physical sciences grade 10 term 2 resource pack...gr10_ps_term2_resource_pack.indb 9 2019/01/03...

PHYSICAL SCIENCESGrade 10TERM 2RESOURCEPACK

Gr10_PS_Term2_Resource_Pack.indb 1 2019/01/03 8:37:49 AM

Contents

Worksheets 3TOPIC 11: Particles Substances Are Made Of 4TOPIC 12: Physical and Chemical Change 12TOPIC 13: Representing Chemical Change 20TOPIC 14: Magnetism 26TOPIC 15: Electrostatics 34TOPIC 16: Electric Circuits 42

Formal Experiment 53Technical Instructions 55Electric Circuits 56Marking Guidelines 66

Assessments 71TOPIC 11: Particles Substances Are Made Of 72TOPIC 12: Physical and Chemical Change 76TOPIC 13: Representing Chemical Change 80TOPICS 14 and 15: Magnetism and Electrostatics 86TOPIC 16: Electric Circuits 94


WORKSHEETS


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4 Grade 10 Physical Sciences

TOPIC 11: Particles Substances Are Made Of

WORKSHEET

1. What is meant by the term:

1.1 compound. (3)

1.2 electrolysis. (2)

2. The following questions are about covalent molecular substances.

2.1 What is meant by a molecule? (2)

2.2 How are molecules formed? (1)

2.3 What types of elements form molecules when they combine? (2)

2.4 Describe the structure of a covalent molecular substance. (3)

3. 3.1 Describe the structure of a covalent network substance. (3)

3.2 Describe two properties of network solids and explain why they have these properties. (4)

4. 4.1 What is meant by the term allotropes? (2)

4.2 Diamond and graphite are allotropes of carbon and they have very different properties. Write down two properties of each allotrope. (4)

4.3 In terms of their structure, explain why diamond and graphite have such different properties. (6)

5. 5.1 Describe how ionic bonding occurs. (2)

5.2 Between what type of elements does ionic bonding occur? (2)

5.3 What type of structure always results from ionic bonding? (1)

5.4 List two properties of ionic compounds and explain why they have these properties. (6)

6. 6.1 What is meant by the term delocalised electrons? (4)

6.2 List three properties of metals that result from metallic bonding. (6)


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TOPIC 11: PARTICLES SUBSTANCES ARE MADE OF

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7. Write down the correct chemical names for the following substances:

7.1 K2Cr2O7

7.2 CS2

7.3 HNO3

7.4 K3PO4

7.5 (NH4)2CO3 (5)

8. Draw circle diagrams to represent the following molecules:

8.1 H2S

8.2 CCℓ4

8.3 NF3 (7)


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CONSOLIDATION EXERCISE

TOTAL: 45 MARKS1. 1.1 Describe the structural differences between covalent molecular substances and

covalent network substances. (4)

1.2 List two differences in the properties of covalent molecular structures and covalent network structures. (4)

2. 2.1 Describe the differences in the structures of ionic compounds and metals. (4)2.2 List two differences between the properties of ionic substances and metals. (4)

3. The table below gives some properties of substances A to D.

SUBSTANCE MELTING POINT (°C)

HARD/SOFT ELECTRICAL CONDUCTIVITY

A −75 Gas Non-conductor

B 1 538 Hard Good conductor

C 1 710 Hard Non-conductor

D 1 087 Hard Conducts when molten

Match the letters A to D to the following substances:lead sulfate sulfur dioxide silicon dioxide iron (8)

4. Write down the chemical formula for each of the following substances:4.1 copper(I) oxide4.2 diphosphorus trioxide4.3 chromium(III) sulphite4.4 nitrogen dioxide4.5 aluminium hydrogen carbonate (10)

5. Draw circle representations for the following molecules:5.1 C2H4 (3)5.2 PCℓ3 (2)


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6. Below is a ball and stick model of a certain molecule. The colour code is: white – hydrogen, small dark grey –carbon, larger dark grey –oxygen.

6.1 Write down the molecular formula of the compound. (2)6.2 Write sown the number of single bonds in the compound. (2)6.3 Write down the number of double bonds in the compound. (2)


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MARKING GUIDELINES

1. 1.1 A compound is made up of atoms of two or more elementsü which are bonded togetherü and which combine in a specific ratio.ü (3)

1.2 Electrolysis is a process by which a compound can be split into its constituent elements üby means of an electric current. ü (2)

2. 2.1 A molecule consists of two or more atomsü which are chemically bonded and behave a single unit.ü (2)

2.2 Molecules are formed when two atoms share a pair of electrons.ü (1)2.3 Molecules are formed when two non-metal atoms combine.üü (2)2.4 A covalent molecular structure consists of individual moleculesü which are held

together by weakü intermolecular forces.ü (3)

3. 3.1 A covalent network structure consists of atoms üwhich are covalently bonded togetherü to form giant molecules.ü (Some covalent network structures consist of molecules which are covalently bonded throughout the structure e.g. silicon dioxide). (3)

3.2 Network solids have high melting points and boiling pointsü because the covalent bonds between the atoms are extremely strong.ü Network solids generally don’t conduct electricityü because there are no delocalised electrons.ü Graphite is an exception. (4)

4. 4.1 Allotropes are different physical formsü of the same element.ü (2)4.2 Diamond has a very high melting point and boiling pointü and is extremely hard.ü Graphite also has a very high melting point and boiling pointü but it is soft.ü (4)4.3 In diamond each carbon atom is bonded to four other atoms üby very strong

covalent bonds ü(this why it is so hard and has a high melting point) and forms a three-dimensional crystal lattice in which there are no delocalised electrons (doesn’t conduct electricity).ü

In graphite each carbon atom is bonded covalently to three other atoms to form large flat molecular sheets. üEvery carbon atom has a delocalised electronü (thus is an electrical conductor), and these electrons hold the sheets together by means of weak intermolecular forcesü (the reason why it is soft). (6)

5. 5.1 Ionic bonding occurs when there is a transfer of electronsü from one atom to another.ü (2)

5.2 Ionic bonding occurs between metals ü and non-metals.ü (2)5.3 A crystal lattice. ü (1)


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5.4 Ionic compounds have high melting points and boiling points.üThis occurs because ionic bonds are very strongü (they are also hard and brittle for this reason). A lot of energy is required to separate the ions. ü Ionic compounds do not conduct electricity when solid, but they do when they are in the molten state. üThis is because when they are in the solid state, the ions occupy fixed positions in a crystal lattice and there are no charges that are free to move,ü When they are molten, or in solution, the ions are free to move and they carry charge (conduct current).ü (6)

6. 6.1 Delocalised electrons are electrons that are spread out over an entire structure üand are not associated with any particular atom.ü They are free to move from valence orbital to valence orbital üand do so in a random way.ü (4)

6.2 Three properties of metals that result from metallic bonding are:• they conduct electricity üü• they are malleable and ductile üü• they have relatively high melting points and boiling points.üü (6)

7. 7.1 Potassium dichromate ü7.2 Carbon disulphide ü7.3 Hydrogen nitrate (nitric acid)ü7.4 Potassium nitrate ü7.5 Ammonium carbonate ü (5)

8. 8.1 (2)

8.2 (3)

8.3 (2)

üü

ü ü

The chlorine atoms (dark grey) are bigger than the carbon atoms.

ü ü

The fluorine atoms and nitrogen atoms are approximately the same size.


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TOTAL: 45 MARKS1. 1.1 Covalent molecular substances are made up of individual moleculesü which are

held together by weak intermolecular forces.ü Covalent network substances are made up of atoms (or molecules)ü which are bonded by strong covalent bonds.ü (4)

1.2 Difference 1: Network substances have very high melting and boiling points while molecular substances have low melting and boiling points.üü Difference 2: Network substances are generally very hard (graphite excepted) while molecular substances are generally soft. üü (4)

2. 2.1 Difference 1: Ionic compounds consist of positive and negative ions while metals consist of positive ions only. üü Difference 2: In ionic compounds the forces holding the substance together are electrostatic forces between positive and negative ions.ü In metals the forces are between delocalised electrons and positive metal ions.ü Difference 3: In metals there are delocalised electrons while in ionic compounds there are no delocalised electrons. üüANY TWO (4)

2.2 Difference 1: Ionic substances are hard and brittle, while metals are generally soft and can be shaped easily.üü Difference 2: Metals conduct electricity in the solid or liquid state, ionic substances conduct electricity only when molten or in aqueous solution.üü (4)

3. Substance A has a low melting point which indicates weak intermolecular forces, so A must be a covalent molecular substance – it must be sulfur dioxide. It is also the only gas out of the four substances. üü

Substance B has a high melting and it is hard, so it has strong interparticle forces. It can be either a covalent network substance, an ionic substance or a metal. The fact that it conducts electricity in the solid state means that it is a metal. B is iron.üü

Substance C has a high melting and it is hard, so it has strong interparticle forces. It can be either a covalent network substance, an ionic substance or a metal. The fact that it is a non-conductor of electricity means that it is a covalent network structure. C is silicon dioxide. üü

Substance D is lead sulfate. üü The reasoning (other than the fact that it is the only one left), is that it has a high melting point, it is hard and it conducts electricity when molten means that it is an ionic substance. (8)


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4. 4.1 Cu2O üü

4.2 P2O3 üü(diphosphorus means 2 P and trioxide means 3 O)

4.3 Cr2(SO3)3 üü

4.4 NO2 üü(dioxide means 2 O)

4.5 Aℓ(HCO3)3 üü (Aℓ forms Aℓ3+ ions, while hydrogen carbonate is HCO3−) (10)

5. 5.1

ü ü ü

(3)

5.2 ü ü

(2)

6. 6.1 C3H6O üü (2)

6.2 There are 8 single bonds. üü (2)

6.3 There is 1 double bond. üü (2)

single bonds

doubleb bond


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TOPIC 12: Physical and Chemical Change WORKSHEET

1. List four observations that could indicate that a chemical change is taking place. (4)

2. Carbon dioxide undergoes sublimation.

2.1 Is this a physical or chemical change? Give a reason for your answer. (3)

2.2 Describe the changes that take place in terms of rearrangement of molecules during this change. (3)

2.3 Describe the energy change that occurs during this change. (3)

3. Write down which of the following are chemical changes and which are physical changes.

3.1 Baking a cake

3.2 Breaking a glass bottle

3.3 Boiling water

3.4 Rusting iron

3.5 Burning wood (5)

4. Write down which of the following are synthesis reactions and which are decomposition reactions.

4.1 Hydrogen gas and oxygen gas combine to form water.

4.2 Ammonia breaks up into nitrogen and hydrogen.

4.3 Ammonium hydroxide is heated and it changes into ammonia and water.

4.4 Sulfur trioxide (SO3) is heated and produces sulfur dioxide (SO2) and oxygen. (8)

5. A piece of magnesium ribbon is placed in a blue solution of copper sulfate. The solution becomes colourless. Magnesium sulfate and a brownish red deposit of copper forms on the bottom of the container. Is this a chemical or physical change? Give reasons for your answer. (4)

6. 6.1 State the law of conservation of mass. (3)

6.2 By using relative atomic masses, show that mass is conserved in the following chemical reaction: Fe + 2HCℓ → FeCℓ2 + H2 (7)


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TOPIC 12: PHYSICAL AND CHEMICAL CHANGE

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7. Use the law of conservation of mass to answer this question. 4,8 g of magnesium metal reacts completely with 12,8 g of oxygen gas. What mass of magnesium oxide produced? Give a reason for your answer. (4)

8. Give the chemical formula for each of the following. A substance in which the ratio of …

8.1 C:H is 1:4. (2)

8.2 C:O is 1:1. (2)

8.3 Na:O is 2:2. (2)


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TOTAL: 60 MARKS1. Water vapour condenses to form water.

1.1 Is this a chemical or physical change? (1)1.2 Describe how the molecules rearrange when this change occurs. (3)1.3 Describe the energy change that occurs during this process. (2)

2. Consider the rusting of iron.2.1 Is this a physical or chemical change? (1)2.2 Give reasons why you chose the answer given in 2.1. (2)

3. Which of the following reactions are synthesis reactions and which are decomposition reactions? Give a reason for each answer.3.1 2H2O → 2H2 + O2 (3)3.2 NH4NO3 → NH3 + HNO3 (3)3.3 2SO2 + O2 → 2SO3 (3)3.4 Mg + O2 → 2MgO (3)

4. Tabulate three differences between physical and chemical changes. (6)

5. 5.1 Which requires more energy:I changing ice to water, orII breaking water down into hydrogen and oxygen? (2)

5.2 Give reasons why this is so. (4)5.3 Which is a physical change and which is a chemical change? (2)

6. Verify the law of conservation of mass by using relative atomic masses for the following chemical reaction: BaCℓ2 + H2SO4 → BaSO4 + 2HCℓ (7)

7. Use circle diagrams to show the rearrangement of atoms in the following chemical reaction: H2 + Cℓ2 → 2HCℓ (4)

8. A reaction is carried out by burning 10 g of carbon completely in oxygen gas. This produces 35,6 g of carbon dioxide. In another experiment 15,2 g of carbon is burned completely in oxygen gas and this produces 55,84 g of carbon dioxide. Use this data to show that the law of constant composition is verified. (9)

9. A sample of calcium oxide (CaO) is found to contain 40 g of calcium and 16 g of oxygen. If another sample of calcium oxide is analysed, it is found to contain 100 g of calcium. Calculate the mass of oxygen that this sample contains. (5)


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MARKING GUIDELINES

1. There could be:• a change in temperature ü• release of a gas ü• a change in colour ü• a precipitate formed ü (4)

2. 2.1 It’s a physical change. ü There is no new substance being formed. ü There is only a rearrangement of the molecules.ü (3)

2.2 In solid carbon dioxide there is a structureü in which the molecules are held in fixed positions. üWhen it becomes a gas, there is no structure in the arrangement of the molecules, they are moving about randomly.ü (3)

2.3 The molecules of the solid absorb energy which enables them to vibrate more strongly. üThe forces between molecules are no longer strong enough to keep them together.ü The forces holding them in place are broken and they are completely free to move.ü (3)

