Chemistry Perfect Score 2011 module answer

43
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

description

Chemistry Perfect Score 2011 module answer

Transcript of Chemistry Perfect Score 2011 module answer

Page 1: 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

Page 2: Chemistry Perfect Score 2011 module answer

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

of each element in a compound. 1

(b) H2SO4 + Zn → ZnSO4 + H2 2

(c) Heating, cooling and weighing are repeated until a constant mass is

obtained. 1

(d) Element Copper Oxygen Mass, g 47.70 – 25.30

=22.40 53.30 – 47.70

=5.60 Mole atom 22.40

64 = 0.35

5.60 16

= 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)

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

Based on the balanced equation; Al2 (CO3)3 : Al2O3 1 : 1 0.3 : 0.3

Mass of Ag = 0.4 x 102 = 30.6 g

1

1 1

(iv) Based on the balanced equation Al2 (CO3)3 : CO2

1 : 3 0.3 : 0.6 Volume of CO2 = 0.9 x 24

= 21.6 dm3

= 21600 cm

3

1

1 1

(b) (i) Zinc carbonate 1

(ii) Zinc oxide and carbon dioxide 1

Heat

Magnesium ribbon

Page 3: Chemistry Perfect Score 2011 module answer

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

(c) P and S // Q and R 1

(d) (i) Q and R 1

(ii) Have same proton number but different nucleon number //

Have same number of protons but different number of neutrons

1

(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

to overcome the forces of attraction between particles

1 1

TOTAL 10

No. Answer Mark

4 (a) Sodium and magnesium // sodium and aluminium // magnesium and

aluminium 1

(b) Halogen 1

(c) 2.8.3 1

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

1

(ii) From left to right : The proton number // the positive charge increases from sodium to argon The forces of attraction by the nucleus on the electrons (nuclei attraction)

in the first three occupied shells become stronger

1 1

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

1

(ii) 2Na + Cl2 → 2NaCl [Formula of reactants and product are correct] [Balanced equation]

1 1

(iii)

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

1

TOTAL 10

Q 33

16

Atomic size decreases

Page 4: Chemistry Perfect Score 2011 module answer

4

No. Answer Mark

5 (a) (i) X 1

(ii) 8 valence electron // electron arrangement 2.8 // achieve octet electron

arrangement 1

(b) Covalent 1

(c)

(a) i

)

(b) (

i

(

VW4 1

(ii)

1+1

(iii) has low melting / boiling point // cannot conduct electricity in molten and solid state . // insoluble in water// soluble in organic

solvent.

1

(d) (i) Ionic compound 1

(ii) Atom U donate one electron to form U+ ion

Atom W accept one electron to form W- ion

U+ ion and W

- ion attracted to each other by strong electrostatic force /

ionic bond.

1 1 1

(iii)

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

1 1

TOTAL 13

V W

W

W

W

U W

(i)

Page 5: Chemistry Perfect Score 2011 module answer

5

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

WX4 molecule / Diagram

Molecules WX4

1 1 1 1

1

1

1

1 1

1

W

W

W

W

W X

X

X

X

Y X

+

Page 6: Chemistry Perfect Score 2011 module answer

6

(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

Page 7: Chemistry Perfect Score 2011 module answer

7

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 to 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

Page 8: Chemistry Perfect Score 2011 module answer

8

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

Page 9: Chemistry Perfect Score 2011 module answer

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

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) MnO4 - + 8 H

+ + 5 e → Mn

2+ + 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

electrode Copper electrode

Page 10: Chemistry Perfect Score 2011 module answer

10

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 + 2Fe2+

→ 2Cl- + 2Fe

3+

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

Page 11: Chemistry Perfect Score 2011 module answer

11

(ii) Test tube P : Cl- ion and Fe

3+ 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

Page 12: Chemistry Perfect Score 2011 module answer

12

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 + 2Fe2+

→ 2Br- + 2Fe

3+

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

Page 13: Chemistry Perfect Score 2011 module answer

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

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+ → Mg

2+ + 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 dm

3

= 0.006 dm3 // 6 cm

3

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 cm

3

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

Page 14: Chemistry Perfect Score 2011 module answer

14

(e) (i) 1. Colourless gas bubbles are released.// efeervesence

1

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 dm

3

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 → 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

