Download MANGA SUB- COUNTY 233/1 CHEMISTRY PAPER 1 (THEORY

MANGA SUB- COUNTY
233/1
CHEMISTRY PAPER 1
(THEORY)
JULY/AUGUST 2015
TIME: 2 HOURS
MARKING SCHEME:
1.
a)
Water that does not lather easily with soap1
b)
Ca 2
and
Mg 2
for
2
Ca 2 aq  CO 3 aq   CaCO 3  s 
c)
OR
Mg 2 aq  CO 3
2
 aq  
MgCO 3  s 
Penalise ½ mk
-
wrong/ missing s.s.
-
Penalise fully for UB
equation/wrong
changes.
2.
3.
4.
a)
b)
a)
b)
c)
Fe  s   S  s  FeS  s 
(i)
(ii)
Iron II chloride/ FeCl2(aq)
Hydrogen sulphide/ H2S(g)
Sodium
Sodium hydroxide
2Na 2 O 2 s  2H 2O l   4NaOH aqO 2  g 
720  360  180  90  45
 No. of half  lifes  4
 Age of the old wood  4  5670 yrs
 22,680 yrs
5.
a)
b)
6.
a)
b)
c)
d)
7.
a)
b)
(i)
Group III/ 3/ Three
(ii)
T3+
T 2 so 3 3
Hydrogen/ H2(g)
It is less dense/ lighter than air.
White solid seen in the boiling tube
Magnesium reacts with steam forming white magnesium oxide.
Mg s  H 2O  g   MgO s   H 2  g 
The mass of a substance produced during electrolysis is directly proportional to the
quantity of electricity passed.
t  32  60   10  1930 sec
dr  It  0.5  1930  165c
If 0.44 g  965c
88
88 g 
 965
0.44
 193,000c  2 F
 ch arg e on Y  2
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8.
a)
b)
c)
9.
a)
b)
c)
10.
(i)
Amphoteric oxides
(ii)
Neutral oxides
Any 2 x ½ mk/ Accept correct names
ZnO, PbO, Al 2 O 3
They have both acidic and basic properties.
(Accept w.t.t.e)
(i)
Dinitrogen tetraoxide/ N2O4(l)
(ii)
Oxygen/ O2(g)
2PbNO3 2 s   2PbOs  4NO 2  g  O 2  g 
-
A brown gas produced.
A residue that is orange when hot and yellow on cooling is formed.
(Any 1x1=1mk)
Mass of the hydrated salt = 28.36 – 25.14 = 3.22 g
Mass of the anhydrous salt = 26.56 – 25.14 = 1.42 g
 Mass of water of crystallization = 3.22 – 1.42 = 1.80 g
Na2SO4
H2O
Mass
mass
1.42 g
1.42
 0.01
No. of moles =
142
0.01
1
Mole ratio
=
0.01
 formula of the salt is Na 2 SO 4 .10H 2O
1.8 g
1 .8
 0.1
18
0.1
 10
0.01
11.
Add sodium hydroxide solution, to the test solutions in separate test- tubes drop by drop until in
excess. The solution containing Fe2+ ions forms a dark green precipitate which is insoluble in
excess sodium hydroxide solution. The solution containing Fe3+ ions forms a reddish- brown
precipitate which is insoluble in excess sodium hydroxide solution.
(NB: Ammonia solution can be used in place of sodium hydroxide.)
12.
a)
13.
14.
Hydrogen is a non- polar molecule/ compound hence insoluble in water which is a polar
solvent.
b)
Hydrogen chloride is a polar molecule/ compound hence dissolves in water which is a
polar solvent.
c)
All the valence electrons in diamond ore used in bounding while in graphite, three valence
electrons are used in bounding leaving and delocalized electron that carries the electric
charge.
(Accept: Diamond does not have delocalized electrons while graphite has delocalized electrons.)
Mass of gas evolved = 4.37 – 4.18 = 0.55 g
If 280 cm3 weight 0.55 g
22400
 0.55 g
Then 22400 cm3 weighs
280
=
44 g
a)
C7H16
b)
Butane
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15.
a)
b)
c)
16.
2- methylpropane
Water gas
H 2O  g  C s   CO g   H 2  g 
-
(penalize ½ mk for wrong/ missing s.s)
Used to reduce metal oxides of some metals.
Used as a fuel.
Na 2 SO 4 aq  BaCl 2 aq  BaSO 4 s  2 NaCl aq
R.M .F of BaSO 4  137  32  16  4   233
6.9
 0.0296
233
 moles of Na 2 SO 4  1  0.0296
 No of moles of BaSO 4 
 0.0296
R.M .F of Na 2 SO 4  23  2   32  16  4   142
 mass of Na 2 SO 4  0.0296  142 g
 4.2032 g
 mass of KCl  10  4.2032 g
 5.7968 g .
17.
a)
b)
18.
19.
a)
b)
(i)
Zn2+
(ii)
SO422
Pb 2 aq   SO 4 aq   PbSO 4  s 
c)
Bauxite
Silicon IV oxide
Iron III oxide
(Reject formula)
Carbon is below aluminium in the reactivity series therefore cannot displace aluminium.
a)
b)
Used as fertilizer
NH 3  g  HNO 3 aq   NH 4NO3 s 
68  1000,000
17
 4,000,000 moles
Moles of ammonia 
4,000,000  80
1,000,000
 320 ions
Mass of NH 4NO 3 in ions  
20.
a)
b)
c)
Copper
Cu s  4HNO 3 aq  CuNO3 2 aq 2 NO 2  g  2H 2O l 
CuOH 2  s  4 NH 3 aq  CuNH 3 4 
2
-
 aq  2OH

 aq 
Penalize ½ mk for wrong/ missing s.s
Penalize fully for U.B equation
Penalize fully for wrong/ incorrect chemical symbols & formulae
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21.
a)
Al s   Al 3 aq  3e 
Cu 2 aq 2e   Cu s 
b)
(i)
2 Al s  3Cu 2 aq  2 Al 3 aq 3Cu s 
(ii)
= Ec 
0
E E
0
0
R
0
=  0.34   1.66v
 2.00v
22.
b)
c)
23.
a)
b)
c)
24.
a)
b)
25.
a)
Polychloroethene
(Reject polyvinyl/ chloride)
5125
No. of monomers =
62.5
= 82 monomers.
Curve B. Reactants have a high concentration of particles at the beginning. As the particles
collide to form products, the concentration of reactant particles decreases.
Equilibrium point
At the start, the concentration of products is zero. The concentration of products increases
as the reaction progresses. When the reacting particles are consumed the reaction comes to
an end hence no further increase in concentration of products.
R. A.M of R  6.0  10 23  3.45  10 22
= 207
3.0  10 22
No. of moles of atoms =
 0.05 moles
6.0  10 23
Mass of Ca  0.05  40  2 g
Heat was absorbed from the surrounding to the solution in the beakers/ Heat was lost from
the body (hand) to the solution.
b)
(Both axes labeled correctly award ½ mk)
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26.
c)
Endothermic reaction.
a)
-
Atoms of the same element which have the same atomic number but different mass
numbers.
OR
-
Atoms of the same element which have the same number of protons but different
number of neutrons. (Any 1x1=1mk)
9  22  1  20
R. A.M 
10
198  20 218


 21.8
10
10
b)
27.
NB:
(i)
(ii)
Marks are awarded only if the axes are labeled correctly.
The curve s MUST become horizontal at different times but at the same maximum
volume.
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