Marking Scheme Paper 2

Section A Question No. 1 (a) (i)

(ii)

Explanation Pure metal : Substance X

1

Alloy

1

: Substance Y

- Pure metal has atoms/element of the same size / same type which are orderly arranged . Reject: particles / substances -Alloy has foreign atoms / two different atoms / different type of atoms / elements R: two different type of metals

(iii)

Mark

Substance Y.

Σ Mark

1 1 1

1

The foreign atoms disrupt/disturbed the orderly arrangement of the metal atoms and reduce the layers of metal atoms from sliding over each other.

(b) (i)

Polymers are large molecules/long chains made up of identical repeating units called monomers which are joined together by covalent bonds.

1

1 (ii)

(iii)

H

H

C

C

Name of monomer: monochloroethene

H

1

Cl

To make drain pipes/ artificial leather/water-proof clothes/shoes /bags //as insulator for electric wiring

1

10

1

Question No. 2(a) (i)

Explanation Melting

BOILING?

Mark 1

(ii)

Molecule

1

(b)

The heat energy absorbed by the particles is used to overcome the forces of attraction between the naphthalene molecules / particles.

1

The particles move faster

1

X : electron a: shell

1

(c) (d) (i)

Y : nucleus

1

(ii)

Electron

1

(e) (i)

W and X

1

(ii)

Has different number of neutrons but same number of protons // has different nucleon number but same proton number

Question No. 3 (a) (i) Solution I // I (ii)

(iii) (b) (i)

Σ Mark

Explanation

1+1

10

Mark

Σ Mark

1

Number of mole = M x V = 25 x 0.01 = 0.00025 1000 1000 = 0.00025 // 2.5 x 10-4 mol

1

Number of H+ = 0.00025 x 6.02 x 1023 = 1.5 x 1020 ions

1

Universal indicator // pH paper // pH meter.

1

Number of mole of NaOH = 25 x 0.1 1000 = 0.0025 mol

1

2

(ii)

HCl + NaOH  NaCl + H2O

1

From the equation, 1 mol of NaOH react with 1 mol of HCl 0.0025 mol of NaOH react with 0.0025 mol of HCl// No. of mole of HCl = 0.0025 mol Volume of HCl = 0.0025 x 1000 0.1 = 25 cm (iii)

1

1

3

Correct diagram of apparatus set-up. Correct label.

Hydrochloric acid / HCl// Solution I

Sodium hydroxide / NaOH solution

2

10

3

Question No. 4 (a)

Explanation

Mark

Electrical to chemical energy

1

Na+, Cl-, H+, OH-

1

R: NaCl  Na+ + Cl- , H2O  H+ + OH-

1

(c) (i)

Oxygen

1

(ii)

Insert/Place/Put/Bring the glowing wooden splinter into the test tube. The glowing wooden splinter will rekindles // Gas will light up the glowing wooden splinter. (If others test of gas given, not accepted)

1

(d) (i)

Chlorine gas

1 1

(ii)

- The concentration of chloride ions/Cl- is higher - Chloride ions / Cl- is selectively discharged

1

(iii)

2Cl-  Cl2 + 2e

1

(e)

Platinum

(b)

Σ Mark

1

10

[ (1-1) = 0 WCR graphite ]

4

Question No. 5(a)

Explanation

Mark

Σ Mark

Boiling points / oC

(i)

Number of carbon atoms per molecule

1. Y axis with boiling point as label and OC as unit , X axis with number of carbon atom per molecule as label. 2. Suitable and consistent scale 3. Smooth line with 6 points transferred correctly (ii)

1 1 1

The size of alkane molecules increases. // The forces of attraction between alkane molecules becomes stronger. / More energy is needed to overcome the stronger forces of attraction between molecules.

1

(iii)

Liquid

1

(b)

Correct structural formulae Correct names of the two isomers

H

H

H

C

C

H

H

H C H

Butane / n-Butane

1 1

H C

H

1

H 1

5

H C

H H H

Question No. 6 (a)

(b) (c) (i) (ii)

(iii)

(d) (i)

1

H

C

C

C

H

H

H

H

2-methylpropane

1

10

Explanation

Mark

Σ Mark

Chlorine water / Acidified Potassium Manganate (VII) solution / Acidified Potassium dichromate (VI) solution.