3. 3.1 Physical ü

3.2 Physical ü

3.3 Physical ü

3.4 Chemical ü

3.5 Chemical ü (5)

4. 4.1 Synthesis üü

4.2 Decomposition üü

4.3 Decomposition üü

4.4 Decomposition üü (8)

5. This is a chemical change. ü The blue colour disappears and the solution becomes colourless. ü This is an indication of a chemical change. üA new, reddish brown substance forms which wasn’t there before – another indication of a chemical change.ü (4)


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6. 6.1 In a chemical reaction, üthe total mass of the reactants üis equal to the total mass of the products.ü (3)

6.2 Fe + 2HCℓ → FeCℓ2 + H2Reactants: Fe: Mr = 56 ü HCℓ: m = 2(1 + 35,5) = 73 ü Total mass = 56 + 73 = 129ü

Products: FeCℓ2: Mr = 56 + (2 × 35,5) = 127 ü H2: Mr = 2 × 1 = 2 ü

Total mass = 127 + 2 = 129 ü

Mass of reactants = Mass of products ü (7)

7. Total mass reactants = 12,8 + 4,8 ü

= 17,6 g ü` Mass of magnesium oxide = 17,6 g ü(applying the law of conservation of mass). ü (4)

8. 8.1 CH4 üü

8.2 CO üü

8.3 Na2O2 üü (6)


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TOTAL: 60 MARKS1. 1.1 Physical change ü (1)

1.2 There are very weak (non-existent) forces between molecules in water vapour. üThe molecules are totally free to move. ü When the vapour turns to liquid, the forces keep the molecules together but they can flow over each other. ü (3)

1.3 The water molecules lose energy üand when the temperature is low enough, they will have lost enough energy for the molecules to hold on to each other and form a liquid.ü (2)

2. 2.1 Chemical change. ü (1)

2.2 The brown crust (rust) that forms on the metal is a new substance. ü There is a change in colour from grey to reddish brown. ü (2)

3. 3.1 Decomposition ü One substance changes into two others. üü (3)

3.2 Decomposition ü One substance changes into two others. üü (3)

3.3 Synthesis ü Two substances combine to from a single substance. üü (3)

3.4 Synthesis ü Two substances combine to form a single substance. üü (3)

4. PHYSICAL CHANGE CHEMICAL CHANGE

No new substances are formedü New substances are formedü

Low change in energy ü High energy change ü

The mass, number of atoms and number of molecules are conservedü

Only the mass and number of atoms are conserved ü

(6)

5. 5.1 II Breaking water down into hydrogen and oxygen üü (2)

5.2 When ice changes to water, weakü intermolecular forces have to be broken slightly.ü When water is broken down into hydrogen and oxygen, strong covalent bonds have to be brokenü and this requires much more energy. ü (4)

5.3 I is a physical change, ü II is a chemical change. ü (2)


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6. BaCℓ2 + H2SO4 → BaSO4 + 2HCℓ

Reactants: BaCℓ2: Mr = 137 + (2 × 35,5) = 208 ü H2SO4: Mr = (2 × 1) + 32 + (4 × 16) = 98 ü

Total reactants = 208 + 98 = 306 ü

Products: BaSO4: Mr = 137 + 32 +(4 × 16) = 233 ü

HCℓ: 2Mr = 2(1 + 35,5) = 73 ü

Total products = 233 + 73 = 306 ü∴ Total mass (m) of reactants = Total mass (m) of products ü (7)

7. H2 + Cℓ2 → 2HCℓ

H H

ü+

Cℓ Cℓ

ü→ H Cℓ

ü

H Cℓü

(4)

8. The law of constant composition states that the ratio of the atoms of elements in a specific compound is always the same.

First compound: mass of carbon = 10 g mass of oxygen = 35,6 – 10 = 25,6 g ü

Mole ratio C:O = :,

1210

3225 6

üü

= 0,83: 0,8 = 1: 1 ü


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We divide the mass of each element by the relative atomic or molecular mass, because we want the mole ratio. The learners do not know about moles at this stage, so they just need to understand that this has to be done.

Second compound: mass of carbon = 15,2 g mass of oxygen = 55,84 – 15,2 = 40,64 g ü

ratio C:O = ,:,

1215 2

3240 64

üü

= 1,27: 1,27 = 1:1 ü ` the ratio of carbon to oxygen is the same for both compounds, which verifies the law

of constant composition. ü (9)

9. Ratio of Ca:O in first sample = :40401616 ü= 1: 1 ü

The ratio in the second sample must also be 1:1ü Ratio of Ca:O in second sample = x:40

10016 ü

` 100 × 16 = 40x x = 40 The second sample contains 40 g of oxygen. ü (5)


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TOPIC 13: Representing Chemical Change

WORKSHEET

1. List four observations that indicate that a chemical reaction has taken place. (4)

2. The representations for two chemical reactions are shown below.

I Fe2O3 + Zn → ZnO + Fe

II potassium hydroxide + sulfuric acid → potassium sulfate + water

2.1 Write down balanced reaction equations for I and II. (9)

2.2 Use equation II to show that mass is conserved during this reaction. (7)

3. Consider the following unbalanced reaction equation:

Aℓ2O3(s) + HNO3(aq) → Aℓ(NO3)3(aq) + H2O(l)

3.1 Write down a balanced reaction equation for this reaction, including phase symbols. (4)

3.2 What does the phase symbol (aq) mean? (2)

3.3 Which chemical law is verified by a balanced reaction equation? (2)

4. Write down chemical formulae for each of the following compounds:

4.1 aluminium oxide

4.2 sodium dichromate

4.3 iron(III) nitrate

4.4 potassium sulfite

4.5 calcium sulfide (5)

5. 5.1 Balance the following chemical equation:

H2(g) + Cℓ2(g) → HCℓ(g) (2)

5.2 Show that the number of atoms is conserved in this reaction. (4)

6. Write down a balanced chemical equation for the reaction between aluminium carbonate powder and dilute sulfuric acid. The products are aluminium sulfate, carbon dioxide and water. (5)


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TOPIC 13: REPRESENTING CHEMICAL CHANGE

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7. Write down a balanced reaction equation for the reaction represented by the following coloured circle diagram: (3)

= oxygen

= sulfur

+ →

8. Draw circle diagrams to represent the following reaction:

2CO(g) + O2(g) → 2CO2(g) (3)


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TOTAL: 37 MARKS1. Consider the following two chemical reactions:

I H2(g) + O2(g) → H2O(l) and

II potassium carbonate is heated to form potassium oxide and carbon dioxide gas.

1.1 Write down a balanced reaction equation for reaction I. (2)

1.2 Write down a balanced reaction equation for reaction II, including phase symbols. (4)

1.3 State which reaction is a decomposition reaction and which is a synthesis reaction. (2)

2. Give the correct chemical formulae for the compounds represented by the following chemical names:

2.1 nitrogen trihydride

2.2 magnesium phosphate

2.3 diphosphorus trioxide

2.4 aluminium fluoride (4)

3. Write down balanced reaction equations for the following:

3.1 methane gas reacts with oxygen to form carbon dioxide and water.

3.2 carbon monoxide gas reacts with oxygen to form carbon dioxide.

3.3 sulfur dioxide reacts with water to form sulphurous acid. (10)

4. 4.1 Balance the following reaction equation:

ZnO(s) + HNO3(aq) → Zn(NO3)2(aq) + H2O(l) (2)

4.2 Show that mass is conserved in this reaction. (7)

5. Write down balanced reaction equations for the following:

5.1 Fe(s) + HCℓ(aq) → FeCℓ3(aq) + H2(g) (4)

5.2 BaCℓ2(aq) + K2SO4(aq) → BaSO4(s) + + KCℓ(aq) (2)


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MARKING GUIDELINES

1. • There is a change in temperature ü• There is a change in colour ü• A gas is given off ü• A precipitate is formedü (4)

2. 2.1 I Fe2O3 + 3Znü → 3ZnOü + 2FeüII 2KOHüü + H2SO4 ü→ K2SO4ü + 2H2Oüü (9)

2.2 Reactants: Fe2O3: m = (2 × 56) + (3 × 16) = 160 ü 3Zn: m = 3 × 65 = 195 üTotal reactants = 355 ü

Products: 3ZnO: m = 3(65 + 16) = 243 ü 2Fe: m = 2 × 56 = 112 üTotal products = 355 üMass of products = Mass of reactants.ü (7)

3. 3.1 Aℓ2O3(s)ü + 6HNO3(aq)ü → 2Aℓ(NO3)3(aq)ü + 3H2O(l)ü (4)3.2 (aq) means ‘in aqueous solution’ i.e. in water. üü (2)3.3 The law of conservation of matter (mass and/or atoms) is verified by a balanced

chemical equation. (2)

4. 4.1 Aℓ2O3 ü4.2 Na2Cr2O7 ü4.3 Fe(NO3)3 ü4.4 K2SO3 ü4.5 CaS ü (5)

5. 5.1 H2(g) + Cℓ2(g) → 2HCℓ(g) ü phases ü (2)5.2 Reactants:

H: 2 atoms üCℓ: 2 atoms üProducts:H: 2 × 1 = 2 atomsüCℓ: 2 × 1 = 2 atomsü (4)


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6. Aℓ2(CO3)3(s)ü + 3H2SO4(aq) ü → Aℓ2(SO4)3(aq)ü + 3CO2(g) ü+ 3H2O(l)ü (5)

7. SO2(g) ü + O2(g)ü → 2SO3(g) ü (3)

8.

+ →ü ü ü

(3)


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TOTAL: 37 MARKS1. 1.1 H2(g) + O2(g) ü → 2H2O(l) ü (2)

1.2 K2CO3(s)ü → K2O(s)ü + CO2(g)ü phases ü (4)

1.3 I is synthesis ü II is decomposition ü (2)

2. 2.1 NH3 ü

2.2 Mg3(PO4)2 ü

2.3 P2O3 ü

2.4 AℓF3 ü (4)

3. 3.1 CH4(g) ü + 2O2(g) ü → CO2(g) ü + 2H2O(l) ü

3.2 CO(g) ü + O2(g) ü → CO2(g) ü

3.3 SO2(g)ü + H2O(l)ü → H2SO3(g) ü (10)

4. 4.1 ZnO(s) + 2HNO3(aq)ü → Zn(NO3)2(aq) + H2O(l)ü (2)

4.2 Reactants: ZnO: Mr = 65 + 16 = 81 ü HNO3: m = 2[1 + 14 + (3 × 16)] = 126 ü

Total reactants: = 81 + 126 = 207ü

Products: Zn(NO3)2: m = [65 + 2(14 + 3 × 16)] = 189 ü H2O: Mr = (2 × 1) + 16 = 18 ü

Total products = 189 + 18 = 207 üMass of products = Mass of reactants ü (7)

5. 5.1 2Fe(s) ü + 6HCℓ(aq) ü → 2FeCℓ3(aq) ü + 3H2(g) ü (4)

5.2 BaCℓ2(aq) + K2SO4(aq)ü → BaSO4(s) + 2KCℓ(aq) ü (2)


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TOPIC 14: Magnetism

WORKSHEET

1.

GRADE 10 TERM 2 RESOURCE PACK

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QUESTION 1

QUESTION 2

PAGE 30

QUESTION 2

PAGE 37

1.1 Will magnet A attract or repel magnet B? (1)

1.2 What will happen to the force of magnet A on magnet B if the distance between them is increased? (1)

1.3 Draw a diagram of the magnetic fi eld between the two magnets. (3)

2. 2.1 What would happen to the poles of a magnet if the magnet were to be cut into two halves (as shown in the diagram below)?


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QUESTION 1

QUESTION 2

PAGE 30

QUESTION 2

PAGE 37

(2)

2.2 Explain what is meant by “a magnetic fi eld”. (2)

2.3 Explain how the magnetic fi eld lines indicate the strength of the magnetic fi eld. (2)

2.4 Study the following information about these three magnets:

A

Pole B attracts Pole P Pole Q REPELS Pole X

B P Q X Y

Would the following poles attract or repel if they were brought close to one another?

2.4.1 Pole B and pole X. (1)

2.4.2 Pole A and pole Q. (1)

2.4.3 Pole B and pole Y. (1)


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TOPIC 14: MAGNETISM

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3. Choose the best option (A, B, C or D) for this question. Which of the following statements is true concerning magnetic fields?

A They always begin on a north pole and travel to the south pole.

B They show both the direction and the strength of the magnetic field.

C They show the direction in which a positive charge will experience force.

D They only cross when multiple magnets are present. (2)

4. The Earth’s magnetic field is shaped like that of a bar magnet. The magnetic poles of the Earth are not aligned with its geographic poles.

4.1 Explain what is meant by the geographic north pole. (2)

4.2 Sketch the magnetic field of the Earth. (3)

4.3 To which pole does the north pole of a compass needle point? (1)

4.4 Comment on this statement: “The north pole is actually a south pole”. Is this statement true or false? Justify your answer. (4)

5. Migratory birds and sea turtles sense variations in the strength and inclination of the Earth’s magnetic field as they travel large distances across the world to mating grounds or beaches each year.

5.1 At which two places on the Earth is its magnetic field the strongest? (2)

5.2 One theory which explain how birds manage to sense the magnetic field is that they contain very small particles of iron oxide (magnetite) in their beaks. Describe how the presence of iron oxide in its beak helps a bird to navigate. (4)


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TOTAL: 20 MARKS

MULTIPLE CHOICE1. Which of the following materials is not ferromagnetic?

A iron

B nickel

C copper

D chromium (2)

2. Which of the following statements about magnetic fields is false?

A The direction of a magnetic field is the direction towards which the north pole of a compass points due to the magnetic force on it.

B Magnetic fields exist inside and outside magnets.

C Magnetic field lines cross one another when there are more than two magnets present.

D The magnetic field strength at the north pole of a magnet has the same size as that at its south pole. (2)

3. The magnetic north pole is …

A located at the south pole.

B is a south pole.

C located at the top of the Earth’s spin axis.

D located at the bottom of the Earth’s spin axis. (2)

LONG QUESTIONS

4. The science class were asked to investigate the relative strength of the magnetic field at the north pole of a bar magnet before it was cut in half, and after it was cut in half. They measured the relative strength by counting the number of paper clips that could be suspended from the north pole of the magnet, and its “half magnet”.

