Chemistry Perfect Score 2011 Module Answer

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

JAWAPAN SET 1 PAPER 2 : STRUCTURED QUESTION Section A No. 1

Answer (a)

Mark

The formula that shows the simplest whole number ratio of atoms of each element in a compound. H2SO4 + Zn → ZnSO4 + H2 Heating, cooling and weighing are repeated until a constant mass is obtained.

(b) (c)

1 2 1

(d) Element Mass, g

Copper 47.70 – 25.30 =22.40 Mole atom 22.40 64 = 0.35 Simplest ratio 1 Empirical formula = CuO

Oxygen 53.30 – 47.70 =5.60 5.60 16 = 0.35 1 4 2 1

H2 + CuO → Cu + H2O To prevent the hot copper from being oxidized again.

(e) (f) (g)

Magnesium ribbon

Heat

2 TOTAL

No. 2

Answer (a)

(i) (ii) (iii)

(iv)

(b)

(i) (ii)

Al2CO3 Al2(CO3)3 Al2O3 + 3CO2 The number number of mole 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 Based on the balanced equation Al2 (CO3)3 : CO2 1 : 3 0.3 : 0.6 Volume of CO2 = 0.9 x 24 3 = 21.6 dm 3 = 21600 cm Zinc carbonate Zinc oxide and carbon dioxide

13 Mark 1 2 1

1 1

1 1 1 1 1

2

JAWAPAN SET 1 PAPER 2 : STRUCTURED QUESTION Section A No. 1

Answer (a)

Mark

The formula that shows the simplest whole number ratio of atoms of each element in a compound. H2SO4 + Zn → ZnSO4 + H2 Heating, cooling and weighing are repeated until a constant mass is obtained.

(b) (c)

1 2 1

(d) Element Mass, g

Copper 47.70 – 25.30 =22.40 Mole atom 22.40 64 = 0.35 Simplest ratio 1 Empirical formula = CuO

Oxygen 53.30 – 47.70 =5.60 5.60 16 = 0.35 1 4 2 1

H2 + CuO → Cu + H2O To prevent the hot copper from being oxidized again.

(e) (f) (g)

Magnesium ribbon

Heat

2 TOTAL

No. 2

Answer (a)

(i) (ii) (iii)

(iv)

(b)

(i) (ii)

Al2CO3 Al2(CO3)3 Al2O3 + 3CO2 The number number of mole 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 Based on the balanced equation Al2 (CO3)3 : CO2 1 : 3 0.3 : 0.6 Volume of CO2 = 0.9 x 24 3 = 21.6 dm 3 = 21600 cm Zinc carbonate Zinc oxide and carbon dioxide

13 Mark 1 2 1

1 1

1 1 1 1 1

2

(iii)

→

ZnCO3

ZnO + CO2 TOTAL

No. 3

Answer (a)

(i) (ii)

(b)

16 (c) (d)

(e)

(i) (ii) (i) (ii)

Mark

The number of  protons  protons found in the nucleus of an atom 7

33

1 12 1 1 1

Q

P and S // Q // Q and R Q and R

1 1 1

Have same proton number but different nucleon number //  Have same number same number of protons but different number different number of neutrons O Melting point : 63 C [values & unit must be correct]

Section

1

Physical state

AB

Solid

DE

Liquid and gas 1

(iii)

1 1

the heat energy absorbed by the particles is used to overcome the forces of attraction between particles TOTAL

No. 4

10

(a)

Answer Sodium and magnesium // sodium and aluminium // magnesium and aluminium

(b)

Halogen

1

(c) (d)

2.8.3 Sodium, magnesium, aluminium, chlorine, argon

1 1

(i)

Mark 1

Atomic size decreases size decreases

(ii)

(e)

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

(i)

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

(ii)

2Na + Cl2 → 2NaCl [ Formula are correct]]  Formula of reactants and product are correct [ Balanced   Balanced equation] has high melting / boiling point // conduct electricity in molten state or aqueous solution // soluble in water

(iii)

1 1 1

1 1 1 TOTAL

10

3

No. 5

Answer (a)

(i) (ii)

(b) (c)

(i)

Mark

X

1

8 valence electron // electron arrangement 2.8 // achieve octet electron arrangement

1

Covalent

1

VW4

1

(ii) 1+1 W

W

W

V

W

(iii)

(d)

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

1

(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) 1 1

U

W

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

13

4

PAPER 2: ESSAY QUESTION Section B No. 6

Answer (a)

Group 17 Period 3 Has seven valence electrons. Has three shells occupied with electron

(b)

(i) Between Y and X

Mark 1 1 1 1

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

1 1 1 1 1

X

Y

1 (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

1

1 1 1

W

Molecules WX4

5

(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

No. 7

Answer (a)

(b)

(i) (ii)

(i)

(ii) (c)

20

Mark

2.8.7, Chlorine 2Fe + 3Cl2 → 2FeCl3 Correct formulae of reactants and product Balanced Z,Y,X Z more reactive than X Atomic size of Z bigger than atomic size X Valence electron become further away from nucl eus 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 1 1 1 1

same/similar Same valence electron 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

TOTAL

1 1 1 1 1 1 20

6

PAPER 2: ESSAY QUESTION Section C No. 8

Answer (a)

(b)

Mark

(i)

Dilute acid: Hydrochloric acid / Sulphuric acid/ Nitric acid Metal N: Magnesium / zinc

1 1

(ii)

Anhydrous calcium chloride To dry the hydrogen gas

1 1

(iii)

Example: Copper(II) oxide Copper ion is reduced// reduction process Because oxidation number of copper decrease from +2 t o 0 Hydrogen is oxidised// oxidation process Because oxidation number of hydrogen increase from 0 to +1 Hydrogen is reducing agent Copper(II) ion// Copper(II) oxide is oxidising agent

1 1

Relative Molecular mass of (CH 2)n = 56 (12 + 2) n = 56 n=4 Molecular formula = C4H8

1

(i)

(ii)

1 1 1 1

1

Unglased porcelain chips

Glass wool soaked in butanol

Heat

Water 2

Procedure:

1. 2. 3. 4. 5. 6.

