BUTERE EAST F3 JOINT EXAM 2015
MARKING SCHEME
CHEMISTRY PP1
1.
a). By opening the air hole completely (l mrk)
b). see diagram (l mrk)
c). Water vapour (steam) (1 mrk)
2. a).35—17=18
b). R.a.m (35 x 90)+ (36 x 4)+(37 x 6) = 3150 +144+222 =35.16 (2mrks)
100
100
3. (a). Ca(OH)2(aq) +CO2(g)  CaCO3(s)+ H2O(l)
(lmrk)
(b). It forms a colourless solution of potassium carbonate hence no visible change
(2mrks)
4. a). -Rusting occurred in tube (I). No rusting in (II).
- In tube (II) anhydrous calcium chloride absorbed moisture “‘but not sodium chloride
(2mrks)
(b).- To improve its appearance /make it beautiful
- To prevent its corrosion or rusting.
(1mrk)
5. Moles of acid= 20 x 0.25 = 0.005
1000
2HNO3(aq) + NaCO3(s) CO2(g) + H2O(l) + NaNO3(aq)
2 moles of acid = 1 mole of Na2CO3
0.005 x1 = 0.0025moles
2
25cm3 = 0.0025
250cm3 = 250x0.0025 = =0.025moles
25 7
Mass ofNa2CO3 = 0.25 x106 = 2.65g
(3mrks)
6. Glucose and Galatose
(2mrks)
7. (a) There’s formation of yellow “solid and white solid. V
Heat from burning magnesium dissociates sulphur (iv) oxide gas to form sulphur” and oxygen gas.
Oxygen supports combustion of magnesium to form magnesium oxide. V (3mrks)
b) 2Mg(s) + SO2(g) 2MgO(s) + S(s) /‘ (lmrk)
8. (a)
C
H
92.3
77
E.F = CH ½
12
1
7.69
7.77 ½(CH)n = 26  13n=26
7.69
7.69
n=2
1 MF=C2H2√½
b).H-C=C-H
(lmrk)
9. a) I. cation Pb2+
II. anion NO3(2mrks)
b). Pb2+(aq)+ Cl-(aq) PbCl2(s)
(lmrk)
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10. a).chlorine has stronger Van der waals forces than argon which is mono-atomic
b)
(lmrk)
c).It has layers in its structures joined by weak ‘van der waals forces which slide over each other.(1 mrk)
11. i). Nitrogen
(lmrk)
ii).4Zn(s) + 2NO2(g) 4ZnO(s) + N2(g)
(lmrk)
-Add water to dissolve lead (II) nitrate to form solution
-Add water to sodium carbonate/ potassium carbonate / (NH4)2CO3 to make solution. Mix with lead
(III) nitrate 1/2 to form precipitate of lead (II) carbonate. Filter ½. Dry the precipitate (3mrks)
13. a). Z. Has the highest tendency to gain electrons due to its smallest radius.
(2mrks)
b)Ionic / electrovalent
(lmrk)
14. (a) (i) Magnesium nitride (Mg3N2)
(lmk)
(ii) Ammonia/NH3
(lmk)
(b) Mg3N2(s) + 6H2O (l)  3Mg (OH)2 + 2NH3(g)
15. a).The solution is colourless
(lmrk)
b)(i).PbO(s) + 2HNO3(aq)  Pb(NO3)2(aq) + H2O(l)
(lmrk)
16. (a) No observation change II copper remains brown .Copper is below hydrogen in the reactivity series
hence cannot displace hydrogen from steam
(b) Gold, Silver. Mercury, platinum
17. (a) 2 — chlorobutane
(b)
(2mks)
(c) Alkene
18. (i) Fractional crystallization
(2mks)
(ii) CaO(s) + H2O  Ca(OH)2(aq)
(iii) Ammonia, Carbon (IV) Oxide
(3mks)
19. Colour change from grey to yellow
Colour change from black to brown.
Zinc oxidized to ZnO//CuO reduced to Cu// Zinc is higher in reactivity, Reject displacement.
20. a) Under the same conditions of temperature and pressure the rate of diffusion of a gas is inversely
proportional to the square root of its density.
21 (a) Thistle funnel should dip into the solutions in the flask.
(b) Hydrogen peroxide H2O2(l)
(c) Insoluble in water slightly soluble in water.
22. (a) - Bulb lights
Brown fumes produced at anode
Grey solid deposited at the cathode molten PbBr2 has mobile ions to conduct electricity current.
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Pb2+ discharged at the anode to produce Br2(g)
Pb2+ discharge at cathode to produce Pb
Br- discharge at the anode to produce Br2(g)
(b) Use fume chamber½ outside I in open air. Reason Br2 vapour is poisonous; ½
23. FM of Ca(NO3)2 =40 + (14x2)+(6x6) - = 104 √½
Mass of N= 28
% of M= 28 x 100
104
26.92308
(2mks)
24. Add water to the mixture ¼ and stir to dissolve sodium chloride
Filter off’ the lead II sulphate residue. √½ Heat the filtrate to dryness √½ to obtain sodium chloride
crystals √½
25. a) Cracking √½
b)
26. a) R,Q,T,U1
b). The solution reaction with fats on the fingers to form soap √. It is this soap that makes them to be
slippery.
27. Due to incomplete combustion CO is produced√½2 This gas is poisonous
28. 1 mole of NaOH => 40g
2g
1x2
40
0.05 mole
Original concentration of H2SO4
1000cm3=> 1 mole
30cm = 30x1
1000
=0.03 moles
Moles of acid reacted with NaOH
Mole ratio of NaOH: H2SO4
2 : 1
But 2 => 0.05 moles
1 =1x 0.05
2
=0.025 moles
Moles of acid reacted with KOH
0.03-0.025 =0.005 moles
Mole ratio KOH: H2SO4
2:1
But I mole => 0.005 moles
2 => 2x0.005
1
=0.01 moles
KOH reacted 0.01 moles
But 0.1 mole of KOH => 1000cm3
0.01 mole => 0.01 x 1000
0.1
=l00cm3
29. a) Sodium carbonate decahydrate
b) Efflorescence
c) Na2CO3.H2O(s)  Na2CO3(s) + H2O(l)
30. P =Sublimation√//sublimate formation
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R = solidification/Freezing √
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