4.1 Draw a neat sketch of the magnetic field of a bar magnet. (3)

4.2 Explain how the paper clips are attracted to the bar magnet. (2)

4.3 Give one reason why this method of using paper clips is a fairly good indicator of the field strength. (3)

4.4 Give one reason why this method could produce faulty results. (2)


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4.5 Predict the outcome of this experiment. Justify your answer. (4)

One group of learners found it very difficult to cut their bar magnet in half. They decided to place it on a stone (outside) and to saw through the magnet with an electrically powered saw. The bar magnet vibrated significantly during this process. When they tested their “half magnets” they found that no paper clips could be suspended from either of the pieces, whereas before cutting the magnet it had suspended 6 paper clips.


RESOURCE PACK


MARKING GUIDELINES

1.1 Attract ü (1)

1.2 Force of magnet A on magnet B decreases. ü (1)

1.3

S N

üCorrect pattern of fi eld lines between magnets

üDirected from north to south pole

üSouth pole on left

(3)

2. 2.1


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QUESTION 1

QUESTION 2

PAGE 30

QUESTION 2

PAGE 37

(2)

2.2 A magnetic fi eld is a region üin which a magnet or magnetic material experiences a magnetic force.ü (2)

2.3 Th e magnetic fi eld strength is greatest in those places where the fi eld lines are closest to one another. üü (2)

2.4 2.4.1 repel ü (1)

2.4.2 attract ü (1)

2.4.3 attract ü (1)

3. B üü

A is incorrect – magnetic fi eld lines are continuous – they exist inside the magnet as well as outside.

C is incorrect – magnetic fi eld lines show the direction of force on a north pole (not on electric charge).

D is incorrect – magnetic fi eld lines never cross. (2)


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4.1 The geographic north pole is the point at the top of the Earth’s spin axis.üü (2)

4.2 The magnetic field axis is about 11° to the left of the Earth’s spin axis. ü

Magnetic field lines point inwards at the magnetic north. ü

Correct pattern (similar to that of a bar magnet). ü

Spin axis

Magneticfield lines

(3)

4.3 towards magnetic north. ü (1)

4.4 The statement is ambiguous (unclear) because it does not tell you whether it is referring to the geographic or the magnetic north pole. ü

It is true that the magnetic north pole is actually a south pole. ü

It is false if it is referring to the geographic north pole ü because that is defined as geographic north. ü (4)

5.1 The magnetic ü north and south poles. ü (2)

5.2 Iron oxide (magnetite) is ferromagnetic. ü It experiences (magnetic) force in the Earth’s magnetic field. ü The bird is able to fly in the correct direction and to the correct position on Earth because it can sense the Earth’s magnetic field ü and the inclination of the field lines. ü (Birds are taught where to fly by their parents and the flock).

NB Direction is linked to the direction of the field – force of attraction / repulsion due to the field.

Position is linked to the inclination of the field. (4)


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TOTAL: 20 MARKS

MULTIPLE CHOICE1. C üü Iron, nickel and chromium are ferromagnetic materials. (2)

2. C üü Magnetic field lines never cross one another. (2)

3. B üü The lines of magnetic field point inwards at the magnetic north pole, therefore it is

a south pole. The north pole of a magnet is actually the north seeking pole. It points towards the magnetic north pole. (2)

LONG QUESTIONS4.1

N

S

(3)

4.2 Paper clips are made of ironü which is ferromagneticü so they are attracted to magnets. (2)

4.3 The more paper clips placed end to end that the magnet is able to attract, the greater the weight of the paper clipsüü and therefore the greater the force of attraction at that pole. ü (3)

4.4 a. The paper clips must all be placed end to end, and not placed randomly at the magnet’s pole. ü We need to measure the force by the amount of weight the magnet’s field can support at the pole. ü

OR

b. Each time a paper clip is used it retains some of its magnetism. ü If the same paper clips are re-used to hang from the poles, you will have an inflated result (making the magnetic field seem stronger than it actually is. ü (2)


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4.5 The halves of the magnet will support fewer paper clips. üü They have less magnetic material in them than the original magnet. ü It would contradict the law of conservation of energy if the half magnets had equal or greater magnetic field intensities. ü (4)


RESOURCE PACK


TOPIC 15: Electrostatics

WORKSHEET

MULTIPLE CHOICE1. A plastic rod and a dry cloth are uncharged. The plastic rod is rubbed with the dry cloth

and they both become charged. The rod becomes negatively charged because some particles move from the cloth to the rod.

What is the charge on the cloth and which particles moved in the charging process?

CHARGE ON ROD PARTICLES THAT MOVED

A Positive Protons

B Positive Electrons

C Negative Protons

D Negative Electrons

(2)

2. A positively charged plastic rod attracts small pieces of paper because ….…

A the paper pieces are negatively charged.

B the paper pieces are neutral.

C the paper pieces are very small.

D the paper pieces become polarised. (2)

LONG QUESTIONS3. A pith ball is a polystyrene sphere coated with metal paint.

A plastic rod is charged by rubbing it with a cloth. It is held next to an uncharged pith ball that is suspended on light cotton thread.

Cotton thread

Postively charged plastic rod

Pith ball


TOPIC 15: ELECTROSTATICS

Term 2 35

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3.1 Describe how the plastic rod becomes positively charged when it is rubbed with a cloth. (2)

3.2 Describe what happens in the metal paint on the pith ball when the positively charged plastic rod is brought near to it. (2)

3.3 The ball is attracted to the rod. Explain why this happens, given that the pith ball is uncharged. (2)

3.4 Predict what you would see if the pith ball touches the positively charged rod. (2)

3.5 Explain your prediction in 3.4. (2)

4. Two identical pith balls are suspended on light, inelastic cotton threads. Pith ball A has a positive charge of 5,4 nC. Pith ball B carries a negative charge of 8,2 nC.

4.1 State the principle of quantization of charge. (2)

4.2 Calculate the number of extra electrons added to pith ball B. (4)

4.3 Describe the type of force that pith ball B exerts on pith ball A. The pith balls are brought up together, and then they are separated again to hang at the same original distance apart. (1)

4.4 State the law of conservation of charge. (2)

4.5 Use the law of conservation of charge to calculate the charge on pith ball B after it has touched pith ball A and is separated, and hangs back at its original position. (4)

4.6 How many electrons were transferred from pith ball B to pith ball A when the pith balls touched each other? (4)

5. Draw a diagram to show why it is dangerous to take shelter under a lone tree in open veld during a thunderstorm. Write a brief description to explain your reasoning. (6)

6. Explain each of the following phenomena using your knowledge of electrostatics.

6.1 When the air is very dry, your hand feels a small sharp electric shock when you touch a metal door knob after walking along the carpet to the door. (3)

6.2 When a charged plastic ruler is brought close to small pieces of paper the paper pieces are attracted to the ruler. (3)


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TOTAL: 34 MARKS

MULTIPLE CHOICE1. A positively charged rod is brought near to the following objects.

i. Small uncharged pieces of paper

ii. A thin stream of water

iii. A positively charged balloon

iv. A negatively charged cloth

Which of these objects will not be attracted to the rod?

A i) ii) and iii)

B i) iii)

C ii) iii)

D iii) (2)

2. When a charged object becomes polarised, it …

A gains electrons.

B loses electrons.

C gains a north and a south pole.

D experiences a shift in charge. (2)

3. A neutral object has …

A no charges.

B no magnetic field.

C more neutrons than protons and electrons.

D no imbalance in charge. (2)



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4. Two identical positively charged pith balls are brought up to touch one another, and then moved apart to their original distance. Initially the fi rst pith ball A carried twice as much charge as the second pith ball B. When they have separated both pith balls each carry a charge of +6 nC. What was the original charge on the pith balls?

PITH BALL A PITH BALL B

A 12 0

B 8 4

C 6 6

D 4 2

(2)

LONG QUESTIONS5. A positively charged plastic rod is placed above a thick metal plate. Th e metal plate is

connected to the Earth by a wire. Th e metal plate rests on an insulator.


PAGE 26

QUESTION 1

QUESTION 2

PAGE 30

QUESTION 2

PAGE 37

A learner disconnects the earthing wire and then removes the positively charged rod.

Th e experiment is repeated. On the second attempt, the learner removes the positive rod, and then disconnects the earthing wire.

5.1 Initially the metal plate is earthed. Explain what happens when the positively charged rod is placed above the metal plate while the earthing wire remains attached to it. (4)

5.2 When the earthing wire is disconnected and then the positively charge rod is removed, what charge (if any) remains on the metal plate? Explain your answer. (4)

5.3 On the second attempt, the positive rod is removed fi rst, and then the earthing wire is disconnected. What is the charge (if any) on the metal plate now? Explain your answer. (4)


RESOURCE PACK


6. Inside a cathode ray tube such as those that were used in the “old fashioned TV sets” the direction of a beam of electrons is controlled by oppositely charged plates. Study the simplified diagram shown below.

+-

+

The positive and negative metalplates provide an electric field tocontrol the direction of the electronbeam.

A beam of electrons is acceleratedthrough a narrow gap.

-

6.1 Explain how the direction of the beam of electrons is controlled by the positive and negative plates. (4)

6.2 These types of TV sets collect dust particles on their screens. Explain why dust particles tend to stick to the screens of these types of TV sets. (4)

7. An electrostatic smoke precipitator collects smoke (carbon) particles and dust from a coal-fired furnace. The diagram below shows how the particles are collected. The flue gases and dust enter through a grid which is negatively charged. They then pass through positively charged plates, and fall into the dust trap. The dust trap is cleaned frequently so that the precipitator does not become clogged up.

plate

positive charge

negativechargedust trap

flue glasses and dust

+ ve

- ve

Explain how the precipitator works in terms of electrostatic principles. (6)



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MARKING GUIDELINES

MULTIPLE CHOICE1. B üü The cloth becomes positively charged when it loses electrons to the rod. (2)

2. Düü The positively charged rod attracts negative charge (electrons) in the paper pieces, causing the molecules inside the paper to become polarised. The negative parts of the molecules are attracted to the rod.

Options B and C are true but each of them is not the reason why the paper pieces are attracted to the rod, therefore they are incorrect answers to this question. (2)

LONG QUESTIONS3. 3.1 Electrons are transferred from the rod to the cloth. ü The rod has less electrons

than protons in it therefore it is positively charged.ü (2)

3.2 The metal paint has free electrons (electrons which are free to move around). üWhen the positive rod is brought near to it, the electrons in the paint move to one side of the pith ball near the positive rod. ü (2)

3.3 The movement of electrons to one side of the pith ball causes the ball to be polarised. ü The negative side of the ball is attracted to the positive rod. ü (2)

3.4 The pith ball will touch the rod and then spring away (be repelled) by it. üü (2)

3.5 When the pith ball touches the rod it will transfer electrons to the rodü(it will share charge with the rod). Both pith ball and rod will have a positive charge (lack of electrons) ü so they repel each other. (2)

4. 4.1 Every charge in the universe consists of integer multiples of the electron charge.üü (2)4.2 Q = nqeü method 8,2 × 10−9 ü= n × 1,6 × 10−19ü substitutions n = 5,13 × 1010ü accuracy (4)

4.3 attraction ü (towards A) (1)

4.4 The net charge of an isolated systemü remains constant during any physical process.ü (2)

4.5 Total charge = 5,4 + (−8,2)ü method = -2,8ü nC accuracy Charge on each pith ball = ½ ü× −2,8 method = −1,4 nC ü (1,4 × 10−9 C) accuracy SI units (nC or C) (4)


RESOURCE PACK


4.6 Q = nqe method

1,4 × 10−9= n × 1,6 × 10−19ü substitutions n = 8,75 × 109ü accuracy

Number of electrons transferred = (5,13 × 1010) – (8,75 × 109)ü method

= 4,24 × 109 ü accuracy (4)

5. Lightning strikes when there is a large imbalance of charge between the Earth and the base of the cloud.ü

Th e negative charge on the base of the cloud repels electrons deeper into the Earth ü causing the surface and the tree tops to be positively charged. ü If you shelter under a tree you are very likely to be struck because the tree attracts the lightning strike due to its height above the ground. ü

Diagram: Clouds charged negatively at baseü

Tree and surface oppositely chargedü

PAGE 39

4.2 𝑄𝑄𝑄𝑄 = 𝑛𝑛𝑛𝑛𝑞𝑞𝑞𝑞𝑒𝑒𝑒𝑒 method

8,2 × 10-9 = 𝑛𝑛𝑛𝑛 × 1,6 × 10-19 substitutions

𝑛𝑛𝑛𝑛 = 5,13 × 1010 accuracy (4)

PAGE 40



𝑛𝑛𝑛𝑛 = 8,75 × 109 accuracy

Number of electrons transferred = (5,13 × 1010) – ( 8,75 × 109) method

= 4,24 × 109 accuracy (4)

PAGE 40

5. INSERT DIAGRAM

PAGE 42

3.

(6)

6. 6.1 When the air is very dry not much water vapour is present, so charge will remain on objects for longer. ü As you walk along charge builds up on your body (and your hand) because your feet (shoes) are rubbing electrons on (or off ) the carpet. üTh e door handle is metal – it conducts the excess charge away from your hand. ü You feel this as an electric shock. (3)

6.2 When a charged plastic ruler is brought close to small pieces of paper the molecules on the surface of the paper are become polarised. üü Opposite charge attracts ü so the paper pieces are attracted to the ruler. (3)



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TOTAL: 34 MARKS

MULTIPLE CHOICE1. D üü Like charges repel each other.

