Chemistry Perfect Score 2011 Module Answer

(1)

BAHAGIAN PENGURUSAN BAHAGIAN PENGURUSAN

SEKOLAH BERASRAMA PENUH DAN SEKOLAH KLUSTER SEKOLAH BERASRAMA PENUH DAN SEKOLAH KLUSTER

JAWAPAN

MODUL

MODUL PERFECT SCORE

PERFECT SCORE

2011

CHEMISTRY

[KIMIA]



Set 1



Set 2



Set 3



Set 4



Set 5

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JAWAPAN SET 1 JAWAPAN SET 1

PAPER 2 : STRUCTURED QUESTION PAPER 2 : STRUCTURED QUESTION Section A

Section A No.

No. AnswerAnswer MarkMark 1

1 (a)(a) The formula that shows the simplest The formula that shows the simplest whole number ratio of atomswhole number ratio of atoms of each element in a compound.

of each element in a compound.

1 1 (b)

(b) HH22SOSO44 + + Zn Zn → → ZnSOZnSO44 + + HH22 22

(c)

(c) Heating, cooling and weighing are repeated until a constant mass isHeating, cooling and weighing are repeated until a constant mass is obtained. obtained. 1 1 (d) (d) Element

Element CopperCopper OxygenOxygen Mass, g Mass, g 47.7047.70 – – 25.3025.30 =22.40 =22.40 53.30 53.30 – – 47.7047.70 =5.60 =5.60 Mole atom Mole atom 22.4022.40 64 64 = 0.35 = 0.35 5.60 5.60 16 16 = 0.35 = 0.35 Simplest ratio Simplest ratio 11 11 Empirical formula = CuO

Empirical formula = CuO

4 4 (e)

(e) HH22 + + CuO CuO → → Cu Cu + + HH22OO 22

(f)

(f) To prevent the hot copper from being To prevent the hot copper from being oxidized again.oxidized again. 11 (g) (g) 2 2 TOTAL TOTAL 1313 No.

2 (a)(a) (i)(i) AlAl22COCO33 11

(ii)

(ii) AlAl22(CO(CO33))33 AlAl22OO33 + 3CO+ 3CO22 22

(iii)

(iii) The numbThe number of mer of mole of ole of AlAl22(CO(CO33))33 = 70.2/ 234= 70.2/ 234

= 0.3 mol = 0.3 mol Based on the balanced equation;

Based on the balanced equation; Al Al22(CO(CO33))33 : : AlAl22OO33 1 1 : 1: 1 0.3 0.3 : : 0.30.3 Mass of Mass of Ag = Ag = 0.4 x 0.4 x 102 102 = 30.6 = 30.6 gg 1 1 1 1 1 1 (iv)

(iv) Based on the balanced equationBased on the balanced equation Al Al22(CO(CO33))33 : : COCO22 1 1 : 3: 3 0.3 0.3 : : 0.60.6 Volume of CO Volume of CO22= 0.9 x 24= 0.9 x 24 = 21.6 dm = 21.6 dm33 = 21600 cm = 21600 cm33 1 1 1 1 1 1 (b)

(b) (i)(i) Zinc Zinc carbonate carbonate 11 Heat Heat Magnesium Magnesium ribbon ribbon

(3)

JAWAPAN SET 1 JAWAPAN SET 1

PAPER 2 : STRUCTURED QUESTION PAPER 2 : STRUCTURED QUESTION Section A

Section A No.

1 (a)(a) The formula that shows the simplest The formula that shows the simplest whole number ratio of atomswhole number ratio of atoms of each element in a compound.

of each element in a compound.

1 1 (b)

(b) HH22SOSO44 + + Zn Zn → → ZnSOZnSO44 + + HH22 22

(c)

(c) Heating, cooling and weighing are repeated until a constant mass isHeating, cooling and weighing are repeated until a constant mass is obtained. obtained. 1 1 (d) (d) Element

Element CopperCopper OxygenOxygen Mass, g Mass, g 47.7047.70 – – 25.3025.30 =22.40 =22.40 53.30 53.30 – – 47.7047.70 =5.60 =5.60 Mole atom Mole atom 22.4022.40 64 64 = 0.35 = 0.35 5.60 5.60 16 16 = 0.35 = 0.35 Simplest ratio Simplest ratio 11 11 Empirical formula = CuO

Empirical formula = CuO

4 4 (e)

(e) HH22 + + CuO CuO → → Cu Cu + + HH22OO 22

(f)

(f) To prevent the hot copper from being To prevent the hot copper from being oxidized again.oxidized again. 11 (g) (g) 2 2 TOTAL TOTAL 1313 No.

2 (a)(a) (i)(i) AlAl22COCO33 11

(ii)

(ii) AlAl22(CO(CO33))33 AlAl22OO33 + 3CO+ 3CO22 22

(iii)

(iii) The numbThe number of mer of mole of ole of AlAl22(CO(CO33))33 = 70.2/ 234= 70.2/ 234

= 0.3 mol = 0.3 mol Based on the balanced equation;

Based on the balanced equation; Al Al22(CO(CO33))33 : : AlAl22OO33 1 1 : 1: 1 0.3 0.3 : : 0.30.3 Mass of Mass of Ag = Ag = 0.4 x 0.4 x 102 102 = 30.6 = 30.6 gg 1 1 1 1 1 1 (iv)

(iv) Based on the balanced equationBased on the balanced equation Al Al22(CO(CO33))33 : : COCO22 1 1 : 3: 3 0.3 0.3 : : 0.60.6 Volume of CO Volume of CO22= 0.9 x 24= 0.9 x 24 = 21.6 dm = 21.6 dm33 = 21600 cm = 21600 cm33 1 1 1 1 1 1 (b)

(b) (i)(i) Zinc Zinc carbonate carbonate 11 (ii)

(ii) Zinc oxide and carbon dioxideZinc oxide and carbon dioxide 11 Heat Heat Magnesium Magnesium ribbon ribbon

(4)

(iii)

(iii) ZnCOZnCO33 →→ ZnO ZnO + + COCO22 11

TOTAL

TOTAL 1212 No.

3 (a)(a) (i)(i) TheThe number of number of protons protons found in thefound in the nucleusnucleus of of an an atom atom 11 (ii)

(ii) 77 11

(b)

(b) 11

(c)

(c) PP andand SS // // QQ andand RR 11 (d)

(d) (i)(i) QQ andand RR 11 (ii)

(ii) _Have_{Have same}_{same proton number}_{proton number but}_{but different}_{different nucleon number}_{nucleon number //}_// Have

Have samesame number of protons number of protons butbut differentdifferent number of neutrons number of neutrons

1 1 (e)

(e) (i)(i) Melting point :Melting point : 6363OOC [[valuesC values && unitunit must bemust becorrect]correct] 11 (ii)

(ii)

Section

Section Physical statePhysical state AB

AB SolidSolid DE

DE Liquid and gasLiquid and gas

1 1 (iii)

(iii) _{the heat energy}_{the heat energy absorbed}_{absorbed by the particles is used}_{by the particles is used} to

to overcomeovercome the forces of attraction between particlesthe forces of attraction between particles

1 1 1 1 TOTAL TOTAL 1010 No.

4 (a)(a) Sodium and magnesium // sodium and aluminium // Sodium and magnesium // sodium and aluminium // magnesium andmagnesium and aluminium aluminium 1 1 (b) Halogen (b) Halogen 11 (c) (c) 2.8.3 2.8.3 11 (d)

(d) (i)(i) Sodium, Sodium, magnesium, magnesium, aluminium, aluminium, chlorine, chlorine, argon argon 11

(ii)

(ii) From left to right From left to right ::

The proton number // the

The proton number // the positive charge increases from sodium to argonpositive charge increases from sodium to argon The forces of attraction by the nucleus on the electrons (

The forces of attraction by the nucleus on the electrons (nuclei attractionnuclei attraction)) in the first three

in the first three occupied shells become strongeroccupied shells become stronger

1 1 1 1 (e)

(e) (i) (i) Sodium Sodium burnt burnt rapidly rapidly and and brightly brightly with with a a yellow yellow flame flame // // White fumes liberated // white solid formed

White fumes liberated // white solid formed

1 1 (ii)

(ii) 2Na + Cl2Na + Cl22 → 2NaCl→ 2NaCl

[[ Formula Formula of of reactantsreactantsandand product product areare correct correct]] [[ Balanced Balanced equationequation]]

1 1 1 1 (iii) (iii)

has high melting / boiling point // conduct electricity has high melting / boiling point // conduct electricity in molten state or aqueous solution // soluble in water in molten state or aqueous solution // soluble in water

1 1 TOTAL TOTAL 1010

Q

33

16

Atomic size

(5)

No.

