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 Chemistry Paper 2,Nov/Dec 2009
 Questions: 1 2 3 4 5 6 7 8 Main
Weakness/Remedies
Strength

Question 1

(i)         What is meant by atomicity?
(ii)        Mention one element in each case which is

1. monatomic,
2. diatomic,
3. tetratomic.                                                           [5 marks]
1. (i)         Write the orbital electron configuration of

I.          20Ca,
II.        9F.

1. In which group does each of the elements belong?
2. How many unpaired electrons are present in 9F?
3. How many electrons are present in 20Ca2+?

[6 marks]

1. (i)         Write a balanced equation for the thermal decomposition of KCƖO3.
1. Mention the catalyst that could be used to increase the rate of reaction in 1(c)(i).
2. If 5.0 g of KCƖO3 was decomposed by heat, determine the volume of

oxygen produced at s.t.p.
[Molar gas volume at s.t.p. = 22.4dm3, K = 39,  Cl = 35.5, O = 16]
[8 marks]

1. (i)         Mention the products formed when each of the following substances is heated strongly:
1. ZnCO3;
2. CuSO4.5H2O.
1. State the colour change observed when each of the residues in 1(d)(i) above is allowed to cool.

_____________________________________________________________________________________________________
OBSERVATION

The question was attempted by most candidates and the performance was good.

In part (a) (i) and (ii), candidates correctly defined atomicity as the number of atoms present in a given molecule (of an element or a compound) and mentioned one element each for question (ii) I – III respectively, thus.
I           -           helium/neon/argon etc
II         -           hydrogen/oxygen/nitrogen/chlorine/fluorine etc
III        -           phosphorus.

In part (b)(i), candidates wrote correct orbital electron configuration of 20Ca and 9F as follows

1. 1s2 2s2 2p6 3s2 3p6 4s2
2. 1s2 2s2 2p5

In part (b)(ii) – (iv), candidates knew that 20Ca and 9F belong to group 2 and 7,  9F has one unpaired electron and that 20Ca 24 has 18 electrons.

In (c)(i), candidates correctly wrote a balanced equation for the thermal decomposition of KCƖO3  thus:
2KCƖO3(s) → 2 KCl(s)  + 3O2)g).

In part (c) (ii), candidates correctly mentioned (Mn O2) manganese (IV) oxide as the catalyst that could be used to increase the rate of reaction.

In part (c) (iii), candidates correctly determined the volume of oxygen produced at s.t.p. as follows:
molar mass of KCƖO3  = 39  + 35.5 + 16X3  / 122.5
number of moles of KCƖO3  =   5.0    = 0.0408 mole
122.5
mole ratio of  KCƖO3: O2  = 2
hence number of moles of O2  evolved = 0.0408 x  3
2
∴ volume of O2  evolved =22.4 x  0.0408 x 3
2
=  1.37 dm3
Alternative methods were also accepted.
In part (d)(i) and (ii) candidates correctly mentioned the products formed when I and II were heated and stated the colour changed observed when the residue in I and II are allowed to cool, respectively, as follows:

(i)         I.          ZnO / CO2
II.        CuSO4 / CuSO4.H2O

1. I.          Yellow to white/grey
2. White / grey colour remained unchanged.