Neutral objects are attracted to charged objects by polarisation of their molecules. (2)

2. D üü Polarisation is a shift in charge inside the molecule (or object). (2)

3. D üü A neutral object has an equal number of positive and negative charges. It has no imbalance of charge. (2)

4. B üü Charge on each = 6 nC Total charge = 6 + 6 = 12 nC Both pith balls were initially positively charges so Option A is incorrect. Option C is incorrect because they were not equally charged. Option D is an impossible answer. (2)

5. 5.1 Extra electronsü are attracted by the positive rodü from the earthü through the wire to the top surface of the metal.ü (4)

5.2 Negative.üü

Extra electronsü moved from earth through the wire onto the metal.ü (4)

5.3 No charge.üü

Th e extra electrons move back to earth through the wireüü when the rod is removed. (4)

6. 6.1 Electrons are negatively charged particles.ü Th ey are attracted to the positively charged plate (repelled by the negatively charged plate)ü so the beam of electrons can be moved up and down by changing the amountü and the type of chargeü on the plates.

6.2 Th e screen of these TV sets is made of glass covered with tiny dots of coloured fl uorescent paint.. Electrons strike the glass ü and activate the fl uorescent paint on the screen so that these colours shine. Negative charge builds up on the screen (glass). ü Th is negative charge attracts particles of dust ü by polarising them. ü (4)

7. Th e dust and smoke particles gain electronsü and become negatively charged when they pass through the negative grid.ü Opposite charges attract each other. ü Th e dust and smoke particles are attracted to the positively charged plates. ü Th ey lose their extra electrons and become positively charged. ü Th e positive plates repel them away ü so they fall into the dust trap.ü (6)


RESOURCE PACK


TOPIC 16: Electric Circuits WORKSHEET

1. A 12 V battery supplies a maximum of 120 J of energy when charge passes through it. Calculate the amount of charge passing through the battery. (4)

2. When three cells are connected in series, the total emf of the battery is 24 V. Two resistors are connected in parallel with switches S1 and S2 as shown in the diagram below, and connected to the battery. An ammeter (A) reads the current passing through the battery and voltmeter(V) reads the potential difference across the 6 Ω resistor. The cells and ammeter have negligible resistance. The voltmeter has a very high resistance.

PAGE 39



𝑛𝑛𝑛𝑛 = 5,13 × 1010 accuracy (4)

PAGE 40



𝑛𝑛𝑛𝑛 = 8,75 × 109 accuracy

Number of electrons transferred = (5,13 × 1010) – ( 8,75 × 109) method

= 4,24 × 109 accuracy (4)

PAGE 40

5. INSERT DIAGRAM

PAGE 42

3.

2.1 Explain what it means when we say: ‘The emf of the battery is 24 V.’ (2)

2.2 What is the reading on the voltmeter when switch S1 is open? (1)

2.3 What is the reading on the voltmeter when S1 is closed? (1)

2.4 What is the reading on the voltmeter when S1 and S2 are closed? (1)

2.5 Calculate the equivalent resistance of the two resistors in parallel. (4)

3. All the resistors in the following diagrams are identical. Each resistor has a resistamnce of 10 Ω. Calculate the effective resistance of each combination of resistors.

3.1 10 10 10 (2)

3.2

10

10 (3)

3.3

10

10

10

10 (3)

3.4

1010

10 (3)


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TOPIC 16: ELECTRIC CIRCUITS

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4. Three identical cells are connected in series to make a battery. The battery is connected to two identical 6 Ω resistors which are in parallel. When the switch S is closed the ammeter reads the current in one of the parallel branches as 600 mA.

PAGE 43

5.

6. PAGE 44

7.

4.1 Determine the current through the other parallel branch. (2)

4.2 Calculate the current passing through the battery. (2)

5. Study the three circuit diagrams shown below.

The cells, light bulbs and resistors are identical in each circuit.

PAGE 43

5.

6. PAGE 44

7.

In which circuit will the light bulb glow the brightest? Show your reasoning. (4)

6. Four circuit diagram are shown below. Each circuit uses an identical cell, and identical light bulbs.

PAGE 44

2.4 1𝑅𝑅𝑅𝑅𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒

= 1𝑅𝑅𝑅𝑅1

+ 1𝑅𝑅𝑅𝑅2

= 16

+ 112

= 312

𝑅𝑅𝑅𝑅𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 123

= 4 Ω (4)

The switch in each of the four circuits is closed.

6.1 In which circuit(s) are the light bulbs brightest? (A, B, C and/or D). Explain briefly. (3)

6.2 Compare the brightness of the bulb in Circuit A with the brightness of the bulbs in Circuit B. (2)

6.3 Compare Circuit C with Circuit D. (2)


RESOURCE PACK


7. 7.1 Explain what is meant by “the current is 2 A”. (4)

7.2 Calculate the charge that passes through a resistor when a steady current of 2 A is maintained for 2 minutes. (4)

7.3 Explain what is meant by “the potential difference across the resistor is 4 V”. (4)

7.4 Calculate the energy transferred when a potential difference of 4 V is connected across a resistor, and a steady current of 2 A passes through it for 2 minutes. (4)

7.5 Explain what is meant by resistance in an electric circuit. (2)


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TOTAL: 40 MARKS

MULTIPLE CHOICE1. The diagram shows a circuit containing three bulbs and three switches.

S1 SBulb 1

Bulb 2

Bulb 3

2

S3

Bulb 1 and Bulb 3 are lit, but Bulb 2 is not lit.

Which switch (or switches) is (are) closed?

A S1 only

B S2 only

C S1 and S3

D S2 and S3 (2)

Refer to the circuit shown below to answer both Question 2 and 3.

A

V

12 V

6.0

The ammeter reads 2,0 A and the voltmeter reads 12 V.

2. How much charge passes through the resistor in 10 seconds?

A 2,0 C

B 20 C

C 12 C

D 120 C (2)


RESOURCE PACK


3. How much energy is transferred by the resistor in 10 seconds?

A 2,4 J

B 14,4 J

C 240 J

D 144O J (2)

4. Th e diagram below shows part of an electrical circuit.

4,0

2,0A

3,0 A

Th e current in the 4,0 Ω resistor is 3,0 A.

What is the current through the ammeter?

A 4,5 A

B 6,0 A

C 9,0 A

D 12,0 A (2)

LONG QUESTIONS

5. Th e electric circuit shown below contains a battery, two resistors, a switch and another component.

GRADE 10 T2 RP 1st proof amendments

PAGE 47

DIAGRAM FOR QUESTION 5

5.6 1𝑅𝑅𝑅𝑅𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒

= 1𝑅𝑅𝑅𝑅1

+ 1𝑅𝑅𝑅𝑅2 = 1

16+ 1

8 = 3

16

𝑅𝑅𝑅𝑅𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 163 = 5,33 Ω

OR

𝑅𝑅𝑅𝑅𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 𝑅𝑅𝑅𝑅1𝑅𝑅𝑅𝑅2𝑅𝑅𝑅𝑅1+𝑅𝑅𝑅𝑅2

= 16×816 +8

= 5,33 Ω (4)

6.3 The light bulb and the resistor divide the current into parts according to their

resistances. More current passes through the resistor than through the light bulb because it has a lower resistance. (It offers less opposition to the flow of charge).

Ratio of resistance = 160 : 240 = 2 : 3 The current splits into 5 parts. The current through the light bulb = 2/5 of 125 mA = 50 mA (4) 6.5 30 mA

The light bulb is connected in series with the resistor and the battery. The current in a series circuit is the same at all points in the circuit. (3) 6.6 A series circuit is a potential divider. The pd divides in the ratio of the resistors: 160 : 240 = 2: 3 Vlight bulb = 240

400× 12 = 7,2 V (3)

5.1 Name component X. (1)

5.2 Which quantity does this instrument measure? (1)

5.3 Th e switch is closed so that current passes through the circuit. What fl ows in the circuit to create the current? Choose one of the following quantities:

charge, potential diff erence, energy or resistance (1)

5.4 Calculate the equivalent resistance of the two resistors when the switch is closed. (3)


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5.5 When the switch is opened, what is the potential diff erence across the 16 Ω resistor? (2)

5.6 If these two resistors are connected in parallel with one another, what would be their equivalent resistance? Show your working. (4)

6. Circuit 1 and Circuit 2 consist of a 12 V battery connected to a 160 Ω resistor and a light bulb with a resistance of 240 Ω.

12V 160

Circuit 1 Circuit 2

16012V

240

240

6.1 Describe the arrangement of the resistor and light bulb in Circuit 1. (1)

6.2 Calculate the equivalent resistance of Circuit 1. (4)

6.3 In Circuit 1, the current through the battery 125 mA. How much current passes through the light bulb? (4)

6.4 Calculate the equivalent resistance of Circuit 2. (2)

6.5 In Circuit 2 the current through the battery is 30 mA. How much current passes through the light bulb? (3)

6.6 Calculate the potential diff erence across the light bulb in Circuit 2. (3)

6.7 In which circuit (1 or 2) does the light bulb shine the brightest? Justify your answer. (3)


RESOURCE PACK


MARKING GUIDELINES

1. 1. V = QW ü method

12ü = Q120 ü substitutions

Q = 12120

= 10 Cü accuracy; SI units (4)

2. 2.1 The work done per unit charge by the battery ü is 24 J.C−1. (1)

2.2 24 Vü accuracy; SI units (1)



2.5 method

substitutionsR R R

R

1 1 1

61121

123

312

4

eq

eq

1 2

X

= +

= +

=

=

=

finding Req by finding the inverse

ccuracy; SI units

ALTERNATIVE METHOD =

Req = R RR R1 2

1 2

+ü method (can be implied)

= x6 126 12+ ü substitutions

= 1872 ü

= 4 Ωü accuracy; SI units (4 )

3. 3.1 Rseries = R1 + R2 + R3ü method = 10 + 10 + 10 = 30 Ωü accuracy; SI units (2)

3.2 R1

eq = R R1 11 2+ ü method

= 101

101+ ü substitutions

= 102

Req = 102

= 5 Ωü accuracy; SI units


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ALTERNATIVE METHOD

Req = R RR R1 2

1 2

+ ü method (can be implied)

= x10 1010 10+ ü substitutions

= 20100

= 5 Ωü accuracy; SI units (3)

3.3 From the answer in 3.2, we know that two 10 Ω resistors in parallel have an equivalent resistance of 5 Ω.ü (Allow c.o.e.)

Req = 5 + 5ü parallel circuits in series = 10 Ωü accuracy; SI units (3)

3.4 Req = 10 + 5 10 Ω in seriesü with parallel resistorsü = 15 Ωü accuracy; SI units (3)

4. 4.1 600 mA (0,6 A)üü (2)

4.2 I = 600 + 600 ü = 1 200 mA (1,2 A) ü (2)

5. Circuit 3 üWhen resistors are added in parallel the total resistance of the circuit decreases.üThe current increases when there is less resistance. üThe light bulb glows brighter when more current passes through it.ü (4)

6. 6.1 Circuits Aü Cü and Dü (3)

6.2 The bulb in Circuit A is brighter than the bulbs in Circuit B. ü Circuit A has less resistance (OR the current in Circuit A is greater than the current in Circuit B) ü (2)

6.3 Circuit C is the same as Circuit D. ü Two light bulbs in parallel with the battery (cell) and switch. ü (2)


RESOURCE PACK


7. 7.1 The rate ü of flow of charge üis 2 Cü per second. ü (4)

7.2 I = tQ3 ü method

2ü = xQ2 60 ü substitutions; conversion of minutes to seconds

Q = 240 Cü accuracy; SI units (4)

7.3 The energy transferredü per unit charge across the ends of the resistorü is 4 Jü per coulombü (OR 4 J.C−1üü). (4)

7.4 V = QW ü method

4ü = W240 ü substitutions; c.o.e. from 7.2

W = 960 Jü accuracy; SI units (4)

7.5 The resistance is the ratio of the potential difference across a component and the current that passes through it. üü (2)


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TOTAL: 40 MARKS

MULTIPLE CHOICE1. Cüü The current must pass through Bulb 1 and Bulb 3, therefore S1 and S3 must be

closed. (2)

2. Büü I= tQ3

2,0 = Q10

Q = 20 C (2)

3. Cüü V= W/Q

12 = W/20

W = 240 J (2)

4. Cüü The current through the 4,0 Ω resistor is 3,0 A. The ratio of the parallel resistors is 4,0: 2,0 = 2 : 1. Therefore the current splits between them in the ratio of 1 : 2. The current through the 2,0 Ω resistor is 2 × 3,0 = 6,0 A. The total current is equal to the sum of these two currents. Itotal = 3,0 + 6,0 = 9,0 A (2)

LONG QUESTIONS

5. 5.1 Ammeter ü (1)

5.2 Current ü (1)

5.3 Charge ü (1)

5.4 Req = R1 + R2ü method

= 16,0 + 8,0ü substitutions

= 24,0 Ωü accuracy; SI units (3)

5.5 Series circuits divide the potential difference in the same proportion as the resistance of their resistors.

Therefore the pd divides as: 16,0 : 8,0 = 2: 1 The pd divides into 3 parts. The 16,0 Ω resistor has , ,V V24

16 12 0 8 0# = (2)


RESOURCE PACK


5.6

,

R R R

R

1 1 1161

81163

316 5 33

eq

eq

1 2= + = + =

= =

OR ,R R R

R R16 816 8 5 33eq

1 2

1 2 #X= + = + = (2)

6. 6.1 in parallelü (1)

6.2 R1

eq= R R1 11 2+ ü method

= 1601

2401+ ü substitutions

= 1/96 inverting the answer

Reqü = 96 Ωü accuracy; SI units

OR ALTERNATIVE METHOD

Req = R RR R1 2

1 2

+ ü method

= 160 240160 240#

+ substitutions

= 96 Ωü accuracy; SI units (4)

6.3 The light bulb and the resistor divide the current into parts according to their resistances. More current passes through the resistor than through the light bulb because it has a lower resistance. (It offers less opposition to the flow of charge). ü

Ratio of resistance = 160 : 240 = 2 : 3 ü The current splits into 5 parts. The current through the light bulb = 5

2 of 125 mAü = 50 mAü (4)

6.4 Req = R1 + R2ü method = 160 + 240 = 400 Ωü accuracy; SI units (2)

6.5 30 mAü The light bulb is connected in series with the resistor and the battery.ü The current in a series circuit is the same at all points in the circuitü (3)

6.6 A series circuit is a potential divider. The pd divides in the ratio of the resistors: 160 : 240 = 2: 3ü ,V V400

240 12 7 2light bulb # = = (3)

6.7 Circuit 1ü The pd across the light bulb is much greater in Circuit 1ü therefore more energy is

transferred per unit charge through the bulb in Circuit 1. ü OR The current through the bulb in Circuit 1 is 50 mAü which is three times

greater than the current in Circuit 2.ü (3)


FORMAL EXPERIMENT


FORMAL EXPERIMENTGRADE 10 TERM 2: PHYSICS

Electric Circuits with Resistors in Series and in

Parallel

75 Marks

This section provides guidance and assessment of the learner’s knowledge and understanding when carrying out a virtual experiment using the NECT

video of the same name.