5 (a)(a) (i)(i) XX 11 (ii)

(ii) 8 valence electron // electron arrangement 2.8 // achieve octet electron8 valence electron // electron arrangement 2.8 // achieve octet electron arrangement arrangement 1 1 (b) (b) CovalentCovalent 11 (c) (c) VWVW44 11 (ii) (ii) 1+1 1+1 (iii)

(iii) has low melting / boiling point // cannot conduct electricityhas low melting / boiling point // cannot conduct electricity

in molten and solid state . // insoluble in water// soluble in organic in molten and solid state . // insoluble in water// soluble in organic solvent.

solvent.

1 1

(d)

(d) (i) (i) Ionic Ionic compoundcompound 11 (ii)

(ii) Atom U donate one electron to form UAtom U donate one electron to form U++ ionion Atom W accept one electron to form W Atom W accept one electron to form W--ionion U

U++ ion and Wion and W--ion attracted to each other by strong electrostatic force / ion attracted to each other by strong electrostatic force / ionic bond. ionic bond. 1 1 1 1 1 1 (iii) (iii)

[Number of electron each shells are correct] [Number of electron each shells are correct] [Number of

[Number of charge charge symbol symbol are correct]are correct]

1 1 1 1 TOTAL TOTAL 1313 V V WW W W W W W W

U

W

(i) (i)

(6)

PAPER 2: ESSAY QUESTION Section B

No. Answer Mark

6 (a) Group 17 Period 3

Has seven valence electrons.

Has three shells occupied with electron

1 1 1 1 (b) (i) Between Y and X

1.Atom Y has 1 valence electron and atom X has 7 valence electron 2. to achieve octet electron arrangement

3. Atom Y loses/donates/transfers 1 electron to form ion Y+ 4. Atom X gains/receives 1 electrons from atom Y to form ion X -5 Y+ ion and X- ion are attracted by a strong electrostatic force / ionic

bond 6. Diagram

(ii) Between W and X

1. Atom W has 4 valence electrons and atom X has 7 valence electrons. 2. Each atom W contributes 4 electrons whereas each atom X contributes

one electron for sharing.

3. to achieve octet electron arrangement

4. Four atoms of X share a pair of electrons with one atom W to form a WX4molecule / Diagram Molecules WX4 1 1 1 1 1 1 1 1 1 1

W

Y

X

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(c) Compound P : ionic bond Compound Q : Covalent bond Melting Point

Compound P

Ions are held by strong electrostatic forces. More energy is needed to overcome these forces. Compound Q

Molecules are held by weak intermolecular forces. Only a little energy is required to overcome the forces. Or

Electrical conductivity Compound P

In molten state or aqueous solution , there are free moving ions Ions carry charge

Compound Q

In molten and solid states , no free moving ions exist as molecule 1 1 1 1 1 1 1 1 1 1 TOTAL 20

No. Answer Mark

7 (a) (i) 2.8.7, Chlorine 1+1 (ii) 2Fe + 3Cl2 → 2FeCl3

Correct formulae of reactants and product Balanced

1 1 (b) (i) Z , Y , X

Z more reactive than X

Atomic size of Z bigger than atomic size X

Valence electron become further away from nucleus Valence electron to be more weakly pulled by the nucleus Valence electron can be released more easily in atom Z

1 1 1 1 1 1 (ii) same/similar

Same valence electron

1 1 (c) X : 2.4

Y : 2.6

to achieve octet electron arrangement

one X atom contributes four electron and each two Y atoms contributes two electrons for sharing

Group 16 Period 2

6 valence electron

2 shells occupied with electrons

1 1 1 1 1 1 1 1 TOTAL 20

(8)

PAPER 2: ESSAY QUESTION Section C

No. Answer Mark

8 (a) (i) Dilute acid: Hydrochloric acid / Sulphuric acid/ Nitric acid Metal N: Magnesium / zinc

1 1 (ii) Anhydrous calcium chloride

To dry the hydrogen gas

1 1 (iii) Example: Copper(II) oxide

Copper ion is reduced// reduction process

Because oxidation number of copper decrease from +2 t o 0 Hydrogen is oxidised// oxidation process

Because oxidation number of hydrogen increase from 0 to +1 Hydrogen is reducing agent

Copper(II) ion// Copper(II) oxide is oxidising agent

1 1 1 1 1 1 (b) (i) Relative Molecular mass of (CH2)n = 56

(12 + 2)n= 56 n = 4 Molecular formula = C4H8 1 1 (ii) Procedure:

1.

A small amount of glass wool soaked in butanol is placed in a boiling tube.

2.

The boiling tube is clamped horizontally

3.

The unglazed porcelain chips are placed in the middle section of the boiling tube.

4.

The boiling tube is closed with a stopper fitted with a delivery tube

5.

The unglazed porcelain chips are heated strongly. Then, the glass wool is warmed gently to vaporize the propanol.

6.

The gas released is collected in a test tube.

2 1 1 1 1 1 1 TOTAL 20 Unglased porcelain chips Heat Glass wool soaked in butanol

Water

(9)

No. Answer Mark 9 (a) (i)

(ii)

Formula that shows the simplest ratio of the number of atoms for each element in the compound.

Copper(II)oxide // lead(II)oxide CuO + H2 Cu + H2O // PbO + H2 Pb + H2O 1 1 1+1 (b) (i) (ii)

Magnesium oxide / zinc oxide Procedure:

1.

Clean magnesium / zinc ribbon with sand paper

2.

Weigh crucible and its lid

3.

Put magnesium ribbon into the crucible and weigh the crucible with its lid

4.

Heat strongly the crucible without its lid

5.

Cover the crucible when the magnesium starts to burn and lift/raise the lid a little at intervals

6.

Remove the lid when the magnesium burnt completely

7.

Heat strongly the crucible for a few minutes

8.

Cool and weigh the crucible with its lid and the content

9.

Repeat the processes of heating, cooling and weighing until a constant mass is obtained

10.

Record all the mass

Description Mass/g Crucible + lid x Crucible + lid + magnesium y Crucible + lid + magnesium oxide z Result:

Calculation:

Element Mg O Mass, g y-x z-y Mole y-x 24 =0.1 z-y 16 =0.1 Simplest ratio 1 1 Empirical formula: MgO

1 1 1 1 1 1 1 1 1 1 1 1 1 1 Max 10 (c) Element C H Mass (%) 84.6 15.4 Number of moles 84.6/12 =7.05 15.4/1 =15.4 Mole ratio 1 2 Empirical formula : CH2 1 RMM of (CH2)n = 70 [ 12 + 2]n = 70 14 n = 70 n = 5 Molecular formula : C5H10 1 1 1 1 1 20

(10)

JAWAPAN SET 2

PAPER 2 : STRUCTURED QUESTION Section A

No. Answer Mark 1 (a) Cell II 1

(b) (i) Magnesium electrode 1 (ii) e

1

(iii) Copper electrode thicker // Brown solid deposited 1 (c) 1. Correct formulae of reactant and product

2. Balanced equation

Cu2+ + 2e → Cu

1 1 (d) (i) Electrical energy to chemical energy 1 (ii) Blue colour remain unchange 1 (iii) 1. Concentration / Number of mole of Cu2+ion remain unchanged

2. Rate of Cu2+ ion discharge at cathode is the same as rate of Cu atom ionize at anode

1 1 TOTAL 10 No. Answer Mark 2 (a) (i) Iodine

r: formula/iodide/iodine gas 1 (ii) _MnO 4 -+ 8 H+ + 5e → Mn2+ + 4 H2O 1 (iii) +7 → +2 reduction 1 1 (iv) Potassium chloride // iron(II) sulphate // [any reducing agent] 1 (b) (i) Zinc 1

(ii) 1. Correct formulae of reactant and product 2. Balanced equation 2 Zn + O2→ 2 ZnO a: 2 J + O2→ 2 JO 1 1 (iii) K,J, L 1 (iv) Predict : no changes

r: no reaction

Reason : L is more reactive than J/zinc r: more electropositive 1 1 TOTAL 11 V Magnesium

(11)

PAPER 2 : ESSAY QUESTION Section B

No. Answer Mark 3 (a) 1. Propanone is a covalent compound

2. Propanone exist as molecule // No freely moving ion in propanone 3. Sodium chloride is an ionic compound

4. Sodium chloride solution has freely moving ion

1 1 1 1 (b) (i) Properties Cell X Cell Y

1. Type of cell Voltaic cell Electrolytic cell 2. Energy

change

Chemical→ electrical Electrical→ chemical 3. Electrodes Anode: A

Cathode: B

Positive terminal: C // Copper Negative terminal: D // Zinc 4. Ions in electrolyte Cu2+, SO42-, H+and OH -ions Cu2+, SO42-, H+ and OH- ions 5. Half equation Anode: Cu → Cu2+ + 2e Positive terminal: Cu2++ 2e→Cu Cathode: Cu2++ 2e→Cu Negative terminal Zn→ Zn2+ + 2e 6. Observation Anode:

