷ 0.24 = 40 % SESSION 1. HOW DALTON’S ATOMIC THEORY CAME INTO BEING 90

For oxygen, 0.24 g of compound contains ⇒ 0.144 g of oxygen 100 g contains = ? ⇒100 × 0.144 ÷ 0.24 = 60 % 1.5 Asking Questions and Making Hypothesis Q1. 15. 9 g of copper sulphate and 10.6 g of sodium carbonate react together to give 14.2 g of sodium sulphate and 12.3 g of copper carbonate. Which law of chemical combination is obeyed? How? [Refer to TB page 73 Q19] A. Copper sulphate + Sodium carbonate = Copper carbonate + Sodium sulphate In this reaction copper sulphate and sodium carbonate are the reactants. Weight of copper sulphate = 15.9 g, Weight of sodium carbonate = 10.6 g Total weight of reactants = 15.9 g + 10.6 g = 26.5 g Sodium sulphate and copper carbonate are the products. Weight of sodium sulphate = 14.2 g Weight of copper carbonate = 12.3 g Total weight of products 14.2 g + 12.3 g = 26.5 g We find that, Total weight of reactants = Total weight of products. Since total mass of reactants is equal to the total mass of products, this data proves law of conservation of mass. Q2. Carbon dioxide is added to 112 g of calcium oxide. The product formed is 200 g of calcium carbonate. Calculate the mass carbon dioxide used. Which law of chemical combination will govern your answer? [Refer to TB page 73 Q20] SESSION 1. HOW DALTON’S ATOMIC THEORY CAME INTO BEING 91

A. Carbon dioxide + Calcium oxide → Calcium carbonate Mass of Carbon dioxide + 112 g → 200 g Mass of carbon dioxide = 200g –112 g = 88 g Mass of CO2 required is 88 g 88 g + 112 g = 200 g The law demonstrated in this reaction is the law of conservation of mass. In this reaction, 2 moles of CO2 combines with 2 moles of CaO to give two moles of CaCO3. 1.6 Communication Through Drawing and Model Making Q1. Draw a diagram to show the experimental setup for the law of conservation of mass. [Refer to TB page 73 Q22] A. SESSION 1. HOW DALTON’S ATOMIC THEORY CAME INTO BEING 92

SESSION 2 ATOMS, MOLECULES AND ELEMENTS 2.1 Mind Map SESSION 2. ATOMS, MOLECULES AND ELEMENTS 93

2.2 Terminology i. Atom – An atom is the smallest particle of an element that can participate in a chemical reaction and retain all its properties. ii. Symbol – Symbol is the shortened form to denote the name of an element. iii. Atomic mass – The atomic mass of an element is defined as the average mass of all the isotopes of the element as compared to 1/12th of the mass of one carbon –12 atom. iv. Atomic mass unit (a.m.u.) – One atomic mass unit (amu) is defined as the mass exactly one twelfth the atomic mass of Carbon–12 isotope. v. Unified mass (µ) – The unified atomic mass unit (symbol: u) or dalton (symbol: Da) is the standard unit that is used for indicating mass on an atomic or molecular scale (atomic mass). vi. Molecule – A molecule is the smallest particle of an element or a compound that is ca- pable of independent existence and retains all the properties of that substance. 2.3 Key Concepts i. An atom is the smallest particle of an element that can participate in chemical reac- tions and retain all its properties. SESSION 2. ATOMS, MOLECULES AND ELEMENTS 94

ii. A molecule is the smallest particle of an element or a compound that is capable of independent existence and retains all the properties of that substance. iii. Symbol is the shortened form to denote the name of element. iv. Scientists use the relative atomic mass scale to compare the masses of different atoms of elements. v. The number of times one atom of an element is heavier than 1/12th of the mass of the carbon –12 atom is called its atomic mass. It is denoted as a.m.u. or simply u. vi. We can write the formula of any compound by applying criss–cross method. 2.4 Conceptual Understanding Q1. In a class, a teacher asks the students to write the molecular formula of oxygen. Of them, Shamita writes the formula as O2 and Priyanka as O. Of these, which one is correct? State the reason. [Refer to TB page 71 Q3] A. The molecular formula of oxygen written by Shamita is correct and formula written by Priyanka is not correct. The correct molecular formula of oxygen is O2. Steps for writing the formula of an element : i. Write the symbol of oxygen: O. ii. The valency of oxygen is ‘2’. iii. So write 2 as subscript on the right side of the symbol. iv. So the correct way of writing the molecular formula is O2. Q2. The formula of a metal oxide is MO. Then write the formula of its chloride. [Refer to TB page 72 Q8] A. The formula of a metal oxide is MO. Since, valency of oxygen is 2 and one atom of oxygen is combining with one atom of metal M, valency of ‘M’ is two. SESSION 2. ATOMS, MOLECULES AND ELEMENTS 95