If your class is carrying out the experiment using laboratory apparatus and taking down their own results, you must set up your classroom appropriately and give the learners the relevant instructions. You may find it useful to refer to the Technical Instructions which precede the Learner’s Instructions

while preparing for this experiment.

If the learners are proceeding with the virtual experiment, then continue with the NECT programme by using the information, handouts and marking

guidelines contained in this section of this Resource Book.


ELECTRIC CIRCUITS WITH RESISTORS IN SERIES AND IN PARALLEL

Term 1 55

EXPERIM

ENT

Formal ExperimentTECHNICAL INSTRUCTIONS

AIM: TO DETERMINE THE EFFECT OF MULTIPLE RESISTORS ON CURRENT AND POTENTIAL DIFFERENCE IN ELECTRIC CIRCUITS: A) CONNECTED IN SERIES (PART 1) AND B) CONNECTED IN PARALLEL (PART 2).

PART 1: APPARATUS4 V DC power supply (or 1,5 V cell)3 × 10 Ω resistors1 switch12 connecting leads1 ammeter 2 voltmeters

0

1020

30

405

15 25

35

V

0

1020

30

405

1525

35

V

0

1020

30

405

1525

35

A

0

1020

30

405

1525

35

A

0

1020

30

405

1525

35

V

0

1020

30

405

15 25

35

V 0

1020

30

405

15 25

35

V

0

1020

30

405

1525

35

V

0

1020

30

405

1525

35

A

Diagram 1 Diagram 2 Diagram 3

Voltmeter T Voltmeter T Voltmeter T

Voltmeter 1 Voltmeter 1 Voltmeter 1Ammeter AmmeterAmmeter

R1 R1 R1

R2 R2

R3

PART 2: APPARATUS4 V power supply (or 1,5 V cell)3 × 10 Ω resistors 1 switch12 connecting leads 2 ammeters 1 voltmeter

0

1020

30

405

1525

35

V

0

1020

30

405

1525

35

V

0

1020

30

405

1525

35

V0

1020

30

405

1525

35

A

0

1020

30

405

1525

35

A

0

1020

30

405

1525

35

A

0

1020

30

405

1525

35

A

0

1020

30

405

1525

35

A

0

1020

30

405

1525

35

A


Ammeter TAmmeter T Ammeter T

Ammeter 1Ammeter 1

Ammeter 1

VV V

R1R1 R1R2 R2 R3


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NAME: GRADE:

Formal ExperimentELECTRIC CIRCUITS

75 MARKS

PART 1: RESISTORS IN SERIES

AIM: TO DETERMINE THE EFFECT OF MULTIPLE RESISTORS CONNECTED IN SERIES ON THE CURRENT AND POTENTIAL DIFFERENCE IN ELECTRIC CIRCUITS.

APPARATUS4 V power supply (or 1,5 V cell)3 × 10 Ω resistors1 switch12 connecting leads1 ammeter 2 voltmeters

SAFETY MEASURES OR PRECAUTIONSThe current and voltages in d.c. circuits using 3 torch cells in series pose no safety risks.

If using a power pack to apply a voltage across the circuit, ensure that that the switch is off when making and breaking connections.

Check that all connecting leads are intact, insulated and in good working order e.g. there is no rust on the connecting points, no loose ends etc.

All connections should made securely to allow current to pass through the circuit.

Moving coil voltmeters and ammeters should be correctly connected to the positive and negative terminals. If using digital ammeters and voltmeters the meters should give positive readings when connected in the circuit.

Open the switch to disconnect the power source when the readings have been taken to keep the temperature of the resistors at room temperature.



Term 1 57

EXPERIM

ENT

VARIABLES1. Identify the variables in this experiment.

• Independent: ________________________________________________ (1)• Dependent: ________________________________________________

________________________________________________ (2)• Controlled: ________________________________________________ (2)

METHOD1. Set up the circuit using one resistor (R1) as shown in diagram 1.

0

1020

30

405

15 25

35

V

0

1020

30

405

1525

35

V

0

1020

30

405

1525

35

A

0

1020

30

405

1525

35

A

0

1020

30

405

1525

35

V

0

1020

30

405

15 25

35

V 0

1020

30

405

15 25

35

V

0

1020

30

405

1525

35

V

0

1020

30

405

1525

35

A


Voltmeter T Voltmeter T Voltmeter T

Voltmeter 1 Voltmeter 1 Voltmeter 1Ammeter AmmeterAmmeter

R1 R1 R1

R2 R2

R3

2. Insert an ammeter (A) into the circuit to measure the current (I) through the resistor (R1).

3. Insert a voltmeter (V1) across the resistor to measure the potential difference (V1) across the resistor.

4. Insert a voltmeter (VT) across the battery (cell) to measure the potential difference (VT) across the terminals of the battery.

5. Close the switch and take readings on the voltmeters and the ammeter.

6. Open the switch.

7. Add another resistor (R2) in series (as shown in diagram 2).

8. Close the switch and take readings on the voltmeters and the ammeter.

9. Open the switch.

10. Repeat instructions 7 to 9 once more (as shown in diagram 3) adding resistor (R3).

ANALYSIS AND INTERPRETATION2. Draw a circuit diagram showing three resistors in series with an ammeter, a cell and a

switch. Draw one voltmeter connected across the first resistor and a second voltmeter connected across the terminals of the battery to measure the potential difference across the battery. (6)


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3. Fill in a title for this table, and fill the SI units into the column headings. Watch the video and record the readings. Leave the last two columns of this table blank while taking down the measurements

Table of results: ____________________________________________________

_________________________________________________________________ (1)

NUMBER OF RESISTORS IN SERIES

V1 ( ) VT ( ) IT ( )

1

2

3

(6)

4. Use the formula R IV= to calculate the resistance R1 of the first resistor in each of the

circuits. Use the same formula to calculate the resistance RT of the series circuits with one, two and three resistors in series. Write your answers in the table. (4)

5. Comment on the value of the resistance of the first resistor R1 as more resistors are added in series to the circuit.

(1)

6. Comment on the value of the resistance of the series circuit as more resistors are added in series to the circuit.

(1)

7. Predict the value of the effective resistance of the series circuit when four identical resistors are connected in series.

(2)



Term 1 59

EXPERIM

ENT

8. Predict the value of the current through the series circuit when four identical resistors are connected in series. Justify your answer with a calculation.

(2)

9. Using the results obtained in this experiment, show that the effective resistance of a series circuit is calculated by Rseries = R1 + R2 + R3 + … (5)

10. Using the values from the table of results in Question 3, complete the following table by filling in the SI units, copying the recorded values and calculating the missing values.

NO. OF RESISTORS IN SERIES

RT R1

R2 OR R2 + R3

VT V1

V2 ORV2 + V3

2

3

(2)


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11. Use the results from the table in Question 10 to show that for a series circuit:

RV

RV

T

T

1

1 = (3)

CONCLUSION12. (2)



Term 1 61

EXPERIM

ENT

PART 2. RESISTORS IN PARALLEL

AIM: TO DETERMINE THE EFFECT OF MULTIPLE RESISTORS CONNECTED IN PARALLEL ON THE CURRENT AND POTENTIAL DIFFERENCE IN ELECTRIC CIRCUITS.

APPARATUS4 V power supply (or 1,5 V cell)3 resistors 1 switch12 connecting leads 2 ammeters 1 voltmeter

SAFETY MEASURES OR PRECAUTIONS:The same safety measures and precautions as noted for Part 1 of this experiment.

VARIABLESThe same variables as listed in Part 1 of this experiment.

METHOD1. Set up the circuit using one resistor R1 as shown in diagram 1 below.

0

1020

30

405

1525

35

V

0

1020

30

405

1525

35

V

0

1020

30

405

1525

35

V0

1020

30

405

1525

35

A

0

1020

30

405

1525

35

A

0

1020

30

405

1525

35

A

0

1020

30

405

1525

35

A

0

1020

30

405

1525

35

A

0

1020

30

405

1525

35

A


Ammeter TAmmeter T Ammeter T

Ammeter 1Ammeter 1

Ammeter 1

VV V

R1R1 R1R2 R2 R3

2. Insert an ammeter A1 into the circuit to measure the current I1 through the resistor.

3. Insert the second ammeter AT into the circuit to measure the current IT passing through the battery.

4. Insert a voltmeter V across the resistor to measure the potential difference V across the resistor.

5. Close the switch and take readings on the voltmeter and the ammeters.

6. Open the switch.


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7. Add another resistor R2 in parallel to the resistor R1 as shown in diagram 2. Be sure to exclude the second ammeter AT in this connection.

8. Close the switch and take readings on the voltmeter and the ammeters.

9. Open the switch.

10. Repeat instructions 7 to 9 once more as shown in diagram 3 and add resistor R3.

OBSERVATIONS AND INTERPRETATION1. Draw a circuit diagram showing a battery connected to an ammeter AT, a switch

and three resistors R1, R2 and R3 in parallel. A voltmeter V is connected across the parallel combination. An ammeter A1 is connected to the first resistor R1 of the parallel combination. (6)



Term 1 63

EXPERIM

ENT

2. Fill in a title for this table, and fill in the SI units in the column headings.

Watch the video and record the readings.

Leave the last two columns of this table blank while taking down the measurements

Table of results: ____________________________________________________

_________________________________________________________________ (1)

NUMBER OF RESISTORS IN SERIES

V1 ( ) VT ( ) IT ( )

1

2

3

(6)

3. Use the formula R IV= to calculate the resistance R of the first resistor and the

resistance RT of the whole circuit for each of the three circuits. Write the answer in the last two columns of the table, and write a column heading with appropriate SI units for each of these columns.

Use the same formula to calculate the resistance RT of the parallel circuit. (4)

4. Comment on the value of the resistance of the first resistor as more resistors are added in parallel.

(1)

5. Comment on the value of the resistance of the parallel circuit as more resistors are added in parallel.

(1)


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6. Predict the resistance of the parallel circuit when four identical resistors are connected in parallel. Briefly justify your answer. (2)

7. Predict the reading on the ammeter when four identical resistors are connected in parallel. (2)

8. Calculate the resistance of the two resistors in parallel R// using the formula:

R R R1 1 1

1 2= =

' (3)

9. Compare the answer to Question 8 with the results of the experiment when using two resistors in parallel.

(1)



Term 1 65

EXPERIM

ENT


NO. OF RESISTORS

IN PARALLEL

RT R1

R2 OR R2 + R3

IT I1 I2 ORI2 + I3

2

3

(2)

11. Use the results from the table (in Question 10) to show that for a parallel circuit

I R I RT T1 1 = (4)

CONCLUSION12. (2)


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MARKING GUIDELINES

75 MARKS

PART 1. RESISTORS IN SERIES

AIM: TO DETERMINE THE EFFECT OF MULTIPLE RESISTORS ON CURRENT AND POTENTIAL DIFFERENCE IN ELECTRIC CIRCUITS.1. Identify the variables in this experiment.

• Independent: Number of resistors connected in series ü (1)• Dependent: Current through the resistors ü Potential difference (voltage) across the resistors ü (2)

• Controlled: Terminal voltage (same battery) ü Identical resistors ü

Same temperature of resistors (room temperature) [ANY TWO VALID ANSWERS] (2)

2.

ü All components presentü Ammeter in seriesü Voltmeter 1 across R1

ü Voltmeter T across batteryü 3 resistors in seriesü Switch (open or closed)

S

R2

R3

R1

VT

V1A

(6)

3 & 4

Table of results: Voltage (Potential difference) and Current (readings for different numbers of) resistors connected in series. ü (1)

Incorrect or no SI units in each column heading: −1 mark

NUMBER OF RESISTORS IN

SERIESV1 (V) VT (V) IT (A) R1 (Ω) RT (Ω)

1 4,02ü 4,0 0,41ü 9,8 9,8ü

2 2,02ü 4,0 0,20ü 10,1ü 20,0ü

3 1,35ü 4,0 0,13ü 10,4 30,8ü

(10)



Term 1 67

EXPERIM

ENT

5. Comment on the value of the resistance of the first resistor R1 as more resistors are added in series.

Its resistance remains (almost) constant (OR is almost 10 Ω).ü (1)

6. Comment on the value of the resistance of the series circuit as more resistors are added in series.

The resistance of the series circuit increasesü as more resistors are added in series. (1)

7. Predict the value of the effective resistance of the series circuit when four identical resistors are connected in series.

Effective resistance with 4 × 10 Ω resistors in series = 40 Ωüü (accuracy; SI units) (2)

8. Predict the value of the current through the series circuit when four identical resistors are connected in series.

R IV= ü (method)

I40 4=

,I 0 1= Aü (accuracy; SI units) (2)

9. From the results obtained in this experiment, show that the effective resistance of a series circuit is calculated by Rseries = R1 + R2 + R3 + …

For two resistors in series RT = 20 Ωü R1 + R2 = 10 + 10ü = 20 Ω Therefore RT = R1 + R2

For three resistors in series RT = 30,8 Ωü R1 + R2 + R3 = 10,4 + 10 + 10ü = 30,4 Ω Therefore RT = R1 + R2 + R3ü (almost equal to RT)

The formula for calculating the effective resistance of resistors in series is correct. (5)


NO. OF RESISTORS IN SERIES

RT

(Ω)

R1

(Ω)

R2 OR R2 + R3

(Ω)

VT

(V)

V1

(V)

V2 ORV2 + V3

(V)

2 20 10,1 9,9 4,0 2,02 1,98

3 30,8 10,4 20,4 4,0 1,35 2,65

üCorrect SI units and values copied correctly (Columns 2, 3, 5 and 6) üValues calculated correctly (Columns 4 and 7) (2)


RESOURCE PACK


11. Use the results from the table in Question 10 to show that for a series circuit:

RV

RV

T

T

1

1 =

For two resistors in series: ,,

,RV

10 12 02

0 21

1 = = ü and ,,

, ,RV

20 44 0

0 196 0 2T

T = = = ü

therefore RV

RV

T

T

1

1 =

For three resistors in series: ,,

,RV

10 41 35

0 131

1 = = and ,,

,RV

30 84 0

0 13T

T = = ü

therefore RV

RV

T

T

1

1 = (3)

CONCLUSION12. A series circuit acts as a potential difference divider. ü

The ratio of the potential difference to resistance is equal to RV

RV

T

T

1

1 = . ü

OR

The current in the series circuit is inversely proportional ü to the number of identical resistorsü connected in series.

