Chemistry Perfect Score 2011 module answer

1. BAHAGIAN PENGURUSAN
SEKOLAH BERASRAMA PENUH DAN SEKOLAH KLUSTER
JAWAPAN
MODUL PERFECT SCORE
2011
CHEMISTRY
[KIMIA]
 Set 1
 Set 2
 Set 3
 Set 4
 Set 5
1

2. JAWAPAN SET 1
PAPER 2 : STRUCTURED QUESTION
Section A
No. Answer Mark
1 (a) The formula that shows the simplest whole number ratio of atoms 1
of each element in a compound.
(b) H2SO4 + Zn → ZnSO4 + H2 2
(c) Heating, cooling and weighing are repeated until a constant mass is 1
obtained.
(d)
Element Copper Oxygen
Mass, g 47.70 – 25.30 53.30 – 47.70
=22.40 =5.60
Mole atom 22.40 5.60
64 16
= 0.35 = 0.35
Simplest ratio 1 1
Empirical formula = CuO 4
(e) H2 + CuO → Cu + H2O 2
(f) To prevent the hot copper from being oxidized again. 1
(g)
Magnesium
ribbon
Heat
2
TOTAL 13
No. Answer Mark
2 (a) (i) Al2CO3 1
(ii) Al2(CO3)3 Al2O3 + 3CO2 2
(iii) The number of mole of Al2 (CO3) 3 = 70.2/ 234
= 0.3 mol 1
Based on the balanced equation;
Al2 (CO3)3 : Al2O3
1 : 1
0.3 : 0.3 1
Mass of Ag = 0.4 x 102 = 30.6 g 1
(iv) Based on the balanced equation
Al2 (CO3)3 : CO2
1 : 3
0.3 : 0.6 1
Volume of CO2 = 0.9 x 24
= 21.6 dm3 1
= 21600 cm3 1
(b) (i) Zinc carbonate 1
(ii) Zinc oxide and carbon dioxide 1
2

3. (iii) ZnCO3 → ZnO + CO2 1
TOTAL 12
No. Answer Mark
3 (a) (i) The number of protons found in the nucleus of an atom 1
(ii) 7 1
(b) 1
33
Q
16
(c) P and S // Q and R 1
(d) (i) Q and R 1
(ii) Have same proton number but different nucleon number // 1
Have same number of protons but different number of neutrons
(e) (i) Melting point : 63 OC [values & unit must be correct] 1
(ii)
Section Physical state
AB Solid
DE Liquid and gas
1
(iii) the heat energy absorbed by the particles is used 1
to overcome the forces of attraction between particles 1
TOTAL 10
No. Answer Mark
4 (a) Sodium and magnesium // sodium and aluminium // magnesium and 1
aluminium
(b) Halogen 1
(c) 2.8.3 1
(d) (i) Sodium, magnesium, aluminium, chlorine, argon 1
Atomic size decreases
(ii) From left to right :
The proton number // the positive charge increases from sodium to argon 1
The forces of attraction by the nucleus on the electrons (nuclei attraction) 1
in the first three occupied shells become stronger
(e) (i) Sodium burnt rapidly and brightly with a yellow flame // 1
White fumes liberated // white solid formed
(ii) 2Na + Cl2 → 2NaCl
[Formula of reactants and product are correct] 1
[Balanced equation] 1
has high melting / boiling point // conduct electricity 1
(iii) in molten state or aqueous solution // soluble in water
TOTAL 10
3

4. No. Answer Mark
5 (a) (i) X 1
(ii) 8 valence electron // electron arrangement 2.8 // achieve octet electron 1
arrangement
(b) Covalent 1
(c) (i) VW4 (
(b) 1
(a) i i
(ii) ) (
1+1
W
W V W
W
(iii) has low melting / boiling point // cannot conduct electricity 1
in molten and solid state . // insoluble in water// soluble in organic
solvent.
(d) (i) Ionic compound 1
(ii) Atom U donate one electron to form U+ ion 1
Atom W accept one electron to form W- ion 1
U+ ion and W- ion attracted to each other by strong electrostatic force / 1
ionic bond.
(iii)
1
1
U W
[Number of electron each shells are correct]
[Number of charge symbol are correct]
TOTAL 13
4

5. PAPER 2: ESSAY QUESTION
Section B
No. Answer Mark
6 (a) Group 17 1
Period 3 1
Has seven valence electrons. 1
Has three shells occupied with electron 1
(b) (i) Between Y and X
1.Atom Y has 1 valence electron and atom X has 7 valence electron 1
2. to achieve octet electron arrangement 1
3. Atom Y loses/donates/transfers 1 electron to form ion Y+ 1
4. Atom X gains/receives 1 electrons from atom Y to form ion X- 1
5 Y+ ion and X- ion are attracted by a strong electrostatic force / ionic
bond 1
6. Diagram
+
Y X
1
(ii) Between W and X
1. Atom W has 4 valence electrons and atom X has 7 valence electrons. 1
2. Each atom W contributes 4 electrons whereas each atom X contributes
one electron for sharing.
3. to achieve octet electron arrangement
1
4. Four atoms of X share a pair of electrons with one atom W to form a
1
WX4 molecule / Diagram
1
X
W
X
X W W
W
X
W
Molecules WX4
5

