MEMO: GRADE 12 MOCK MATRIC SEPTEMBER 2012
Question One:
1.1
catalyst
1.2
endothermic
1.3
alkyl-chlorides/halides or halo-alkanes
1.4
electrolyte
1.5
eutrophication
5
Question Two:
2.1 C
2.2 C
2.3 B
2.4 D 2.5 D 2.6 B
2.7 A 2.8 A 2.9 A 2.10 C x 10 = 20
Question Three:
3.1
Einput
= 2 x 3 x (C-C) + 2 x 10 x (C-H) + 13 x (O=O)
= 6 x 348 + 20 x 412 + 13 x 496
= 16 776kJ
3.2
Eoutput
4
= 8 x 2 x (C=O) + 10 x 2 x (O-H)
= 16 x 743 + 20 x 463
= 21 148kJ
3.3
∆E
3
= 16 776 – 21 148
= -4 372kJ
Since ∆E is negative – reaction is exothermic
2
Any correctly explained answer without the calculation may also score 2
Question Four:
4.1.1
The rate at which reactants are used up or products are produced.
2
4.1.2
nCaCO3
3
= 2g/(40 + 12 + 48) = 0,02 moles of CaCO3
4.1.3 According to the mole ratio of 2:1 HCl to CaCO3 the number of moles of HCl required to
react with the CaCO3 is 2 x 0,02moles or 0,04 moles. Since we have 0,3 moles of HCl - it is in
excess and CaCO3 is the limiting reagent.
2
4.2.1
On graph paper.
5
4.2.2
Aprox 0,6 grams (mark off graph allowing some leeway)
4.2.3
Since only 0,04 moles of HCl react according to the balanced chemical equation, its
concentration only drops to 0,026 mol•dm-3 which is still close to 0,3 mol•dm-3. 2
4.2.4
The graph flattens out as the CaCO3 gets used up. / Slope gets less / any reasonable
description and explanation gets 2.
2
4.2.5
See graph.
4.2.6
Volume of acid used / Temperature / state of division / each x 2
4.2.7
Greater concentration gives greater number of collisions between particles this results
in more effective collisions per unit time and this gives rise to an increase in the rate
of reaction.
3
2
2
Question Five:
[NO]2/[N2]•[O2]
5.1.1
kc =
2
5.1.2
The equilibrium lies almost completely to the left of this equation: i.e. very little product
is formed and almost all the reactants remain unreacted.
2
5.2.1
For any closed system which is at equilibrium, a disturbance to that equilibrium will
result in the system reacting in such a way as to overcome that disturbance. 2
5.2.2
kc = (2 x 10-6)2 / (2 x 10-2) x (2.5 x 10-3) = 8 x 10-8
3
5.3.1* YES - NO is a greenhouse gas and by removing NO from exhaust fumes we are
removing a potentially harmful gas from the environment.
3
5.3.2
Replace fossil fuel driven cars with electric cars / hydrogen driven cars / steam driven
cars etc. (Any reasonable suggestion)
2
*
If the answer to 5.3.1 is NO then they must give a plausible reason as to why the device
is not cost effective.
Question Six:
6.1.1
Mn3+ because it has the most positive / highest E value
6.1.2
Co2+
6.1.3
Co2+ → Co 3+ and Mn3+ → Mn2+
6.1.4
Ecell = Ecathode - Eanode
2
1
If they choose the wrong equations then mark
2
positively in 6.1.4
= 1,49 – (-0,28)
= 1,77V
4
6.2.1
Fe has a greater oxidizing potential than Cu and so will be oxidized in preference to
Cu. (boys may say that Fe lies above Cu in the table 4B and so will act as the
anode with respect to Cu. )
3
6.2.2
It completes the circuit by allowing ions to flow in the solutions.
6.2.3
The damp soil which would naturally carry some salts and hence form an electrolyte. 1
1
Question Seven:
7.1.1
A homologous series is one which is characterized by a general formula.
2
7.1.2
Unsaturated hydrocarbons contain at least one double or triple bond.
2
7.1.3
a)
alkyl-halides or halo-alkanes.
1
b)
carboxylic acids
1
c)
ketones
1
d)
alcohols / alkanols.
1
2
7.1.4
(boys will need to put in the C-H bonds – I have
left them out because I couldn’t find good pics!)
7.1.5
When marking assume 2 marks and subtract for each mistake – 2 mistakes – no marks!
2
7.1.6
pent-1-ene
2
7.1.7
1-chlorobutane
2
7.2.1
A molecule with the same molecular formula as another but with a different structural
formula.
2
7.2.2
propan-2-ol or 2-propanol.
7.2.3
Heated alcohol reacted with hot conc H2SO4 or Heated alcohol passed over a
catalyst - Al2O3
7.2.4
1
Dehydration/ elimination of water/ dehydrolysis
2
1
7.2.5
2-bromopropane
7.3.1
CH3COOH + C4H9OH ⇌ CH3COOC4H9 + H2O
7.3.2
7.3.3
4
3
2
Concentrated sulphuric acid. / H2SO4(conc) if formula given must have (conc)
1
Question Eight:
8.1.1
Electrolytic cells convert electrical energy into chemical change/ chemical potential
energy
2
8.1.2
Concentrated sodium chloride solution / salt solution.
8.1.3
Chlorine gas – used in plastics and the paper /pulp industry / insecticides.
2
Sodium hydroxide – soap manufacture / bleaching / tanning industry
2
8.2.1
2Cl- → Cl2 + 2e-
2
8.2.2
Mercury is a poisonous metal which can leak into the environment and poison
ecological systems.
3
1
Question Nine:
9.1.1
Nitrogen
1
9.1.2
1:
The production of H2 which is a primary feedstock for the process is generally
produced from fossil fuels.
2:
Fossil fuels are used to heat the gasses prior to reaction in the Haber process.
Or used in the generation of electricity to power the plant
2
9.1.3
N2 + 3H2 ⇌ 2NH3
9.1.4
Increased yields of crops per hectare of land / quicker crop rotation in fields without
them having to lie fallow / better quality of crops as a result of better nutrients. 1
9.1.5
It could poison underground water resources which may be scarce in farming areas. It
could leak into natural water supply and result in eutrophication / algal bloom / blue
baby syndrome etc.
2
(-1 for each mistake)
3
Question Ten:
10.1.1 +2
2
10.1.2 Pb(s) + PbO2(s) + 2H2SO4 ⇌ 2PbSO4 + 2H2O
3
10.1.3 Pb - it is the only element oxidized in the process.
2
10.2.1 2H2O ⇌ O2 + 4H+ + 4e-
2
10.2.2 If both the H2 and the O2 are produced at the time of charging they could potentially
form an explosive mixture which could be detonated by a spark.
2
Graph for Question Four:
0.8
Mass lost in grams
0.7
4.2.2
0.6
4.2.1
0.5
4.2.5
0.4
0.3
0.2
0.1
0
0
100
200
300
Time in seconds
400
500
0