Copper ecomes thinner

Positive terminal:

Copper plate becomes thicker Cathode:

Copper becomes thicker

Negative terminal: Zinc becomes thinner

1 1 1 1 1 1 1 1….6 (c) 1. Ag, M, L

2. L is more electropositive than silver

3. L displace silver from silver nitrate solution 4. M is more electropositive than silver

5. M displace silver from silver nitrate solution 6. M is less electropositive than L

7. M cannot displace L from L nitrate solution

1 1 1 1 1 1 1 (i i) Copper // Cu 1 TOTAL 20 No. Answer Mark 4 (a) (i) 1. Correct formulae of reactant and product

2. Balanced equation

Zn + 2e → Zn2+

3. Correct formulae of reactant and product 4. Balanced equation Pb2+ + 2e → Pb 1 1 1 1 (ii) 1. Zinc is oxidized

2. Zinc atom donates / losses electrons 3. Lead(II) nitrate / Pb2+is reduced

4. Lead(II) nitrate / Pb2+receives electrons

1 1 1 1 (b) (i) 1. Green colour of iron(II) sulphate change to brown

2. Correct formulae of reactant and product 3. Balanced equation

Cl2 + 2Fe 2+

→ 2Cl- + 2Fe3+ 4. Colourless solution of potassium iodide change to brown 5. Correct formulae of reactant and product

6. Balanced equation Cl2 + 2I- → 2Cl- + I2 1 1 1 1 1 1

(12)

(ii) Test tube P : Cl- ion and Fe3+ion Test tube Q : Cl- ion and I2

1 + 1 1 + 1 (iii) 1. Add starch solution

2. Dark blue precipitate formed

1 1 TOTAL 20 PAPER 2 : ESSAY QUESTION

Section C

No. Answer Mark 5 (a) (i) 1. Cu2+ // copper(II) ion

Equation

2. Correct formula of reactant and product 3. Balance Cu2+ + 2e → Cu 4. Copper 1 1 1 1 (ii) 1. Oxygen

2. Insert glowing splinter into the test tube 3. Glowing splinter relights

1 1 1 (iii) 1. NO3 -// nitrate ion 2. Oxygen 3. OH- ion is discharge

4. OH- ion is place lower than NO3

-ion in the electrochemical series Equation

5. Correct formula of reactant and product 6. Balance 4 OH- → 2 H2O + O2 + 4 e 1 1 1 1 1 1 (b) Diagram 1. Functional apparatus 2. Label

3. Pour [50 – 200 cm3] copper(II) sulphate solution into a beaker 4. Connect pure copper as cathode and impure copper as anode 5. Dip both pure and impure copper into copper(II) sulphate solution 6. Anode : Cu → Cu2+ + 2e 7. Cathode : Cu2+ + 2e → Cu 1 1 1 1 1 1 1 TOTAL 20 Impure copper

Pure

copper

Copper(II)

sulphate solution

(13)

No. Answer Mark 6 (a) (i) Metal P : Tin // Lead // Copper

Metal Q : Magnesium // Aluminium // Zinc

1 1 (ii) Exp I

1. Metal P is less electropositive than iron 2. Iron is oxidized

3. Iron losses electron // Fe → Fe2+ + 2e

4. Dark blue precipitate indicates the presence of Fe2+ion Exp II

5. Metal Q is more electropositive than iron 6. Metal Q is oxidized // Metal Q losses electron

7. Water and oxygen receive electron // 2H2O+O2+ 4 e → 4OH

-8. Pink colouration indicates the presence of OH- ion

1 1 1 1 1 1 1 1 (b) (i) 1. Bromine is reduced

2. Bromine molecule receives electron // Oxidation number of bromine

decrease / 0 → -1

3. Iron(II) sulphate / Fe2+is oxidized

4. Fe2+ losses electron // Oxidation number of iron increases/ +2→ +3

5. Correct formula of reactant and product 6. Balanced equation

Br2 + 2Fe 2+

→ 2Br - + 2Fe3+ 7. Brown colour of bromine decolourise

8. Green colour of iron(II) sulphate change to brown

1 1 1 1 1 1 1 1 (ii) 1. Add sodium hydroxide solution

2. Brown precipitate formed

1 1 TOTAL 20

(14)

JAWAPAN SET 3

No.

Answer

Mark

1 (a) (i) Solution in test tube C 1 (ii) Solution in test tube A 1 (b) 1. Higher than pH value of 0.1 moldm-3HCl // The pH is 3/4/5/6

2. Ethanoic acis is a weak acid// Etanoic acid ionizes partially in water to produce low concentration oh hydrogen ion

3. The lower the concentration, the lower the pH value

1 1 1 (c ) (i) Magnesium chloride 1

(ii) Mg + 2 H+ → Mg2+ + H2

1. Correct formula of reactant and product 2. Balanced equation

1 1 (iii) No of mole, HCl = 0.1 x 5 / 1000

= 0.0005 mol Based on balanced equation,

2 mol of HCl : 1 mol of H2

0.0005 mol of HCl : 0.00025 mol of H2// mol of H2= 0.005/2 = 0.0025

Volume of hydrogen gas = 0.00025 x 24 dm3 = 0.006 dm3// 6 cm3 1 1 1 (d) White precipitate 1 TOTAL 12

No.

Answer

Mark

2 (a) (i) Solvent P: Water

Solvent Q: methyl benzene / propanone / suitable organic solvent

1 1 (ii) Effervescence / gas released // magnesium ribbon dissolved 1 (iii) 1. In the presence of solvent P/water , ethanoic acid ionize to form H+ion.

2. H+ ion causes the ethanoic acid to show its acidic properties

3. In solvent Q, ethanoic acid exist as molecule// hydrogen ion does not present

1 1 1 (b) (i) 1. pH value increase / bigger

2. The lower the concentration of acid the higher the pH value

1 1 (ii) (0.5)(V) = (0.04)(250) // V = 20 cm3 1 1 3 (a)

Alkali that ionize/dissociate completely in water to produce high concentration of hydroxide ions.

1 1 (b) Alkaline / alkaline solution 1 1 (c) P: ion

Q: molecule

1

1 2 (d) No

Because there are no hydroxide ions in the solution// ammonia exist in the form of molecule.

1

1 2

 

   

(15)

(e) (i)

1. Colourless gas bubbles are released.// efeervesence

1

(ii) Mg + 2HCl  MgCl2 + H2 1. Correct formula 2. Balanced equation 2. Mol of Mg = 2.4/24 // 0.1 mol 3. Volume of H2= 0.124 dm3= 2.4 dm3 1 1 1 1

4

Total 11 NO ANSWER MARK

4 (a) (i) Green

(ii) Double decomposition reaction

1 1 (b) (i) carbon dioxide

(ii) CuCO3 → CuO + CO2

1. Reactants and products are correct 2. Equation is balanced (iii) - Labelled diagram - Functional 1 1+ 1 2 (c)

1 mol CuCO3 = 12.4/124 = 0.1 mol

Mol of CuCl2 = 0.1 x 135g

Mass = 13.5g 3

10

No.

Answer

Mark

5

(a)

_{Mg + 2HCl}

→

MgCl

2

+ H

2

1+1

(b) (i) 0.4/24 = 0.0167 mol 1 (ii

)

The number of mole of HCl = MV/1000 = 1x 50/1000 = 0.05 mol 1 (c) From the chemical equation 1 mol of magnesium produce 1 mol hydrogen

If 0.0167 mol produce 0.0167 mol hydrogen

Volume of hydrogen = 0.0167 x 24 dm3= 0.4 dm3 / 400 cm3 1 1 (d) I 400 /100 =4 cm3s-1 II 400 /60 = 6.67 cm3s-1 1 1 (e) As catalyst 1 (f) The temperature of hydrochloric acid

The concentration of hydrochloric acid

1 1 TOTAL 11 Copper(II) carbonate Lime water Heat

(16)

No.

Answer

Mark 6 (a) The heat released when 1 one mole of copper is displaced from copper (II) sulphate

solution by zinc.

1 (b) Cu2+ + Zn → Cu + Zn2+ 1 (c) The blue colour of the solution become colourless//Brown deposit is formed//

The polystyrene cup become hot//The reading of the thermometer increase

1 1 (d) (i) Heat release = 50 x 4.2 x 10

= 2100 J

(ii) The number of moles = 50 x 0.5 = 0.025 mol 1000

(iii) Heat of displacement = 2100 = -84000 J 0.025

H = 84.0 kJ/mol

1

1 (e) To ensure all the copper(II) sulphate solution reacted completely 1 (f)

1+1

TOTAL 10

No.