A small amount of glass wool soaked in butanol is placed in a boiling tube. The boiling tube is clamped horizontally The unglazed porcelain chips are placed in the middle section of the boiling tube. The boiling tube is closed with a stopper fitted with a delivery tube The unglazed porcelain chips are heated strongly. Then, the glass wool is warmed gently to vaporize the propanol. The gas released is collected in a test tube. TOTAL

1 1 1 1 1 1 20

7

No. 9

Answer (a)

(b)

Mark

(i)

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

(ii)

Copper(II)oxide // lead(II)oxide CuO + H2  Cu + H2O // PbO + H2  Pb + H2O

(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

1

1 1+1 1

1 1 1 1 1 1 1 1 1 1

Result: Description Crucible + lid Crucible + lid + magnesium Crucible + lid + magnesium oxide Calculation: Element Mass, g Mole

Mg y-x y-x 24 =0.1 1

Simplest ratio

Mass/g x y z O z-y z-y 16 =0.1 1

Element Mass (%) Number of moles Mole ratio

1

1 Max 10

Empirical formula: MgO (c)

1

C 84.6 84.6/12 =7.05 1

Empirical formula : CH2 RMM of (CH 2)n = 70 [ 12 + 2]n = 70 14 n = 70 n = 5 Molecular formula : C5H10

H 15.4 15.4/1 =15.4 2

1 1 1

1

1 1 20

8

JAWAPAN SET 2 PAPER 2 : STRUCTURED QUESTION Section A No. 1 (a) Cell II (b) (i) Magnesium electrode (ii) e

Answer

Mark 1 1 V

Magnesium electrode

(iii) (c)

1

Copper electrode

Copper electrode thicker // Brown solid deposited 1. Correct formulae of reactant and product 2. Balanced equation

1 1 1

2+

(d)

2

(i) (ii) (iii)

No. (a) (i) (ii) (iii)

(b)

(iv) (i) (ii)

(iii) (iv)

Cu + 2e → Cu Electrical energy to chemical energy Blue colour remain unchange 2+ 1. Concentration / Number of mole of Cu ion remain unchanged 2+ 2. Rate of Cu ion discharge at cathode is the same as rate of Cu atom ionize at anode TOTAL Answer

-

+

2+

1 10 Mark 1

Iodine r: formula/iodide/iodine gas MnO4 + 8 H + 5 e → Mn

1 1 1

+ 4 H2O

1 1 1 1 1 1 1

+7 → +2 reduction Potassium chloride // iron(II) sulphate // [any reducing agent] Zinc 1. Correct formulae of reactant and product 2. Balanced equation 2 Zn + O2 → 2 ZnO a: 2 J + O2 → 2 JO K,J, L Predict : no changes r: no reaction Reason : L is more reactive than J/zinc r: more electropositive

1 1 1 TOTAL

11

9

PAPER 2 : ESSAY QUESTION Section B

3

No. (a)

(b)

1. 2. 3. 4. (i)

Answer Propanone is a covalent compound Propanone exist as molecule // No freely moving ion in propanone Sodium chloride is an ionic compound Sodium chloride solution has freely moving ion Cell X Voltaic cell Chemical → electrical

4. Ions in electrolyte 5. Half  equation

Cu , SO4 , H and OH ions Anode: 2+ Cu → Cu + 2e Cathode: 2+ Cu + 2e → Cu Anode: Copper ecomes thinner

Cu , SO4 , H and OH ions

1

Positive terminal: 2+ Cu + 2e → Cu Negative terminal 2+ Zn → Zn + 2e Positive terminal: Copper plate becomes thicker

1

Cathode: Copper becomes thicker

Negative terminal: Zinc becomes thinner

1. 2. 3. 4. 5. 6. 7. (i i)

1 1

Properties 1. Type of cell 2. Energy change 3. Electrodes

6. Observation

(c)

Mark 1 1

Cell Y Electrolytic cell Electrical → chemical

Anode: A Cathode: B 2+

2-

Positive terminal: C // Copper Negative terminal: D // Zinc +

-

2+

2-

+

-

Ag, M, L L is more electropositive than silver L displace silver from silver nitrate solution M is more electropositive than silver M displace silver from silver nitrate solution M is less electropositive than L M cannot displace L from L nitrate solution

Copper // Cu

4

1. 2. 3. 4.

(ii)

(b)

(i)

1. 2. 3. 4. 1. 2. 3. 4. 5. 6.