6. (c) Compound P : ionic bond 1
Compound Q : Covalent bond 1
Melting Point
Compound P
Ions are held by strong electrostatic forces.
More energy is needed to overcome these forces. 1
Compound Q 1
Molecules are held by weak intermolecular forces.
Only a little energy is required to overcome the forces.
Or
1
Electrical conductivity
1
Compound P
In molten state or aqueous solution , there are free moving ions
1
Ions carry charge
1
Compound Q
1
In molten and solid states , no free moving ions
1
exist as molecule
TOTAL 20
No. Answer Mark
7 (a) (i) 2.8.7, Chlorine 1+1
(ii) 2Fe + 3Cl2 → 2FeCl3
Correct formulae of reactants and product 1
Balanced 1
(b) (i) Z,Y,X 1
Z more reactive than X 1
Atomic size of Z bigger than atomic size X 1
Valence electron become further away from nucleus 1
Valence electron to be more weakly pulled by the nucleus 1
Valence electron can be released more easily in atom Z 1
(ii) same/similar 1
Same valence electron 1
(c) X : 2.4 1
Y : 2.6 1
to achieve octet electron arrangement
one X atom contributes four electron and each two Y atoms 1
contributes two electrons for sharing 1
Group 16 1
Period 2 1
6 valence electron 1
2 shells occupied with electrons 1
TOTAL 20
6

7. PAPER 2: ESSAY QUESTION
Section C
No. Answer Mark
8 (a) (i) Dilute acid: Hydrochloric acid / Sulphuric acid/ Nitric acid 1
Metal N: Magnesium / zinc 1
(ii) Anhydrous calcium chloride 1
To dry the hydrogen gas 1
(iii) Example: Copper(II) oxide 1
Copper ion is reduced// reduction process 1
Because oxidation number of copper decrease from +2 to 0
Hydrogen is oxidised// oxidation process 1
Because oxidation number of hydrogen increase from 0 to +1 1
Hydrogen is reducing agent 1
Copper(II) ion// Copper(II) oxide is oxidising agent 1
(b) (i) Relative Molecular mass of (CH2)n = 56 1
(12 + 2) n = 56
n=4
Molecular formula = C4H8 1
(ii)
Unglased
porcelain chips
Glass wool
soaked in
butanol
Heat
Water
2
Procedure:
1. A small amount of glass wool soaked in butanol is placed
in a boiling tube. 1
2. The boiling tube is clamped horizontally 1
3. The unglazed porcelain chips are placed in the middle
section of the boiling tube. 1
4. The boiling tube is closed with a stopper fitted with a
delivery tube 1
5. The unglazed porcelain chips are heated strongly. Then,
the glass wool is warmed gently to vaporize the propanol. 1
6. The gas released is collected in a test tube. 1
TOTAL 20
7

8. No. Answer Mark
9 (a) (i) Formula that shows the simplest ratio of the number of atoms for each 1
element in the compound.
(ii) Copper(II)oxide // lead(II)oxide 1
CuO + H2  Cu + H2O // PbO + H2  Pb + H2O 1+1
(b) (i) Magnesium oxide / zinc oxide 1
(ii) Procedure:
1. Clean magnesium / zinc ribbon with sand paper 1
2. Weigh crucible and its lid 1
3. Put magnesium ribbon into the crucible and weigh the crucible with
its lid 1
4. Heat strongly the crucible without its lid 1
5. Cover the crucible when the magnesium starts to burn and lift/raise 1
the lid a little at intervals
6. Remove the lid when the magnesium burnt completely 1
7. Heat strongly the crucible for a few minutes 1
8. Cool and weigh the crucible with its lid and the content 1
9. Repeat the processes of heating, cooling and weighing until a constant 1
mass is obtained
10. Record all the mass 1
Result:
Description Mass/g 1
Crucible + lid x
Crucible + lid + magnesium y
Crucible + lid + magnesium oxide z
Calculation:
Element Mg O
Mass, g y-x z-y
Mole y-x z-y
24 16 1
=0.1 =0.1
Simplest ratio 1 1
1
Empirical formula: MgO Max 10
(c)
Element C H
Mass (%) 84.6 15.4
Number of moles 84.6/12 15.4/1
1
=7.05 =15.4
1
Mole ratio 1 2
Empirical formula : CH2 1
RMM of (CH2)n = 70 1
[ 12 + 2]n = 70
14 n = 70
n = 5 1
Molecular formula : C5H10 1
20
8

