Grade 12
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
2.1
Paper 2 (Chemistry)
MEMO
September 2014










B
B
A
B
A
A
D
D
C
D
H
H
H C
C
(2 x 10 = 20)
O
O
H
H
H
C
C
C
C
H
H
H
H
H


H

(3)
2.2
They are volatile and evaporate  quickly.
(1)
2.3
ethanol and butanoic acid
(2)
2.4
hydrolysis/elimination/esterification/condensation
2.5
heat and sulphuric acid (H2SO4)
3.1
Methane is a non-polar substance and is therefore insoluble in water. Methanol is polar
and contains and –OH group which allows it to make hydrogen bonds.
(4)
3.2
Methane and Methanal
3.3
As the mass of the carboxylic acid increases, the boiling point of the carboxylic acids also
increases. This is because larger molecules have more intermolecular forces than
smaller molecules.
(2)
4.1.1
Hydration
4.1.2
No because it needs a catalyst and heat to take place.
(2)
4.1.3
propan-1-ol
(2)
4.1.4
primary 
(1)
4.1.5
No , also propan-2-ol 
(2)
4.2.1
hydrohalogenation

(1)
4.2.2
no water present

(1)
(any one)




(1)
(2)
(2)
(1)
H
4.2.3
H
H
H
C
C
C
H
H
H
Br


(2)
4.3
propene

(2)
5.1.1
B = butanal

(2)
5.1.2
F = 2,2-dibromo-3,3-dimethylpentane


 

(6)
5.1.3
J = pent-1-yne
(2)
5.2
C&E

(1)
5.3
G & H or C & E
(1)
5.4.1
Condensation polymerization
H
C
C
C
H
O
C
H O C
5.4.2
(1)
H
C
H

C
H
C
O
C
C
C
C
O
H

C
H
O
H
C
H

(2)
6.1.1
sufficient energy & correct orientation
6.1.2
A substance that increases the rate of a chemical reaction without itself undergoing a
permanent change.
(2)
6.2
A: If the amount of the catalyst increases, then the rate of the reaction will also
increase. OR
B: If the catalyst is present, then the rate will increase. 

(2)
(2)
6.3.1
volume of O2 produced
(1)
6.3.2
A: amount of catalyst added OR B: presence of catalyst 
(1)
6.3.3
amount of H2O2 added OR conc of reactants
(1)
6.4
(can be either endo or exo thermic as question doesn’t specify)
Energy of
Reactants
Potenttial Energy 
Activation Energy
Energy of the
Products
Activation
Energy with
Catalyst 
Reaction Progress 
(6)
6.5
7.1.1
As the slope  of the graph increases, one can see that the hypothesis was correct. As
more catalyst is added, the rate at which O2 is produced increases.
(3)
𝐾𝑐 =
[𝐴𝐵2 ]2
[𝐴2 ][𝐵2 ]2
(numerator) (denominator)
(3)
7.1.2 i)
when equilibrium in a closed system is disturbed , the system will
re-instate a new equilibrium by favouring the reaction that will
opposethe disturbance.
(3)
ii)
This is because the forward reaction is exothermic. By removing heat, one
will favour the forward reaction, thus increasing the concentration of the
products (2AB2).
(3)
7.2
N2
1.5
-0.5
1
c=n/v
=1/0.5
2 mol.dm-3
Start
Change
Equilibrium
Concentration
2NH3
0
+1
1
=1/0.5
1 mol.dm-3
=2 mol.dm-3


[𝐴𝐵2 ]2
[𝐴2 ][𝐵2 ]2
𝐾𝑐 =
=


3H2
2
-1.5
0.5
=0.5/0.5
22
(2)(13 )
(numerator, denominator)
=2 at 470°C
(8)

7.3.1
remains the same
7.3.2
Only temperature changes Kc. the gases will stay in the same proportions with a change
in volume.
(2)
8.1
It is a strong acid because it ionizes completely in water to form a high concentration of
H3O+ ions.
(2)
8.2.1
Mg(OH)2 + 2HCl → MgCl2 + 2H2O
8.2.2
n=m/M
=
8.3
0.25
=
58
(1)

(2)
0.0043 mol Mg(OH)2 OR 4.3x10-3mol 
(3)
pH= -log[H3O+] 
1.5= -log[H3O+] 
[H3O+] = 0.032 mol.dm-3
8.4
𝑐𝑎 𝑣𝑎
𝑐𝑏 𝑣𝑣
𝑛𝑎
=
0.032∗1
109∗𝑥
𝑛𝑏
2
=1
(0.03 mol.dm-3 )
(3)


x = 1.47x10-4dm3 of Gaviscon

OR
0.03∗1
109∗𝑥
2
= 1 
= 1.38 x 10-4dm3

Or
c=n/v 
0.032=n/1
n = 0.032 mol HCl

HCl:Mg(OH)2
1:2

0.032:0.016
c=n/v
109 = 0.016/v 
v=1.47 x 10-4 dm3

(5)
9.1.1
copper

(2)
9.1.2
K+ ions will move towards the silver solution (right or cathode) to make it more neutral.
NO3- ions will move towards the copper solution (left or anode) making it more neutral.
(2)
9.1.3
flow of electrons
9.1.4
EΘcell = EΘcathode - EΘanode

(1)
=0.80 – 0.34
=0.46 V 
9.1.5
(4)
Cu(s)/Cu2+(aq) (1mol.dm-3)//Ag+(aq)(1mol.dm-3)/Ag(s) at 298K (25⁰C)




(3)
9.1.6
INCREASE
(1)
9.1.7
Primary
9.2.1
2Cl- → Cl2(g) + 2e-

(2)
9.2.2
NaOH or sodium hydroxide

(1)
9.2.3
formation of soaps; making paper; purification of bauxite; making rayon, textiles, dyestuffs,
or pharmaceuticals
(any one)

(1)
9.2.4
the membrane is cationic – meaning that cations (like Na+) can go through but not Cl-. (1)
10.1
Nitrogen

(1)
10.2
fractional distillation

(1)

(1)
10.3
4NH3 + 5O2  (in presence of Pt & 1000°C) → 4NO + 6H2O

(3)
10.4.1 ammonium sulphate

(1)
10.4.2 growth of leaves

(1)
10.5
Ostwald process

(1)
10.6
NO2 OR nitrogen dioxide

(1)
10.7
The extra fertilizer has run-off her land causing eutrophication in the dam water. By
using less fertilizer or a system of crop rotation, she can avoid this in the future.
(3)
END of PAPER