Question 2 | ||||||||||||||||
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ANSWER
This question was attempted by majority of the candidates and the performance was good.
In (a)(i) and (ii); most candidates correctly define diffusion, state Charles’ law and draw the
graph of Charles’ law. However some candidates did not correctly labelled their axes and
some candidates left out the word absolute in their definition of Charles’ law.
In (vi) most candidates were able to calculate the temperature correctly. In (v) most
candidates were able to arrange the three states of matter in order of increasing kinetic
energy and forces of cohesion.
In b(I), most candidates correctly stated Le Chatelier’s principle but only a few candidates
could correctly show by calculation that the figures agree with the law of multiple
proportion.
In (c) most candidates could not describe how mixture of A,B and C could be separated but
correctly calculated the percentage by mass of C in the mixture.
The expected answers were:
(a) (i) Diffusion is the movement of particles from the region of higher
concentration to the region of lower concentration.
(ii) Charles’ law states that the volume of a given mass of gas is directly
proportional to its absolute temperature at a constant pressure.
Accept = K at constant pressure with each term defined.
OR
V µ T at constant pressure with each term defined.
(iv) Applying Charles’s equation
V1 = V2 T2 = T1V2
T1 T2 V1
T2 = 300 x 300
150 = 600K
(v) (i) Solid liquid gas /Solid, liquid, gas.
(ii) Gas liquid solid/Gas, liquid, solid.
(b) (i) In a reversible reaction at equilibrium, if any of the factors affecting the equilibrium is altered, the position of equilibrium will change so as to annul the effect of the change.
(iii) Oxide II
% of O2 = 20%
% by mass of metal = 80%
20 of O2 combined 80 of metal \ 100 of O2 = 100 x 80
20
= 400 g
= 400.9
(ii) Oxide I
% of O2 = 11.1 %
% by mass of metal = 88. 9 %
11.1 of O2 combined 88.9 of metal
\ 100 of O2 = 100 x 89 .9 = 800.9 g
11 .1
Ratio of metal in both oxides = 800.9: 400 = 2 : 1 [7 marks]
(ii) ALTERNATIVE
Consider 100g of each oxide:
Oxide I Oxide II
Mass of O2………………….. 11.1g 20.0g
Mass of metal ………………. 100 – 11.1 = 88.9 100 – 20.0 = 80.0
Mass of metal combined with 11.1g of O2 88.9 80.0 x 11.1=44.4
20.0
Ratio of masses of metal = 88.9 : 44.4
= 2:1
(ii) ALTERNATIVE
For sample A 11.1 g of O2 combined with 88.9 g of metal
For sample B 20 g of O2 combined with 80 g of metal
\ 11.1g O2 → x
= 44.4 g
\ Ratio of metal I – A and B = 88.9 : 44.1
= 2 : 1
Note: The mass of any of the elements can be fixed.
(c) (i) – Add water to mixture of A, B and C and stir;
– C dissolves;
– Filter to obtain C as filtrate and A and B as residue;
– Evaporate filtrate to dryness to obtain solid C;
– Heat residue until melted;
– Separate by simple distillation;
– Pure A is collected at 1170C and pure B at 160oC.
(ii) Mass of mixture = 25.25 g
Mass A+ B = 7.52 g + 8.48 g
= 16.00 g
- Mass of C = 25.25 – 16.00 g
= 9.25 g
- % of C in mixture = 9.25 x 100
25.25
= 36.6 %
Question 3 | ||||||||
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ANSWER
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Question 4 |
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ANSWER
The expected answers were:
(a) (i) Structural isomers are compounds with the same molecular formula but different
arrangement of constituent atoms/or different structures.
(ii) CH3 CH2 CH2 CH2 OH, CH3 CHCH2 OH
CH3 CHCH2 CH3 CH3
OH (CH3) 3- C- OH
(iii) (CH3)3 C-OH
(b) (i) C5 H12 + Cl2 light C5H11 Cl + HCl
(ii) H H H H H
H C C C C C Cl
H H H H H
(iii) Light acts as catalyst.
(iv) Pentane C5H12 would come out first.
Pentane has a lower molar mass (72) as compared to chloropentane
(C5H11 Cl) (86.5).
Hence weaker inter molecular forces/van der Waals’ forces.