9. JAWAPAN SET 2
PAPER 2 : STRUCTURED QUESTION
Section A
No. Answer Mark
1 (a) Cell II 1
(b) (i) Magnesium electrode 1
(ii) e
V 1
Magnesium Copper electrode
electrode
(iii) Copper electrode thicker // Brown solid deposited 1
(c) 1. Correct formulae of reactant and product 1
2. Balanced equation 1
Cu2+ + 2e → Cu
(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 1
2. Rate of Cu2+ ion discharge at cathode is the same as rate of Cu
atom ionize at anode 1
TOTAL 10
No. Answer Mark
2 (a) (i) Iodine 1
r: formula/iodide/iodine gas
(ii) MnO4 - + 8 H+ + 5 e → Mn2+ + 4 H2O 1
(iii) +7 → +2 1
reduction 1
(iv) Potassium chloride // iron(II) sulphate // [any reducing agent] 1
(b) (i) Zinc 1
(ii) 1. Correct formulae of reactant and product 1
2. Balanced equation 1
2 Zn + O2 → 2 ZnO
a: 2 J + O2 → 2 JO
(iii) K,J, L 1
(iv) Predict : no changes 1
r: no reaction
Reason : L is more reactive than J/zinc 1
r: more electropositive
TOTAL 11
9

10. PAPER 2 : ESSAY QUESTION
Section B
No. Answer Mark
3 (a) 1. Propanone is a covalent compound 1
2. Propanone exist as molecule // No freely moving ion in propanone 1
3. Sodium chloride is an ionic compound
4. Sodium chloride solution has freely moving ion 1
1
(b) (i) Properties Cell X Cell Y
1. Type of cell Voltaic cell Electrolytic cell
2. Energy Chemical → electrical Electrical → chemical
change 1
3. Electrodes Anode: A Positive terminal: C // Copper 1
Cathode: B Negative terminal: D // Zinc
4. Ions in Cu2+, SO42-, H+ and OH- Cu2+, SO42-, H+ and OH- ions 1
electrolyte ions
5. Half Anode: Positive terminal: 1
equation Cu → Cu2+ + 2e Cu2+ + 2e → Cu
Cathode: Negative terminal 1
Cu2+ + 2e → Cu Zn → Zn2+ + 2e
6. Observation Anode: Positive terminal: 1
Copper ecomes thinner Copper plate becomes thicker
1
Cathode: Negative terminal: 1….6
Copper becomes Zinc becomes thinner
thicker
(c) 1. Ag, M, L 1
2. L is more electropositive than silver 1
3. L displace silver from silver nitrate solution 1
4. M is more electropositive than silver 1
5. M displace silver from silver nitrate solution 1
6. M is less electropositive than L 1
7. M cannot displace L from L nitrate solution 1
(i Copper // Cu 1
i)
TOTAL 20
No. Answer Mark
4 (a) (i) 1. Correct formulae of reactant and product 1
2. Balanced equation 1
Zn + 2e → Zn2+
3. Correct formulae of reactant and product 1
4. Balanced equation 1
Pb2+ + 2e → Pb
(ii) 1. Zinc is oxidized 1
2. Zinc atom donates / losses electrons 1
3. Lead(II) nitrate / Pb2+ is reduced 1
4. Lead(II) nitrate / Pb2+ receives electrons 1
(b) (i) 1. Green colour of iron(II) sulphate change to brown 1
2. Correct formulae of reactant and product 1
3. Balanced equation 1
Cl2 + 2Fe2+ → 2Cl- + 2Fe3+
4. Colourless solution of potassium iodide change to brown 1
5. Correct formulae of reactant and product 1
6. Balanced equation 1
Cl2 + 2I- → 2Cl- + I2
10

11. (ii) Test tube P : Cl- ion and Fe3+ ion 1+1
Test tube Q : Cl- ion and I2 1+1
(iii) 1. Add starch solution 1
2. Dark blue precipitate formed 1
TOTAL 20
PAPER 2 : ESSAY QUESTION
Section C
No. Answer Mark
2+
5 (a) (i) 1. Cu // copper(II) ion 1
Equation
2. Correct formula of reactant and product 1
3. Balance 1
Cu2+ + 2e → Cu
4. Copper 1
(ii) 1. Oxygen 1
2. Insert glowing splinter into the test tube 1
3. Glowing splinter relights 1
(iii) 1. NO3- // nitrate ion 1
2. Oxygen 1
3. OH- ion is discharge 1
4. OH- ion is place lower than NO3- ion in the electrochemical series 1
Equation
5. Correct formula of reactant and product 1
6. Balance 1
4 OH- → 2 H2O + O2 + 4 e
(b) Diagram
1. Functional apparatus 1
2. Label 1
Impure Pure
copper
copper
Copper(II)
sulphate solution
3. Pour [50 – 200 cm3] copper(II) sulphate solution into a beaker 1
4. Connect pure copper as cathode and impure copper as anode 1
5. Dip both pure and impure copper into copper(II) sulphate solution 1
6. Anode : Cu → Cu2+ + 2e 1
7. Cathode : Cu2+ + 2e → Cu 1
TOTAL 20
11

