SKEMA PEMARKAHAN KERTAS KIMIA 2 PERCUBAAN SPM 2014

Question Description Marks

1 (a) (i) [Able to give a definition correctly]

The temperature at which solid turns to liquid 1......1

(ii) [Able to state the boiling point correctly]

136 0C 1 ......1

(iii) [Able to explain the reasons correctly]

Heat energy is absorbed and

used to overcome the force of attraction between particles

// Heat is absorbed to overcome the intermolecular force between molecule

r: ion/atom

1

1......2

(b) (i) [Able to state the process correctly ]

Diffusion

1….1

(ii) [Able to state the type of particle correctly ]

Ion1

....1

(iii) [Able to explain the observations correctly]

- Potassium manganate(VII)is made up of tiny particles- The spaces between particles in gel are smaller than in water//vice versa- Potassium manganate (VII) particles diffuse slower in gel // vice versa

1

1

1….3

9

No2

Explanation Mark ∑ Mark

(a) Able to state the type of compound correctlySample answer: Ionic / salt / inorganic

1 1

(b) Able to write the formula correctlySample Answer:XY2 / MgCl2

1 1

(c)(i) Able to write the electron arrangement correctlyAnswer:2. 8 . 2 / 2 , 8 , 2

1

(ii) 2 . 7 / 2 , 7 1 2

(d)(i) Able to show:1. nucleus for both atoms & all shells filled with correct number of electrons.2. sharing one pair of electrons

1

1

2(ii) Able to state the bond correctly

Answer:Covalent

1 1

(e) Able to compare both melting points correctly.Sample answer:Compound/(a) is higher than substance/(d)

Able to give reasons correctly

Sample Answer:1.(Electrostatic) forces between particles/ ions is stronger.2. more heat/energy required to overcome the forces

1

11 3

Total 10

YY

Question No Mark Scheme Sub Mark

Total Mark

3 (a) HCl 1 1

(b) (i) No changes // blue litmus paper remain unchanged 1 1

(ii) Consist of molecule // no hydrogen ion / H+ 1 1

(c) (i) 1. Helps acid to ionizes / dissociate 2. Produce hydrogen ion / H+ // HCl H+ + Cl-

[ award pt 2 if write correct equation]

1

1 2

(ii) 2H+ + CO32- CO2 + H2O

1. Correct reactants and products 2. Balanced equation

11 2

(iii) 1. Effervescence // bubbles of gas2. `Pop` sound

11 2

(iv) Acid only shows its acidic properties when dissolve in water // H+/ hydrogen ions are responsible for acid to show their properties.

1 1

Total 10

No 4 Explanation Mark ∑ mark

(a)(i) Able to state the energy conversion

Answer

Chemical Electrical 1 1

(ii) Able to state the negative terminal & reason

Answer

Magnesium / Mg

Magnesium / Mg is more electropositive // higher position in Electrochemical Series // higher tendency to release electron

1

1 2

(iii) Able to write the half equation

Answer

Mg Mg 2 + + 2e 1 1

(iv) Able to state the observation

Sample answer

Brown solid deposited // copper electrode becomes thicker 1 1

(b)(i) Able to state the ions present

Answer

Na +, Cl - , H + , OH - 1 1

(ii) Able to name the gas collected

Answer

Chlorine 1 1

(iii) Able to calculate the volume of gas collected

1. No. of mole of gas = 12 / 24 000 // 0.0005

2. Mass of gas = 0.0005 x 71 g

= 0.0355 g

1

1

1

3

Total 10

Question No Mark Scheme Sub Mark

Total Mark

5 (a) C3H8O [ r : C3H7OH ]

1 1

(b) 1. No. of mole of carbon dioxide = 0.1 X 3 // 0.3

2. Volume of carbon dioxide = 0.3 x 24 dm3 // 7.2 dm3 // 7200 cm3

1

1 2

(c) (i) Compound that contain carbon and hydrogen only 1 1

(ii)

water1. Functional diagram 2. Label : water

11 2

(d) (i) Oxidation 1 1

(ii) Propanoic acid

[ r : formula ]

1 1

(e) (i) H O H H H | || | | | H – C – C – O – C – C – C – H | | | | H H H H

[ a: CH3COOC3H7 ]

1 1

(ii) To increase rate of reaction // As a catalyst 1 1

Total 10

SOALAN 6

6 a The amount of heat energy released when one mol of precipitate formed from its ion

1

b(i) Heat released= mcQ

= 100 x 4.2 x 3.5 show working ...1 m = 14700 J

11.....2

ii No of moles of Ag+ = 0.5 x 50 1000

= 0.025 mol

No of moles of Cl- = 0.5 x 50 1000

= 0.025 mol

1

1............2

iii Heat of precipitation = 14700 J 0.025 = - 588 kJ/mol

1

1.........2

c Energy Ag+ + Cl-

AgCl

Label energy,anak panah .........................12 aras yg betul utk exothermic rection .....................13 formula bahan dan hasil yang betul 3

d Ag+ + Cl- AgCl 1

11

BAHAGIAN B

7 (a) 1.Smaller pieces of charcoal has larger/bigger total surface total area

2.Smaller pieces of charcoal is easier to burn when exposed to oxygen

3.More heat is produced by smaller pieces of charcoal than big pieces

4.More heat is absorbed by the food

1

1

1

1……… 4

(b) (i) ) 40/160 // 0.25 cm3s -1 1

(ii) ) Zn + H2SO4 ZnSO4 + H2

1. Correct formula od reactants2. Correct formula of products3. Mol of H2SO4 = 0.5 X 50/1000 // 0.025

From equation, 1 mol of H2SO4 1 mol of H2

4. If 0.025 mol of H2SO4 0.025 mol of H2

5. Volume of H2 = 0.025 x 24 dm3 //0.6 dm3 //

0.025 x 24000//600 cm3

1+1

1

1

1

5

(iii)