Answer Correct formulae of reactant and product Balanced equation 2+ Zn + 2e → Zn Correct formulae of reactant and product Balanced equation 2+ Pb + 2e → Pb Zinc is oxidized Zinc atom donates / losses electrons 2+ Lead(II) nitrate / Pb is reduced 2+ Lead(II) nitrate / Pb receives electrons Green colour of iron(II) sulphate change to brown Correct formulae of reactant and product Balanced equation 2+ 3+ Cl2 + 2Fe + 2Fe → 2Cl Colourless solution of potassium iodide change to brown Correct formulae of reactant and product Balanced equation Cl2 + 2I + I2 → 2Cl

1 1 1 1….6

1 1 1 1 1 1 1 1 TOTAL

No. (a) (i)

1 1

20 Mark 1 1 1 1 1 1 1 1 1 1 1 1 1 1

10

-

3+

(ii)

Test tube P : Cl ion and Fe Test tube Q : Cl ion and I2

ion

(iii)

1. Add starch solution 2. Dark blue precipitate formed

1+1 1+1 1 1 20

TOTAL PAPER 2 : ESSAY QUESTION Section C No. 5

(a)

(b)

Answer (i)

Mark

2+

1. Cu // copper(II) ion Equation 2. Correct formula of reactant and product 3. Balance 2+ Cu + 2e → Cu 4. Copper

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

-

Diagram 1. Functional apparatus 2. Label

1

1 1 1 1 1 1

1 1

Impure copper

3. 4. 5. 6. 7.

1 1

Pure copper Copper(II) sulphate solution

3

Pour [50 – 200 cm ] copper(II) sulphate solution into a beaker Connect pure copper as cathode and impure copper as anode Dip both pure and impure copper into copper(II) sulphate solution 2+ Anode : Cu → Cu + 2e 2+ Cathode : Cu + 2e → Cu TOTAL

1 1 1 1 1 20

11

6

No. (a) (i) (ii)

(b)

(i)

(ii)

Answer Metal P : Tin // Lead // Copper Metal Q : Magnesium // Aluminium // Zinc Exp I 1. Metal P is less electropositive than iron 2. Iron is oxidized 3. Iron losses electron // Fe → Fe 2+ + 2e 2+ 4. Dark blue precipitate indicates the presence of Fe 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 → 4OH8. Pink colouration indicates the presence of OH ion 1. Bromine is reduced 2. Bromine molecule receives electron // Oxidation number of bromine decrease / 0 → -1 2+ 3. Iron(II) sulphate / Fe is oxidized 4. Fe2+ losses electron // Oxidation number of iron increases / +2→ +3 5. Correct formula of reactant and product 6. Balanced equation 2+ 3+ Br2 + 2Fe + 2Fe → 2Br  7. Brown colour of bromine decolourise 8. Green colour of iron(II) sulphate change to brown 1. Add sodium hydroxide solution 2. Brown precipitate formed TOTAL

Mark 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 20

12

JAWAPAN SET 3 PAPER 2 : STRUCTURED QUESTION Section A

No. 1

(a)

Answer (i) (ii)

(b)

(c )

(i) (ii)

(iii)

(d)

Mark

Solution in test tube C Solution in test tube A -3 1. Higher than pH value of 0.1 moldm 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 1

Magnesium chloride + 2+ Mg + 2 H → Mg + H2 1. Correct formula of reactant and product 2. Balanced equation

1

1

1 1

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 3 Volume of hydrogen gas = 0.00025 x 24 dm 3 3 = 0.006 dm // 6 cm

1

1 1 1 12

White precipitate TOTAL

No. 2

(a)

Answer (i)

Solvent P: Water Solvent Q: methyl benzene / propanone / suitable organic solvent

1 1

(ii)

Effervescence / gas released // magnesium ribbon dissolved

1

(iii)

(b)

Mark

(i)

+

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

1. pH value increase / bigger 2. The lower the concentration of acid the higher the pH value

(ii)

1 1

  

(0.5)(V) = (0.04)(250) // 



1



3

V = 20 cm

3 (a) (b) (c) (d)

Alkali that ionize/dissociate completely in water to produce high concentration of hydroxide ions. Alkaline / alkaline solution P: ion Q: molecule No Because there are no hydroxide ions in the solution// ammonia exist in the form of molecule.

1

1

1

1 1 1 1 1

1

13

2 2

(e)

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

1

1

Mg + 2HCl  MgCl 2 + H2 1. Correct formula 2. Balanced equation

(ii)

2.

1 1

Mol of Mg = 2.4/24 // 0.1 mol

1

3

3

3. Volume of H2 = 0.1  24 dm = 2.4 dm

4

1 Total

NO 4

ANSWER

11

MARK 1

(a) (i) Green (ii) Double decomposition reaction

1

(b) (i) carbon dioxide (ii) CuCO3 → CuO + CO2 1. Reactants and products are correct 2. Equation is balanced

1 1+ 1

(iii) Copper(II) carbonate Heat Lime water

- Labelled diagram - Functional

2

(c) 1 mol CuCO3 = 12.4/124 = 0.1 mol Mol of CuCl2 = 0.1 x 135g Mass = 13.5g

3 10

No.

5

(a) (b)

(c)

(d) (e) (f)

Mg + 2HCl (i) (ii )

→

Answer MgCl2 + H2

Mark 1+1

0.4/24 = 0.0167 mol The number of mole of HCl = MV/1000 = 1x 50/1000 = 0.05 mol

1 1

From the chemical equation 1 mol of magnesium produce 1 mol hydrogen If 0.0167 mol produce 0.0167 mol hydrogen 3 3 3 Volume of hydrogen = 0.0167 x 24 dm = 0.4 dm  / 400 cm

1 1

3 -1

I 400 /100 =4 cm s 3 -1 II 400 /60 = 6.67 cm s As catalyst The temperature of hydrochloric acid The concentration of hydrochloric acid TOTAL

1 1 1 1 1 11

14

No. 6

Answer

Mark

(a)

The heat released when 1 one mole of copper is displaced from copper (II) sulphate solution by zinc.