12. No. Answer Mark
6 (a) (i) Metal P : Tin // Lead // Copper 1
Metal Q : Magnesium // Aluminium // Zinc 1
(ii) Exp I
1. Metal P is less electropositive than iron 1
2. Iron is oxidized 1
3. Iron losses electron // Fe → Fe2+ + 2e 1
4. Dark blue precipitate indicates the presence of Fe2+ ion 1
Exp II
5. Metal Q is more electropositive than iron 1
6. Metal Q is oxidized // Metal Q losses electron 1
7. Water and oxygen receive electron // 2H2O+O2 + 4 e → 4OH- 1
8. Pink colouration indicates the presence of OH- ion 1
(b) (i) 1. Bromine is reduced 1
2. Bromine molecule receives electron // Oxidation number of bromine 1
decrease / 0 → -1
3. Iron(II) sulphate / Fe2+ is oxidized 1
4. Fe2+ losses electron // Oxidation number of iron increases / +2→ +3 1
5. Correct formula of reactant and product
6. Balanced equation 1
Br2 + 2Fe2+ → 2Br- + 2Fe3+ 1
7. Brown colour of bromine decolourise
8. Green colour of iron(II) sulphate change to brown 1
1
(ii) 1. Add sodium hydroxide solution 1
2. Brown precipitate formed 1
TOTAL 20
12

13. JAWAPAN SET 3
PAPER 2 : STRUCTURED QUESTION
Section A
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-3 HCl // The pH is 3/4/5/6 1
2. Ethanoic acis is a weak acid// Etanoic acid ionizes partially in water to produce 1
low concentration oh hydrogen ion
3. The lower the concentration, the lower the pH value 1
(c ) (i) Magnesium chloride 1
(ii) Mg + 2 H+ → Mg2+ + H2
1. Correct formula of reactant and product 1
2. Balanced equation 1
(iii) No of mole, HCl = 0.1 x 5 / 1000
= 0.0005 mol 1
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 1
= 0.006 dm3 // 6 cm3
1
(d) White precipitate 1
TOTAL 12
No. Answer Mark
2 (a) (i) Solvent P: Water 1
Solvent Q: methyl benzene / propanone / suitable organic solvent 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. 1
2. H+ ion causes the ethanoic acid to show its acidic properties 1
3. In solvent Q, ethanoic acid exist as molecule// hydrogen ion does not present 1
(b) (i) 1. pH value increase / bigger 1
2. The lower the concentration of acid the higher the pH value 1
(ii)
(0.5)(V) = (0.04)(250) // 1
V = 20 cm3 1
3 Alkali that ionize/dissociate completely in water to produce high concentration of hydroxide 1 1
(a) ions.
(b) Alkaline / alkaline solution 1 1
(c) P: ion 1
Q: molecule 1 2
(d) No 1
Because there are no hydroxide ions in the solution// ammonia exist in the form of molecule. 1 2
13

14. (e) (i) 1
1. Colourless gas bubbles are released.// efeervesence 1
(ii) Mg + 2HCl  MgCl2 + H2
1. Correct formula 1
2. Balanced equation 1
2. Mol of Mg = 2.4/24 // 0.1 mol 1 4
3. Volume of H2 = 0.1  24 dm3 = 2.4 dm3 1
Total 11
NO ANSWER MARK
4 (a) (i) Green 1
(ii) Double decomposition reaction 1
(b) (i) carbon dioxide 1
(ii) CuCO3 → CuO + CO2
1. Reactants and products are correct 1+ 1
2. Equation is balanced
(iii)
Copper(II)
carbonate
Heat
Lime water
- Labelled diagram 2
- Functional
(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 → MgCl2 + H2 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 1
Volume of hydrogen = 0.0167 x 24 dm3= 0.4 dm3/ 400 cm3 1
(d) I 400 /100 =4 cm3s-1 1
II 400 /60 = 6.67 cm3s-1 1
(e) As catalyst 1
(f) The temperature of hydrochloric acid 1
The concentration of hydrochloric acid 1
TOTAL 11
14

15. No. Answer Mark
6 (a) The heat released when 1 one mole of copper is displaced from copper (II) sulphate 1
solution by zinc.
(b) Cu2+ + Zn → Cu + Zn2+ 1
(c) The blue colour of the solution become colourless//Brown deposit is formed// 1
The polystyrene cup become hot//The reading of the thermometer increase 1
(d) (i) Heat release = 50 x 4.2 x 10 1
= 2100 J
(ii) The number of moles = 50 x 0.5 = 0.025 mol
1000 1
(iii) Heat of displacement = 2100 = -84000 J
0.025
H = 84.0 kJ/mol 1
(e) To ensure all the copper(II) sulphate solution reacted completely 1
(f)
Energy
Zn + Cu2+
H= - 84.0 kJ/mol
Zn2+ + Cu 1+1
TOTAL 10
No. Answer Mark
7 (a) Graph : Axes labeled with units 1
All points plotted correctly 1
& Shape of graph correct 1
(i) 50 cm3 ( marked on the graph) 1
(ii) NaOH + HCl NaCl + H2O
Mol of NaOH = 50 x 1 = 0.05 1
1000
From the equation : 1 mol NaOH : 1 mol HCl
0.05 mol NaOH : 0.05 mol HCl 1+1
Concentration HCl = 0.05 x 1000 = 1 moldm-3
50
(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 1
Therefore for 1 mole of NaOH reacted, 5.6/0.1 = 56 kJ heat energy released
15

