memory

Topics: Sodium chloride, Chemistry, Hydrogen Pages: 2 (388 words) Published: October 14, 2014
NAROK FORM FOUR DISTRICT MOCK – 2007
233/3
CHEMISTRY
PAPER 3

MARKING SCHEME.

1. Table I :
School value - 1mk 8cm3 – 9cm3 1mk
Decimal / Accuracy  0.1 ( 1mk) Max ( 2mks)
Penalise for unrealistic values
a) Average volume 8.5 + 8.5 + 8.5  ½ = 8.5 cm3  ½ penalise for missing units
3
b) Moles of solution A 8.5 x 0.25  ½ = 0.002125  ½ moles
1000
c)i) Moles of HCl in 25cm3 of solution D:
HCl + NaOH NaCl + H2O

1 : 1  ½

Moles of HCl in 25cm3 = 0.002125  ½ ( Mole ratio)
ii)Moles of HCl on 100cm3 of solution D
25 0.002125
100x 0.002125  ½ = 0.0085 moles  ½
25
iii) Moles of HCl in 100cm3 Solution B:
0.5 moles 1000cm3
0.5 x 100cm3 = 0.05moles ½
1000
iv) Moles reacting with x2CO3
X2CO3 + 2HCl 2xCl + CO2 + H2O
1 : 2
Moles of HCl reacting = 0.05 – 0.0085  ½
= 0.0415 moles X2CO3 ½
d) mole ratio 1 : 2 ½
Moles of X2CO3 = 0.0415  2 = 0.02075 Moles that reacted e) i) Moles of X2CO3 = 0.02075
0.02075 2.2 g
1 mole x 2.2 ½ = 106.024g ½
0.02075 = ~ 106 g ½

ii)RAM of x X2CO3 =
2x + 60 = 106 ½
2x = 46 ½
2x 2

x = 23 ½

2.Table II
Initial
Final
DT.C
13.0C
9.0C
- 4.0C

a) H = Mc T  ½
= mass = 20g ( 1 g/cm3 )  ½
C = 4.2
DT = 4.0C
= 20 x 4.2 x 4 = + 336J  ½ Deny ½ for lack of +ve sign b) Mole of KCl RFM = 39 + 35.5
= 74.5  ½
 2 = 0.02685 moles
74.5
0.02685 336
1 x 336 = 12514 J Mol-1  ½
0.02685  ½
12514
1000
= + 12.514 KJ Mol-1 Insist on units as KJ Mol-1 J-1
Hsoln = Hhyd + Hlattice ½

12.514 =  ½ Hhyd + 690  ½
= -677.5 KJ mol-1 ½ deny ½ mark for of sign (-ve) 3.
OBSERVATION
INFERENCES
a)i) Colourless...
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