[Allow any similar conclusion as either of the above – if it is incomplete, but appropriate, award one mark; if complete award two marks.] (2)

PART 2.1.

ü All components presentü Ammeter in seriesü Voltmeter 1 across R1

ü Voltmeter T across batteryü 3 resistors in seriesü Switch (open or closed)

S

AT

A1

R1

R2 R3

V

(6)

2&3. Table of results: Voltage (Potential difference) and Current (readings for different numbers

of ) resistors connected in parallel ü (1)

NUMBER OF RESISTORS

IN PARALLELV (V) IT (A) I1 (A) RT (Ω) R1 (Ω)

1 2ü 0,19ü 0,194ü 10,5ü 10,3ü

2 2 0,39ü 0,189 5,1ü 10,6

3 2 0,58ü 0,194 3,45ü 10,3

ü Correct SI units (10)



Term 1 69

EXPERIM

ENT

4. Comment on the value of the resistance of the first resistor as more resistors are added in parallel.

The resistance of R1 remains (fairly) constant (is about 10,3 Ω)ü (1)

5. Comment on the value of the resistance of the parallel circuit as more resistors are added in parallel.

The resistance of the circuit decreases as more resistors are added in parallel.ü (1)

6. Predict the resistance of the parallel circuit when four identical resistors are connected in parallel.

,

R 410

2 5

T

X

=

= [Learners can calculate the value using the formula] (2)

7. Predict the reading on the ammeter when four identical resistors are connected in parallel.

0,8 Aüü [Learners can calculate the value using R V1= .] (2)

8. Calculate the resistance of two resistors in parallel R// using the formula:

R R R1 1 1

1 2= +

'.

R R R

R

1 1 1

101

101

102

210

5

1 2

X

= +

= +

=

=

=

'

'

(3)

9. Compare the answer to Question 8 with the results of the experiment when using two resistors in parallel.

The result is (almost) the same as that found by experiment. ü (1)


NUMBER OF RESISTORS IN

PARALLEL

RT

(Ω)R1

(Ω)R2 ORR2 + R3

(Ω)

IT

(A)I1

(A)I2 ORI2 + I3

(A)

2 5,1 10,6 10,6 0,39 0,19 0,20

3 3,45 10,3 5,1 0,58 0,19 0,39

ü Correct SI units and values copied correctly (Columns 2, 3, 5 and 6) ü Values calculated correctly (Columns 4 and 7) (2)


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11. Use the results from the table (in question 10) to show that for a parallel circuit

I1R1 = ITRT

For 2 resistors I1R1 = (0,19)(10,6) = 2,0 Vü

ITRT = (0,39) (5,1) = 1,989 = 2,0 Vü therefore I1R1 = ITRT

For 3 resistors I1R1 = (0,19)(10,6) = 2,0 Vü

ITRT = (0,58) (3,45) = 2,0 V ü therefore I1R1 = ITRT (4)

CONCLUSION12. Resistors in parallel are current dividers. ü The current divides in a ratio according to the resistance of each branch so that

I1R1 = ITRT ü

OR

The potential difference (voltage) across each branch of a parallel circuit is the same.ü The current in the circuit is equal to the sum of the currents through all the branches.

OR (IT = I1 + I2 + I3) ü (2)


ASSESSMENTS


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TOPIC 11: Particles Substances Are Made OfQUESTIONS

MULTIPLE CHOICE1. Which one of the following has a covalent network structure?

A Ar B NH3 C graphite D lithium chloride (2)

2. Which one of the following does not have a crystal lattice structure at room temperature? A diamond B sodium fluoride C potassium nitrate D neon (2)

3. What type of substances only conduct electricity when they are molten or in aqueous solution?A ionic substancesB covalent molecular substancesC metalsD covalent network structures (2)

4. The following is a model of a chemical substance. Which one of the following chemical formulae could possibly be correct for this model?

carbon large blackhydrogen small light greyoxygen medium darker grey

A CH4 B CH O2 2 C CHO D CO2 (2)


Term 2 73


ASSESSM

ENTS

5. How many atoms, in total, are there in the molecule represented by H PO3 4 ? A 1 B 3 C 8 D 12 (2)

6. What type of crystal structure is present in ice? A ionic B metallic C covalent network D covalent molecular (2)

LONG QUESTIONS 7. 7.1 What types of solids are diamond and graphite? (2)

7.2 Give an explanation for the fact that diamond is very hard yet graphite is a soft, powdery substance. (3)

7.3 Give an explanation of the fact that diamond does not conduct electricity, while graphite is a good conductor in the solid state. (4)

8. 8.1 What types of solids are potassium chloride and lead? (2)8.2 Give an explanation for the fact that potassium chloride is a hard, brittle

solid and lead is a soft solid that can be shaped as necessary. (6)8.3 Give an explanation for the fact that potassium chloride does not conduct

electricity in the solid state, but conducts when molten or in solution and lead conducts electricity in both the solid and molten states. (6)

9. Covalent bonding occurs in both covalent molecular solids and in covalent network solids. Give explanations for the following facts.9.1 Molecular solids are soft, while network solids are usually very hard. (4)9.2 Molecular solids have low melting points, while network solids have very

high melting points. (4)

10. Fluorine is a gas at room temperature, while sodium is a soft metal with a relatively low melting point. However; when they react, they form a hard crystalline solid. 10.1 Explain in detail how sodium and fluorine combine when they react. (4)10.2 Explain why the product formed is a hard, crystalline solid with a high

melting point. (4)

11. State which one of each of the following pairs of substances will have the higher melting point and give a brief explanation.11.1 Ice and aluminium. (4)11.1 Carbon dioxide and silicon dioxide (SiO2 ). (5)


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MARKING GUIDELINES

MULTIPLE CHOICE 1. C Graphite contains atoms bonded together covalently to form large

flat sheets. It has a high melting point and boiling point. [CL 2] (2) Ar is a noble gas and does not form covalent bonds. NH3 is a

covalent molecular substance and lithium chloride is ionic.

2. D Neon is a gas at room temperature (noble gas). Sodium fluoride and potassium nitrate have ionic crystal lattices and diamond has a covalent network crystal lattice. [CL 2] (2)

3. A Ionic substances do not conduct electricity when in the solid state, but they do conduct when molten or in aqueous solution. Metals conduct electricity in both the solid state and the molten state. [CL 2] (2)

4. B There are three different coloured spheres in the illustration. That means that there are three elements represented. There is one black sphere and two light grey ones and two darker grey ones, so one element has one atom present and the other two have 2 atoms present. The answer can only be B. [CL 3] (2)

5. C There are 3 hydrogen atoms, 1 phosphorus atom and 4 oxygen atoms. [CL 2] (2)

6. D Ice is made up of molecules of water held together by weak intermolecular forces, so it is a covalent molecular solid. [CL 3] (2)

LONG QUESTIONS7. 7.1 They both have covalent network structures. [CL 1] (2)

7,2 Diamond is very hard because every carbon atom in diamond is bonded to four others by very strong covalent bonds.

Graphite is soft because it doesn’t have only covalent bonds but there are also weak intermolecular forces present between flat sheets of carbon atoms and these make graphite soft. [CL 2] (3)

7.3 In diamond all electrons are paired up in strong covalent bonds, there are no free charges of any type that are free to move and become charge carriers.

In graphite there is one unpaired electron on every carbon atom. These electrons become delocalised and are free to move and act as charge carriers. [CL 2] (4)

8. 8.1 Potassium chloride is an ionic solid. Lead is a metallic solid. [CL 1] (2)8.2 In potassium chloride there are cations and anions which are held in place

in a crystal lattice by very strong ionic bonds.This makes it hard. If


Term 2 75


ASSESSM

ENTS

any force is exerted on the lattice, ions with like charges can come into contact and repel each other, causing the crystal to shatter.

In a metal there are forces between delocalised electrons and positive ions. There is always a layer of delocalised electrons between cations and this prevents the ions to come into contact and repel. This makes the structure flexible and makes it possible to be shaped. [CL 3] (6)

8.3 When potassium chloride is solid, the ions occupy fixed positions in the crystal lattice. There are no charges which are free to move and become charge carriers. When the substance is in the molten state (or in aqueous solution), the ions are free to move and become charge carriers.

In lead there are delocalised electrons which are present in both the solid and molten states. They are free to move and are the charge carriers. [CL 3] (6)

9. 9.1 Molecular solids consist of individual atoms which are held together by weak intermolecular forces and so they are soft.

Network solids consist of atoms (or molecules) which are bonded together by strong covalent bonds and hence they are hard. [CL 3] (4)

9.2 Because molecular solids have weak intermolecular forces, relatively small amounts of energy are needed to separate the molecules and so they have low melting points.

In network solids, the forces between atoms are strong covalent bonds and so large amounts of energy are needed to separate the molecules, so they have high melting points. [CL 3] (4)

10. 10.1 When sodium and fluorine react, the sodium atom, being a metal donates electrons and they are transferred to the fluorine. The sodium atoms become cations and the fluorine atoms become anions. The cations and anions attract each other (electrostatically). [CL 2] (4)

10.2 The attraction between cations and anions results in a crystal lattice being formed in which the oppositely charged ions attract each other with strong ionic bonds. A lot of energy is required is required to separate the ions and so sodium fluoride has a high melting point. [CL 3] (4)

11. 11.1 Aluminium has the higher melting point. The metallic bonds between aluminium ions and the delocalised electrons are stronger than the weak intermolecular forces between ice molecules. [CL 4] (4)

11.2 Silicon dioxide has the higher melting point. Carbon dioxide is a molecular solid, held together by weak intermolecular forces . Silicon dioxide is a network solid and the molecules are held together by strong covalent bonds. [CL 4] (5)


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TOPIC 12: Physical and Chemical ChangeQUESTIONS

MULTIPLE CHOICE 1. Which one of the following is NOT applicable to a physical change?

A The number of molecules is conserved. B Mass is conserved. C A new substance is formed. D The number of atoms is conserved. (2)

2. Which one of the following involves the greatest energy change? A Ice melts.B Carbon dioxide is split into carbon and oxygen.C Iodine sublimates.D Water is heated until it boils. (2)

3. Which one of the following is an indication that a chemical change is taking place? A Salt dissolves in water and becomes invisible. B Chocolate is heated until it melts. C The temperature remains constant during the change. D A gas is released while the change is occurring. (2)

4. Select the correct statement. During a physical change …A matter can change shape. B matter changes from being one substance to another. C there is a very large change in temperature. D new substances are formed. (2)

5. The law of constant composition states that: A molecules combine in a specific ratio in any chemical reaction. B atoms combine in a fixed ratio to form a specific compound. C atoms combine in a specific mass ratio when they react. D the number of molecules in a specific reaction is conserved. (2)

6. The law of conservation of mass states that: A the mass of the reactants is equal to the mass of the products. B the number of molecules of reactants is equal to the number of molecules

of products. C the number of atoms of reactants is equal to the number of atoms in the

products. D the number of atoms of each element in the reactants is equal to the

number of atoms of each element in the product. (2)


Term 2 77


ASSESSM

ENTS

LONG QUESTIONS7. A chemical reaction is carried out in the laboratory. The chemical equation for

the reaction is shown below.$( ) ( ) ( ) ( )AgNO aq KI aq AgI s KNO aq3+ +

What mass of silver nitrate reacts completely with potassium iodide, when 16,6 g of KI reacts to form 23,5 g of AgI and 10,1 g of KNO3 ? (3)

8. Consider the following chemical reaction: ( ) ( ) ( ) ( )NH g O g NO g H O l4 5 4 62 24 $+ +

8.1 Show, by using relative atomic masses that mass is conserved during the reaction. (7)

8.2 Show that the number of atoms of each element is conserved. (7)

9. Explain why the number of molecules is not conserved during a chemical change. (3)

10. Sulfur consists of S8 molecules. Describe the rearrangement of molecules that occurs in each of the following changes.10.1 Sulfur is heated until it melts. (3)10.2 Liquid sulfur is heated until it boils. (4)

11. 11.1 For each of the changes mentioned in Question 10, state whether a physical or chemical change is happening. (2)

11.2 Compare the energy required to bring about these changes with the energy needed to break the covalent bonds between sulfur atoms in S8 molecules. Explain your answer. (4)

12. Describe each of the following reactions as synthesis or decomposition reactions. 12.1 $( ) ( ) ( )Na s Cl g NaCl s2 22+ 12.2 $( ) ( ) ( )K CO s K O s CO g2 3 2 2+ 12.3 $( ) ( ) ( )Fe s S s FeS s+ (3)


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MARKING GUIDELINES

MULTIPLE CHOICE1. C New substances are only formed by means of a chemical change.