1

Cooking oil has no free moving ions / is a covalent compound / contain molecular only

1

green to brown

1

Oxidation reaction is a reaction where iron (II) ions release one electron to become iron (III) ions. // Oxidation is loss of electrons.

1

Sodium hydroxide solution / / potassium hexacyanoferate (II)/(III) solution / potassium thiocyanate solution.

1

Zinc sulphate,

1

(ii)

Zn

(iii)

+2 to 0

(iv)

H

+

CuSO4



ZnSO4

No. of mole of CuSO4 =

+

Cu

1 1

25 × 0.2 = 0.005 mol 1000

1 10

6

(v)

mass of ZnSO4=0.005 x 161=0.8 g//0.805 g

1

Section B Question Explanation No. 7 (a) (i) - To allow the oxygen / air to enter the crucible - for the complete combustion of magnesium (ii)

Element Mass(g) Number of moles of atom Simplest ratio of moles Empirical formula

(b)(i)

Magnesium, Mg 30.64 – 28.24 = 2.4 2.4 24 = 0.1

Mark

∑Mark

1 1

2

Oxygen, O 32.24 – 30.64 = 1.6 1.6 16 = 0.1

1

1 MgO

Diagram 7.1 Diagram 7.2 It involves heating It involves Similarity heating The reaction is The reaction is between a gas between a gas and a solid and a solid Metal oxide is Metal is formed formed Metal is reacted Metal oxide is Difference with oxygen gas reacted with hydrogen gas The mass of the The mass of the solid increases solid decreases

2 1 1 1

5

Criteria

1 1 1 1 1

Max.4

7

(ii)

-

Dilute hydrochloric acid // or any dilute acid

1

-

Zinc // or any reactive metal

1

2 1

(iii) (iv)

2HCl + Zn  ZnCl2 + H2 Precaution Dry hydrogen gas is passed through the combustion tube for a few minutes / throughout the experiment to remove all the air in the tube. During cooling, the flow of hydrogen is continued. The heating, cooling and weighing processes are repeated until a constant mass is obtained. The combustion tube must be slanted slightly towards the tiny hole.

Explanation A mixture of hydrogen and air can cause an explosion when lighted.

To ensure the oxygen from the air does not oxidise the hot copper to copper(II) oxide. To ensure that all of the copper(II) oxide has been reduced into copper. To prevent the water formed during the reaction from flowing towards the hot porcelain dish. Total marks

1

2

2

2 Max.6 2

20

8

Question Explanation No 8 (a) (i) Lead(II) chloride // Silver chloride // Mercury chloride (ii)

(b)

(c)(i)

Mark

∑Mark

1

1

Chemicals needed: Lead(II) nitrate and sodium chloride // any suitable answer

2

Name of reaction: precipitation reaction // double decomposition reaction

1

Black powder X: Copper(II) oxide

1

Blue solution Y: Copper(II) chloride

1

Cation: Copper(II) ion // Cu2+

1

Anion: Chloride ion // Cl-

1

3

4

Number of moles of lead(II) ions, Pb2+ = number of moles of lead(II) nitrate, Pb(NO3)2 5 1000 = 0.005 mol = 1.0 x

1 1

Number of moles of chromate(VI) ions, CrO42= number of moles of potassium chromate(VI), K2CrO4 5 1000 = 0.005 mol

1

1 mol of chromate(VI) ions, CrO42- reacted completely with 1 mol of lead(II) ions, Pb2+.