(b)

Cu

(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 1000 (iii) Heat of displacement = 2100 0.025

1

2+

1

2+

+ Zn →

Cu + Zn

1

= 0.025 mol 1 = -84000 J H = 84.0 kJ/mol

(e)

1 1

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

(f)

Energy Zn + Cu

2+

H= - 84.0 kJ/mol 2+

Zn

+ Cu

1+1

TOTAL

No.

7

Answer

(a)

Mark

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

1 1 1

3

50 cm ( marked on the graph) NaOH

+

HCl

NaCl

10

1 +

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 -3 Concentration HCl = 0.05 x 1000 = 1 moldm 50

1

1+1

(c)

To ensure uniform temperature of mixture in t he polystyrene cup

1

(d)

All the sodium hydroxide has reacted completely

1

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

(e)

(i)

15

(ii)

Energy NaOH + HCl H= - 56.0 kJ/mol 1+1

H2O + NaCl

(i) (ii)

Less than 5.6 kJ

-

1 1

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

TOTAL

14

PAPER 2 : ESSAY QUESTION Section B

No. 8

(a)

Answer (i)

(ii) (iii)

Mark

Label axes with units All points are transferred correctly Shape of the graph is smooth and correct 3 2.5 cm 2+ moles of Pb ions = 2.5 x 1.0 / / 0.0025 1000 moles of I ions = 5 x 1.0 // 0.005 1000 2+

1 1 1 1 1

1

-

Pb : I 0.0025 : 0.0005 1 : 2 (b)

1 1

Test tube 1: + 1. Ion exist : K , I and NO3 2. All lead(II) nitrate reacts completely 3. Excess of potassium iodide 4. Solution contains soluble salt of potassium iodide and pota ssium nitrate

1 1 1 1

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

1 1 1

Test tube 7: 8. 9. 10. 11.

+

2+

-

Ion exist : K , Pb and NO3 All potassium iodide reacts completely Excess of lead(II) nitrate Solution contains soluble salt of lead(II) nitrate and potassium nitrate

1 1 1 1 Max 10

16

No. 9

(a)

(b)

Answer

Mark

(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)

(i)

Temperature : 450-550 C Catalyst : iron Pressure : 200 atm 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 1 1 1

(ii)

Average rate of reaction for experiment I =

1

o

26.0 210 3 -1 = 0.12 cm s Average rate of reaction for experiment II = 26.0 150 3 -1 = 0.17 cm s

1 1 [correct unit ]

(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

TOTAL

PAPER 2 : ESSAY QUESTION Section C

10

No. (a)

(i)

(ii)

(iii)

Answer Experiment I  – hydrochloric acid or Experiment II – sulphuric acid Mg + 2HCl MgCl 2 + H2 → 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 The rate of reaction is the change of volume of hydrogen gas per unit time

Mark 1 1+1

1

1

17

1 1 1 1 1 1

20

(b)

(i)

Volume of hydrogen/ cm3

Experiment II

Experiment I

1 1 1. 2.

(ii)

Curve with label Axis with title and correct unit

Time/s

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 -3 3. 50 cm of 0.2 moldm hydrocloric acid / sulphuric acid is measured and poured into a conical flask  4. 5 cm of magnesium ribbon are added into the conical flask  5. close conical flask immediately with the stopper fitted with delivery tube. 6. At the same time the stopwatch is started shake the conical flask. 7. The burette readings are recorded at 30 second intervals for 5 minutes Time/s 0 30 60 90 120 150 180 3 Volume of gas / cm TOTAL

11

No. (a) (b)

Answer Water on the wet shirt evaporated Evaporation absorbs heat energy from body (i) C2H5OH + 3 O2  2 CO2 + 3 H2O

1 1…..2

1 1 1 1 1 1 1 7 max 5 ……1 20

Mark

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 1 1

18

than ethanol 3. No. of mole of CO 2 and H2O produced during combustion of  propanol is more than ethanol 4. Formation of CO2 and H2O releases heat energy

1 1..4

1 1….2

(ii) Diagram – labelled and functional Material : Water , ethanol Apparatus : spirit lamp. weighing balance, copper can, clay-pipe triangle, thermometer, wind shield

1 1…..2

Procedure : 3 1. Measure (100  – 250) cm 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 = a60b = y mol

12.

The released heat = mc   = 100 x 4.2 x t = xJ

13.

(c)

1

1 1 1

The heat of combustion of propanol = -

1

x y

-1

1

J mol or

-Z 1 13 max 8

-1

TOTAL

(b)

1 1



kJ mol

No. (a) 12

1

20

Answer 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 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 mole of silver nitrate solution produces 1 mole of Ag ion 2. 1 mole of sodium chloride solution produces 1 mole of Cl ion

Mark 1 1 1 1 1 1 1 1 1 1 1

19

-

(d)

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

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 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 : o

Initial temperature of Ca(NO3)2  /  C o Initial temperature of Na2CO3 /  C o Average initial temperature /  C o Lowest temperature of the mixture /  C o Change in temperature /  C

Ө1 Ө2 (Ө1 + Ө2)/2  Ө3 Ө4 Ө3- Ө4

1 Max 4

1 1 1 1

1

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

1

heat change  mc(Ө 4 – Ө3) = x kJ -1 heat of reaction = + x kJmol 0.05 -1 = + y kJmol

1

TOTAL

13

No. a (i)

Answer 1. 2.