All the other statements are applicable to physical changes. [CL 1] (2)

2. B Splitting carbon dioxide into carbon and oxygen means that the covalent bonds between carbon and oxygen have to be broken. This requires much more energy than a physical change. The other changes described are physical changes. [CL 3] (2)

3. D Release of a gas indicates that a chemical change is occurring. During a chemical change there is always a change in temperature. [CL 2] (2)

4. A A change of shape is a physical change. All the other changes described are chemical changes. [CL 2] (2)

5. B The law of constant composition states that atoms of elements combine in fixed ratios to produce a specific compound. Note that there is not a specific mass ratio in which elements combine. [CL 3] (2)

6. A The number of molecules being conserved is not what occurs in a chemical change. The number of atoms is conserved but this is not stated by the law of conservation of mass. [CL 4] (2)

LONG QUESTIONS7. By the law of conservation of mass, the total mass of reactants and products must

be the same. So, reactants products X + 16,6 = 23,5 + 10,1

X = 17,0 g [CL 3] (3)8. $NH O NO H O4 5 4 63 2 2+ +

8.1 Reactants: : ( )NH m 4 14 1 33 #= + = 68 : ( )O m 5 2 162 #= = 160 Total reactants = 228 Products: : ( )NO m 4 14 16= + = 120 : ( )O m 6 2 1 16#= +H2 = 108


Term 2 79


ASSESSM

ENTS

Total products = 228 ` Mass of reactants = Mass of products [CL 2] (7)

8.2 Reactants: N = 4 atoms H 4 3 12#= = atoms O 5 2 10#= = atoms Products: N 4 1 4#= = atoms O 4 1 6 1 10# #= + = atoms H 6 2 12#= = atoms Number of atoms of each element is the same in reactants and products

` conserved. [CL 2] (7)

9. The number of molecules of each substance is not conserved in a chemical change because the molecules of the reactants each break up when they react and form different molecules in the products. The number of molecules may increase or decrease. [CL 3] (3)

10. 10.1 In solid sulfur, the molecules are held in place in a crystal lattice by intermolecular forces. When the sulfur melts, the molecules become disordered because the intermolecular forces can no longer keep them in the crystal lattice arrangement. [CL 2] (3)

10.2 In liquid sulfur there are still intermolecular forces in place so the molecules are not totally free to move. When the sulfur absorbs more energy, the intermolecular forces are broken completely and the sulfur molecules are totally free to move. [CL 2] (4)

11. 11.1 Both changes described in 10.1 and 10.2 are physical changes. [CL 2] (2)11.2 The energy needed to break covalent bonds between sulfur atoms is much

greater than the energy needed for either change described in 10.1 or 10.2. This is because covalent bonds are much stronger than intermolecular

forces, so more energy is needed to separate the atoms than to separate the molecules. [CL 3] (4)

12. 12.1 This is a synthesis reaction. (Two substances react to form one substance). 12.2 This is a decomposition reaction. (One substance breaks up into two

other substances). 12.3 This is a synthesis reaction. (Two substances react to form one

substance). [CL 2] (3)


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TOPIC 13: Representing Chemical ChangeQUESTIONS

MULTIPLE CHOICE 1. Magnesium carbonate is heated and reacts according to the following equation:

$( ) ( ) ( )MgCO s MgO s CO g3 2+

40 g of the powder is heated in a closed container. Which of these statements is correct? A The mass of MgO is 40 g.B The total mass of the products is 40 g. C The total mass of the products is less than 40 g. D The total mass of the products is greater than 40 g. (2)

2. Hydrogen gas reacts with oxygen gas to produce water. The correct balanced reaction equation for this reaction is: A $( ) ( ) ( )H g O g H O l2 2+ B $( ) ( ) ( )H g O g H O l2 2 2+ C $( ) ( ) ( )H g O g H O l2 22 2 2+ D $( ) ( ) ( )H g O g H O l4 2 2 2+ (2)

3. Which balanced reaction equation can represent the reaction shown in the diagram?

+ +

A $CO H CO H O2 2 2+ + B $CH O CO H O4 2 2 2+ + C $CH O CO H O2 22 2 24 + + D $C H O CO H O2 24 2 2 2+ + (2)

4. Which one of the following is the correct balanced reaction equation for: $( ) ( ) ( )Fe s O g Fe O s2 3 4+

A $( ) ( ) ( )Fe s O g FeO s2 2+ B $( ) ( ) ( )Fe s O g Fe O s3 4 3 4+ C $( ) ( ) ( )Fe s O g Fe O s23 2 3 4+ D $( ) ( ) ( )Fe s O g Fe O s3 2 2 3 4+ (2)


Term 2 81


ASSESSM

ENTS

5. Consider the following statements about a balanced reaction equation. From a balanced equation we can verify … I the law of conservation of mass. II the fact that atoms are conserved. III the fact that molecules are conserved.The correct statement(s) is/are …A I and II only.B I and III only.C I only.D all of I, II and III. (2)

6. Which one of the following statements is NOT correct?

A reaction equation is balanced when:A the number of molecules of reactants equals the number of molecules of

products.

B the mass of the reactants is equal to the mass of the products.

C the number of atoms of each element in the reactants and products is equal. D the mass of the atoms of each element in the reactants and products is equal. (2)

LONG QUESTIONS 7. Solid potassium chlorate is heated and breaks up into potassium chloride powder

and oxygen gas. 7.1 Write down a balanced reaction equation for this reaction. (3)7.2 Verify that mass is conserved during this reaction. (6)

8. Consider the chemical reaction: $( ) ( ) ( ) ( )NO g H O l HNO aq NO g2 2 3+ +

8.1 Write down a balanced reaction equation for this reaction. (2)8.2 Represent the reaction by means of a diagram using coloured circles. (4)

9. Potassium metal reacts with water to form potassium hydroxide and hydrogen gas. 9.1 Write down a word equation for this reaction. (2)9.2 Write down a balanced reaction equation (in symbols) to represent this

reaction. (2)9.3 Verify that the number of atoms is conserved during this reaction. (6)

10. Calcium hydrogen carbonate reacts with nitric acid to produce calcium nitrate, carbon dioxide and water. 10.1 Write down a balanced reaction equation for this reaction. (5)10.2 40,5 g of calcium hydrogen carbonate reacts with 31,25 g of nitric acid. The

mass of carbon dioxide produced is 22,0 g, while 9,0 g of water is produced at the same time. Calculate the mass of calcium nitrate produced. (3)


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11. The following three word equations are given 1. $hydrogen gas chlorine gas hydrogen+ chloride gas 2. $silver nitrate magnesium sulfate silver sulfate magnesium+ + nitrate 3. $calciumcarbonate hydrochloric acid calciumchloride carbondioxide water+ + +

For each of these equations write down …11.1 a symbol equation (12)11.2 a balanced reaction equation. (7)

12. Use the balanced equation in Question 11.2.2 to verify that atoms are conserved during this reaction. (10)


Term 2 83


ASSESSM

ENTS

MARKING GUIDELINES

MULTIPLE CHOICE

1. B Mass is conserved during a chemical reaction, so the mass of the reactants is equal to the mass of the products. [CL 3] (2)

2. C Remind learners that both hydrogen and oxygen are diatomic gases and do not occur in nature in the atomic form. [CL 2] (2)

3. C The molecule on the left can only be CH4 (there is no such thing as C F4 ). This is followed by 2 molecules of a diatomic gas. [CL 3] (2)

4. D Remind learners that Fe3 does not mean 3 separate atoms of iron, but rather a single molecule of iron containing 3 atoms. [CL 2] (2)

5. A The number of molecules is not conserved during a chemical reaction but atoms are and mass is. [CL 2] (2)

6. A In a chemical reaction the numbers of molecules in the reactants and products does not have to be the same. Remind learners of synthesis and decomposition reactions where there are different numbers of molecules in reactants and products. [CL 2] (2)

LONG QUESTIONS 7. 7.1 $( ) ( ) ( )KC O s KC s O gl l2 2 33 2 + [CL 2] (3)

7.2 Reactants: : [ , ( )]KC O ml 2 39 35 5 3 163 #= + + = 245 Total reactants = 245 Products: : ( , )KC ml 2 39 35 5= + = 149 : ( )O m 3 2 162 #= = 96 Total products = 245 Mass reactants = mass products [CL 3] (6)


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8. 8.1 $( ) ( ) ( ) ( )NO g H O l HNO aq NO g3 22 2 3 + + [CL 2] (2)8.2

+ +→

One mark for each correct substance, and the correct number of molecules. [CL 4] (4)

9. 9.1 $potassium water potassium hydroxide hydrogen + + [CL 2] (2)9.2 $( ) ( ) ( ) ( )K s H O l KOH aq H g2 2 22 2 + + [CL 2] (2)9.3 Reactants: :K atoms2= :H 2 2#= = 4 atoms :O atoms2= Products: :K atoms2= :O atoms2= :H 2 2= + = 4 atoms [CL 3] (6)

10. 10.1 $( ) ( ) ( ) ( ) ( ) ( ) ( )Ca HCO s HNO aq Ca NO aq CO g H O l2 2 23 2 3 3 2 2 2 + + +

[CL 3] (5)10.2 Mass reactants = 40,5 + 31,25 = 71,75 g Mass products = x + 22,0 + 9,0 = x + 31 = 71,75 x = 71,75 – 31 = 40,75 g = mass of calcium nitrate [CL 3] (3)

11. 11.1 11.1.1 $( ) ( ) ( )H g C g HC gl l2 2 + 11.1.2 $( ) ( ) ( ) ( ) ( )AgNO aq MgSO aq Ag SO s Mg NO aq3 4 2 4 23 + + 11.1.3 $( ) ( ) ( ) ( ) ( )CaCO s HC aq CaC aq CO g H O ll l3 2 2 2 + + +

[CL 2] (12)11.2 11.2.1 $( ) ( ) ( )H g C g HC gl l22 2+ 11.2.2 $( ) ( ) ( ) ( ) ( )AgNO aq MgSO aq Ag SO s Mg NO aq2 3 4 2 4 3 2 + +

11..3 $( ) ( ) ( ) ( ) ( )CaCO s HC aq CaC aq CO g H O ll l23 2 2 2 + + + [CL 2] (7)


Term 2 85


ASSESSM

ENTS

12. Reactants: Ca = 1 atom C = 1 atom O = 3 atoms H = 2 atoms Cl = 2 atoms

Products: Ca = 1 atom C = 1 atom O = 2 + 1 = 3 atoms H = 2 atoms Cl = 2 atoms [CL 2] (10)


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TOPICS 14 and 15: Magnetism and ElectrostaticsQUESTIONS

MULTIPLE CHOICE1. Which of the following lists of materials contains ONLY non-ferrous materials?

A copper, zinc, brass B copper, iron, brass C nickel, iron, cobalt D aluminium, iron, nickel (2)

2. Diagram 1 shows two thin, uncharged strips of plastic. Diagram 2 shows the same strips after they have been rubbed with a dry cloth.

diagram 1 diagram 2

stripsof plastic

Which row correctly describes the charge on the strips after rubbing, and

the force between the strips after rubbing?

Charge on strips Force between strips

A The same Attraction

B The same Repulsion

C Opposite Attraction

D Opposite Repulsion (2)

3. A rubber balloon possesses a positive charge. If brought near and allowed to touch a wooden door, it sticks to the door. This does not occur with an uncharged balloon. These observations can lead one to conclude that the wooden door is …… A electrically neutral. B negatively charged. C lacking electrons. D a conductor. (2)


Term 2 87

TOPICS 14 AND 15: MAGNETISM AND ELECTROSTATICS

ASSESSM

ENTS

4. Which of the diagrams below best represents the charge distribution on an insulated metal sphere when a positively charged plastic tube is placed nearby?

+__

__ ___

______+ +

+ ++++++

+++

A B C D

(2)

5. Three bar magnets, of differing magnetic strength, are placed close to each other. Magnet A and Magnet C are fixed in their positions. Magnet B which is the strongest of the magnets, is free to move on a smooth frictionless surface. Magnet B takes up the arrangement shown in the diagram below and remains at rest.

N NS SXMagnet AMagnet A Magnet B Magnet C

(2)

Which row correctly describes the relative strengths of magnets A and C, and names the pole X?

Magnetic strength of magnets A and C

Type of pole at X

A A is weaker than C North

B A is stronger than C North

C A is stronger than C South

D A is weaker than C South

(2)


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LONG QUESTIONS 6. Two pointers are suspended in a magnetic field. One pointer is made of brass and

the other is a magnet. The pointers are initially held in the positions shown in the upper circles of the

diagram below, and then they are released.

N S

brass pointer N pole ofmagnet

S pole ofmagnet

arrows show direction of

strongmagnetic

field

draw final position of brasspointer in this circle

draw final position of magnet in this circle

6.1 Copy the diagram and draw the settled final positions of the two pointers

in the lower circles. (2)6.2 Explain the final position of the brass pointer. (2)6.3 Explain the final position of the magnet. (2)6.4 Suggest a material from which the magnet is made. (1)

7. The Earth has a magnetic field which is shaped like that of a bar magnet. 7.1 Draw a diagram of the magnetic field of this bar magnet.

N S

(3)7.2 Draw a circle to represent the Earth. On this circle draw and label 7.2.1 the Earth’s axis of rotation (1) 7.2.2 the geographic north pole (1) 7.2.3 the magnetic north (1) 7.2.4 the Earth’s magnetic field (3)


Term 2 89


ASSESSM

ENTS

8. A positively charged plastic rod is held above a metal plate. The metal plate rests on an insulator, and it is connected to the earth with an earthing wire. This apparatus is shown in the diagram below.

positively charged rod metal plate

insulator

earthing wire+ + + + + +

8.1 What is an insulator? (2)8.2 Explain how a plastic rod becomes positively charged when it is rubbed

with a soft cloth. (2)8.3 Is the metal plate charged when the apparatus is arranged as shown in the

diagram? Explain briefly. (2)8.4 The earthing wire is disconnected, and then the positively charged rod is

removed. What happens to the metal plate? Explain briefly. (3)8.5 The experiment is repeated from the beginning again. This time the

positively charged rod is removed first, and then the earthing wire is disconnected. What happens to the metal plate? Explain briefly. (3)

9. The diagram below shows a learner, standing on a polystyrene box, while touching the negatively charged dome of a Van de Graaff generator.

Van de Graaff generator

Earthed metal dome

Polystyrene box

Explain why the learner’s hair is sticking up. (4)


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10. Static electricity can be both useful and a nuisance. 10.1 The diagram shows a method of producing sandpaper using static electricity.

Roll ofpaper

Positively charged metal plate

Earthed metal plate

Sand

Conveyor belt

Glue droplets

Glue spray gun

+ + + + +

Glue is sprayed onto a moving strip of paper. As the glue leaves the spray

gun, the glue breaks up into tiny negatively charged droplets which coat the paper. The sticky paper passes between two metal plates. Sand moving on a conveyor belt also passes between the metal plates.