16. (ii)
Energy
NaOH + HCl
H= - 56.0 kJ/mol
1+1
H2O + NaCl
(i) Less than 5.6 kJ 1
(ii) - Hydrochlolric acid is strong acid dissociates completely in water ; ethanoic 1
acid is a weak acid dissociates in partially water
- Part of the heat released during neutralisation is absorbed to ionise further 1
ethanoic acid molecules, therefore heat released will be less than 5.6 kJ
TOTAL 14
PAPER 2 : ESSAY QUESTION
Section B
No. Answer Mark
8 (a) (i) Label axes with units 1
All points are transferred correctly 1
Shape of the graph is smooth and correct 1
(ii) 2.5 cm3 1
(iii) moles of Pb2+ ions = 2.5 x 1.0 / / 0.0025 1
1000
moles of I- ions = 5 x 1.0 // 0.005
1000 1
Pb2+ : I-
0.0025 : 0.0005 1
1 : 2 1
(b) Test tube 1:
1. Ion exist : K+, I- and NO3-
2. All lead(II) nitrate reacts completely 1
3. Excess of potassium iodide 1
4. Solution contains soluble salt of potassium iodide and potassium nitrate 1
1
Test tube 5:
5. Ion exist : K+ and NO3-
6. All lead(II) nitrate reacts completely and all potassium iodide reacts 1
completely 1
7. Solution contains soluble salt of potassium nitrate
1
Test tube 7:
8. Ion exist : K+, Pb2+ and NO3- 1
9. All potassium iodide reacts completely 1
10. Excess of lead(II) nitrate 1
11. Solution contains soluble salt of lead(II) nitrate and potassium nitrate 1
Max 10
16

17. No. Answer Mark
9 (a) (i) Size of the reactant/the total surface area of the reactant 1
Concentration of the reactant 1
Temperature of the reactant 1
Catalyst 1
(ii) Temperature : 450-550oC 1
Catalyst : iron 1
Pressure : 200 atm 1
(b) (i)  The axes are labeled together with its unit 1
 The scale is correct 1
 The points are transferred correctly 1
1
 The curve is smooth
(ii) Average rate of reaction for experiment I = 26.0 1
210
= 0.12 cm3 s-1 1
Average rate of reaction for experiment II = 26.0
150 1
= 0.17 cm3 s-1
[correct unit] 1
(iii) 1. The rate of reaction for Experiment II is higher than in Experiment I 1
2. The concentration of HCl in Experiment II is more/higher than in 1
Experiment I
3. The number of hydrogen ion/ H+ per unit volume of the solution in 1
1
Experiment II is more than in Experiment I
4. The frequency of collisions between hydrogen ion and calcium carbonate 1
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
TOTAL 20
PAPER 2 : ESSAY QUESTION Section C
No. Answer Mark
10 (a) (i) Experiment I – hydrochloric acid or 1
Experiment II – sulphuric acid
Mg + 2HCl → MgCl2 + H2 1+1
(ii) The number of mole of HCl = MV/1000
= 1.0 x 50 = 0.05 mol
1000 or 1
The number of mole of H2SO4 = MV/1000
= 1.0 x 50 = 0.05 mol
1000
(iii) The rate of reaction is the change of volume of hydrogen gas per unit time 1
17

18. (b) (i) Volume of hydrogen/ cm3
Experiment II
Experiment I
1
1
Time/s
1. Curve with label
2. Axis with title and correct unit
(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 1
Label correctly 1…..2
Procedure :
1. A burette is filled with water and inverted over a basin containing
water. The burette is clamped vertically to the retort stand. 1
2. The water level in the burette is adjusted and the initial burette
reading is recorded.
1
3. 50 cm3 of 0.2 moldm-3 hydrocloric acid / sulphuric acid is
measured and poured into a conical flask
4. 4. 5 cm of magnesium ribbon are added into the conical flask 1
5. 5. close conical flask immediately with the stopper fitted with 1
delivery tube. 1
6. At the same time the stopwatch is started shake the conical flask. 1
7. The burette readings are recorded at 30 second intervals for 5 1
minutes 7 max 5
Time/s 0 30 60 90 120 150 180
……1
Volume of gas / cm3
TOTAL 20
No. Answer Mark
11 (a) Water on the wet shirt evaporated 1
Evaporation absorbs heat energy from body 1
(b) (i) C2H5OH + 3 O2  2 CO2 + 3 H2O H = - 1,376 kJ / mol 1+ 1
1. Heat of combustion for propanol is higher than ethanol 1
2. No. of carbon and hydrogen atoms per molecule propanol is higher 1
18