Answer

Mark

7

(a) Graph : Axes labeled with units All points plotted correctly & Shape of graph correct

1 1 1 (i) 50 cm3 ( marked on the graph) 1 (ii) NaOH + HCl NaCl + H2O

Mol of NaOH = 50 x 1 = 0.05

1000

From the equation : 1 mol NaOH : 1 mol HCl

0.05 mol NaOH : 0.05 mol HCl

Concentration HCl = 0.05 x 1000 = 1 moldm

-3

50

1

1+1

(c) To ensure uniform temperature of mixture in the polystyrene cup 1 (d) All the sodium hydroxide has reacted completely 1 (e) (i) 0.1 mole of NaOH when reacted releases 5.6 kJ

Therefore for 1 mole of NaOH reacted, 5.6/0.1 = 56 kJ heat energy released

1

Energy

H= - 84.0 kJ/mol

Zn + Cu

2+

(17)

PAPER 2 : ESSAY QUESTION Section B

No.

Answer

Mark

8

(a) (i) Label axes with units

All points are transferred correctly

Shape of the graph is smooth and correct

1 1 1 (ii) 2.5 cm3 1 (iii) moles of Pb2+ions = 2.5 x 1.0 / / 0.0025

1000 moles of I-ions = 5 x 1.0 // 0.005 1000

Pb

2+

: I

-0.0025 : 0.0005

1 :

2

1 1 1 1

(b)

Test tube 1:

1. Ion exist : K+, I-and NO3

-2. All lead(II) nitrate reacts completely 3. Excess of potassium iodide

4. Solution contains soluble salt of potassium iodide and pota ssium nitrate

Test tube 5:

5. Ion exist : K+ and NO3

-6. All lead(II) nitrate reacts completely and all potassium iodide reacts completely

7. Solution contains soluble salt of potassium nitrate

Test tube 7:

8. Ion exist : K+, Pb2+and NO3

-9. All potassium iodide reacts completely 10. Excess of lead(II) nitrate

11. Solution contains soluble salt of lead(II) nitrate and potassium nitrate

1

1 Max 10

(ii) 1+1

(i) Less than 5.6 kJ 1 (ii)

-

Hydrochlolric acid is strong acid dissociates completely in water ; ethanoic

acid is a weak acid dissociates in partially water

-

Part of the heat released during neutralisation is absorbed to ionise further ethanoic acid molecules, therefore heat released will be less than 5.6 kJ

1 1 TOTAL 14

Energy

H= - 56.0 kJ/mol

NaOH + HCl

H

2

O + NaCl

(18)

No.

Answer

Mark

9 (a) (i) Size of the reactant/the total surface area of the reactant Concentration of the reactant

Temperature of the reactant Catalyst 1 1 1 1 (ii) Temperature : 450-550oC Catalyst : iron Pressure : 200 atm 1 1 1 (b) (i) _ _{The axes are labeled together with its unit}

 The scale is correct

 The points are transferred correctly  The curve is smooth

1 1 1 1 (ii) Average rate of reaction for experiment I = 26.0

210

= 0.12 cm3s-1 Average rate of reaction for experiment II = 26.0

150 = 0.17 cm3s-1 [correct unit ] 1 1 1 1 (iii) 1. The rate of reaction for Experiment II is higher than in Experiment I

2. The concentration of HCl in Experiment II is more/higher than in Experiment I

3. The number of hydrogen ion/ H+ per unit volume of the solution in Experiment II is more than in Experiment I

4. The frequency of collisions between hydrogen ion and calcium carbonate in Experiment II is more than in Experiment I

5. The frequency of effective collisions hydrogen ion and calcium carbonate in Experiment II is more than in Experiment I

1 1 1 1 1 TOTAL 20

PAPER 2 : ESSAY QUESTION Section C

No. Answer Mark 10 (a) (i) Experiment I – hydrochloric acid or

Experiment II – sulphuric acid

Mg + 2HCl → MgCl2 + H2

1 1+1 (ii) The number of mole of HCl = MV/1000

= 1.0 x 50 = 0.05 mol 1000 or The number of mole of H2SO4 = MV/1000

= 1.0 x 50 = 0.05 mol 1000

1

(19)

(b) (i)

1. Curve with label

2. Axis with title and correct unit

1 1

(ii) 1. Sulphuric acid in experiment II is diprotic acid, hydrochloric acid in experiment I is monoprotic acid//Concentration of hydrogen ion, H+ in experiment II is higher than experiment I

2. The number of hydrogen ion per unit volume in experiment II is higher than experiment I

3. Frequency of collisions between hydrogen ions and magnesium atoms in experiment II is higher than experiment I

4. Frequency of effective collisions between hydrogen ions and magnesium atoms in experiment II is higher than experiment I 5. Rate of reaction in experiment II is higher than experiment I

…5

(c) Diagram : Functional apparatus set-up Label correctly

Procedure :

1. A burette is filled with water and inverted over a basin containing water. The burette is clamped vertically to the retort stand.

2. The water level in the burette is adjusted and the initial burette reading is recorded.

3. 50 cm3of 0.2 moldm-3 hydrocloric acid / sulphuric acid is measured and poured into a conical flask

4. 5 cm of magnesium ribbon are added into the conical flask 5. close conical flask immediately with the stopper fitted with

delivery tube.

6. At the same time the stopwatch is started shake the conical flask. 7. The burette readings are recorded at 30 second intervals for 5

minutes Time/s 0 30 60 90 120 150 180 Volume of gas / cm3 1 1…..2 1 1 1 1 1 1 1 7 max 5 ……1 TOTAL 20

No. Answer Mark 11 (a) Water on the wet shirt evaporated

Evaporation absorbs heat energy from body

1 1 (b) (i) C2H5OH + 3 O2  2 CO2 + 3 H2O - 1,376 kJ / mol

1. Heat of combustion for propanol is higher than ethanol

2. No. of carbon and hydrogen atoms per molecule propanol is higher

1+ 1 1 1 Experiment II Time/s H = Experiment I Volume of hydrogen/ cm3

(20)

than ethanol

3. No. of mole of CO2and H2O produced during combustion of

propanol is more than ethanol

4. Formation of CO2and H2O releases heat energy

(ii) Diagram – labelled and functional Material : Water , ethanol

Apparatus : spirit lamp. weighing balance, copper can, clay-pipe triangle, thermometer, wind shield

Procedure :

1. Measure (100 – 250) cm3of water and pour into the copper can and initial temperature is recorded after 5 minutes

2. Weigh the spirit lamp filled with ethanol

3. Light the spirit lamp to heat the water in the can and stir

4. Extinguish the spirit lamp when the temperature increase reaches

30˚C, record the maximum temperature of water reached

5. Weigh the spirit lamp with its remain. Result :

7. The initial mass of the spirit lamp + ethanol = a g The final mass of the spirit lamp + ethanol = b g 8. The mass of ethanol burnt = (a-b) g

9. The initial temperature of water = t1˚C

The maximum temperature of water = t2˚C

10. Increase in temperature of the water = (t2 – t1)  t˚C

Calculation :

RMM of ethanol C2H5OH = 46

11. The no. of mol of ethanol burnt =a₆₀b

= y mol 12. The released heat = mc

= 100 x 4.2 x t = x J

13. The heat of combustion of propanol =

-y

x

J mol-1 or - Z kJ mol-1 1 1..4 1 1….2 1 1…..2 1 1 1 1 1 1 1 1 1 1 13 max 8 TOTAL 20

No. Answer Mark 12 (a) Exothermic reaction is a reaction that releases heat to the surrounding

The total energy content of the products is lower than the total energy content of the reactants

Endothermic reaction is a reaction that absorbs heat from the surrounding

The total energy content of the products is higher than the total energy content of the reactants

1 1 1 1 (b) A reacts with B to form C and D

A and B are the reactants while C and D are the products

Heat energy is absorbed from surrounding //It is an endothermic reaction

Total energy content of C and D/ product is higher than total energy content of A and B/ reactants

When reaction occurs, the temperature of mixture of solutions increases / becomes hot (any 4 of the above)

1 1 1 1 1 (c) 1. 1 mole of silver nitrate solution produces 1 mole of Ag+ ion 1

(21)

3. One e mole of potassium chloride produces 1 mole of Cl- ion

4. The heat of precipitation of silver chloride is heat that released when 1 mole of AgCl is formed from Ag+ ion and Cl-ion // Ag+ + Cl-  AgCl

5. Number of mole of AgCl produced in bothe reactions are the same, heat released are the same.

1 1 1 Max 4 (d) Materials : calcium nitrate solution, sodium carbonate solution

Procedures :

- measure 50 cm3of 1.0 mol/ dm3Ca(NO3)2 solution and 50 cm 3

of 1.0 mol / dm3 Na2CO3solution separately and poured into a plastic cup

- measure and record the initial temperature of both solutions after 5 minutes - pour quickly and carefully Ca(NO3)2 solution into the plastic cup that contains

Na2CO3solution and stir continuously

- measure and record the lowest temperature reached Tabulation of data :

Calculation :

No. of moles of CaCO3 = No. of moles of Ca(NO3)2 = mv/1000 = 1.0(50)/1000 = 0.05

heat change  mc(Ө4 – Ө3)

= x kJ

heat of reaction = + x kJmol-1 0.05

= + y kJmol-1

Initial temperature of Ca(NO3)2 / oC Ө1

Initial temperature of Na2CO3 / oC Ө2

Average initial temperature / oC (Ө1 + Ө2)/2  Ө3

Lowest temperature of the mixture / oC Ө4

Change in temperature / oC Ө3- Ө4 1 1 1 1 1 1 1 1 TOTAL 20

No.