10.1.1 Explain the advantage of having all the droplets of glue negatively charged. (3)

10.1.2 Explain why the sand moves towards the sticky paper. (3)

10.2 People often experience an electric shock when getting out of a car on a dry warm day. This happens because charge is separated as they move around on the car seats.

On a dry warm day, the potential difference between the driver and the car is 10 000 V as he gets out of the car. When the driver touches the body of the car, 20 J of energy are transferred by the electric shock.

Calculate the amount of charge transferred between the car and the driver. (3)


Term 2 91


ASSESSM

ENTS

MARKING GUIDELINES

MULTIPLE CHOICE1. A All the other options include iron, which is ferromagnetic

(ferrous). [CL 2] (2)

2. B The strips are being repelled by each other, therefore they both have the same charge. [CL 2] (2)

3. B The positively charged balloon is attracted to a negatively charged object. It would also attract an electrically neutral object, but it may not stick to a neutral object. [CL 3] (2)

4. D The electrons in the metal sphere are free to move towards the positively charged rod. [CL 3] (2)

5. B Magnet B is repelled by both magnets A and C. We can conclude this because if magnet B was attracted to magnet C, it would attach itself to that magnet. Magnet B is repelled most strongly by magnet A. Therefore X is the north pole, and magnet A has a stronger magnetic field than magnet C. [CL 4] (2)

LONG QUESTIONS 6.1

N S

brass pointer N pole ofmagnet

S pole ofmagnet

arrows show direction of

strongmagnetic

field

draw final position of brasspointer in this circle

draw final position of magnet in this circle

N

S

(2)6.2 Brass is not a ferrous (ferromagnetic) material therefore the position of the

pointer is not affected by the magnetic field. [CL2] (2)6.2 The magnet aligns itself with the magnetic field (which points from N to S in

the direction of the arrows.) [CL2] (2)6.3 iron (nickel, cobalt). ANY of these DO NOT accept ferromagnetic material, as this answer is too vague. [CL1] (1)


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7.1 symmetrical pattern

direction of field lines from N to S

N S field lines run through the magnet [CL2] (3)

7.2

symmetrical pattern direction of field lines from Earth’s S to N

axis of rotation

Magnetic N N

[CL2] (6)

8.1 An insulator is material (or an object) which does not conduct electric current. [CL1] (2)

8.2 The soft cloth removes electrons from the surface of the rod. This leaves the rod with more positive than negative charge (therefore it is positively charged). [CL2] (2)

8.3 No (It is uncharged). The wire that connects it to the Earth maintains the metal with an equal number of positive and negative charges. [CL3] (2)

8.4 The metal plate is negatively charged. Before the wire was disconnected the positive rod attracts excess electrons from the Earth. When the wire is disconnected while the positive rod is held in place, these extra electrons remain in excess on the metal plate because they have no path to return to Earth. [CL4] (3)

8.5 The metal plate is not charged. When the positively charged rod is moved away, the excess electrons which gathered on the plate, return to Earth. There are no excess electrons on the metal plate. [CL4] (3)

9. The dome of the generator is negatively charged, therefore when the learner touches it he also becomes charged. The hairs on his head are negatively charged (have the same charge) therefore each hair repels the hairs around it. This causes all the hairs to stand up on end repelling each other. [CL3] (4)


Term 2 93


ASSESSM

ENTS

10.1 10.1.1 The negatively charged droplets repel each other, therefore they spray outwards and cover the strip of paper evenly. The negatively charged droplets polarise the molecules of paper causing charge to separate inside the molecules. They can then attract the positive sides of the molecules so they stick firmly to the paper. [CL4] (3)

10.1.2 The droplets of glue are still negatively charged. The sand particles are positively charged. The attraction between positively charged sand particles and negatively charged glue droplets causes the sand particles to move towards the paper. [CL4] (3)

10.2 W VQ= (method) Q20 10000 #= (substitutions) ,Q C0 002= (accuracy; SI units) [CL3] (3)


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6.0 V

4.0 Ω

8.0 Ω

5.0 Ω

5.0 Ω

12.0 V

10 Ω

20 Ω V

TOPIC 16: Electric CircuitsQUESTIONS

MULTIPLE CHOICE 1. The circuit diagram shows a ,4 0X resistor and an ,8 0X

resistor connected to a , V6 0 battery. What is the potential difference across the ,4 0X resistor?

A , V0 5B , V2 0C , V4 0D , V6 0 (2)

2. Two ,5 0X resistors are connected as shown in the diagram. What is the total resistance of this combination?

A less than ,5 0XB ,5 0XC more than ,5 0X and less than ,10 0XD ,10 0X (2)

3. The diagram shows a 10X and a 20X resistor connected in a potential divider circuit.

What is the reading on the voltmeter?A , V4 0 B , V6 0 C , V8 0 D , V12 0 (2)

4. In the circuit shown, only one of the fuses has blown, but none of the lamps is lit. Which fuse has blown?

A

B CD

powersupply


Term 2 95


1 Ω

3 Ω

2 Ω

1 Ω

3 Ω

2 Ω

ASSESSM

ENTS

5. The resistance of a wire depends on its length and on its cross-sectional area. Which of the following statements correctly describes the relationship between resistance of a wire and its length, and resistance of a wire and its cross-sectional area?

Length Cross-sectional area

A Resistance is directly proportional to the length.

Resistance is directly proportional to the cross-sectional area.

B Resistance is directly proportional to the length.

Resistance is inversely proportional to the cross-sectional area.

C Resistance is inversely proportional to the length.

Resistance is directly proportional to the cross-sectional area.

D Resistance is inversely proportional to the length.

Resistance is inversely proportional to the cross-sectional area.

(2)

6. Which of the following resistors has the largest potential difference across it? A 1 XB 2 XC 3 XD They all have the same potential

difference across them. (2)

7. Which of the following resistors has the largest current passing through it? A 1 XB 2 XC 3 XD They all have the same current passing

through them. (2)

LONG QUESTIONS 8. A cell has an emf of 4,5 V.

8.1 What does this statement mean? (2)8.2 Two 180 Ω resistors are connected in parallel across a battery with an emf

of 4,5 V. (2) 8.2.1 Draw the circuit diagram for this circuit. (3) 8.2.2 Calculate the resistance of the parallel combination of resistors. (3) 8.2.3 The current supplied by the battery is 0,05 A. Calculate the current

in each of the 180 Ω resistors. (3)


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9. A thermistor is a temperature-dependent device. Its resistance changes with temperature as shown in the graph below.

Graph of resistance against temperature for a thermistor.

resistancein kΩ

40

30

20

10

00 10 20 30 40 50 60 70

temprature in ºC

The thermistor is connected to a potential divider circuit as shown in the circuit diagram below. The voltmeter reads the potential difference across the thermistor.

V

9.0 V

12 kΩ

9.1 A series circuit acts as a potential difference divider. What does this statement mean? (2)

9.2 The voltmeter reads 6,0 V. 9.2.1 What is the potential difference across the 12 kΩ resistor? (2) 9.2.2 Determine the resistance of the thermistor. (3) 9.2.3 Refer to the graph, and determine the temperature of the thermistor. (2)9.3 The temperature of the thermistor is increased. Refer to the circuit

diagram and the graph of resistance against temperature above to answer these questions.

How does an increase in temperature of the thermistor affect … 9.3.1 its resistance? (1) 9.3.2 the reading on the voltmeter? (1)


Term 2 97


A

B

C

M

Fuse

ASSESSM

ENTS

10. A circuit consists of a battery of emf V12 and a heater with a resistance of ,6 0X .

A6.0 Ω

12 V

The reading on the ammeter is 2,0 A.10.1 Define an electric current. (2)10.2 Calculate the amount of charge that passes through the resistor in 2 minutes. (3)10.3 Calculate the amount of energy transferred by the charge in 2 minutes. (3)10.4 Name the charge carriers in the circuit. (1)10.5 Which reference to the circuit diagram, in which direction do these charge

carriers (named in 10.4) travel through the resistor? Answer: To the left OR To the right (1)

11. The diagram shows a simplified electrical circuit for a household heater.

It consists of a fan driven by the motor (M), three identical heating elements (resistors), and a lamp.

A fuse is included to protect the device from a surge in current.11.1 Name the components that work when only

switch A is closed. (3)11.2 The heating elements are controlled with the

two switches, B and C. The heater operates at three different settings: high, medium and low heat. The lower the resistance of the combination of the heating elements, the higher the temperature setting.

Copy the table below, and fill in a tick for each switch that is CLOSED at that setting.

Heater settings Switch B Switch C

High

Medium

Low (6)


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12. In the circuit diagram shown below, two resistors of 10 Ω and 40 Ω are connected in parallel. A resistor R of unknown resistance is connected in series with the parallel combination. The ammeter reads 0,6 A. The battery has an emf of 9,0 V, and it has no internal resistance. The voltmeter connected across resistor R reads 4,2 V.

AV

R

10 Ω

40 Ω

0,6 A

9,0 V

4,2 V

12.1 Calculate the effective resistance of the parallel resistors. (3)12.2 Explain what is meant by: "A parallel circuit is a current divider." (2)12.3 Using the fact that the parallel circuit is a current divider, calculate the

current through the 10 Ω resistor. (4)12.4 Explain what is meant by: "A series circuit is a potential divider." (2)12.5 Hence, calculate the resistance of the unknown resistor R. (4)12.6 How is the reading on the ammeter affected when the 10 Ω resistor is

removed from the circuit? Just answer: increases, decreases, or remains the same. (1)


Term 2 99


ASSESSM

ENTS

MARKING GUIDELINES

MULTIPLE CHOICE

1. D The potential difference across the branches of a parallel circuit is the same ( V6 ). [CL 2] (2)

2. A The resistance of the parallel combination is always less than the resistance of the smallest resistor in the combination OR

R R R1 1 1 1 1 2

5 5 5// 1 2= + = + =

,R 5 2 52// X= = [CL 2] (2)

3. C VV

2012 10 20= + applying the ratios for the potential divider

,V V8 0= [CL 4] (2)

4. A Fuse A provides current to all the other fuses because it is attached to the leads from the power supply. [CL 2] (2)

5. B Resistance increases as length of conductor increases.

Resistance decreases as thickness (cross-sectional area) increases. [CL 3] (2)

6. C The potential difference divides up in direct proportion to the resistance. [CL 2] (2)

7. D The current through a series circuit is the same through all components (resistors). [CL 2] (2)

LONG QUESTIONS8.1 The total amount of energy per unit charge supplied by the battery is 4,5 J.C−1 OR The potential difference across the battery when no current passes through it (on

open circuit) is 4,5 V. [CL1] (2)8.2 8.2.1

battery two resistors connected in parallel

[CL1] (3)


RESOURCE PACK


8.2.2 R R R1 1 1

1 2= + (method)

1 1180 180= + (substitutions)

2180=

R 180 902// X= = (accuracy; SI units) [CL3] (3) 8.2.3 The parallel circuit acts as a current divider, so the current splits up

according to the ratio of the resistors connected in parallel. In this case there are two resistors of equal resistance so the current splits

up equally. The current through each resistor is 1/2 of 0,05 A = 0,025 A. [CL 3] (3)

9.1 The potential difference across each resistor (component) in a series circuit divides according to the ratio of their resistances. [CL2] (2)

9.2 9.2.1 , , ,V V9 0 6 0 3 0= - = (method) (accuracy; SI units) [CL2] (2)9.2.2 The voltage across the thermistor is 6,0 V, as compared to the voltage

across the 12 kΩ resistor being 3,0 V. Therefore, the resistance of the thermistor is k2 12 24# X= . [CL4] (3)9.2.3 (Approximately) C8 o (accuracy; SI units; c.o.e.) [CL3] (2)

9.3 9.3.1 R decreases This is determined from the graph which shows that as temperature

increases, resistance decreases. [CL4] (1)9.3.2 V decreases If the resistance decreases, the pd across it will also decrease (because the

two resistors act as a potential divider in proportion to their resistances.) [CL4] (1)

10.1 The rate of flow of charge [CL1] (2)10.2 Q I tD= (method) ( )20 2 60# #= (substitutions;

conversion of minutes to second) C240= (accuracy; SI units) [CL2] (3)10.3 ALTERNATIVE 1 V W

Q= W VQ= (method) 12 240#= (substitutions; c.o.e) J2880= (accuracy; SI units) OR W VI tD= (method) ( )12 2 2 60# # #= (substitutions; conversion of minutes to second) J2880= (accuracy; SI units) [CL2] (3)10.4 electrons [CL2] (1)10.5 to the left [CL2] (1)


Term 2 101


ASSESSM

ENTS

11. 11.1 The fuse, the motor and the lamp [CL2] (3) 11.2

Heater settings Switch B Switch C

High

Medium

Low

2 marks for each correct row [CL4] (6)

12. 12.1 R R R1 1 1

1 2= + (method)

1 110 40= + (substitutions)

540=

R 40 85// X= = (accuracy; SI units) [CL3] (3)

12.2 The current splits up according to the ratio of the resistances in the parallel branches e.g. if the ratio is 10:40 then the current will split up in the ratio of 40:10. [CL 2] (2)

12.3 The main current is 0,6 A. The ratio of the resistances is 10:40 therefore the current splits into 50 parts. The current through the 10 Ω resistor = 50

40 × 0,06 = 0,48 A [CL 3] (4)12.4 The potential difference across each resistor (component) in a series circuit

divides according to the ratio of their resistances. [CL 1] (2)12.5 Ratio of pd = 4,8 V ; 4,2 V = 8 : 7 Ratio of resistances = 8 ; R = 8 : 7 R = 7 Ω [CL 3] (4)12.6 Decreases [CL 4] (1) Removing the 10 Ω resistor causes resistor R and the 40 Ω resistor to be

connected in series. This increases the effective resistance of the whole circuit, and therefore there is more opposition to the flow of charge i.e. the current decreases.


physical sciences grade 10 term 2 resource pack...gr10_ps_term2_resource_pack.indb 9 2019/01/03...

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