Download Final - SKH Lam Kau Mow Secondary School

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S.K.H. LAM KAU MOW SECONDARY SCHOOL
FINAL EXAMINATION (12-13)
F.4 Chemistry
Marking Scheme
Section A
1
1-10
B
11-20 C
21-25 C
2
B
C
B
3
A
D
C
4
C
C
B
5
D
D
C
6
D
B
7
A
B
8
A
D
9
A
B
10
A
D
Section B
1.
(a)
Zn(s)  Zn2+(aq) + 2e
[1]
(b)
(i)
It acts as an oxidizing agent to oxidize any hydrogen produced to water.(not remove)
[1]
(ii)
Carbon powder acts as an electrical conductor in the inner part of the cell.
[1]
-
Zinc case in used cells is thinner and corroded.
[1]
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Paste in new cell is dry, but paste in used cell is moist.
[1]
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Paste in new cell is acidic but less acidic or even alkaline in used cell.
[1]
(c)
(Any two)
(d)
The ammonium ions in the paste undergo reduction to form hydrogen gas. When it covers the carbon rod,
it prevents the flow of electric circuit.
[1]
or 2NH4+(aq) + 2e  2NH3(aq) + H2(g)
The manganese(IV) oxide surrounding the carbon rod is used to oxidize any hydrogen gas produced from
ammonium ions to water.
[1]
or 2MnO2(s) + H2(g)  Mn2O3(s) + H2O(l)
However, if the zinc-carbon cells discharges rapidly, there is too much hydrogen gas produced. It cannot
be converted to water fast enough. Hence, the voltage drops.
[1]
The voltage resumes to a high value when the cell stands idle for several minutes because there is enough
time for manganese(IV) oxide to oxidze the hydrogen gas produced.
[1]
Effective communication [1]
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2.
(a)
Very dilute nitric acid only behaves as a typical acid. Therefore, copper is not reactive enough to react
with it.
[1]
On the other hand, concentrated nitric acid acts as an oxidizing agent. Therefore, it can react with both
reactive metals and unreactive metals.
(b)
[1]
Zn + 2H+  Zn2+ + H2
[1]
Oxidizing agent: H+
(c)
[1]
Cu(s) + 4H (aq) + 2NO3 (aq)  Cu (aq) + 2NO2(g) + 2H2O(l)
+

2+
(not accept dilute acid)
[1]
Reducing agent: Zn
[1]
Zinc reacts with hydrogen ions while copper does not.
[1]
From this, we can deduce that zinc is higher and copper is lower than hydrogen in the electrochemical
3.
(a)
(b)
series. (i.e. reactivity Zn> H2 > Cu)
[1]
(i)
+7
[1]
(ii)
+6
[1]
(i)
The reaction is a redox reaction.
Oxidizing agent: MnO4(aq),
(ii)
4.
(a)
[1]
reducing agent: Fe2+
The reaction is not a redox reaction.
[1]
[1]
A, B and C
[1]
(b)
D and E
[1]
(c)
A is sulphuric acid, [1] while B is nitric acid [1] and C is ethanoic acid. [1]
(d)
Solution A/ sulphuric acid is a dibasic acid.[1]
(e)
D is sodium hydroxide solution [1] and E is aqueous ammonia. [1]
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5.
(a)
X: pipette†
[1]
Y: burette†
[1]
† correct spelling
(b)
Erroneous step:
The pipette should be rinsed with the solution it is about to contain (orange juice).
[1]
The error will lead to:
(c)
The amount of sodium hydroxide solution titrated is less than the actual value.
[1]
The concentration of the acid determined will be lower than the actual value.
[1]
Taking into account of the orange colour of the juice, the apparent colour change is from orange to pinkish
orange.
[1]
(d)
To dilute the orange juice by 10 times can make the colour change of the indicator more obvious. [1]
(e)
The student’s result is likely to be higher.
[1]
Reason:
There may be acids (e.g. sulphurous acid, Vitamin C) other than citric acid in the juice which can also be
6.
(a)
(b)
(c)
(d)
(e)
neutralized by sodium hydroxide.
[1]
To ensure all aluminium sulphate has been reacted.
Ba2+(aq) + SO42(aq)  BaSO4(s)
3.20
No. of moles of BaSO4 used = 137.3  32.1  16.0  4
= 0.013 7 mol ( not accept 0.01)
No. of moles of SO42 in Al2(SO4)3 = No. of moles of BaSO4
= 0.013 7 mol
3BaCl2(aq) + Al2(SO4)3(aq)  3BaSO4(s) + 2AlCl3(aq)
Mole ratio of Al2(SO4)3 : BaSO4 = 1 : 3
[1]
[1]
0.013 7
No. of moles of Al2(SO4)3 = 3
= 0.004 57 mol
(f)
(g)
Mass of aluminium sulphate present in the spray
= 0.004 57  [27.0  2 + (32.1 + 16.0 × 4)  3] = 0.004 57  342.3
= 1.56 g
Percentage by mass of aluminium sulphate in the spray
1.56
10
= .00  100% = 15.6 %
[1]
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