Answer

Mark

13 a

(i)

1. Zinc nitrate, zinc sulphate 2. Zinc carbonate

1

1 (ii)

I :Sodium carbonate solution/ potassium carbonate solution / ammonium

carbonate solution

II : Sulphuric acid

1

1 (iii)

1.

50 cm

3

of 1 mol dm

-3 magnesium

nitrate solution is

measured and

poured into a beaker

2.

50 cm

3

of 1 mol dm

-3

Sodium carbonate solution/ potassium carbonate

solution / ammonium carbonate solution solution is measured and

poured into the beaker.

3.

The mixture is stirred with a glass rod and a white solid,

magnesium carbonate

is formed.

4.

The mixture is filtered

5.

and the residue is rinsed with distilled water

6.

The white precipitate is dried by pressing it between filter papers.

1

1 1…6

(22)

c

(i)

1. nitrate ion / NO3-ion

2. Add dilute sulphuric acid followed by iron(II) sulphate solution into test tube containing salt X solution

3. Add a few drops of concentrated sulphuric acid through the wall of test tube 4. A brown ring is formed.

1

1 (ii)

1. Zn2+ , Pb2+, Al3+

2. Add ammonia solution into test tube containing salt X solution until excess 3. White precipitate dissolves in excess ammonia solution showing t he

presence of Zn2+ions

4. White precipitate insoluble in excess ammonia solution showing the presence of Pb2+and Al3+ions.

5. Add potassium iodide solution into test tube containing salt X solution Yellow precipitate formed showing the presence of Pb2+ions // 6. No change showing the presence of Al3+ions.

1 1 1 1 1 1

20

(23)

JAWAPAN SET 4

No Answer Mark

1 (a) Compound that contains only carbon and hydrogen Has double bonds between carbon – carbon atoms

1 1

(b) Alkene 1

(c) Propene 1

(d) (i) Hydrogenation / Addition reaction 1 (ii)

1 (e) (i) C3H6+ 9/2 O2 →3CO2 + 3H2O or

2C3H6+ 9O2 →6CO2 + 6H2O 2 (ii) No. of mole of C3H6 = 42 1 . 2 = 0.05 Volume of gas CO2 = 0.05 x 3 x 24 = 3.6 dm3 1 1 TOTAL 10 No Answer Mark 2 (a) Ethanol 1 (b) Hydroxyl group 1 (c) (i) (ii) (iii) Oxidation

Orange colour of potassium dichromate (VI) solution turns to green

1 1 1 (d) (i) (ii) (iii) (iv) Esterification Ethyl ethanoate Pleasant smell CH3COOH + C2H5OH → CH3COOC2H5 + H2O 1 1 1 2 TOTAL 10

H

O

H



C



C



O



H

(24)

No. Explanation Mark 3 (a) (i) Haber process 1

(ii) N2 + 3H2 2NH3

Correct formula

Balanced 1 1 (iii) 450oC --- 550oC Vanadium(V) oxide 1 1 (iv) As a fertiliser 1 (b)

(i) Polyvinyl chloride // polychloroethene 1

(ii) 1 (c) Correct arrangement Correct label 1 1 TOTAL 10

No. Explanation Mark 4 (a) (i)

(ii) (iii)

glycerol

saponification / alkaline hydrolysis to cause precipitation of soap

1 1 1 (b) (i) X: detergent Y :soap 1 1 (ii) magnesium stearate or calcium stearate 1 (iii) Mg2+and Ca2+ 1+1 (iv) causes water pollution / non-biodegradable 1

TOTAL 9

Tin atom

(25)

PAPER 2 : ESSAY QUESTION Section B No Answer Mark 5 (a) (i) 14.3 % 1 (ii) Element C H Mass/ % 85.7 14.3 1 No. of moles

12

7 .

85

_{= 7.14} 1 3 . 14 = 14.3

2

Ratio of moles/ Simplest ratio

₇

_.

₁₄

14 .

7

= 1

14 .

7

3 .

14

= 2 3 Empirical formula = CH2 RMM of (CH2)n = 56 ...1 [(12 + 1(2)]n = 56 14n = 56 n = 14 56 = 4 ………..1 Molecular formula : C4H8 ………..1 6 max 5 (iii) 1+1 1+1 Max 4

(iv) Compound M (Butene, C4H8) has a higher percentage of carbon atom in

their molecule than butane, C4H10 ……….1

% of C in C4H8 =

8 )

12 (

4 )

12 (

4

 x 100% =

56

48

x 100% = 85.7% …………1 % of C in C4H10= 10 ) 12 ( 4 ) 12 ( 4  x 100% =

58

48

x 100% = 82.7% ………..1

...3

(b) (i) Starch Protein 1 1 (ii) H H CH3 H I I I I C = C – C = C I I H H 2-methylbut-1,3-diene or isoprene 1 1..2 (c) (i) Rubber that has been treated with sulphur 1

But-2-ene But-1-ene

(26)

(ii)

In vulcanised rubber sulphur atoms form cross-links between the rubber molecules

These prevent rubber molecules from sliding too much when stretched

1 1 TOTAL 20 No. Explanation Mark 6 (a) Examples of food preservatives and their functions:

 Sodium nitrite – slow down the growth of microorganisms in meat  Vinegar – provide an acidic condition that inhibits the growth of

microorganisms in pickled foods

1+1 1+1 (b) (i) Paracetamol Codeine 1 1 (ii) To follow the instructions given by the doctor concerning the dosage and

method of taking the medicine

To visit the doctor immediately if there are symtoms of allergy or other side effects of thye medicine

1 1 (iii) If the correct dosage is not given by the doctor, it will cause abuse of the

medicine. For instance, if the child is given a overdose of codeine, it may lead to addition.

If the child is given paracetamol on a regular basis for a long time, it may cause skin rashes, blood disorders and acute inflammation of the pancreas.

1 1 (c) Type of food additives Examples Function

Preservatives Sugar, salt To slow down the growth of microorganisms

Flavourings Monosodium glutamate, spice, garlic

To improve and enhance the taste of food

Antioxidants Ascorbic acid To prevent oxidation of food

Dyes/ Colourings Tartrazine Turmeric

To add or restore the colour in food

Disadvantages of any two food additives:

Sugar – eating too much can cause obesity, tooth decay and diabetes Salt – may cause high blood pressure, heart attack and stroke. Tartrazine – can worsen the condition of asthma patients

- _{May cause children to be hyperactive}

MSG – can cause difficult in breathing, headaches and vomiting.

2 2 2 2 1 1 TOTAL 20 PAPER 2 : ESSAY QUESTION

Section C

Questions Marking criteria Mar ks 7 (a) (i) 1. Sulphur is burnt in air to produce sulphur dioxide //

2. Burning of metal sulphides/zinc sulphide / lead sulphide produce sulphur dioxide

3. Sulphur dioxide is oxidised to sulphur trioxide in excess oxygen

4. Sulphur trioxide is dissolved in concentrated sulphuric acidto form oleum. 5. The oleum is diluted with water to produce concentrated sulphuric acid

1 1 1 1 1

(27)

(ii) H2SO4+ 2NH3→ (NH4)2SO4

Formula for reactants and product correct Balanced

1 1 (b) 1. Bronze is harder than copper

2. Atoms of pure copper are same size and arrange in layers 3. when force applied the layers will slide.

4. In bronze tin atom has different size compare to pure copper

5. _{and interrupt the orderly arrangement of pure copper.}

1 1 1 1 1 max4 (c) Procedure:

1.

Iron nail and steel nail are cleaned using sandpaper.

2.

Iron nail is placed into test tube A and steel nail is placed into test tube B.