Volume of hydrogen = 0.0167 x 24 dm3= 0.4 dm

3/ 400 cm

3

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

Page 15: Chemistry Perfect Score 2011 module answer

15

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 + Zn

2+ 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

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 + Cu2+

Zn2+

+ Cu

Page 16: Chemistry Perfect Score 2011 module answer

16

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 potassium 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

+, Pb

2+ 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

1

1

1

1

1

1

1

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

H2O + NaCl

Page 17: Chemistry Perfect Score 2011 module answer

17

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 cm

3 s

-1

Average rate of reaction for experiment II = 26.0

150 = 0.17 cm

3 s

-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

(iii) The rate of reaction is the change of volume of hydrogen gas per unit time 1

Page 18: Chemistry Perfect Score 2011 module answer

18

(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 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

5. 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 H = - 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

Experiment I

Volume of hydrogen/ cm3

Page 19: Chemistry Perfect Score 2011 module answer

19

than ethanol

3. No. of mole of CO2 and H2O produced during combustion of

propanol is more than ethanol

4. Formation of CO2 and 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) cm3 of 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 =60

ba = 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

2. 1 mole of sodium chloride solution produces 1 mole of Cl- ion

1

1

Page 20: Chemistry Perfect Score 2011 module answer

20

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 cm3 of 1.0 mol/ dm

3 Ca(NO3)2 solution and 50 cm

3 of 1.0 mol / dm

3

Na2CO3 solution 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

Na2CO3 solution 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 cm3 of 1 mol dm

-3 magnesium nitrate solution is measured and

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

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

1

1

1

1…6

Page 21: Chemistry Perfect Score 2011 module answer

21

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

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 the

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

Page 22: Chemistry Perfect Score 2011 module answer

22

JAWAPAN SET 4

PAPER 2 : STRUCTURED QUESTION

Section A

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 dm

3

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

H

Page 23: Chemistry Perfect Score 2011 module answer

23

No. Explanation Mark 3

(a) (i)

Haber process

1

(ii)

N2 + 3H2 2NH3

Correct formula Balanced

1 1

(iii)

450 oC --- 550

oC

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

Copper atom

Page 24: Chemistry Perfect Score 2011 module answer

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

12

7.85 = 7.14

1

3.14 = 14.3

2 Ratio of moles/ Simplest ratio 14.7

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 2-methylpropene

Page 25: Chemistry Perfect Score 2011 module answer

25

(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

Page 26: Chemistry Perfect Score 2011 module answer

26

(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 (VI) solution) to

form carboxylic acid

1 1

(iii) Procedure: 1. Place glass wool in a boiling tube

2. Pour 2 cm3 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. Using test tube collect the gas given off

Diagram:

6 max 5

Page 27: Chemistry Perfect Score 2011 module answer

27

[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

...2

Total 20

Heat Heat

Glass wool

soaked with

ethanol

Porous pot chips

Water

Page 28: Chemistry Perfect Score 2011 module answer

28

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 1.70 1.75 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

No response or wrong response 0

Page 29: Chemistry Perfect Score 2011 module answer

29

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

Page 30: Chemistry Perfect Score 2011 module answer

30

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 alloy 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

Page 31: Chemistry Perfect Score 2011 module answer

31

2. (a)

EXPLANATION SCORE [Able to state 4 inferences correctly]

Test tube Inference A Iron (II) /Fe

2+ 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) /Fe

2+ ions are not formed / produced // iron / Fe does not rusted /

oxidized D Iron (II) /Fe

2+ 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

No response given / wrong response 0

Page 32: Chemistry Perfect Score 2011 module answer

32

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

Page 33: Chemistry Perfect Score 2011 module answer

33

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.

1

No response given / wrong response 0 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.

1

No response given / wrong response 0

Page 34: Chemistry Perfect Score 2011 module answer

34

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.

1

No response given / wrong response 0

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 30

oC.