19. than ethanol
3. No. of mole of CO2 and H2O produced during combustion of 1
propanol is more than ethanol
1..4
4. Formation of CO2 and H2O releases heat energy
(ii) Diagram – labelled and functional 1
1….2
Material : Water , ethanol
Apparatus : spirit lamp. weighing balance, copper can, clay-pipe 1
triangle, thermometer, wind shield 1…..2
Procedure :
1. Measure (100 – 250) cm3 of water and pour into the copper can
and initial temperature is recorded after 5 minutes 1
2. Weigh the spirit lamp filled with ethanol
3. Light the spirit lamp to heat the water in the can and stir 1
4. Extinguish the spirit lamp when the temperature increase reaches 1
30˚C, record the maximum temperature of water reached
5. Weigh the spirit lamp with its remain. 1
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 1
9. The initial temperature of water = t1˚C
The maximum temperature of water = t2˚C 1
10. Increase in temperature of the water = (t2 – t1) t˚C
Calculation : 1
RMM of ethanol C2H5OH = 46
11. The no. of mol of ethanol burnt = ab = y mol
60
12. The released heat = mc 
= 100 x 4.2 x t 1
= xJ
x 1
13. The heat of combustion of propanol = - J mol-1 or -Z
y 1
kJ mol-1 13 max 8
TOTAL 20
No. Answer Mark
12 (a) Exothermic reaction is a reaction that releases heat to the surrounding 1
The total energy content of the products is lower than the total energy content of the 1
reactants
Endothermic reaction is a reaction that absorbs heat from the surrounding 1
The total energy content of the products is higher than the total energy content of the 1
reactants
(b) A reacts with B to form C and D 1
A and B are the reactants while C and D are the products 1
Heat energy is absorbed from surrounding //It is an endothermic reaction 1
Total energy content of C and D/ product is higher than total energy content of A and B/ 1
reactants
When reaction occurs, the temperature of mixture of solutions increases / becomes hot 1
(any 4 of the above)
(c) 1. 1 mole of silver nitrate solution produces 1 mole of Ag+ ion 1
2. 1 mole of sodium chloride solution produces 1 mole of Cl- ion 1
19

20. 3. One e mole of potassium chloride produces 1 mole of Cl- ion 1
4. The heat of precipitation of silver chloride is heat that released when 1 mole of AgCl 1
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
1
the same. Max 4
(d) Materials : calcium nitrate solution, sodium carbonate solution 1
Procedures :
- measure 50 cm3 of 1.0 mol/ dm3 Ca(NO3)2 solution and 50 cm3 of 1.0 mol / dm3 1
Na2CO3 solution separately and poured into a plastic cup
- measure and record the initial temperature of both solutions after 5 minutes 1
- pour quickly and carefully Ca(NO3)2 solution into the plastic cup that contains 1
Na2CO3 solution and stir continuously
- measure and record the lowest temperature reached 1
Tabulation of data :
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 1
Change in temperature / oC Ө3- Ө4
Calculation :
No. of moles of CaCO3 = No. of moles of Ca(NO3)2 1
= mv/1000 = 1.0(50)/1000 = 0.05
heat change mc(Ө4 – Ө3) 1
= x kJ
heat of reaction = + x kJmol-1
0.05
= + y kJmol-1
TOTAL 20
No. Answer Mark
13 a (i) 1. Zinc nitrate, zinc sulphate 1
2. Zinc carbonate 1
(ii) I :Sodium carbonate solution/ potassium carbonate solution / ammonium 1
carbonate solution 1
II : Sulphuric acid
(iii) 1. 50 cm3 of 1 mol dm-3 magnesium nitrate solution is measured and 1
poured into a beaker
2. 50 cm3 of 1 mol dm-3 Sodium carbonate solution/ potassium carbonate
solution / ammonium carbonate solution solution is measured and 1
poured into the beaker. 1
3. The mixture is stirred with a glass rod and a white solid, magnesium 1
carbonate is formed.
1
4. The mixture is filtered
5. and the residue is rinsed with distilled water 1
6. The white precipitate is dried by pressing it between filter papers. 1…6
20

21. c (i) 1. nitrate ion / NO3- ion 1
2. Add dilute sulphuric acid followed by iron(II) sulphate solution into test tube
1
containing salt X solution
3. Add a few drops of concentrated sulphuric acid through the wall of test tube 1
4. A brown ring is formed.
1
(ii) 1. Zn2+ , Pb2+ , Al3+ 1
2. Add ammonia solution into test tube containing salt X solution until excess 1
3. White precipitate dissolves in excess ammonia solution showing the
1
presence of Zn2+ ions
4. White precipitate insoluble in excess ammonia solution showing the 1
presence of Pb2+ and Al3+ ions. 1
5. Add potassium iodide solution into test tube containing salt X solution
1
Yellow precipitate formed showing the presence of Pb2+ ions //
6. No change showing the presence of Al3+ ions.
20
21

22. JAWAPAN SET 4
PAPER 2 : STRUCTURED QUESTION
Section A
No Answer Mark
1 (a) Compound that contains only carbon and hydrogen 1
Has double bonds between carbon – carbon atoms 1
(b) Alkene 1
(c) Propene 1
(d) (i) Hydrogenation / Addition reaction 1
(ii)
1
(e) (i) C3H6 + 9/2 O2 → 3CO2 + 3H2O or 2
2C3H6 + 9O2 → 6CO2 + 6H2O
(ii) 2 .1 1
No. of mole of C3H6 =
42
= 0.05
Volume of gas CO2 = 0.05 x 3 x 24
= 3.6 dm3 1
TOTAL 10
No Answer Mark
2 (a) Ethanol 1
(b) Hydroxyl group 1
(c) (i) Oxidation 1
(ii) Orange colour of potassium dichromate (VI) solution turns to green 1
(iii) 1
H O
H CC O H
H
(d) (i) Esterification 1
(ii) Ethyl ethanoate 1
(iii) Pleasant smell 1
(iv) CH3COOH + C2H5OH → CH3COOC2H5 + H2O 2
TOTAL 10
22