3.

Pour the agar-agar solution mixed with potassium hexacyanoferrate(III) solution into test tubes A and B until it covers the nails.

4.

Leave for 1 day.

5.

Both test tubes are observed to determine whether there is any blue spots formed or if there are any changes on the nails.

6.

The observations are recorded Results:

Test tube The intensity of blue spots A High

B Low Conclusion:

Iron rust faster than steel.

1 1 1+ 1 1 1 1 1 1 1 TOTAL 20 No Answer Mark

8 (a) (i) X - any acid – methanoic acid

Y - any alkali – ammonia aqueous solution

1 1 (ii) 1. Methanoic acid contains hydrogen ions

2. Hydrogen ions neutralise the negative charges of protein membrane 3. Rubber particles collide,

4. Protein membrane breaks

5. Rubber polymers combine together

1 1 1 1 1 5 max 4 (iii) Ammonia aqueous solution contains hydroxide ions

Hydroxide ions neutralise hydrogen ions (acid) produced by activities of bacteria

1 1 (b) (i) Alcohol 1

(ii) Burns in oxygen to form carbon dioxide and water

Oxidised by oxidising agent (acidified potassium dichromate (V I) solution) to form carboxylic acid

1 1 (iii) Procedure:

1. Place glass wool in a boiling tube

2. Pour 2 cm3of ethanol into the boiling tube

3. Place pieces of porous pot chips in the boiling tube

4. Heat the porous pot chips strongly 5. Heat ethanol gently

6. Using test tube collect the gas given off

Diagram:

(28)

[Functional diagram] ….1

[Labeled – porous pot, water, named alcohol, heat]….1

Test:

Put a few drops of bromine water ...1 Brown colour of bromine water decolourised ...1

...2

Total

20

Heat Heat Glass wool soaked with ethanol

Porous pot chips

(29)

JAWAPAN SET 5 PAPER 3 SET 1

EXPLANATION SCORE 1. (a)

(i)

[Able to record all reading accurately with unit] Sample answer

Experiment Metal X Metal Y I 1.70 cm 1.40 cm II 1.75 cm 1.45 cm III 1.75 cm 1.45 cm

3

[Able to record all reading correctly without unit] 2 [able to record three to five reading correctly 1 No response or wrong response 0 EXPLANATION SCORE 1(a)

(ii)

[Able to construct a table to record the diameter of the dents and average diameters for material X and Y that contain:

1. correct title

2. Reading and unit Sample answer:

Material Diameter of the dents(cm) Average diameter,(cm)

1 2 3

X 1.701.75 1.75 1.73

Y 1.40 1.45 1.45 1.43 3 [Able to construct a table to record the diameter of the dents and average

diameters for material X and Y that contain 1. title

2. Reading

2

[Able to construct a table with at least one title / reading 1 No response or wrong response 0

EXPLANATION SCORE 1.(b) [Able to state correct observation]

Sample answer:

The diameter of dents made on material Y is smaller than material X// The diameter of dents made on material X is bigger than material Y

3

[Able to state correct observation, incompletely] Sample answer:

The diameter of dents made on material Y is smaller// The diameter of dents made on material X is bigger

2

[Able to state an idea of the observation] Sample answer:

The diameter of dents for Y is small// The diameter of dents for X is big

1

(30)

EXPLANATION SCORE 1.(c) [Able to state the inference correctly]

Sample answer:

Material Y is harder than material X// Material X softer than material Y

3 [Able to state the inference correctly/

Sample answer:

Material Y is harder // Material X softer

2 [Able to state an idea of inference.

Sample answer:

Material Y is hard// Material X is soft

1 No response or wrong response 0

EXPLANATION SCORE 1.(d) [Able to state the correct operational definition for alloy]

1. what should be done and

2. what should be observe correctly Sample answer:

When the weight of 1 kilogram is dropped at height of 50 cm to hit the ball bearing which is taped onto the alloy block using cellophane tape a smaller dent is formed.

3

[Able to state the meaning of alloy, incompletely] Sample answer:

Material that form small dent is hard

2

[Able to state an idea of alloy] Sample answer:

Alloy form dent//alloy is hard

1

No response or wrong response 0 KK0508

EXPLANATION SCORE 1.(e) [able to give all three explanations correctly]

Sample answer:

1. atoms in material X are in orderly arrangement 2. atoms in material Y are not in orderly arrangement

3. layer of atoms in material Y difficult to slide on each other

3

[able to give any two explanations ] 2 [able to give any one explanations ] 1 No response given / wrong response 0

(31)

EXPLANATION SCORE 1.(f) [Able to state any alloy for material Y and its major pure metal for materials X correctly]

Sample answer:

Material X: copper // iron// any suitable metal

Material Y: bronze/ brass//stainless steel// any suitable al loy for pure metal given.

3

[Able to state any alloy for material Y and its major pure metal for materials X correctly] Sample answer:

Material X: tin/ zinc// chromium / nickel // any suitable metal

Material Y: bronze// brass//stainless steel// any suitable alloy for pure metal given.

2

[Able to state any alloy for material Y and its major pure metal for materials X correctly] Sample answer:

Material X: magnesium // aluminium//zinc // any metal Material Y: pewter // bronze // stainless steel //any alloy

1

No response given / wrong response 0 EXPLANATION SCORE 1.(g) [Able to state the relationship correctly between the manipulated variable and responding

variable with direction] Sample answer:

The harder/ softer the material, the smaller / bigger the diameter of the dent.

3

[Able to state the relationship correctly between the manipulated variable and responding variable with direction]

Sample answer:

Alloy/ pure metal will form smaller/ bigger dent than pure / alloy //

The smaller / bigger the diameter of the dent, the harder/softer the material

2

[able to state the idea of hypothesis] Sample answer:

Y is harder // X is softer // alloy is harder

1 No response given / wrong response 0 1.(h)

EXPLANATION SCORE [Able to state all the three variables and all the three actions correctly]

Sample answer:

Names of variables Action to be taken (i) manipulated :

Type of materials / material X and Y

(i) the way to manipulate variable:

Change pure metal/ alloy with alloy /pure metal

(ii) responding: Diameter of dent

(ii) what to observe in the responding variable: The diameter of the dent formed on material X and Y.

(iii) controlled:

Mass of the weight // height of the weight // size of steel ball bearing.

(iii) the way to maintain the controlled variable:

Uses same mass of weight // same height of weight // same size of steel ball bearing

3

[able to state any two variables and any two actions correctly] 2 [able to state any one variablesand any two action correctly] 1 No response given / wrong response 0

(32)

2. (a)

EXPLANATION SCORE

[Able to state 4 inferences correctly] Test tube Inference

A Iron (II) /Fe2+ions formed / produced // iron / Fe rusted / oxidized

B Iron (II) /Fe2+ions are not formed / produced // iron / Fe does not rusted / oxidized

C Iron (II) /Fe2+ions are not formed / produced // iron / Fe does not rusted / oxidized

D Iron (II) /Fe2+ions formed / produced // iron / Fe rusted / oxidized

3

[Able to state 3 inferences correctly] 2 [Able to state 1 inferences correctly] 1 No response given / wrong response 0

2.(b)

EXPLANATION SCORE [able to explain a difference in observation correctly between test tube 1 and 2]

Sample answer:

Iron / Fe in test tube A rust / oxidized because iron is in contact with less electropositive metal, but iron in test tube B does not rust / oxidized because iron is in contact with less electropositive metal.

3

[able to explain a difference in observation correctly between test tube A and B incompletely] Sample answer:

Iron / Fe in test tube A rust / oxidized but iron in test tube B does not rust / oxidized

2 [able to explain a difference in observation correctly between test tubeA1 and B]

Sample answer:

Iron / Fe / nail / metal rust / oxidized // iron/ Fe/ nail/ metal does not rust / oxidized

1 No response given / wrong response 0

2.(c)

EXPLANATION SCORE [Able to state the hypothesis correctly]

Sample answer:

When a more/ less electropositive metal is in contact with iron / Fe, the metal inhibits/ speed up rusting of iron.//

When a more / less electropositive metal is in contact with iron/ Fe, rusting of iron is faster / slower// The higher /lower the metal in contact with iron/ Fe in electrochemical series than iron /Fe ,the

rusting of iron/ Fe is slower / faster

3

[Able to state the hypothesis less correctly]

When a more/ less electropositive metal, the metal inhibits/ speed up rusting of iron.//

The rusting of iron/ Fe is slower / faster if a more / less electropositive metal in contact with iron/ Fe .

2

[Able to give an idea of hypothesis ] Sample answer:

Different metal in contact with iron, will cause iron to rust// metal can cause iron rust.

