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(a) Polythene is produced from ethene by polymerization. (b) Ethene can be produced from kerosine fraction of petroleum by cracking process. (c) Soap is produced from vegetable oil by alkaline hydrolysis of the oil (ester) (d) Magarine is produced from vegetable oil by hydrogenation of the oil

(a)(i) Carbon(IV) oxide is evolved. (ii) Sulphur (V) oxide is evolved. (iii) Hydrogen sulphide is evolved. (b) Sulphur (IV) oxide turns potassium tetraoxomanganate (VII) solution colourless without any precipitation. Sulphuric (IV) oxide turns acidified potassium heptaoxodichromate (IV) solution from golden yellow to green.

(a) The time taken for half amount of \(^{56}Mn\) to decompose is the half life which is 9.3 x 10\(^{3}\)s (b) They are the Alpha, the Beta and Gamma particles.

(a) Compounds A and B are geometrical isomers. (b) Compound A = trans-1, 2-dichloroethene, B =Cis-1, 2-dichloroethene (c) The Chemical properties of A and B will be the same because they contain the same functional group.

(a)(i) Producer gas = \(\frac{1}{3}CO_{(g)}\) + \(\frac{2}{3}N_{2(g)}\) (ii) Water gas = \(\frac{1}{2}H_{2(g)}\) + \(CO_{(g)}\) (b) The components of water gas are all combustible gases whereas the Nitrogen content of producer gas is not combustible.

(a) Le Chateller's principles states that a system at equilibrium will change in such a way as to nullify any change about any factors affecting the equilibrium, e.g temperature. (b) For the Haber process, 3H2(g) + N2\(_{(g)}\) 2NH\(_{3(g)}\) \(\Delta\)H. Increase in pressure favours the yield of ammonia because the reaction flows in the direction of the less number of moles. Decrease in temperature favours the exothermic process as the Haber process. Increase in the number of moles of the reactants will favour the production of ammonia. (c) Ammonia form white fumes of ammonium chloride, when it is brought in contact with hydrogen chloride solution on rod dipped into it. (d) Test Observation (i) Ammonia solution is added to a clear solution of zinc chloride (ii) Ammonia solution is added to a clear solution of solution of copper(II) tetraoxosulphate (VI) solution White gelatinous precipitate which became soluble in excess of the ammonia solution. Precipate is soluble forming deep blue solution (e) The electron cloud formed by the unsharedl pair in NH\(_3\) spread out over a greater volume than that of the three pair connected to hydrogen atoms. This would tend to force the bonding pairs closer to one another and thereby reduce the bond angle to less than the bond angle in the tetrahedral methane. (f) Ammonia is used to make trioxonitrate (V) acid. Ammonia solution is used in laundry work. It removes temporary hardness by precipitating the Calcium ion as CaCO\(_3\).

(a) Members of a homologous series can be prepared by the same method. Members exhibit the same general chemical reaction. The molecular formula of each member of a homologous series can be derived from a general formula. (b)(i) Hexane (CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)CH\(_2\)CH\(_3\))CH = CH(ethyne) (c)(i) moles H = \(\frac{7.7}{1.0}\) = 7.7 moles moles C = \(\frac{92.3}{12}\) = 7.7 moles i.e \(\frac{7.7}{7.7}\) H ; \(\frac{7.7}{7.7}\) C = 1 : 1 Empirical formula of the hydrocarbon is CH (CH)x = 78; (13)x = 78; x = \(\frac{78}{13}\) = 6 Molecular formula is C\(_6\)H\(_6\) (ii) The molecule is Benzene. The structure is (d)(i)Y undergoes substitution reaction only. (ii) x (iii) x H L (e)(i) Suitable catalyst must be applied when reacted directly with air. In another method, acidified suitable oxidising agent must be applied. (ii) Oxidation reaction CH\(_3\)CH\(_2\)OH\(_{(g)}\) + CH\(_2\)COOH\(_{(l)}\) +H\(_2\)O\(_{(g)}\)

(a) A conductor is a solid metal which allows electricity to pass through it while an electrolyte is a compound in the molten or in solution which allows electric current to pass through it. (b)(i) Faraday's first law of electrolysis states that the amount of element deposited at the electrodes is directly proportional to the quantity of electricity applied in the electrolysis. (ii) Different quantity of electricity at different time is allowed to pass through a fixed concentration of CuSO\(_4\) solution using copper electrodes. It could be observed that the higher the quantity of electricity, the greater the amount of copper element deposited at the cathode for example. (c) Quantity of electricity = Current x tim i.e 0.45 ampere x 1500 secs. 0.45 Coulomb x 15000 Sec. 0.45 \(\frac{coulomb}{Sec}\) x \(\frac{15000}{1}\) = 675 coulombs If 675 coulombs give 0.222g of x\(^{2+}\) i.e. 1 coulomb will give \(\frac{0.222}{675}\)g x\(^{2+}\) 2 For 2 x 965000 coulombs = \(\frac{0.222}{675}\) x 2 x \(\frac{965000}{1}\) = 63.5g (d) Electrolysis is used to convert electrical energy to chemical energy. It can also be used for electroplating