(v) C4 H6 Cl2
(c) (i) H H H H
H C C C C H
H
OH H
H C H
H
3-methyl butan–2–ol/ 3- methyl-2-butanol
(ii) CH3 – CH – COOH
OH
2–hydroxypropanonic acid
(d) (i) I CH3 – CH COOH
+ NH3
II CH3 – CH –COO-
NH2
(ii) % N = 0. 0313 x 100
0. 309
= 10. 1 %
% O = 100 – (60.5 + 6.5 + 10.1)
= 22. 9
C H N O
60.5 6.5 10.1 22.9
12 1 14 16
5.04 6.5 0.72 1 43
5.04 6.5 0.72 1.43
0.72 0.72 0.72 0.72
7 9.027 1 1.99
Empirical formula
C7 H9 NO2
Question 7 |
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ANSWER
In (a)(i)(ii) and (iii) most candidates correctly define standard electrode potential, state factors that affect standard electrode potential and gave uses of the values of standard potential. However, some candidates could not properly define standard electrode potential.
In (a)(iv) most candidates could not draw and label correctly electrochemical cell.
In (b) (ii) and (III) only a few candidates could define oxidation in terms of electron transfer. Also majority of the candidates could not balance redox reaction.
In (c) most candidates correctly classified oxides as basic, amphoteric, acidic or neutral.
In (d) most candidates could not correctly define hydrogen bonding.
The expected answers were:
(a) (i) Is a measure of the tendency of an element to form ions in solution relative to
the tendency of hydrogen atoms to form ions in solution at standard conditions
(at 298k and 1 atm. pressure and 1 molar concentration)
(ii) Condition, temperature, pressure.
(iii) – To predict the direction/ feasibility of a chemical reaction.
- To calculate the e. m. f. of a cell.
- To predict the standard potential of unknown elements
(v) Emf = Eo (Right) – Eo ( left)
+ 0 .34V – (- 0. 76V)
= + 1. 10V
(b) (i) I . Oxidation – a process of election loss.
II. Oxidation agent – Is an electron acceptor.
(ii) Separate into half reactions
2I- ® I2 + 2e
5e+ 8H++ MnO4- ® Mn2+ + 4H2O
Balancing 10I- ® 5I2 x 10e
10e+ 16H+ + 2MnO4- ® 2Mn2+ + 8H2O
10I- +16H+ + 2MnO4- ® 5I2 +2Mn2+ + 8H2O
(c) (i) Neutral
(ii) Acidic
(iii) Amphotenic
(iv) Basic
(d) Hydrogen bonding is an intermolecular force between molecules in which hydrogen is
attached to a very electronegative atom which has a small size e.g N, O, F
Question 8 |
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ANSWER
The expected answers were:
(a) (i) I. Biotechnology is the use of biological processes/living organisms to
make products for human use.
II. Biogas is a fuel obtained from organic waste through the process of anaerobic decomposition.
(ii) Production of drugs (medical) fertilizer, waste management, production of kenkey, fufu, garri, burukutu/any food or beverage produced from fermentation (domestic), industrial application etc.
ALTERNATIVE
(b) (i) Yeast is added to sugar cane juice. Left to stand (at room temperature)
(for few days) to ferment. The mixture is distilled to obtain ethanol.
(ii) Carbon (IV) oxide
(iii) – in aerated drinks;
– as fire extinguisher;
– coolant/refrigerant;
– solvay process;
– synthesis of Na2CO3.
(iv) Ethanol from cane sugar is renewable because it is from plants which can be cultivated continuously.
Ethanol from petroleum is non-renewable because it is from a source which can get exhausted.
(c) Heavy chemicals are chemicals used extensively in industries and are
produced in very large quantities.
Fine chemicals are chemicals produced in small quantities for specific purposes
and to a very high degree of purity.
Heavy chemical – H2SO4, NaOH, NH3, Ca CO3, HNO3, HCl, NaCO3, bleaching
powder, metals (iron, tin, aluminum, copper, zinc).
Fine chemical
- laboratory/analytical reagents, drugs, dyes, perfumes, photographic reagents, additives, cosmetics.
(d) O2, Cl2 HCl Increasing order to deviation
The extent of deviation depends on intermolecular forces.
HCl is polar and will have a greater intermolecular attraction than oxygen and chlorine. The molar mass of chlorine is greater than that of oxygen hence stronger
vander waals forces than oxygen.