1

(33)

2.(d)

EXPLANATION SCORE [able to state all the variable in this experiment correctly]

Sample answer:

(i) manipulated variable: Type/different metal (ii) responding variable:

Rusting // presence of blue colour (iii) constant variable:

Size/mass of iron nail // type of nail // medium in which iron nail are kept// temperature

3

[able to state any two the variable in this experiment correctly] 2 [able to state any one the variable in this experiment correctly] 1 No response given / wrong response 0

2.(e)

EXPLANATION SCORE [able to state the operational definition for the rusting of iron nail correctly ]

1. What should be done and

2. what should be observe correctly Sample answer:

When iron nail is in contact with copper/tin/less electropositive metal and immersed in potassium hexacyanoferrate (III) solution, blue colouration is formed

3

[able to state the operational definition for the rusting of iron nail less correctly ] Sample answer:

Rusting of iron is the formation of blue colouration when iron nail is in contact with different metal.

2

[able to state the operational definition for the rusting of iron nail correctly ] Sample answer:

Rusting of iron is the formation of blue colouration.

1

No response given / wrong response 0 2.(f)

EXPLANATION SCORE

[able to classify all the three metals correctly] Metal that can provide sacrificial

protection to iron

Metal that cannot provide sacrificial protection to iron

Y Z

X

3

[able to classify any two metals correctly] 2 [able to classify any one metal correctly] 1 No response given / wrong response 0

(34)

2.(g)

EXPLANATION SCORE

[Able to compare the intensity of blue colour and relate the intensity of blue colour with the concentration of Fe2+accurately ]

Sample answer:

The intensity of blue colouration after two days is higher. The concentration of iron (II) ion is higher.

3

[Able to compare the intensity of blue colour and relate the intensity of blue colour with the concentration of Fe2+correctly]

Sample answer:

The intensity of blue colouration after two days is higher. The number of iron (II) ion is higher.

2

[able to state an idea of the intensity of blue colour and relate the intensity of blue colour with the concentration of Fe2+correctly]

Sample answer:

The intensity of blue colouration after two days is higher // The number of iron (II) ion is higher.

1

No response given / wrong response 0 3 (a) KK051021 – Statement of problem

EXPLANATION SCORE [Able to make a statement of the problem accurately and must be in question form]

Suggested answer:

Does a different type of alcohols have different heat of combustions? //

How does the number of carbon atom per molecule of alcohol affect the heat of combustion ?

3 [Able to make a statement of the problem but less accurate//Accurate statement of the problem but

not in question form. ] Suggested answer:

How does the number of carbon per molecule of alcohol affect the heat of combustion?//Does the increase in the number of carbon per molecule of alcohol increases the heat of combustion?

2 [Able to state an idea of statement of the problem]

Suggested answer:

Alcohols have different heat of combustion.

3(b) KK051202 – Stating variables

EXPLANATION SCORE [Able to state all the three variables correctly]

Suggested answer:

Manipulated variable: Different types of alcohols//Different alcohols such as ethanol, propanol and butanol.

Responding variable: Heat of combustion//Increase in temperature Fixed variable: Volume of water // type of container/ size of container

3

[Able to state any two of the variables correctly] 2 [Able to state any one of the variables correctly] 1 No response given / wrong response 0

3 (c) KK051202 – Stating hypothesis

EXPLANATION SCORE [Able to state the relationship between manipulated variable and responding variable correctly]

Suggested answer:

When the number of carbon per molecule of alcohol increases, the heat of combustion increases.

3 [Able to state the relationship between manipulated variable and responding variable but in reverse

direction]

Suggested answer:

The heat of combustion increases when the number of carbon per molecule of alcohol increases.// Different types of alcohols has different heat of combustion.

2

[Able to state an idea of the hypothesis] Suggested answer:

Alcohols have different heat of combustion.

(35)

3(d) KK051205 – List of substances and apparatus

EXPLANATION SCORE [Able to state the list of substances and apparatus correctly and completely]

Suggested answer:

Ethonol, propanol, butanol, water, [metal] beaker, spirit lamp, thermometer, weighing balance, wooden block, tripod stand, wind shield, measuring cylinder.

3

[Able to state the list of substances and apparatus correctly but not complete] Suggested answer:

Ethanol, propanol, butanol, water, [metal] beaker, spirit lamp, thermometer, weighing balance.

2 [Able to state an idea about the list of substances and apparatus]

Suggested answer:

Ethanol/propanol/butanol/water, beaker, thermometer.

3(e) KK051204 – Procedures

EXPLANATION SCORE [Able to state a complete experimental procedure]

Suggested answer:

1.

[200 cm3] of water is poured into a [copper] beaker.

2.

Initial temperature of the water is recorded.

3.

A spirit lamp is half filled with ethanol.

4.

Weight the spirit lamp with ethanol and record the mass

5.

The spirit lamp is put under the copper can and ignites the wick immediately.

6.

The water is stirred and the flame is put off after the temperature has increased by 30oC.

7.

The highest temperature of the water is recorded

8.

Immediately weight the spirit lamp and record the mass.

9.

The experiment is repeated t by replacing ethanol with propanol and butanol.

3

[Able to state the following procedures] 1, 2, 4, 5,7,8

2 [Able to state the following procedures]

2, 4, 5, 7

3(f) Tabulation of data

EXPLANATION SCORE [Able to exhibit the tabulation of data correctly with suitable headings and units ]

Types of alcohols Initial temperature/ oC Highest temperature/ oC

Initial mass of spirit lamp/g

Final mass of spirit lamp/g Ethanol

Propanol

Butanol 3

[Able to exhibit the tabulation of data less accurately with suitable headings without units ] Types of alcohols Initial temperature Highest temperaturer Initial mass of spirit lamp

Final mass of spirit lamp

2

[Able state an idea about the tabulation of data]

Alcohol Temperature Mass

1

(36)

PAPER 3 SET 2

Question Rubric Score

1(a)

Able to record the burette readings accurately with 2 decimal places. Experiment I II III Initial burette reading 1.00 cm3 13.50 cm3 26.00 cm3 Final burette reading 13.50 cm3 26.00 cm3 38.50 cm3 3

Able to record the burette reading correctly with 1 decimal place//any 5 readings correctly

2 Able to record any 4 burette readings correctly 1 Wrong response or no response 0 Question Rubric Score

1(b)

Able to construct a table with the following information: 1. Accurate titles and units:

2. Burette readings and volume of acid used/cm3 Sample answer: Experiment I II III Initial burette reading/cm3 1.00 13.50 26.00 Final burette reading/cm3 13.50 26.00 38.50 Volume of acid used/cm3 12.50 12.50 12.50 3

Able to construct a with correct titles and burette readings and volume of acid used (without units)

2 Able to construct a table with a least a title and a burette reading. 1 Wrong response or no response 0 Question Rubric Score

1(c)

Able to calculate correctly the molarity of acid with the following steps: Step 1: MaVa = 1 MbVb 1 Step 2: Ma= 1.0 x 25 12.5 Step 3: 2.0 mol dm-3 3

Able to show any 2 steps correctly. 2 Able to show any 1 step correctly. 1 Wrong response or no response 0 Question Rubric Score

1(d))

Able to state the operational definition for neutralization accurately. Sample answer:

When 12.5 cm3of hydrochloric acid 1.0 mol dm-3is added to 25 cm3 sodium hydroxide 1.0 mol dm-3with a few drops of phenolphthalein, colourless solution turns pink.

3

Able to state the operational definition less accurately. Sample answer:

When hydrochloric acid is added to sodium hydroxide solution, the solution turns pink

2

Able to state the idea for neutralisation. Sample answer

Acid react with alkali

1 Wrong or no response 0

(37)

1(e) Able to give the volume and explaination correctly with following aspects: 1. 6.25 cm3

2. Sulphuric acid is a diprotic acid 3. Concentration of H+ ions is double

3

Able to give any two of the above aspects 2 Able to give aby one of the above aspects 1 Wrong response or no response 0 Question Rubric Score 1(f) Able to state the three variables correctly.

Sample answer

Manipulated variable: Type of acids//Hydrochloric acid, ethanoic acid Responding variable: pH values

Fixed variable: Concentration of acids

3

Able to give any two variable correctly 2 Able to give one variable correctly 1 Wrong response or no response 0 Question Rubric Score

1(g)

Able to state the hypothesis accurately. Sample answer

When the concentration of hydrogen ion in acid is higher, , the pH value is lower// The higher the concentration of hydrogen ion, the lower the pH value

3

Able to state the hypothesis less accurately. Sample answer;

The strong acid has lower pH value // The pH value of weak acid is higher.