Valency of chlorine is one. Write symbols of metal and chlorine with the numbers of their valencies on the right- hand top of them: M2 Cl1. Crisscross the valencies and write: M1 Cl2. Ignore the number ‘1’ and write the formula of the metal chloride: MCl2. Q3. Mohith said “H2 differs from 2H”. Justify. [Refer to TB page 71 Q4] A. i. Since the element hydrogen is diatomic, H2 represents a molecule of hydrogen. ii. But 2H represents two atoms of hydrogen. The symbol by itself indicates one atom of the element. So 2H is two atoms of hydrogen. Q4. Write the valencies of Fe in FeCl2 and FeCl3 . [Refer to TB page 72 Q14] A. i. Valency is the combining capacity of an element or group of atoms. ii. The valency of Cl (Chloride) is 1. iii. Cl2 indicates 1 x 2 = 2. iv. In FeCI2 , one atom of iron (Fe) is combining with two chloride ion. So, from FeCl2 the valency of Fe is 2. v. In FeCI3, one atom of iron (Fe) is combining with three chloride ions. So from FeCl3 the valency of Fe is 3. Q5. Which has more number of atoms –100 g of sodium or 100 g of iron? Justify your answer. (Atomic mass of sodium = 23 u, atomic mass of iron = 56 u) [Refer to TB page 72 Q16] A. 23 g of Na is 1 mole and has 6.022 × 1023 atoms 100 g of Na =? 100X6.022X1023 23 = 602.2 × 1023 23 = 26.182 × 1023 = 2.6182 × 1024 atoms SESSION 2. ATOMS, MOLECULES AND ELEMENTS 96

56 g of iron has 6.022 × 1023 atoms. 100 g of iron =? 100X6.022X1023 56 = 602.2 × 1023 56 = 10.753 × 1023 = 1.075 × 1024 100 g of Na contain 2.618 × 1024 atoms 100 g of Fe contain 1.075 × 1024 atoms So, 100 g of sodium contains more atoms than 100 g of iron. Q6. Complete the following table: [Refer to TB page 72 Q17] A. Chloride Hydroxide Nitrate Sulphate Carbonate Phosphate Anions Cations Sodium NaCl NaOH NaNO3 Na2SO4 Na2CO3 Na3PO4 Mg(OH)2 Mg(NO3)2 MgSO4 MgCO3 Mg3(PO4)2 Magnesium MgCl2 Ca(OH)2 Ca(NO3)2 CaSO4 CaCO3 Ca3(PO4)2 Al(OH)3 Al(NO3)3 Al2(SO4)3 Al2CO3 AlPO4 Calcium CaCl2 NH4OH NH4NO3 (NH4)2SO4 (NH4)2CO3 (NH4)3PO4 Aluminium AlCl3 Ammonium NH4Cl SESSION 2. ATOMS, MOLECULES AND ELEMENTS 97

Q7. Lakshmi gives a statement “CO and Co both represents elements”. Is it correct? State the reason. [Refer to TB page 72 Q5] A. It is wrong. CO = Carbon monoxide Co = Cobalt CO is carbon monoxide, where C and O represent the different elements present in the compound carbon monoxide. This can be identified with the help of both C and O, which are capital(upper case) letters. Co represents an element, where the first letter is capital(upper case) and the second letter is small(lower case). 2.5 Asking Questions and Making Hypothesis Q1. Imagine what would happen if we do not have standard symbols for elements? [Refer to TB page 73 Q21] A. i. If we do not have standard symbols for elements, we have to write the complete names of elements and compounds which would be time consuming and labori- ous. ii. So, the standard symbols have enabled students of science to write the names of chemical substances and represent chemical reactions in a precise manner. 2.6 Information Skills and Projects Q1. Make playcards with symbol and valencies of the atoms of the elements separately. Each student should hold two playcards, one with the symbol in the right hand and the other with the valency in the left hand. Keeping the symbol in place, student should criss–cross their valencies to form the formula of a compound. [Refer to [Refer to TB page 74 Q1, Try These] A. Students’ Activity. SESSION 2. ATOMS, MOLECULES AND ELEMENTS 98