Expt I and II

1.Rate of reaction of expt I is higher

2.The size of zinc in Expt I is smaller

3.Total surface area of zinc in Expt I is bigger/larger

4.The frequency of collision between zinc atom and hydrogen ion/H + in Expt I is higher

5. The frequency of effective collision between particles in Exp I is higher

Expt II and III

1. Rate of reaction in Expt II is higher

2.The concentration of sulphuric acid in Exp II is higher

3. The no. of H+ per unit volume in Expt II is higher/greater in Expt II// the concentration of hydrogen ion in Expt II is higher

4. The frequency of collision between zinc atom and H+ in Expt II is higher

5. The frequency of effective collision in Expt II is higher

1

1

1

1

1

1

1

1

1

1…… 10

Total 20

8 (a) (i) A mixture of two or more elements with a certain fixed composition in which the major component is a metal.

11 2

(ii) 1. Improve the appearance2. Improve the strength and hardness3. Increase the resistance to corrosion

[Any two corrections]

1+1

2

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

(ii) 1. Pure copper is made up of same type of atoms and are of the same size.

2. The atoms are arranged in an orderly manner.3. The layer of atoms can slide over each other.4. Bronze is made up of atoms of different size//

In bronze, tin atoms and copper atoms are of different size.5. The atoms are not orderly arranged// The presence of tin atoms

disturb the orderly arrangement of copper atoms.6. This reduces/prevents the layer of copper atoms from sliding.

1

11

1

11Max5

(iii) Pure copper:

[minimum 3 3 layers ]

Bronze:

1

1+1

3(c) (i) Sulphur trioxide is dissolved in concentrated sulphuric acid to form oleum.

Oleum is diluted with water to produce sulphuric acid.11

(ii) SO3 + H2SO4 H2S2O7

H2S2O7 + H2O 2H2SO4

11

(iii) Moles of S = moles of sulphur = 48 / 32 =1.5Volume of SO2 = 1.5 24 dm3

= 36 dm3

1

1 7

Total 20

Tin atom

Copper atom

Copper atom

9 (a (i)) Oxidation number of Aluminium : +3

Oxidation number of Copper : +1

1

1 2

(ii) Aluminium oxide

Copper (I) Oxide

1

1 2

(iii) Oxidation number is not written as Al has only one oxidation number.

Oxidation number of copper is written as Copper has more than one oxidation number

1

1 2

1. Example :

CuSO4 + Zn ZnSO4 + Cu

2. Copper ion undergoes reduction

3. Oxidation number of copper decreases from +2 to 0.

4. Zinc atom undergoes oxidation5. Oxidation number of zinc increases from

0 to +2

6. NaOH + HCl NaCl + H2O

7. no change in oxidation number8. The reaction is not a redox reaction

1

1

1

1

1

1

1

1 …..8

(a) (i) 1. potassium hexacyanoferrat (III)- to detect Fe2+ ion

2. Phenolfthalein - to detect OH-

1

1

2

(ii) Fe → Fe2+ + 2e 1 1

(iii) Test tube B.

Iron/Ferum is more electropositive than P.

1

1 2

(iv) Q, Fe, P 1

Total 20

10 (a)(i) Lead(II) nitrate: Acid reacts with base/ metal oxide/ metal

hydroxide/ metal carbonate/ metal

Lead(II) sulphate: Precipitation reaction

1

1

(a)(ii) Lead(II) nitrate solution and [any suitable solution which

contains sulphate ion, SO42-]

1+1

(b) 1. Pour [20 -100] cm3 nitric acid to a beaker

2. Add PbO/PbCO3 in small portion

3. Stirring after each addition

4. Until some solid remains unreacted

5. Filter off excess

6. Evaporate until a 1/3 of volume of solution.

7. Leave to cool until crystal form

8 .Filter and dried between filter paper

9.+ 10 . PbO + 2HNO3 Pb(NO3)2 + H2O

PbCO3 + 2HNO3 Pb(NO3)2 + CO2 + H2O

1

1

1

1

1

1

1

1

1+1

(c) 1. Add barium chloride solution to each solution

2. Sodium sulphate and sulphuric acid produce a white

precipitate

3. The one that does not produce a precipitate is hydrochloric

acid

4. Add sodium carbonate to the two remaining solutions

5. Sulphuric acid produces gas bubbles

6. Sodium sulphate does not produce gas bubbles

Note: There are other methods that can be used here.

1

1

1

1

1

1

Total 20