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Chemistry Question Paper

Chemistry

Course:Secondary Level

Exam Year:2012

EMPOWERMENT CONCEPTS – 2012
233/1
CHEMISTRY
Paper 1
JULY/ AUGUST 2012

1. (i) Saponification. (1 mk)
(ii) Vulcanisation (1 mk)

2. (a) Deliquescence (1 mk)
(b) Drying of gases / drying agent. (1 mk)

3. (i) Using vanadium ? oxide / pt catalyst so that to lower the activation ?energy?
- Lowering the temperature since the reaction is exothermic ? to increase the rate of production.
- Increasing pressure – favour ? the side with fewer moles. (2 mks)
(ii) = Activation energy ?1
= Energy changes ?1

4. (a) M – Sodium Chloride crystals. ( ½ mk)
P – Concentrated Sulphuric VI acid ( ½ mk)

5. (i) Condensation (1 mk)
(ii) Oxidation – reduction (Redox) / displacement (1 mk)
(iii) Substution reaction (1mk)

6.

No of half-life is 4
No of years = 440X4?1 = 1760yrs?1 (3 mks)

7. Add? ½ water to the mixture stir ? ½ the mixture for all Sodium Carbonate to dissolve. Filter ? ½ the mixture to obtain calcium carbonate as residue and sodium carbonate as filtrate. Heat ? ½ the filtrate to evaporate ? ½ excess water and leave it to cool slowly for sodium carbonate to crystallize ? ½ . Finally filter the product and obtain pure crystals of sodium carbonate. (3 mks)

8. (i) High temperature of over 5000C and pressure less latin. (1 mk)
(ii) Burn in air rich of oxygen / Burn in pure oxygen. (1 mk)
(iii) Presence of hot platinum catalyst and oxygen. (1 mk)

9. (a) (i) S16 = 2.8.6 (1 mk)
(ii) S12 = 2.8.2 (1 mk)

(b) (i) Neutron – 14 ( ½ mk)
(ii) Electron - 10 ( ½ mk)

10. (i) At constant temperature the volume is inversely proportional to the pressure Formula
Vd (1 mk)
(ii) P1V1 = P2V2 ? ½
12 X1 = 2.5 X V2 ? ½
V2 = ? ½ = 4.8 litres ? ½ (2 mks)

11. H2SO4 + 2 WOH W2SO4 + 2H2O
Moles of H2SO4 = =0.04? ½
Mole ratio = 1:2 ? ½
Moles of WOH = ? ½ = 0.08 ? ½
RFM of WOH = =40 ? ½
RAM of W = W +10+1 = 40
= 40 -17 = 23 ? ½ (3 mks)

12. (a) (I) methyl benzene/Toluene/ Accept any organic solvent. ( ½ mk)
(II) Water ( ½ mk)

(b) B – Red litmus remained red? ½
- The non-polar / methylbenzene with ammonia does not form polar ions ? ½ e.g. NH4+ and OH- thus no effective to litmus paper/ Ammonia remains in a molecular state / exists as a molecule.? ½
C – Red litmus paper turned blue water being polar solvent when react with ammonia it form NH+ ions and OH- which affect the litmus paper. (2 mk)

13. (i) Slip a piece of manila paper / wooden ? ½ splint into region A and ? ½ quickly remove before it catches fire. Inner part remains unburnt / Not charred ? ½ or Hold a match stick on a pin and let the head on the head of the match stick in zone A does not light. ( 1 ½ mk)
(ii) Unburnt carbon ? ½ particles form soot other carbon particles glow ? ½ white hot emitting light ? ½ . ( 1 ½ mk)

14. (a)
Penalize for condemned structure (1mk)
/ CH3CH2CH2CH2CH2ONa /C4H10ONa

(b) Alkoxide (1 mk)
(c) 2C4H10O (l) + 2Na (s) 2C4H9ONa (l) + H2 (s) (1 mk)
- Balanced equation ½
- State symbols ½
Penalize
- Joining of letters
- Mixing small and capital letters.
- Wrong state symbols
- Penalize fully for unbalanced equations

15. (a) Ca(OH)2 (aq) + CO2(s) CaCO3(s) + H2O(l)? ½

- Lime water forms white ? ½ ppt due to the formation of calcium carbonate but in excess calcium carbonate forms colourless solution due to the formation of soluble ? ½ calcium hydrogen carbonate.
CaCO3(s) + H2O(l) + CO3 (g) Ca(HCO3)2(aq) (2 mks)
(b) In diamond each carbon atom is bonded to four other carbon atoms ? ½ arranged in a regular tetrahedron shape the whole structure of diamond extends all directions forming a rigid ? ½ mass of atoms. (1 mk)

16. 2M Sulphuric acid has a higher ? ½ than 2M ethanoic acid because it is a stronger ? ½ acid and fully ? ½ ionize in water than ethanoic acid that only partially ionizes. ? ½ (2 mks)
17. Piece covered with copper ? 1 Iron is more reactive than? copper and will lose electron more readily in the presence of air and water. (2 mks)

18. (i) When forward and reverse reaction are taking place at the same rate the reaction has not stopped but is still continuing.
(ii) B – CrO42- & H+ - Is the reactant since the start of the reaction (1 mk)

(b)

19. (a) (i) Key X - P
- H

(ii)

(b) - Dative bond / co- ordinate
- Covalent bond.

20. (i) Orange ? ½ Potassium chromate VII turns green ? ½ due to reduction process SO2 is a reducing agent where it reduces chlomate VII ions to chloromium III ions ? ½ (1 ½ mk)

(ii) Brown ? ½ iron III sulphate solution turns green ? ½ due to reduction of Fe3+ to Fe2+ ions ? ½ (1 ½ mk)
21. (a) It is a strong acid?1
- The PH value decreases ?1 from 10+2 shows that as the acid is added to urea which is a base the PH decreases.
(b) P & R
- Zinc hydroxide is amphoteric and it reacts with both acids and bases. (1 mk)

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