(a) Zinc reacts with dilute hydrochloric acid to produce hydrogen as but with trioxonitrate (VI) acid. the hydrogen initially produced is at once oxidized by the powerful oxidizing ability of the trioxonitrate (V) acid to give water and the reduction products of the acid. (b) Step I 4NH\(_{3(g)}\) + SO\(_{2(g)}\) --> 4NO\(_{(g)}\) + 6H\(_2\)O\(_{(l)}\) Step II 2NO\(_{(g)}\) + O\(_{2(g)}\) —> 2NO\(_{2(g)}\) Step III 2H\(_2\)O\(_{(l)}\) + 4NO\(_{2(g)}\) + O\(_{2(g)}\) —> 4HNO\(_{3(aq)}\) (c) NaOH(\(_{(aq)}\) + HNO\(_{3(aq)}\) —> NaNO\(_{3(aq)}\) + H\(_2\)O\(_{(l)}\) 30g NaOH x \(\frac{1 mole}{g}\) Since mole = concentration x volume Mole HNO\(_3\) = 1.00 mole x 100dm\(^3\) = 0.1 mole dm\(^3\) From the balanced neutralization reaction, \(\frac{NaOH}{HNO_ 3}\) = \(\frac{1}{1}\) i.e mole NaOH = mole HNO\(_3\). The limiting reagent in the above rxn, is 0.1 mole HNO\(_3\) 0.1 mole HNO\(_3\) will produce 85 x 0.1 = 8.5g NaNO\(_3\) (d)(i) 2NaNO\(_{3(g)}\) \(\to\) 2NaNO\(_{2(g)}\) + O\(_{2(g)}\) (ii) 2Cu(NO\(_3\))\(_{2(g)}\) \(\to\) 2CuO\(_{(g)}\) + 4NO\(_2\) + O\(_{2(g)}\) (iii) Hg(NO\(_3\))\(_{2(g)}\) -> Hg\(_{(l)}\) + 2NO\(_{2(g)}\) + O\(_{2(g)}\)

Solution D is 2M CH\(_3\)COOH. and solution E of solution E is 2M KOH (a) Temp. T\(_1\) of solution D = 31\(^o\)C (b) Temp. T\(_2\) of solution E = 32\(^o\) (c) Average Temperature T\(_3\) = \(\frac{T_1 + T_2}{2} = \frac{31 + 32}{2}\) = 31.50\(^o\)C (d) The highest temp. T\(_4\) = 40\(^o\) (e)(i) Rise in temp. T\(_4\) - T\(_3\) = 40 - 31.50 = 8.50\(^o\)C (ii) 50cm\(^3\) of soln D + 50cm\(^3\) of soln. E = 100cm\(^3\) of mixture.1cm\(^3\) of solution weighs 1g. (given) 100cm\(^3\) of the mixture will weigh 1 x 100 = 100g (f)(i) heat evolved = mass x specific heat capacity x rise in temperature = 100 x 4.2 x 8.5 joules = 3570 joules (ii) nos. of moles = \(\frac{volume \times mol}{100}\) for the acid; \(\frac{50 \times 2}{1000}\) = 0.10 moles for the base = \(\frac{50 \times 2}{1000}\) = 0.10 moles Equation for the reaction; CH\(_3\)COOH\(_{(aq)}\) + KOH\(_{(aq)}\) \(\to\) H\(_2\)O\(_{(l)}\) From the aqueous neutralisation above, 1 mole of CH\(_3\)COOH + 1 mole of KOH produces one mole of water 0.1 mole of CH\(_3\)COOH will produce 0.1 mole water. Hence 0.10 mole of H\(_2\)O is produced with the evolution of 3570 joules of heat energy 1 mole of H\(_2\)O will be produced with the evolution of \(\frac{350}{0.1}\) J mol\(^{-1}\) = 35700 Jmol\(^{-1}\) = 35.70KJ mol\(^{-1}\) The heat of neutralisation 35.70KJ water formed, (g) Sources of errors. (i) Heat is lost (ii) Density of solutions not exactly 1 gcm\(^{-3}\) (iii) Thermometer may not be accurately calibrated (iv) Loss of soln. during transfer and stirring. To minimize error; (i) Use lagged calorimeter (ii) Find water equivalent of calorimeter (iii) Measure mass of solution and mixture accurately. (iv) Careful string and transfer

Test Observation Inference (a) F + water and filtered The residue is black and filtrate is colourless (b)(i) (ii) Filtrate + litmus paper Filtrate + fehlings solution No effect Brick - red ppt. Neutral solution (c) Residue + dil. HCl A colourless, odourless gas evolved. The gas gives a pop sound with a lighted splint, light green solution formed The gas is hydrogen The residue is a metal above hydrogen in the activity series. Fe\(^{2+}\) suspected (d) Clear solution + NaOH\(_{aq}\) Dirty green ppt., insoluble in excess NaOH\(_{(aq)}\) Fe\(^{2+}\) confirmed (e) From the result above, F is confirmed to be a mixture. The residue is grey and metal, that is iron filings. The other component is soluble in water and reduces fehling's solution, which is either glucose or fructose.