1

= 1.0 x

(ii)

1

4

The ionic equation for the reaction is: Pb2+ + CrO42-

PbCrO4

1

2

9

(iii)

•

•

•

- The height of precipitate formed increases for the first 4 test tubes because as the volume of potassium chromate(VI) increases, more lead(II) chromate(VI) is formed - The height of precipitate formed becomes constant when all Pb2+ have reacted completely. - colourless to yellow - Presence of chromate(VI) ions give the yellow colour to the solution // Chromate(VI) ions in the first 5 test tubes are all reacted // In the last 3 test tubes, chromate(VI) ions are in excess -To ensure the height of precipitate represents the amount of precipitate formed - because diameter of the test tubes are the same Total marks

1 1 1 1 6 1 1 20

10

Section C Questio n No 9 (a)

(i)

Explanation

Compound formed between X and Y Ionic bond is formed because X atom donates electrons and Y atom receives electrons to achieve stable electron arrangement // X is metal and Y is non-metal

Types of chemical bonds

(ii) Boiling point and melting point

High because a lot of energy needed to overcome the strong electrostatic forces between ions

Mar k

Molecule formed between Z and Y Covalent bond is formed because Z and Y atoms share the electrons to achieve stable electron arrangement // Y and Z are nonmetal Low because less energy is needed to overcome the weak forces of attraction between molecules

(b)

X

X

X X

X X

X

2

4

2-

X

X

X X X

2

1 1

Correct electron arrangement of 2 ions Correct charges and nuclei are shown 2+ XX XX X

∑Mar k

X

X XX

X2+

X

X X

X X

X X

X X

X X XX

X X

X

X

Y2-

11

-

-

-

X atom with an electron arrangement of 2.8.2 donates 2 valence electrons to achieve the stable octet electron arrangement, 2.8. X2+ ion is formed // X X2+ + 2eY atom with an electron arrangement of 2.6 accept 2 electrons to achieve the stable octet electron arrangement, 2.8. Y2- ion is formed // Y + 2eY2The oppositely-charged ions, X2+ and Y2- are attracted to each other by a strong electrostatic force. An ionic compound XY is formed

1

1 1 1

6

1 (c)

Battery + − +

−

1 1 з

Bulb

1 1

Carbon (Anode) Crucible

Carbon (Cathode)

1 P / naphthalene

Clay pipe triangle Tripod stand

1

Bunsen burner

1 1 1

10

1. A crucible is filled with solid P until it is half full. 2. Two carbon electrodes are dipped in the solid P and connected to the batteries. 3. Switch is turned on and observation is recorded. 4. The solid P is then heated until it melts completely. 5. The switch is turned on again and observation is recorded. 6. Steps 1 to 5 are repeated using solid Q to replace solid P. 7. Observations: Compound Observation P P does not light up the bulb in 12

both solid and molten states. lights up the bulb in molten state only.

Q

P does not light up the bulb in both solid and molten states. Q lights up the bulb in molten state only. P: naphthalene // any suitable answer Q: lead(II) bromide // any suitable answer

Question No. 10 (a)

Explanation

Mark

CH4 + 2O2  CO2 + 2H2O

1

RMM of methane, CH4 = 12 + 4 = 16 16 g of methane releases 898 kJ heat

1

Σ Mark

No. of mol CH4 = 1 / 16 = 0.0625 mol Thus 1 g of methane releases = 898 x 1 /16 kJ g-1 = 56.125 kJ g-1 (Correct answer and unit) (b)

Item

Diagram 10.1

1 1

4

Diagram 10.2

13

Temperatur e Total Energy content

Heat

Increase

Decrease

A and B / reactants have higher energy content than its products / (C and D) Released

R and S / products have higher energy content than its reactants / (P and Q)

2 2

2

6

1 1 1 1

4

Absorbed

(c) Plotted graph 1. correct axis and labelled 2. all 4 points/values transferred correctly 3. straight line graph 4. Consistent scale -1

Heat of combustion of alcohols (kJ mol )

Number of carbon atoms

Heat of combustion increases from

1

14

methanol → ethanol → propanol → butan-1-ol// methanol, ethanol, propanol, butan-1-ol Heat of combustion increases because : 1. Number of carbon atom per molecule / mole increases from methanol to butanol.

1

2. Number of hydrogen atoms per molecule / mole increases from methanol to butanol.

1

3. More carbon dioxide / CO2 and water / H2O molecules are formed.

1

4. More covalent bonds are formed in the products. 5. More heat is given out.

1

Total marks

1

6

20

END OF MARKING SCHEME

15