(ii)

(iii)

Zinc nitrate, zinc sulphate Zinc carbonate

I :Sodium carbonate solution/ potassium carbonate solution / ammonium carbonate solution II : Sulphuric acid 3 -3 1. 50 cm of 1 mol dm magnesium nitrate solution is measured and

20

Mark 1 1

1 1 1

poured into a beaker 2.

3. 4. 5. 6.

3

-3

50 cm of 1 mol dm Sodium carbonate solution/ potassium carbonate solution / ammonium carbonate solution solution is measured and poured into the beaker. The mixture is stirred with a glass rod and a white solid, magnesium carbonate is formed. The mixture is filtered and the residue is rinsed with distilled water The white precipitate is dried by pressing it between filter papers.

1 1 1 1 1 1…6

20

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

1

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

(ii)

2+

2+

1

3+

1. Zn , Pb , Al 2. Add ammonia solution into test tube containing salt X solution until excess 3. White precipitate dissolves in excess ammonia solution showing t he 2+ presence of Zn ions 4. White precipitate insoluble in excess ammonia solution showing the 2+ 3+ presence of Pb and Al ions. 5. Add potassium iodide solution into test tube containing salt X solution 2+ Yellow precipitate formed showing the presence of Pb ions //  3+ 6. No change showing the presence of Al ions.

1 1 1 1 1 1

20

21

JAWAPAN SET 4 PAPER 2 : STRUCTURED QUESTION Section A

1

No (a) (b) (c) (d)

(i)

Answer Compound that contains only carbon and hydrogen Has double bonds between carbon – carbon atoms Alkene Propene Hydrogenation / Addition reaction

Mark 1 1 1 1 1

(ii)

1 (e)

(i)

C3H6 + 9/2 O2 → 3CO2 + 3H2O or

2

2C3H6 + 9O2 → 6CO2 + 6H2O (ii)

1

2 .1

No. of mole of C3H6

=

Volume of gas CO2

= 0.05 x 3 x 24 3 = 3.6 dm

42 = 0.05

1 TOTAL

2

No (a) (b) (c)

Answer

(i) (ii) (iii)

Ethanol Hydroxyl group Oxidation Orange colour of potassium dichromate (VI) solution turns to green

H

10 Mark 1 1 1 1 1

O

H    C    C    O    H H (d)

(i) (ii) (iii) (iv)

Esterification Ethyl ethanoate Pleasant smell CH3COOH + C2H5OH → CH3COOC2H5 + H2O

1 1 1 2 TOTAL

10

22

3

No. (a) (i) (ii)

Explanation

Mark 1

Haber process N2 + 3H2

2NH3

Correct formula

1 1

Balanced o

o

(iii)

450 C --- 550 C Vanadium(V) oxide

1 1

(iv)

As a fertiliser

1

(i)

Polyvinyl chloride // polychloroethene

1

(b)

1

(ii)

(c)

Tin atom

Correct arrangement Correct label

1 1

Copper atom TOTAL

4

No. (a) (i) (ii) (iii) (b)

Explanation

10

Mark 1 1 1

glycerol saponification / alkaline hydrolysis to cause precipitation of soap

(i)

1 X: detergent

(ii) (iii) (iv)

1

Y :soap magnesium stearate or calcium stearate Mg2+ and Ca2+ causes water pollution / non-biodegradable TOTAL

1 1+1 1 9

23

PAPER 2 : ESSAY QUESTION Section B

5

No (a) (i)

Answer

Mark 1

14.3 %

(ii) 1

2

Element Mass/ % No. of moles

C 85.7

H 14.3

85 .7 = 7.14 12

Ratio of moles/  Simplest ratio

14.3

7.14 =1 7.14

1

= 14.3

14 .3 =2 7.14

3 Empirical formula RMM of (CH 2)n [(12 + 1(2)]n 14n

= CH2 = 56 .............1 = 56 = 56 56 n = 14 = 4 ………..1 Molecular formula : C4H8 ………………..1

6 max 5

(iii)

1+1

1+1 But-1-ene

But-2-ene

Max 4

2-methylpropene (iv)

Compound M (Butene, C 4H8) has a higher percentage of carbon atom in their molecule than butane, C4H10 …………….1 % of C in C 4H8

4(12) x 100% 4(12)  8 48 = x 100% 56 =

= 85.7% …………1 4 (12 ) % of C in C 4H10 = x 100% 4 (12 )  10 =

48 x 100% 58

= 82.7% (b)

(c)

(i)

Starch Protein

(ii)

H H CH3 I I I C = C –  C = I H

(i)

………..1

.....3 1 1

H I C I H

2-methylbut-1,3-diene or isoprene Rubber that has been treated with sulphur

1 1..2

1

24

(ii)

In vulcanised rubber sulphur atoms form cross-links between the rubber molecules These prevent rubber molecules from sliding too much when stretched TOTAL

6

No. (a)

(b)

Explanation 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 20 Mark 1+1 1+1

(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

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

(iii)

1

1

(c) Type of food additives Preservatives

Examples

Function

Sugar, salt

2

Flavourings

Monosodium glutamate, spice, garlic Ascorbic acid

To slow down the growth of microorganisms To improve and enhance the taste of food To prevent oxidation of  food To add or restore the colour in food

2

Antioxidants Dyes/ Colourings

Tartrazine Turmeric

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

1 1

TOTAL

20

PAPER 2 : ESSAY QUESTION Section C Questions 7

(a)

(i)

Mar ks 1 1 1 1 1

Marking criteria 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

25

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

(ii)

 Results: Test tube A B

1 1 1+ 1 1 1 1

1 1

The intensity of blue spots High Low

1 Conclusion: Iron rust faster than steel.

TOTAL

8

No (a) (i)

(b)

Answer X - any acid – methanoic acid Y - any alkali – ammonia aqueous solution

(ii)

1. 2. 3. 4. 5.