23. No. Explanation Mark
3 (a) (i) Haber process 1
(ii) N2 + 3H2 2NH3
Correct formula 1
Balanced 1
(iii) 450 oC --- 550oC 1
Vanadium(V) oxide 1
(iv) As a fertiliser 1
(b)
(i) Polyvinyl chloride // polychloroethene 1
(ii) 1
(c) Correct arrangement 1
Tin atom Correct label 1
Copper atom
TOTAL 10
No. Explanation Mark
4 (a) (i) glycerol 1
(ii) saponification / alkaline hydrolysis 1
(iii) to cause precipitation of soap 1
(b) (i) 1
X: detergent
1
Y :soap
(ii) magnesium stearate or calcium stearate 1
(iii) Mg2+ and Ca2+ 1+1
(iv) causes water pollution / non-biodegradable 1
TOTAL 9
23

24. 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 85.7 14.3
= 7.14 = 14.3
12 1
2 Ratio of moles/ 7.14 14.3
Simplest ratio =1 =2
7.14 7.14
3 Empirical formula = CH2
RMM of (CH2)n = 56 .............1
[(12 + 1(2)]n = 56
14n = 56
56 6 max 5
n =
14
= 4 ………..1
Molecular formula : C4H8 ………………..1
(iii) 1+1
1+1
But-1-ene But-2-ene Max 4
2-methylpropene
(iv) Compound M (Butene, C4H8) has a higher percentage of carbon atom in
their molecule than butane, C4H10 …………….1
4(12)
% of C in C4H8 = x 100%
4(12)  8
48
= x 100%
56
= 85.7% …………1
4(12)
% of C in C4H10 = x 100%
4(12)  10
48
= x 100%
58
= 82.7% ………..1
.....3
(b) (i) Starch 1
Protein 1
(ii) H H CH3 H
I I I I
C = C– C = C
I I 1
H H 1..2
2-methylbut-1,3-diene or isoprene
(c) (i) Rubber that has been treated with sulphur 1
24

25. In vulcanised rubber sulphur atoms form cross-links between the rubber 1
(ii) molecules
These prevent rubber molecules from sliding too much when stretched 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 1+1
 Vinegar – provide an acidic condition that inhibits the growth of
microorganisms in pickled foods 1+1
(b) (i) Paracetamol 1
Codeine 1
(ii) To follow the instructions given by the doctor concerning the dosage and 1
method of taking the medicine
To visit the doctor immediately if there are symtoms of allergy or other side 1
effects of thye medicine
(iii) If the correct dosage is not given by the doctor, it will cause abuse of the 1
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
(c)
Type of food Examples Function
additives
Preservatives Sugar, salt To slow down the growth 2
of microorganisms
Flavourings Monosodium To improve and enhance
glutamate, spice, the taste of food 2
garlic
Antioxidants Ascorbic acid To prevent oxidation of 2
food
Dyes/ Colourings Tartrazine To add or restore the
Turmeric colour in food 2
Disadvantages of any two food additives:
Sugar – eating too much can cause obesity, tooth decay and diabetes
1
Salt – may cause high blood pressure, heart attack and stroke.
1
Tartrazine – can worsen the condition of asthma patients
- May cause children to be hyperactive
MSG – can cause difficult in breathing, headaches and vomiting.
TOTAL 20
PAPER 2 : ESSAY QUESTION
Section C
Mar
Questions Marking criteria
ks
7 (a) (i) 1. Sulphur is burnt in air to produce sulphur dioxide // 1
2. Burning of metal sulphides/zinc sulphide / lead sulphide produce sulphur 1
dioxide 1
1
3. Sulphur dioxide is oxidised to sulphur trioxide in excess oxygen
1
4. Sulphur trioxide is dissolved in concentrated sulphuric acidto form oleum.
5. The oleum is diluted with water to produce concentrated sulphuric acid
25

