F.6 / Chemistry I/16-17 Ans/ Page 1 Of8

F.6 / Chemistry I/16-17 Ans/ Page 1 of8

S.K.H.LAMKAUMOWSECONDARY SCHOOL

F.6 MOCK EXAMINATION (16-17)

CHEMISTRY (Paper I)

Marking Scheme

Section A

1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10
1-10 / C / B / C / D / D / D / B / D / A / C
11-20 / D / A / B / A / B / A / B / D / C / B
21-30 / C / A / C / D / A / B / C / C / D / B
31-36 / A / A / A / D / C / C

Section B

Part I

1.i)

(1 mark for showing the diatomic molecules of iodine; 1 mark for showing the regular arrangement of the molecules) (2)

ii)

There are weak van der Waals’forces between iodine molecules. (1)

In sand, the attractive forces between atoms are strong covalent bonds.(1)

Much less heat is required to separate the iodine molecules. Hence the boiling point of iodine is much lower.

iii)

Purple vapour is given off.(1)

Dark solid is formed at the bottom of the evaporating dish.(1)

2a)

i)Halogens(1)

ii)

An atom of a needs one more electron to obtain the stable electronic arrangement of a neon atom. (1)

Each atom of a can obtain the electronic arrangement of a neon atom by sharing one of its outermost shell electrons with another atom. (1)

iii)

y = 18z = 7(1)

b) i)

(1)

ii)

(1)

iii)

P has a higher melting point.(1)

P has a giant ionic structure and the ions are held by strong ionic bonds.(1)

Q has a simple molecular structure and the molecules are held by weak van der Waals’forces. (1)

More heat is required to break the strong ionic bonds between the ions in P. Hence the melting point of P is higher.

3. a)

2CuO(s) + C(s)  2Cu(s) + CO2(g)(1)

A reddish brown powder would be observed.(1)

c) i)

Number of moles of CuO

= 0.120 mol(1)

According to equation, 2 moles of CuO produce 2 moles of Cu.

∴number of moles of Cu produced

= 0.120 mol

Theoretical yield of Cu

= number of moles of Cu x molar mass of Cu

= 0.120 mol x 63.5 g mol–1

= 7.62 g(1)

ii)

Mass of Cu produced

= 7.62 g x 82.0%

= 6.25 g(1)

The position of magnesium is high in the reactivity series. / Magnesium oxide is stable. (1)

Hence carbon is not strong enough to reduce magnesium oxide.(1)

a)A monobasic acid is an acid that can produce one hydrogen ion per molecule.(1)

b)pH of solution= –log10 (1.23 x 10–3)(1)

= 2.91(1)

(1 mark for showing equivalence point at 25 cm3; 1 mark for curve finishing between pH 12 – 13, extending to 50 cm3) (2)

d)Phenolphthalein is a suitable indicator(1)

because the indicator changes colour within the pH range of the vertical part of the titration curve. (1)

5. a)i)Ag+(aq) + e– Ag(s)(1)

ii)Tl(s) + Ag+(aq)  Tl+(aq) + Ag(s) (1)

iii)Thallium(1)

Oxidation occurs at the thallium electrode.(1)

iv)From the thallium electrode to the silver electrode.(1)

b)Pt and Cd(1)

They are furthest apart in the electrochemical series.(1)

c)No reaction would occur.(1)

The position of Ag in the electrochemical series is higher than that of Pt, i.e. Ag is more reactive than Pt. (1)

Hence Pt cannot displace Ag from AgNO3(aq).

a)i)It is produced by the incomplete combustion of petrol in the car engine.(1)

ii)It is poisonous. / It reduces the oxygen carrying capacity of blood and thus leads to dizziness and even death. (1)

b)At high temperatures, nitrogen and oxygen in the air combine to give oxides of nitrogen inside the car engine. (1)

c)i)Irritate the respiratory system.(1)

ii)Reduce visibility.(1)

d)i)Hydrocarbons(1)

ii) Cause cancer(1)

*e)Sustainability

‧Crops that produce oil can be grown in most places.(1)

‧Biodiesel can be made from many vegetable oils.It is renewable.(1)

‧Use less fossil fuels.(1)

Low pollution

‧Emission of pollutants is lower in amount.(1)