(iii)

Ammonia aqueous solution contains hydroxide ions Hydroxide ions neutralise hydrogen ions (acid) produced by activities of  bacteria

(i) (ii)

Alcohol Burns in oxygen to form carbon dioxide and water Oxidised by oxidising agent (acidified potassium dichromate (V I) solution) to form carboxylic acid 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. Using test tube collect the gas given off 

(iii)

Methanoic acid contains hydrogen ions Hydrogen ions neutralise the negative charges of protein membrane Rubber particles collide, Protein membrane breaks Rubber polymers combine together

20

Mark 1 1 1 1 1 1 1 5 max 4 1 1 1 1 1

6 max 5

Diagram:

26

Porous pot chips Glass wool soaked with ethanol

Heat

Heat

Water

[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

27

JAWAPAN SET 5 PAPER 3 SET 1 1. (a) (i)

EXPLANATION [Able to record all reading accurately with unit] Sample answer Experiment I II III

Metal X 1.70 cm 1.75 cm 1.75 cm

Metal Y 1.40 cm 1.45 cm 1.45 cm

[Able to record all reading correctly without unit] [able to record three to five reading correctly  No response or wrong response

1(a) (ii)

1.(b)

EXPLANATION [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

SCORE

3

2 1 0 SCORE

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 [Able to construct a table with at least one title / reading

2

 No response or wrong response

0

EXPLANATION [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

1

SCORE 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 [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

2

 No response or wrong response

0

1

28

1.(c)

EXPLANATION [Able to state the inference correctly] Sample answer: Material Y is harder than material X// Material X softer than material Y [Able to state the inference correctly/  Sample answer: Material Y is harder // Material X softer [Able to state an idea of inference. Sample answer: Material Y is hard// Material X is soft  No response or wrong response

SCORE 3

EXPLANATION [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.

SCORE 3

1.(d)

2

1

0

[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 1.(e)

EXPLANATION [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

SCORE 3

[able to give any two explanations ]

2

[able to give any one explanations ]

1

 No response given / wrong response

0

29

1.(f)

1.(g)

EXPLANATION [Able to state any alloy for material Y and its major pure metal for materials X correctly] Sample answer: Material X: copper // iron// any suitable metal Material Y: bronze/ brass//stainless steel// any suitable al loy for pure metal given. [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. [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  No response given / wrong response

SCORE 3

2

1

0

EXPLANATION [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. [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 [able to state the idea of hypothesis] Sample answer: Y is harder // X is softer // alloy is harder  No response given / wrong response

1.(h) EXPLANATION [Able to state all the three variables and all the three actions correctly] Sample answer:

SCORE 3

2

1

0

SCORE 3

Names of variables (i) manipulated : Type of materials / material X and Y

Action to be taken (i) the way to manipulate variable: 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] [able to state any one variablesand any two action correctly]  No response given / wrong response

2 1 0

30

2. (a) EXPLANATION [Able to state 4 inferences correctly] Test tube A B C D

SCORE 3

Inference 2+ Iron (II) /Fe ions formed / produced // iron / Fe rusted / oxidized 2+ Iron (II) /Fe ions are not formed / produced // iron / Fe does not rusted /  oxidized 2+ Iron (II) /Fe ions are not formed / produced // iron / Fe does not rusted /  oxidized 2+ Iron (II) /Fe ions formed / produced // iron / Fe rusted / oxidized

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

2 1 0

2.(b) EXPLANATION [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. [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 [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  No response given / wrong response

SCORE 3

2

1

0

2.(c) EXPLANATION SCORE 3 [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 [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

31

2.(d) EXPLANATION [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

SCORE 3

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

2 1 0

2.(e) EXPLANATION [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

SCORE 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 [able to classify all the three metals correctly] Metal that can provide sacrificial protection to iron Y Z

SCORE

Metal that cannot provide sacrificial protection to iron X

3

2 1 0

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

32

2.(g) EXPLANATION SCORE 3 [Able to compare the intensity of blue colour and relate the intensity of blue colour with the 2+ concentration of Fe accurately ] Sample answer: The intensity of blue colouration after two days is higher. The concentration of iron (II) ion is higher. 2 [Able to compare the intensity of blue colour and relate the intensity of blue colour with the 2+ concentration of Fe correctly] Sample answer: The intensity of blue colouration after two days is higher. The number of iron (II) ion is higher. [able to state an idea of the intensity of blue colour and relate the intensity of blue colour with the 1 2+ concentration of Fe correctly] Sample answer: The intensity of blue colouration after two days is higher  // The number of iron (II) ion is higher. 0  No response given / wrong response 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: 3 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 ? [Able to make a statement of the problem but less accurate//Accurate statement of the problem but  not in question form. ] Suggested answer: 2 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? [Able to state an idea of statement of the problem] Suggested answer: 1 Alcohols have different heat of combustion. 0  No response given / wrong response 3(b) KK051202 – Stating variables EXPLANATION SCORE 3 [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 2 [Able to state any two of the variables correctly] 1 [Able to state any one of the variables correctly] 0  No response given / wrong response 3 (c) KK051202 – Stating hypothesis EXPLANATION SCORE [Able to state the relationship between manipulated variable and responding variable correctly] 3 Suggested answer: When the number of carbon per molecule of alcohol increases, the heat of combustion increases. [Able to state the relationship between manipulated variable and responding variable but in reverse 2 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. [Able to state an idea of the hypothesis] Suggested answer: Alcohols have different heat of combustion.  No response given / wrong response