26. (ii) H2SO4 + 2NH3→ (NH4)2SO4
Formula for reactants and product correct 1
Balanced 1
(b) 1. Bronze is harder than copper 1
2. Atoms of pure copper are same size and arrange in layers 1
3. when force applied the layers will slide. 1
1
4. In bronze tin atom has different size compare to pure copper
1
5. and interrupt the orderly arrangement of pure copper. max4
(c) Procedure:
1. Iron nail and steel nail are cleaned using sandpaper. 1
2. Iron nail is placed into test tube A and steel nail is placed into test tube B. 1
3. Pour the agar-agar solution mixed with potassium hexacyanoferrate(III) 1+ 1
solution into test tubes A and B until it covers the nails.
4. Leave for 1 day. 1
5. Both test tubes are observed to determine whether there is any blue spots
formed or if there are any changes on the nails. 1
6. The observations are recorded 1
Results:
Test tube The intensity of blue spots 1
A High 1
B Low
1
Conclusion:
Iron rust faster than steel.
TOTAL 20
No Answer Mark
8 (a) (i) X - any acid – methanoic acid 1
Y - any alkali – ammonia aqueous solution 1
(ii) 1. Methanoic acid contains hydrogen ions 1
2. Hydrogen ions neutralise the negative charges of protein membrane 1
3. Rubber particles collide, 1
4. Protein membrane breaks 1
5. Rubber polymers combine together 1
5 max 4
(iii) Ammonia aqueous solution contains hydroxide ions 1
Hydroxide ions neutralise hydrogen ions (acid) produced by activities of 1
bacteria
(b) (i) Alcohol 1
(ii) Burns in oxygen to form carbon dioxide and water 1
Oxidised by oxidising agent (acidified potassium dichromate (VI) solution) to 1
form carboxylic acid
(iii) Procedure:
1. Place glass wool in a boiling tube
3
2. Pour 2 cm of 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 max 5
6. Using test tube collect the gas given off
Diagram:
26

27. Porous pot chips
Glass wool
soaked with
ethanol
Heat Heat Water
[Functional diagram] ….1 ...2
[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
27

28. JAWAPAN SET 5
PAPER 3 SET 1
EXPLANATION SCORE
1. (a) [Able to record all reading accurately with unit]
(i) Sample answer
Experiment Metal X Metal Y 3
I 1.70 cm 1.40 cm
II 1.75 cm 1.45 cm
III 1.75 cm 1.45 cm
[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) [Able to construct a table to record the diameter of the dents and average
(ii) diameters for material X and Y that contain:
1. correct title
2. Reading and unit
Sample answer:
Material Diameter of the dents(cm) Average
1 2 3 diameter,(cm)
X 1.70 1.75 1.75 1.73
3
Y 1.40 1.45 1.45 1.43
[Able to construct a table to record the diameter of the dents and average 2
diameters for material X and Y that contain
1. title
2. Reading
[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] 3
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
[Able to state correct observation, incompletely] 2
Sample answer:
The diameter of dents made on material Y is smaller// The diameter of dents
made on material X is bigger
[Able to state an idea of the observation] 1
Sample answer:
The diameter of dents for Y is small// The diameter of dents for X is big
No response or wrong response 0
28

29. EXPLANATION SCORE
1.(c) [Able to state the inference correctly] 3
Sample answer:
Material Y is harder than material X// Material X softer than material Y
[Able to state the inference correctly/ 2
Sample answer:
Material Y is harder // Material X softer
[Able to state an idea of inference. 1
Sample answer:
Material Y is hard// Material X is soft
No response or wrong response 0
EXPLANATION SCORE
1.(d) [Able to state the correct operational definition for alloy] 3
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.
[Able to state the meaning of alloy, incompletely] 2
Sample answer:
Material that form small dent is hard
[Able to state an idea of alloy] 1
Sample answer:
Alloy form dent//alloy is hard
No response or wrong response 0
KK0508
EXPLANATION SCORE
1.(e) [able to give all three explanations correctly] 3
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
[able to give any two explanations ] 2
[able to give any one explanations ] 1
No response given / wrong response 0
29

30. EXPLANATION SCORE
1.(f) [Able to state any alloy for material Y and its major pure metal for materials X correctly] 3
Sample answer:
Material X: copper // iron// any suitable metal
Material Y: bronze/ brass//stainless steel// any suitable alloy for pure metal given.
[Able to state any alloy for material Y and its major pure metal for materials X correctly] 2
Sample answer:
Material X: tin/ zinc// chromium / nickel // any suitable metal
Material Y: bronze// brass//stainless steel// any suitable alloy for pure metal given.
[Able to state any alloy for material Y and its major pure metal for materials X correctly] 1
Sample answer:
Material X: magnesium // aluminium//zinc // any metal
Material Y: pewter // bronze // stainless steel //any alloy
No response given / wrong response 0
EXPLANATION SCORE
1.(g) [Able to state the relationship correctly between the manipulated variable and responding 3
variable with direction]
Sample answer:
The harder/ softer the material, the smaller / bigger the diameter of the dent.
[Able to state the relationship correctly between the manipulated variable and responding 2
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
[able to state the idea of hypothesis] 1
Sample answer:
Y is harder // X is softer // alloy is harder
No response given / wrong response 0
1.(h)
EXPLANATION SCORE
[Able to state all the three variables and all the three actions correctly] 3
Sample answer:
Names of variables Action to be taken
(i) manipulated : (i) the way to manipulate variable:
Type of materials / material X and Y Change pure metal/ alloy with alloy /pure
metal
(ii) responding: (ii) what to observe in the responding variable:
Diameter of dent The diameter of the dent formed on material X
and Y.
(iii) controlled: (iii) the way to maintain the controlled
Mass of the weight // height of the weight variable:
// size of steel ball bearing. Uses same mass of weight // same height of
weight // same size of steel ball bearing
[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
30