2 Able to give an idea of hypothesis

Sample answer

Different acid has different pH value

1 No response or wrong response 0 Question Rubric Score 1(h) Able to classify all the substances correctly.

Sample answer:

Substances with pH less than 7 Substances with pH more than 7 Ethanoic acid

Nitric acid

Ammonia solution Barium hydroxide

3

Able to classify any 3 substances correctly 2 Able to classify any two substances correctly 1 Wrong response or no response 0

(38)

Question Rubric Score 2(a) Able to state the inference accurately

Sample answer

When alcohol react with ethanoic acid, ester is formed//Esters have sweet pleasant smell property

3

Able to state the inference less accurately Sample answer

Reaction between alcohol and ethanoic acid produced sweet pleasant smell product

2

Able to give an idea of making an inference Sample answer

Ester formed

1 Wrong response or no response 0 Question Rubric Score 2(b) Able to construct a table correctly with the following information:

1. Columns with titles for alcohol, carboxylic acid, Ester 2. Name of all alcohols, carboxylic acid and ester

Alcohol Carboxylic acid Ester

Methanol Ethanoic acid Methyl ethanoate Ethanol Propanoic acid Ethyl propanoate Propanol Methanoic acid Propyl methanoate

3

Able to construct a table correctly with 2 esters named correctly 2 Able to construct a table correctly with 1 ester named correctly 1 Wrong response or no response 0 Question Rubric Score 2(c) Able to name the alcohol and carboxylic acid correctly.

Alcohol: Propanol

Carboxylic acid: Butanoic acid

3 Able to name alcohol or carboxylic acid correctly 2 Able to name any alcohol or any carboxylic 1 Wrong response or no response 0 Question Rubric Score 2(d)(i) Able to state the three variables correctly.

Sample answer

Manipulated variable: Hexane and hexene

Responding variable: Colour change of bromine water

// colour change of potassium manganate (VII) solution Fixed variable: Bromine water//acidified potassium manganate(VII)

solution

3

Able to state any two variable correctly 2 Able to state any one variable correctly 1 Wrong response or no response 0 Question Rubric Score 2(d)(ii) Able to state the hypothesis accurately

Sample answer:

Hexane declourised the brown colour of bromine water, hexane does not// Hexane declourised the purple colour of acidified potassium manganate(VII)

solution, hexane does not

3

Able to state the hypothesis but less accurately Sample answer

Brown bromine water decolourised with hexene but no change with hexane.

2 Able to state an idea of hypothesis 1

(39)

Question Rubric Score 2(d)(iii) Able to state the operational definition accurately

Sample answer

When bromine water //acidified potassium manganate(VII) solution is added to hexane/alkene brown bromine water //purple colour potassium manganate(VII) solution decolourised

3

Able to state the operational definition less accurately Sample answer

Alkene decolourised brown bromine water//Alkene decolourised purple colour potassium manganate(VII) solution

2

Able to state an idea of operational definition

Alkene decolourised bromine water//acidified potassium manganate(VII) solution

1 No response or wrong response 0 Question Rubric Score 2(d)(iv) Able to predict and make explainations accurately

Answer

1. Hexene

2. Percentage of carbon atoms per molecule hexene is higher than hexane

3

Able to give anyone of the above answer 2 Able to give an idea of prediction/explanation

Alkene//more carbon atoms

1

No response or wrong response 0 Question Rubric Score 3(a) Able to state the aim accurately

Sample answer

To compare the effectiveness of cleaning agents A and B on cleansing action in hard water .

2 Able to state the problem statement less accurately

Sample answer

Cleansing action of cleaning agent B is more effective

1

No response or wrong response 0 Question Rubric Score

3(b) Able to state the three variables accurately. Answer

Manipulated variable: Celaning agent A and B

Responding variable: Effectiveness of cleansing action Fixed variable: Type of water//hard water

3

Able to state any two variables accurately 2 Able to state any one variable accurately 1 No response or wrong response 0 Question Rubric Score 3(c) Able to state the hypothesis accurately with direction

Sample answer

Cleaning agent B is more effective than clening agent A in hard water

3

Able to state the hypothesis less accurately Sample answer

Cleansing action of c

leaning agent B

is better /more effective

(40)

Able to state an idea of hypothesis Sample answer

Cleansing action in hard water

1 No response of wrong response 0 Question Rubric Score 3(d) Able to state the complete list of apparatus and material as follows

Hard water, cleaning agent A and B,2 beakers, 2 pieces of cloths stained with oil, galss rod

3

Able to state a list of apparatus and materials as follows

Hard water, cleaning agent A and B, 2 beakers, 2 pieces of cloths stained with oil,

2

Able state one apparatus one material 1 No response or wrong response 0 Question Rubric Score 3(e) Able to state procedures correctly as follows

1. [50 - 200] cm3of hard water is poured into a beaker 2. Cleaning agent A is added into the beaker

3. A piece of cloth stained with oil is immersed in the solution 4. The cloth is shaken/rubbed/stirred

5. Observation is recorded

6. Repeat steps 1 – 5 by using cleaning B .

3

Able to state steps 2, 3, 5 and 6 2 Able to state steps 2 and 3 1 No response or wrong response 0

Question Rubric Score 3(f) Able to tabulate the data correctly

Sample answer

Type of cleaning agentr Observation Cleaning agent A

Cleaning agent A

3

Type of cleaning agentr Observation

2

Able to give an idea of tabulation of data 1 No response or wrong response 0

Marking Scheme Paper 3 Set 3

Question Rubric Score 1(a)(i) Able to record the thermometer reading accurately to 1 decimal place

with unit. Answer

Initial temperature = 30.0oC Highest temperature = 60.0oC

3

Able to record the thermometer reading correctly without unit Sample answer:

Initial temperature = 30

(41)

Highest temperature = 60

Able to record one thermometer reading correctly Sample answer

30 // 60

1 No response or wrong answer 0 (a)(ii) Able to state one observation accurately

Sample answer

Thermometer reading increases//Temperature increases

3 Able to state the observation less accurately

Sample answer

Temperature reading increases/higher//Temperature rises

2 Able to state the idea of observation

Sample answer

Thermometer reading change//Temperature change

1 Wrong or no response 0 (a)(iii) Able to state the inference accurately.

Sample answer

Heat energy is released //The reaction is exothermic reaction

3 Able to state the inference correctly

Sample answer

Heat energy is absorbed by water

2 Able to state an idea of inference

Temperature of water increases

1 Wrong or no response 0 (b)(i) Able to state all the mass of alcohols accurately to 2 decimal places

with unit Answers

1.55g, 2.23g, 3.56g, 4.01g

3

Able to state at least 3 mass of alcohols accurately or 4 mass of alcohol correctly to 4 decimal places

Answer

1.5536, 2.2309, 3.5601, 4.0101

2

Able to state at least 2 mass of alcohol correctly 1 Wrong or no response 0 (b)(ii) Able to tabulate the initial mass, final mass and mass of alcohols

accurately with units Sample answer:

Alcohol Initial mass/g Final mass/g Mass of alcohol/g Methanol 354.9548 353.4012 1.55 Ethanol 342.0201 339.7892 2.23 Propanol 364.4303 360.8702 3.56 Butanol 332.9891 328.9790 4.01 3

Able to tabulate the initial mass, final mass and mass of alcohols correctly without unit

2 Able to tabulate at least 2 readings of the initial mass, final mass and

mass of alcohols with correctly

(42)

(c) Able to calculate the heat of combustion of methanol correctly with the following steps: 1. Heat change = 200 x 4.2 x 30 J = 25200 J 2. No of mole of methanol = 1.55 ÷ 16 = 0.1 3. Heat of combustion = - 252 kJ mol-1

3

Able to calculate with at least 2 steps correctly 2 Able to calculate at least 1 step correctly 1 Wrong or no response 0 (d)(i) Able to state the variables correctly

Sample answer

Manipulated variable

Type of alcohols// Methanol, Ethanol, Propanol, Butanol Responding variable

Heat of combustion Fixed variable Volume of water

3

Able to state any two variables correctly 2 Able to state any one variable correctly 1 Wrong or no response 0 (d)(ii) Able to state the hypothesis accurately with the manipulated variable

related to responding variable Sample answer

When the number of carbon atoms per molecule alcohol increases, the heat of combustion increases

3

Able to state the hypothesis correctly with RV to MV Sample answer

Heat of combustion increases when the number of carbon atoms per molecule increases

2

Able to state an idea of hypothesis. Sample answer

Heat of combustion is different for different alcohol

1

Wrong or no response

(e) Able to predict the heat of combustion for pentanol correctly Sample answer

2350 kJ mol-1 // (2300 – 2400)kJ mol-1

3 Able to give the heat of combustion with the following value

Sample answer

More than 1860 kJ mol-1

2 Able to give an idea to predict a value of heat of combustion:

more than 2350 kJ mol-1