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S.K.H. LAM KAU MOW SECONDARY SCHOOL
FINAL EXAMINATION (2013-2014)
F.4 Chemistry
Marking Scheme
Section A
1
2
3
4
5
6
7
8
9
10
0+
B
D
C
C
D
A
C
D
D
C
10+
B
B
A
C
A
D
B
-
A
C
20+
A
B
C
D
D
Section B
1.
(a)
[2]
(b)
Relative molecular mass of NH3 = 17
No. of mole of NH3 used = 8500/17 = 500
[1]
Max. no. of mole of HCN produced = 500
[1]
Relative molecular mass of HCN = 27
(c)
Max. mass of HCN produced = 500 x 27 = 13500g or 13.5 kg
[1]
Hydrocyanic acid is only slightly ionised in aqueous solution
[1]
+
and the H concentration of hydrocyanic acid is low.
[1]
Hence, hydrocyanic acid is a weak acid and has a higher pH value as compared to nitric
acid. On the hand, nitric acid is almost completely ionised in aqueous solution and the H+
concentration of nitric acid is high.
[1]
Hence, nitric acid is a strong acid and has a low pH value.
2.
(a)
(b)
Burette†
The pH meter is more sensitive than indicators./ gives accurate pH value
[1]
[1]
(c) (i) Beaker: aqueous sodium hydroxide;
[1]
X/Burette: dilute hydrochloric acid.
[1]
(ii) Aqueous sodium chloride / NaCl solution
[1]
(salt solution or water not accepted)
(† correct spelling required)
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(d)
(i)
Beaker: dilute ethanedioic acid;
[1]
X/Burette: aqueous sodium hydroxide.
[1]
(ii) Ethanedioic acid is a dibasic acid./ There are two dissociable hydrogen atoms in each
ethanedioic molecule.
[1]
The reaction between ethanedioic acid and sodium hydroxide takes place in two stages, as
shown.
[1]
HOOC-COOH + NaOH  HOOC-COONa + H2O
and then HOOC-COONa + NaOH  NaOOC-COONa + H2O
or /It forms acid salt & normal salt during neutralization.
/[1]
Thus, the pH curves show two end points which correspond to the two stages.
3.
+ 3 HCl  AlCl3 + 3 H2O
(a)
Al(OH)3
(b)
no. of mole of Al(OH)3 required per day = 0.24/3 =0.08
[1]
[1]
Formula mass of Al(OH)3 = 27 + (16+1)x3 = 78
Mass of Al(OH)3 required = 0.08 x 78 = 6.24g
Volume of medication required = 6.24/2.6x100 =240 cm
4.
[1]
3
[1]
(c)
CaCO3 + 2 HCl  CaCl2 + CO2 + H2O
[1]
(a)
FeO42-
[1]
(b)
Let the oxidation no. of iron be x
2 + x + 4(-2) = 0
x = +6
(c)
with ‘+’ sign [1]
Yes, a reaction will take place.
Potassium iodide is a reducing agent
[1]
and it will reduce the iron (VI) ions present in BaFeO4 to either iron(III) or iron(II) ions,
(the more stable oxidation states of iron.)
5.
(a)
[1]
Firstly, recycling saves these resources. (The amount of metal ores which can be mined is
limited.)
[1]
Secondly, recycling saves energy/fuel because the energy required to extract metals from
ore is higher than the energy required to melt the metals.
(† correct spelling required)
[1]
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or/environmental impact can be reduced because the metal waste and air pollutants are
reduced
/[1]
{no mark for just stating e.g. ‘environmental friendly’ }
(b)
The steel and aluminium cans may be separated by large electro-magnets.
[1]
The electro-magnets only attract steel which is magnetic but not aluminium which is
non-magnetic.
(c)
[1]
The thin layer of tin protects the steel from rusting.
[1]
6.
(a)
Fe
O
Mass of element /g
0.69
0.952-0.69 = 0.262
No. of moles
0.69/55.8=0.0124
0.262/16=0.0164
Simplest ratio
1
1.32
Simplest whole no. ratio
3
4
Therefore, the empirical formula of compound is Fe3O4
(b)
Fe3O4 + 4 CO  3 Fe + 4 CO2
(c)
Let the average oxidation no. of Fe be x
[1]
[1]
[1]
3 (x) + 4 (-2) = 0
with ‘+’ sign [1]
x = + 2.666
7.
(a)
Distilled water is considered a non-conductor of electricity, and is unsuitable as an
electrolyte.
[1]
Distilled water contains only negligible amounts of H + and OH- ions. There are no (other)
ions present.
[1]
(b) (i)
Most
Least
reactive
reactive
C
B
E
A
D
[2]
(if show B/C > E or E > A/D max. [1])
(ii)
Metals C and D
(† correct spelling required)
[1]
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8.
9.
Nitrogen disproportionates
[1]
from +4 in NO2 to +5 in HNO3 and +3 in HNO2
[1]+[1]
(a)
Water should be added into the beaker to create an aqueous solution.
(b) (i)
[1]
At electrode 1: Cu(s)  Cu2+(aq) + 2 e-
with correct state symbols [1]
At electrode 2: Cu2+(aq) + 2 e-  Cu(s)
with correct state symbols [1]
(ii)
copper electrode 1
21.5
18.5
copper
electrode 2
[1]+[1]
(iii) The colour of the solution remains blue as the conc./amount of Cu2+ remains unchanged.
[1]+[1]
(† correct spelling required)