(a) Heat the mixture. Ammonium chloride is sublime and collected in the cold upper part of the test-tube. The resulting solid is cooled and stirred with water to dissolve sodium chloride. Filter, wash, and dry the residue to obtain lead (lI) tetraoxosulphate (VI). Evaporate the filtrate to dryness to obtain sodium chloride crystals. (b)(i) Methyl orange (or phenolphthalein) (ii) Any two of the following; - Ensure that there are no air bubbles in the burette - Read off burette at eye level - Ensure that burette is not leaking - Rinse the conical flask with distilled water - Take three concordant readings. (ii) with methyl orange: yellow \(\to\) pink/orange with phenolphthalein: pink \(\to\) colourness.

No expanded working has been attached for this question yet. The verified answer is Option C.

(a) (1) Fast decline in the state of the economy is one of the reasons for the wide spread corruption in Nigeria. (2) Widespread corruption in Nigeria is also due to poor awareness of the rights of the citizens. (b) The common pattern that corruption takes in Nigeria is that people can only get what they .want through. bribery and personal relationships. (c) (1) Our leaders must live a life that is corruption - free (2) Emphasis must be laid on justice, fairness and selflessness in teaching moral instructions. (3) Less emphasis should be placed on acquisition of material wealth.

CO-EDUCATION IN SECONDARY SCHOOL SHOULD NOT BE ABOLISHED Good day, Chairman, Panel of Judges, my co-debaters, Ladies and Gentlemen. I am here to speak against the motion which states: Co-education in secondary schools should be abolished. Co-educational school or mixed school is a school where both male and female students learn together under the same roof and condition. A few years back, some people believed that mixed schools should be abolished. Nowadays, mixed or co-educational schools are very common and I strongly believe that the system should not be abolished, rather it should be encouraged. The greatest fear and the strongest point of the disciples of the old system is that immorality is common in mixed schools. I want to disagree with this point because there are cases of immorality occurring in purely girls' schools and one wonders who is to blame for this. It is therefore, hypocritical to have separate schools for boys and girls. The basic cause of immorality may even be the separation of boys from girls because when the two sexes attend the same school, study and play together, there is the tendency that they will take one another for granted. However, any opportunity for boys and girls to come together when they attend separate schools is often abused because from experience, boys continually nurse fanciful ideas about girls and vice-versa when they attend separate schools. Co-education encourages unity that is supposed to exist between male and female students. When boys and girls attend the same school, they learn about the psyche of one another. They will be able to understand the motives behind every behaviour and change of attitude. Over the years. they will gather numerous experiences which will enable them to live together in unity and harmony with the opposite sex even after leaving school. Therefore, co-education naturally gives practical and balanced orientation which make it possible for the young boys and girls to exist together under the same conducive atmosphere. Chairman, Panel of Judges, Ladies and Gentlemen you will agree with me that our young boys and girls need to be encouraged to study together so that they can know one another better. Co-education is one of the ways through which we can ensure peaceful marriages and homes in future. I am aware that my opponents would want to argue that juvenile delinquency and indiscipline are common in mixed schools. But I want to ask them a question: Are there no cases of juvenile delinquency and flagrant indiscipline in purely boys' or girls' schools? There have been cases of students flagrantly disobeying rules in girls' schools as well as in boys' schools. In co-educational schools, behaviours and comportment of boys and girls are finely regulated by the co-existence of the two sexes in schools. Boys will not reckon with girls who usually disobey their teachers and vice versa. Co-education promotes a healthy academic rivalry between boys and girls. No boy likes to be relegated to the background by a girl in his class. Therefore. it is common in co-educational schools that boys will want to dominate the girls academically and conversely, the girls always struggle hard to prove that whatever boys can do, they can do it better. In this way, a progressive and healthy academic and sporting rivalries always exist among boys and girls. Finally, it is cheaper to have boys and girls learning in the same schools than to have separate schools for the two sexes. Building separate schools for the different sexes will amount to wastage of public funds and duplication of schools and resources which can be conserved if we encourage co-education. Chairman, Panel of Judges Ladies and Gentlemen. why are we being hypocritical and wasteful by abolishing co-education? Co-education should be encouraged and promoted in our society. Thank you very much for listening.