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

1

No response given / wrong response 0

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

No response given / wrong response 0

Page 35: Chemistry Perfect Score 2011 module answer

35

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 cm

3 13.50 cm

3 26.00 cm

3

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 cm

3 of hydrochloric acid 1.0 mol dm

-3 is added to 25 cm

3

sodium hydroxide 1.0 mol dm-3

with 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 Question Rubric Score

Page 36: Chemistry Perfect Score 2011 module answer

36

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

3

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

Page 37: Chemistry Perfect Score 2011 module answer

37

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 No response or wrong response 0

Page 38: Chemistry Perfect Score 2011 module answer

38

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 cleaning agent B is better /more effective

2

Page 39: Chemistry Perfect Score 2011 module answer

39

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] cm3 of 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 = 30.0 oC

Highest temperature = 60.0 oC

3

Able to record the thermometer reading correctly without unit

Sample answer:

Initial temperature = 30

2

Page 40: Chemistry Perfect Score 2011 module answer

40

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

1

Wrong or no response 0

Page 41: Chemistry Perfect Score 2011 module answer

41

(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

1

Wrong or no response 0

Page 42: Chemistry Perfect Score 2011 module answer

42

(f) Able to state the operational definition of heat of combustion of methanol

in this experiment correctly

Sample answer

When 1 mole of ethanol is burnt to heat 200 cm3 of water, 252 kJ of heat

energy is released

3

Able to state the operational definition but less accurately

Sample answer

When 1 mole of ethanol is burnt, 252 kJ of heat energy is released

2

Able to state an idea of operational definition

Sample answer

When ethanol is burnt heat energy is relaeased

1

Wrong or no response 0

(g) Able to state the three reasons correctly

Sample answer

1. Some of the heat energy is released to the surrounding

2. Some of the heat energy is absorbed by the copper can

3. Incomplete combustion of ethanol

3

Able to state any two reasons correctly 2

Able to state any one reason correctly 1

Wrong or no response 0

(h) Able to classify all the compounds correctly

Sample answer

Hydrocarbon Non-hydrocarbon

Hexene Propanoic acid

Methane Ethanol

3

Able to classify at least three compounds correctly 2

Able to classify at least two compounds correctly 1

Wrong or no response 0

Question 2

2(a) Able to state the aim of experiment accurately

Sample answer

To determine the effect of type of electrode on the selection of ions to be

discharged at the anode/ on the product formed at the anode.

2

Able to state the aim of experiment less accurately

Sample answer

To determine type of electrode affects the product formed at the anode

1

Wrong or no response 0

2(b) Able to state all the variables

Sample answer

Manipulated variable

Type of electrodes//Carbon electrodes and copper electrodes

Responding variable

Product formed at anode

Fixed variable

Electrolyte

3

Able to state any two variables correctly 2

Able to state any one variable 1

Wrong or no response 0

Page 43: Chemistry Perfect Score 2011 module answer

43

2(c) Able to state the hypothesis accurately

Sample answer

When carbon electrodes are used, bubbles/oxygen gas released at anode,

when copper electrodes are used, anode becomes thinner/ionised

3

Able to state the hypothesis less accurately

Sample answer

When different type of electrodes are used, different product is formed at the

anode

2

Able to state an idea of hypothesis

Sample answer

Type of electrodes affect products formed

1

Wrong or no response 0

2(d) Able to list completely the materials and apparatus as the following

Copper(II) sulphate solution (0.5 – 2.0) mol dm-3

, copper rods, carbon rods,

electrolytic cell, battery, connecting wires, test tube

3

Able to list the following apparatus and materials:

Copper(II) sulphate solution, carbon rods/ copper rods, beaker, battery,

connecting wire

2

Able to give at least one material, one apparatus

Copper(II) sulphate solution//carbon rods//copper rods

Beaker

1

Wrong or no response 0

2(e) Able to state the all the following procedures

1. Half filled the electrolytic cell/beaker with copper(II) sulphate

solution

2. A test tube filled with copper(II) sulphate solution is inverted over

the anode carbon electrode

3. Both electrodes are connected to the batteries using connecting

wires//Complete the circuit

4. Record the observations at the anode

5. Repeat steps 1-4 by using copper electrodes

3

Able to state steps 1, 3, 4 and 5 2

Able to state steps 1 and 3 1

Wrong or no response 0

2(f) Able to draw a table for tabulation of data correctly with the following

Type of electrodes Observation at anode

Carbon

Copper

3

Able to draw a table with the following

Type of electrodes Observation

2

Type of electrodes

1

Wrong or no response 0

END OF MARKING SCHEME