YEAR 10 CHEMISTRY MARKING SCHEME

DEPARTMENT FOR CURRICULUM,

LIFELONG LEARNING AND EMPLOYABILITY

Directorate for Learning and Assessment Programmes

Educational Assessment Unit

Track 3

Annual Examinations for Secondary Schools 2019

YEAR 10

CHEMISTRY

Item

1

1

1

a

b

c

1

d

1

1

1

1

1

1

e

f

g

h

i

j

MARKING SCHEME

Answer and guidelines

SECTION A

Marks

Rubidium.

Fluorine.

Beryllium.

Iron.

1

1

1

Accept other suitable metals.

Iodine.

Helium or Neon or Argon or Krypton or Xenon.

Nitrogen.

Magnesium.

Sulfur.

Sodium.

Total

2

2

a

b

2

c

2

d

i

2

2

2

2

d

d

d

d

ii

iii

iv

v

3

a

No reaction.

Reactivity increases down the group.

As atoms get bigger, outer electrons experience less attraction by the

nucleus and become easier to lose.

Ca(s) + 2H2O(l) ¡ú Ca(OH)2 (s) + H2 (g)

1 mark for formulae, 1 mark for balancing.

White precipitate; or gas bubbles produced; or heat generated.

Slaked lime.

Carbon dioxide, CO2.

Brick red.

Total

Very dilute NaCl: OH-, oxygen O2, H+, hydrogen H2

Conc. NaCl: Cl-, chlorine Cl2, H+, hydrogen H2

Copper(II) sulfate solution: OH-, oxygen O2, Cu2+, copper

? mark for each answer.

Chemistry ¨C Marking Scheme ¨C Year 10 ¨C Track 3 ¨C 2019

1

1

1

1

1

1

1

10

1

1

1

1

2

1

1

1

1

10

2

2

2

Page 1 of 6

Item

3

b

3

3

c

c

3

3

3

d

d

d

4

a

4

b

4

4

4

b

c

c

4

d

4

1 mark for assembly.

1 mark each for labelling charge on

electrode.

1 mark for labelling electrolyte.

a

5

b

b

Page 2 of 6

Marks

4

i

ii

Anode slowly disintegrates (gets smaller).

Purification of copper or electroplating.

1

1

2

i

Na+ + e- ¡ú Na

1 mark for balancing; 1 mark for correct addition of electrons.

Reduction.

(a) Q = It = 5 ¡Á 32 ¡Á 60 = 9600 C

1

ii

(b) 96500 C ¡Á 2 give 1 mole Cl2

9600 C

give ?

1

1

iii

(c) 0.05 ¡Á 22.4 = 1.12 dm3

Turns damp blue litmus red then bleaches it.

No marks awarded if bleaching action is not mentioned.

Total

i

ii

i

ii

e

5

5

Answer and guidelines

i

ii

9600 / (96500¡Á2) = 0.05 moles Cl2

1

1

1

20

K2SO3

2.5 moles in 1000 cm3

? moles in 100 cm3

1

1

2.5 ¡Á 100/1000 = 0.25 moles

0.25 ¡Á 158 = 39.5 g

Sulfur dioxide is denser than air.

Conc. sulfuric acid absorbs any water vapour .

Oxidising agent.

Cu is oxidised as oxidation number increases from 0 to +2 or Cu loses

electrons.

Potassium dichromate paper or solution.

Turns from orange to green.

Accept also the test using potassium permanganate.

Total

1

1

1

1

1

1

Forms dense white cloud if exposed to hydrogen chloride gas.

Ca(OH)2(s) + 2NH4Cl(s) ¡ú CaCl2(s) + 2H2O(l) + 2NH3(g)

1 mark for formulae, 1 mark for balancing.

It would react and neutralise the ammonia.

1

1

10

1

2

1

Chemistry ¨C Marking Scheme ¨C Year 10 ¨C Track 3 ¨C 2019

Item

5

b

iii

Answer and guidelines

(a) A white precipitate formed.

(b) White precipitate, insoluble in excess NaOH.

Total

Marks

1

1

6

6

a

magnesium nitrate, Mg(NO3)2

1

6

b

sodium sulfate, Na2SO4

1

6

c

sulfuric acid, H2SO4

1

6

d

copper(II) bromide, CuBr2

1

Total

4

SECTION B

7

a

i

7

a

ii

7

a

iii

7

a

iv

7

b

i

7

b

ii

Haematite.

Fe2O3(l) + 3CO(g) ¡ú 2Fe(l) + 3CO2(g)

1

1 mark for formulae.

1 mark for balancing.

1 mark for state symbols.

CaCO3 ¡ú CaO + CO2

7

b

iii

7

b

iv

7

b

v

7

c

i

7

c

ii

7

c

iii

2

1 mark for formulae.

1 mark for balancing.

Sand reacts with the CaO produced by the thermal decomposition of

CaCO3 to form slag which is removed separately.

Cryolite reduces considerably the amount of heat required to melt bauxite.

Ions must be mobile for them to be able to carry electrical energy. They

cannot do this in the solid state.

Al3+ + 3e- ¡ú Al

1 mark for formulae.

1 mark for number of electrons added.

