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CROSS COUNTRY EXAM 2014
CHEMISTRY 233/2
MARKING SCHEME
MAY/JUNE 2014
1.
(a) Halogen
(b) L-2.8.8
P- 2.8.8
(c) Hydrogen A
- weather balloon
- Hydrogenatation
- Rocket fuel
- Welding /ox hydrogen
- Manufacture of ammonia (any 1mk)
M- Chlorine
 Main bleaching agent
 P.V.C manufacture /plastics
 HCl
”
 Water treatment
(any 1mk)
(d) G has a larger atomic radius than C hence √1mk the metallic bonds in C is stronger than in G
√1mk
T.T 2mks
(e) Giant atomic structure / molecular structre√1mk
(f) P has a smaller atomic radius than O due to a larger nuclear charge which pulls electrons strongly √1mk
since added electrons enter the same energy level.
(g) E
(i) Smallest atomic radius hence readily gains electrons √1mk
(ii) O√ ½mk - It loses electrons√ ½mk readily since it has the largest atomic radius√ ½ mk
2.
(a)
-
cost
availability
effect on environment
method of storage
heating value (Any 2mks each 1mk)T.T 2mks
(b) (i) Mass = Density x volume
√ ½ mk
-3
3
= 1gcm x 450cm
= 450g
= 450 = 0.45Kg
1000
∆T = 46.5 – 25 = 21.5oC
√ ½ mk
Heat evolved = MC∆T √ ½ mk
= (0.45Kg x 4.2KjKg-1K-1 x 21.5OK)
= 40.635KJ
√ ½ mk
(ii) Mass of ethanol = (125.5 -124)g= 1.5g
√ ½ mk
R.M.M of ethanol = (24+16+6) = 46g
Moles of ethanol = 1..5/46
= 0.0326moles √ ½ mk
If 0.0326moles = 40.635KJ
1mole = 1mole x 40.635KJ√ ½ mk
0.0326moles
Exothermic reaction = -1246.472KJmol-1
√ ½ mk
-1
(iii) C2H5OH(l) + 7O2(g) → 2CO2(g) + 3H2O(l)
∆H=-1246.472KJmol √ 1 mk
No heat change 0mks
(iv)
- Heat loss to surrounding /heat absorbed by apparatus
- Error in reading temperature /mass
- Incomplete ethanol combustion
(any 2 =
√ 2 mks)
1
(d)
∆H=-1246.472KJmol-1
Energy (KJ)
√½mk
√½mk
√½mk
√½mk
Reactants
Prod
Enthalpy
E
C2H5OH(aq) + 7O2 (g)
2CO2 + 3H2
Reaction path √½mk
3.
(a) (i) Solid A – Magnesium Oxide / MgO
√ ½mk
(ii) D – Hydrogen gas / H2
√ ½mk
(iii) X-Ammonia / NH3
√ ½ mk
(iv) B – Magnesium Nitride/ MO3N2
√ ½ mk
(b) Oxygen is more reactive than nitrogen hence reacts faster and completely with
Magnesium/Nitrogen
√ 1mk
(i) Less reactive, little nitrogen reacts with Magnesium
√ ½mk
(c) Mg3N2(s) + 6H2O(l) →3Mg(OH)2(aq) + 2NH3(g)
√ 1mk
(d) (i) Passing gas X/NH3 through heated
√ 1mk Copper (III) oxide (CuO) where ammonia is √
1mk oxidize to nitrogen/(Y)
(ii) 2NH3(g) + 3CuO (s) → N2(g) + 3Cu(s) + 3H2O(l) √ 1mk
(e) Carbon (IV) oxide/CO2
√ 1mk
(f) Any noble gases (He, Ne, Ar)
√ 1mk
(g) (i) Haber process
√ 1mk
(ii) More of gas X formed √ 1mk lower temperature favours forward reaction √ 1mk in an
exothermic reaction .
4.
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2O
Volume of
gas produced
(Cm3)
Scale - √ ½mk
Labeling √ ½mk
Plotting 1mk 6 correct plots
- √ ½mk at least three
Shape – 1mk
Shown above
Total mks = 3mks
600
500
400
300
200
100
0
10
5. (a)
20
30
40
50
60
70
80
90
100
Volume of acid used
(c) (i) 300cm3± 2
√ 1mk
3
(ii) 600cm
(d) (i) Rate of reaction will be slow since the surface area could be reduced as compared to zinc
powder. √ 1mk
T.T 2mks
(ii) Rate of reaction will increase√ ½mk since the number of hydrochloric acid molecules will
increase√ ½mk thereby increasing the frequency of collision
(i) B- Ammonia gas /formula√ ½mk
(ii) C-Nitrogen (IV) oxide / NO2 √ ½mk
(iii) E-Water/formula √ ½mk
(iv) F-Air and Nitrogen (IV) oxide mixture
√ ½mk
2
6.
(b) Ammonia is oxidized to form Nitrogen (II) oxide√ 1mk in the presence of a catalyst √ 1mk to
form nitrogen (II) oxide
(c) Cooling of the gas
(d) Concentrated acid is heated, Nitric acid being √ ½mk more volatile than water distills of fast√
½mk and is condensed to form a more concentrated acid√ ½mk
(e) Nitric (V) acid is a strong oxidizing agent√ 1mk and therefore reacts as a strong oxidizing agent
and not as an acid √ 1mk
(f) Prevent decomposing by light
√ 1mk
(i) Ƶ2/Z
√ 1mk
e-
X
X √ ½ mk
Y√ ½ mk
Salt bridge√½ mk
Y2+(aq) √ ½ mk
X2+(aq) √ ½ mk
Correct half cell
√ ½ mk
Spoon
Silver
metal
Silver
nitrate
(b) (i) Magnesium ions/formula√ ½mk
sulphate ions/formula√ ½mk
hydroxyl ions/formula√ ½mk
hydrogen ions/ formula √ ½mk
(ii) 4H+(aq) + 4e-(aq) →2H2(g)
√1mk
(iii) 4OH-(aq) →2H2O(l) + O2(g) + 4e-(aq)
√1mk
4OH (aq) →2H2O(l) + O2(g) + 4e
√1mk
4e = 4F = 4 x 96500C
√1mk
If 4 x 96500C = 24dm3 = 1.2dm3
1.2dm3 x 4 x 96500C √1mk = 19300 C √1mk
24dm3
7.
(a) (i) 2-methyl pentane √1mk
(ii) 3-chloro2-methylbut-1-ene √1mk
(b) (i) Water/H2O
(ii) Nickel catalyst
√1mk
(iii) H – C
C – H √1mk
(iv) Polymerization
√1mk
(v) Chloroethene/C2H3Cl
√1mk
(vi)
H H
C= C
H Cl
(viii) Non-biodegradable hence causes pollution
3
- releases poisonous gases when burnt(any 2 = 2mks)
4