‧The photosynthesis of plants uses carbon dioxide from the atmosphere.
Biodiesel releases carbon dioxide when burnt.
Hence biodiesel is carbon neutral. Less new carbon dioxide is produced. (1)

Effective communication(1)

(Chemical knowledge = 0 to 2: communication mark = 0

Chemical knowledge = 3 to5: communication mark = 0 or 1

incomplete answer / difficult to understand, communication mark = 0)

a)The standard enthalpy change of combustion of a substance is the enthalpy change when one mole of the substance (1)
is completely burnt in oxygen under standard conditions.(1)

(1)

ΔHO c[C2H4(g)]= ΔHO r + ΔHO c[C2H5OH(l)]

ΔHO r= ΔHO c[C2H4(g)] –ΔHO c[C2H5OH(l)]

= [(–1 411) – (–1 368)] kJ mol–1(1)

= –43 kJ mol–1(1)

∴the standard enthalpy change of the reaction is –43 kJ mol–1.

Part II

a)Number of moles of Br2

= 0.0200 mol(1)

According to the equation, 1 mole of Br2 reacts with methanoic acid to produce 1 mole of CO2.

i.e. number of moles of CO2 produced

= 0.0200 mol

Volume of CO2 (at room temperature and pressure)

= 0.0200 mol x 24.0 dm3 mol–1

= 0.480 dm3(1)

b)Measure the absorbance of (Br2)the reaction mixture using a colorimeter at regular time intervals. (1) +(1)
From a calibration curve, read off concentrations of bromine at regular time intervals according to the absorbance recorded.

Measure the volume change of (CO2)the reaction mixture using agas syringe at regular time intervals. (1) +(1)
or

Measure the change in mass of (CO2)the reaction mixture using an electronic balance at regular time intervals. (1)+(1)

c)i)Instantaneous rate of consumption of bromine

= – (1)

= 8.25 x 10–6 mol dm–3 s–1(1)

ii)Instantaneous rate of formation of H+(aq) ions

= 2 xInstantaneous rate of consumption of bromine

= 2 x8.25 x 10–6 mol dm–3 s–1

= 1.65 x 10–5 mol dm–3 s–1(1)

a)Kc= (1)

b)Initial concentration of Fe3+(aq) ions

= mol dm–3

= 6.00 x 10–3mol dm–3(1)

c)1.00 x 10–3 mol dm–3(1)

e)The forward reaction is exothermic.(1)

When the temperature is increased, the concentrations of Fe3+(aq) ions and SCN–(aq) ions increase while the concentration of [Fe(SCN)]2+(aq) ions decreases. (1)

It can be deduced that when the temperature is increased, the system will undergo a net backward reaction so as to lower the temperature. (1)

Thus, the backward reaction should be an endothermic reaction.

10.

11.

Property / Compounds
(i) CANNOT be oxidized by acidified potassium dichromate solution / BE / (1)
(ii) are chiral / AD / (1)

b)CH3CH2CH2CH2OH (1)
butan-1-ol(1)

12.

Similarities

•Both main group metals and transition metals are metals. They are both lustrous, good conductors of electricity and heat, malleable and ductile. (1)

•Both main group metals and transition metals form oxides when heated in air.(1)
Calcium forms calcium oxide while vanadium forms vanadium(V) oxide / vanadium(IV) oxide / vanadium(III) oxide.

Differences

•Main group metals exhibit only one oxidation state while transition metals exhibit variable oxidation states.
Calcium exhibits only one oxidation state (+2) while vanadium exhibits variable oxidation states (e.g. +1, +2, +3, +4 and +5). (1)

•Main group metals form colourless compounds while transition metals form coloured compounds.
Calcium oxide is not coloured while vanadium(V) oxide is orange in colour (comparing the colours of calcium ions and vanadium ions are also acceptable). (1)

•Main group metals do not demonstrate catalytic activity while transition metals do.
Calcium is not used as catalyst while vanadium(V) oxide is used as a catalyst in the Contact process. (1)

Effective communication(1)

(Chemical knowledge = 0 to 2: communication mark = 0

Chemical knowledge = 3 to5: communication mark = 0 or 1

without examples/incomplete answer / difficult to understand, communication mark = 0)

 correct spellings required