Oxygen is released at the anodes.

This reacts with the graphite anodes to form carbon dioxide.

Aluminium is more reactive than iron so it forms more stable compounds.

Al2O3 is therefore harder to break than Fe2O3 and electrolysis is a more

aggressive process than reduction in a blast furnace.

Finely divided iron.

Accept Iron.

Moles N2 : moles NH3 = 1 : 2

20 moles N2 produce 40 moles NH3

40 moles ¡Á 22.4 dm3 = 896 dm3

P1V1/T1 = P2V2/T2

(1 ¡Á 896) / 273 = (200 ¡Á V2) / 723

V2 = (1 ¡Á 896 ¡Á 723) / (200 ¡Á 273)

= 11.86 dm3

Total

Chemistry ¨C Marking Scheme ¨C Year 10 ¨C Track 3 ¨C 2019

3

1

1

1

2

1

1

1

1

1

1

1

1

1

20

Page 3 of 6

Item

Answer and guidelines

Marks

8

8

a

b

1

1

8

c

Volumetric flask.

Yellow to red.

Any two of:

? There should not be any leakage from the burette during titration.

? Keep your eye in level with the liquid surface while taking the burette

reading or while reading the pipette or measuring flask etc.

? Do not blow through the pipette to expel the last drop of solution from

it. Simply touch the inner surface of the titration flask with the nozzle

of the pipette for this purpose.

? Shaking of the titration flask during titration should be continuous.

? Use a white tile under the flask to make colour change at end point

more clearly visible.

? Carry out a ¡®rough¡¯ titration first, followed by repeated accurate

titrations.

Accept other suitable precautions related to accuracy of procedure.

Award 1 mark for each precaution.

Na2CO3(aq) + 2HCl(aq) ¡ú 2NaCl(aq) + CO2(g) + H2O(l)

8

2

e

1 mark for formulae.

1 mark for balancing.

Moles of HCl used:

1000 cm3 HCl(aq) contain 0.25 moles

20 cm3

contain ?

8

f

i

Answer: 0.005 moles HCl

Moles of Na2CO3 reacting:

From equation: Na2CO3 : HCl = 1 : 2

So, moles of Na2CO3 reacting = 0.005/2 = 0.0025 moles Na2CO3

Moles of Na2CO3 in 1000 cm3 of solution:

Moles in 25 cm3 solution = 0.0025 moles

Moles in 1000 cm3 solution = ? Answer: 0.1 moles Na2CO3

Accept answer presented as concentration in mol dm-3.

RFM Na2CO3 = (23 ¡Á 2) + (12 ¡Á 1) + (16 ¡Á 3) = 106

8

f

Page 4 of 6

3

1 mark for formulae.

1 mark for balancing.

1 mark for state symbols.

d

CO32- + 2H+ ¡ú CO2 + H2O

8

2

ii

Mass of anhydrous Na2CO3 in 1000 cm3 solution = 0.1 moles ¡Á 106 =

10.6 g

1

1

1

1

1

1

Chemistry ¨C Marking Scheme ¨C Year 10 ¨C Track 3 ¨C 2019

Item

8

8

9

Answer and guidelines

Mass of water in 28.6g of hydrated Na2CO3 = 28.6 g ? 10.6 g = 18.0 g

RMM H2O = (1 ¡Á 2) + (16 ¡Á 1) = 18

Moles H2O in 28.6 g hydrated Na2CO3 = 18/18 = 1.0 moles

g

h

a

i

Marks

1

1

1

Ratio Na2CO3 : H2O = 0.1 : 1.0 = 1 : 10

Answer: ¡®x¡¯ = 10

Sodium carbonate is heat stable, so it will not be affected by the excess

heat once all the water is evaporated.

Sodium hydrogen carbonate can decompose on heating, so a slower

evaporation process is required.

Total

20

Magnesium; Aluminium; Zinc.

1

1

1

1 mark for each test-tube

drawn and labelled.

9

9

a

a

ii

iii

2

3Mg(s) + 2AlCl3(aq) ¡ú 3MgCl2(aq) + 2Al(s)

OR

2Al(s) + 3Zn(NO3)2(aq) ¡ú 2Al(NO3)3(aq) + 3Zn(s)

1 mark for formulae.

1 mark for balancing.

1 mark for state symbols.

3Mg + 2Al3+ ¡ú 3Mg2+ + 2Al

9

9

a

a

iv

v

3

1 mark for formulae.

1 mark for balancing.

Mg loses electrons (or oxidation state increases from 0 to +2) ¨C Oxidised

Al3+ gains electrons (or oxidation state decreases from +3 to 0) - Reduced

OR

2Al + 3Zn2+ ¡ú 2Al3+ + 3Zn

1 mark for formulae.

1 mark for balancing.

Al loses electrons (or oxidation state increases from 0 to +3) ¨C Oxidised

Zn2+ gains electrons (or oxidation state decreases from +2 to 0) - Reduced

Abrasive paper is used to remove any oxide layer which may prevent the

underlying metal to react as expected.

Chemistry ¨C Marking Scheme ¨C Year 10 ¨C Track 3 ¨C 2019

2

1

1

OR

2

1

1

1

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