1

0

33

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. [Able to state an idea about the list of substances and apparatus] Suggested answer: Ethanol/propanol/butanol/water, beaker, thermometer.  No response given / wrong response

2

1

0

3(e) KK051204 – Procedures EXPLANATION [Able to state a complete experimental procedure] Suggested answer: 1. [200 cm3] of water is poured into a [copper] beaker. 2. Initial temperature of the water is recorded. 3. A spirit lamp is half filled with ethanol. 4. Weight the spirit lamp with ethanol and record the mass 5. The spirit lamp is put under the copper can and ignites the wick immediately. 6. The water is stirred and the flame is put off after the temperature has increased by 30oC. 7. The highest temperature of the water is recorded 8. Immediately weight the spirit lamp and record the mass. 9. The experiment is repeated t by replacing ethanol with propanol and butanol.

SCORE 3

[Able to state the following procedures] 1, 2, 4, 5,7,8 [Able to state the following procedures] 2, 4, 5, 7  No response given / wrong response

2 1 0

3(f) Tabulation of data EXPLANATION [Able to exhibit the tabulation of data correctly with suitable headings and units ] Types of   Initial  Highest   Initial mass of  Final mass of  o o alcohols temperature/  C  temperature/  C  spirit lamp/g spirit lamp/g  Ethanol Propanol  Butanol

SCORE

3

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

2

[Able state an idea about the tabulation of data] Alcohol Temperature

1 Mass 0

 No response given / wrong response

34

Question

1(a)

Question

1(b)

Question

1(c)

Question

1(d))

Question

PAPER 3 SET 2 Rubric Able to record the burette readings accurately with 2 decimal places. Experiment I II III 3 3 3 Initial burette 1.00 cm 13.50 cm 26.00 cm reading 3 3 3 Final burette 13.50 cm 26.00 cm 38.50 cm reading Able to record the burette reading correctly with 1 decimal place//any 5 readings correctly Able to record any 4 burette readings correctly Wrong response or no response

Score

3

2 1 0

Rubric 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 1.00 13.50 26.00 3 reading/cm Final burette 13.50 26.00 38.50 3 reading/cm Volume of acid 12.50 12.50 12.50 3 used/cm Able to construct a with correct titles and burette readings and volume of  acid used (without units) Able to construct a table with a least a title and a burette reading. Wrong response or no response

Score 3

Rubric 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 -3 Step 3: 2.0 mol dm Able to show any 2 steps correctly. Able to show any 1 step correctly. Wrong response or no response

Score 3

Rubric Able to state the operational definition for neutralization accurately. Sample answer: 3 -3 3 When 12.5 cm of hydrochloric acid 1.0 mol dm is added to 25 cm -3 sodium hydroxide 1.0 mol dm with a few drops of phenolphthalein, colourless solution turns pink. Able to state the operational definition less accurately. Sample answer: When hydrochloric acid is added to sodium hydroxide solution, the solution turns pink  Able to state the idea for neutralisation. Sample answer Acid react with alkali Wrong or no response Rubric

Score 3

2 1 0

2 1 0

2

1

0 Score

35

1(e)

Question 1(f)

Question

1(g)

Question 1(h)

Able to give the volume and explaination correctly with following aspects: 3 1. 6.25 cm 2. Sulphuric acid is a diprotic acid + 3. Concentration of H ions is double Able to give any two of the above aspects Able to give aby one of the above aspects Wrong response or no response

3

2 1 0

Rubric 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 Able to give any two variable correctly Able to give one variable correctly Wrong response or no response

Score 3

Rubric 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 Able to state the hypothesis less accurately. Sample answer; The strong acid has lower pH value // The pH value of weak acid is higher. Able to give an idea of hypothesis Sample answer Different acid has different pH value No response or wrong response

Score 3

Rubric Able to classify all the substances correctly. Sample answer: Substances with pH less than 7 Ethanoic acid Nitric acid

2 1 0

2

1

0 Score 3

Substances with pH more than 7 Ammonia solution Barium hydroxide

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

2 1 0

36

Question 2(a)

Question 2(b)

Question 2(c)

Question 2(d)(i)

Question 2(d)(ii)

Rubric Able to state the inference accurately Sample answer When alcohol react with ethanoic acid, ester is formed//Esters have sweet pleasant smell property Able to state the inference less accurately Sample answer Reaction between alcohol and ethanoic acid produced sweet pleasant smell product Able to give an idea of making an inference Sample answer Ester formed Wrong response or no response

Score 3

2

1

0

Rubric 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 Able to construct a table correctly with 2 esters named correctly Able to construct a table correctly with 1 ester named correctly Wrong response or no response

Score 3

Rubric Able to name the alcohol and carboxylic acid correctly. Alcohol: Propanol Carboxylic acid: Butanoic acid Able to name alcohol or carboxylic acid correctly Able to name any alcohol or any carboxylic Wrong response or no response

