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PAHANG TRIAL 2009 EXAMINATION

CHEMISTRY PAPER 2 MARKING SCHEMES

SECTION A - Structural Questions:

Question 1.

(a)(i) The presence of isotopes 1M

(ii) Let the abundance of 63X be a %.
The % abundance of 65X. = ( 100 – a ) 1M Relative atomic mass = ( 62.93 x a) + ( 64.93 x ( 100 -a) ) 1M 100 63.55 = 62.93a + 6493 -64.93a 100

6355 = -2a + 6500
a = 69.0% 1M
The % abundance of 65X = 100- 69.0
= 31.0 %

Relative abundance 63X : 65X
1 : 2 1M

(iii)

Relative
Abundance

63 64 65 Relative mass /m/e 2M

SpeciesprotonsneutronsElectrons

20 Ne
10

10
10
10
16O2-
8

8
8
10
2 M The species have same number of electrons or isoelectronic. 1M

----------------
10M

2. (a) (i) H2O2 + 2H+ + 2 I- → 2H2O + I21M
(ii) Rate = k [H2O2] [I-]1M
(iii) 0.21M
0.11M
(iv) second order1M

(b) (i) 121M
(ii) 1s2 2s2 2p6 3s2. 1M
(iii) +2 , X has two valence electrons2M (iv) X is a better electricity conductor.1M
----------------
10M

3.(a) Atomic size increases, screening effect increases with more inner shells of
electrons 1M
effective nuclear charge decreases, ionisation energy lowered, valence
electrons are more easily removed.1M

(b) i.Be2+ (aq) + 4H2O (l) → [ Be (H2O)4 ]2+ (aq)1M

ii.It is acidic, acting as a Bronsted-Lowry acid1M

© The Be2+ ion has a high charge density 1M
and can strongly polarise large anions due to its smaller size.1M The ions of other Group 2 elements have larger sizes and charge densities and
weaker polarising power

(d)i.platinum and rhodium1M

ii.4NH3(g) + 5O2(g)  4NO(g) + 6H2O (g)1M

iii.low temperature1M
low pressure1M
( Note : The reaction is exothermic reaction. According to le Chatelier
principle, a low temperature will favour the formation of NO. For gaseous
equilibrium, a decrease in pressure will favour the reaction which produces
more gaseous molecules. Thus in the above equilibrium a low pressure will
favour the formation of NO.)

________
10M

4.(a) i. A is...
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