Score 3

Rubric 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 Able to state any two variable correctly Able to state any one variable correctly Wrong response or no response Rubric 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 Able to state the hypothesis but less accurately Sample answer Brown bromine water decolourised with hexene but no change with hexane. Able to state an idea of hypothesis No response or wrong response

2 1 0

2 1 0 Score 3

2 1 0 Score 3

2

1 0

37

Question 2(d)(iii)

Question 2(d)(iv)

Question 3(a)

Question 3(b)

Question 3(c)

Rubric 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 Able to state the operational definition less accurately Sample answer Alkene decolourised brown bromine water//Alkene decolourised purple colour potassium manganate(VII) solution

Score 3

Able to state an idea of operational definition Alkene decolourised bromine water//acidified potassium manganate(VII) solution No response or wrong response

1

2

0

Rubric Able to predict and make explainations accurately Answer 1. Hexene 2. Percentage of carbon atoms per molecule hexene is higher than hexane Able to give anyone of the above answer Able to give an idea of prediction/explanation Alkene//more carbon atoms

Score 3

No response or wrong response

0 Rubric

Able to state the aim accurately Sample answer To compare the effectiveness of cleaning agents A and B on cleansing action in hard water . Able to state the problem statement less accurately Sample answer Cleansing action of cleaning agent B is more effective No response or wrong response

2 1

Score 2

1 0

Rubric 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 Able to state any two variables accurately Able to state any one variable accurately No response or wrong response

Score 3

Rubric Able to state the hypothesis accurately with direction Sample answer

Score 3

2 1 0

Cleaning agent B is more effective than clening agent A in hard water Able to state the hypothesis less accurately Sample answer Cleansing action of cleaning agent B is better /more effective

2

38

Able to state an idea of hypothesis Sample answer Cleansing action in hard water No response of wrong response

1

Question Rubric  Able to state the complete list of apparatus and material as follows 3(d) Hard water, cleaning agent A and B,2 beakers, 2 pieces of cloths stained with oil, galss rod

Question 3(e)

Question 3(f)

0 Score 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 No response or wrong response

1 0

Rubric 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 . Able to state steps 2, 3, 5 and 6 Able to state steps 2 and 3 No response or wrong response

Rubric Able to tabulate the data correctly Sample answer Type of cleaning agentr Cleaning agent A Cleaning agent A Type of cleaning agentr

Score 3

2 1 0

Score 3 Observation

Observation

2

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

Marking Scheme Paper 3 Set 3 Question Rubric 1(a)(i) Able to record the thermometer reading accurately to 1 decimal place with unit. Answer o Initial temperature = 30.0 C o Highest temperature = 60.0 C Able to record the thermometer reading correctly without unit Sample answer: Initial temperature = 30

1 0

Score 3

2

39

Highest temperature = 60 Able to record one thermometer reading correctly Sample answer 30 // 60 No response or wrong answer (a)(ii)

(a)(iii)

(b)(i)

(b)(ii)

Able to state one observation accurately Sample answer Thermometer reading increases//Temperature increases Able to state the observation less accurately Sample answer Temperature reading increases/higher//Temperature rises Able to state the idea of observation Sample answer Thermometer reading change//Temperature change Wrong or no response  Able to state the inference accurately. Sample answer Heat energy is released //The reaction is exothermic reaction  Able to state the inference correctly Sample answer Heat energy is absorbed by water Able to state an idea of inference Temperature of water increases Wrong or no response  Able to state all the mass of alcohols accurately to 2 decimal places with unit  Answers 1.55g, 2.23g, 3.56g, 4.01g 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 Able to state at least 2 mass of alcohol correctly Wrong or no response 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 Able to tabulate the initial mass, final mass and mass of alcohols correctly without unit Able to tabulate at least 2 readings of the initial mass, final mass and mass of alcohols with correctly Wrong or no response

1

0 3

2

1

0 3

2

1 0 3

2

1 0 3

2 1 0

40

(c)

(d)(i)

(d)(ii)

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 -1 3. Heat of combustion = - 252 kJ mol Able to calculate with at least 2 steps correctly Able to calculate at least 1 step correctly Wrong or no response 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 Able to state any two variables correctly Able to state any one variable correctly Wrong or no response 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 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 Able to state an idea of hypothesis. Sample answer Heat of combustion is different for different alcohol

3

2 1 0 3

2 1 0 3

2

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 Able to give the heat of combustion with the following value Sample answer -1 More than 1860 kJ mol Able to give an idea to predict a value of heat of combustion: -1 more than 2350 kJ mol Wrong or no response

3

2

1 0

41

(f)

(g)

(h)

Question 2 2(a)

2(b)

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 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 Able to state an idea of operational definition Sample answer When ethanol is burnt heat energy is relaeased Wrong or no response 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 Able to state any two reasons correctly Able to state any one reason correctly Wrong or no response

Able to classify all the compounds correctly Sample answer Hydrocarbon Non-hydrocarbon Hexene Propanoic acid Methane Ethanol Able to classify at least three compounds correctly Able to classify at least two compounds correctly Wrong or no response

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. Able to state the aim of experiment less accurately Sample answer To determine type of electrode affects the product formed at the anode Wrong or no response 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 Able to state any two variables correctly Able to state any one variable Wrong or no response

3

2

1

0 3

2 1 0

3

2 1 0

2

1

0 3

2 1 0

42