WAEC Chemistry 2012 Theory Past Questions

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1(a)(i) Rutherford proposed that an atom comprises a positively charged nucleus (situated at the centre) with electrons revolving around it. (ii) The atomic number of an element is the number of protons present in the nucleus of its atom, while the mass number is the sum of the proton number and the neutron number. (iii) The relative atomic mass of chlorine is not a whole number because chlorine is an isotopic mixture; the conventional relative atomic mass for an isotopic element is the average weight of the various isotopes. (b)(i) The first ionization energy is the energy required to remove the most loosely bound electron from a gaseous atom to form a unipositive gaseous ion. (ii) Electrovalent compounds are composed of ions, rather than molecules; they conduct electricity in molten or aqueous form; they have high melting and boiling points. (iii) The elements in each pair have similar chemical properties because they have equal number of valence electrons. (c)(i) A homologous series is a family of organic compounds which are closely related. Features: members have a general molecular formula; they have the same functional group; they have a general method of preparation; they have similar chemical properties; they show gradation in physical properties; consecutive members differ in their molecular formula by a CH2 group and in their relative molecular mass by 14. (ii) Heterolytic fission refers to a process by which bond breaking is done in such a way that the electron pair is completely transferred to one of the atoms, leaving the other atom electron-deficient. (d)(i) For dilute HCl and NaOH, any indicator (methyl orange, litmus, phenolphthalein etc.) can be used. (ii) For HCl and NH4OH, methyl orange is used. (iii) For CH3COOH and NaOH, phenolphthalein is used. (e) E is ammonium chloride, F can be NaOH, KOH, Ca(OH)2, CaO etc., G is ammonia. 2(a) Diffusion is defined as the movement of particles from the region of higher concentration to the region of lower concentration, until even distribution is attained.

Model answer

(iv) P1=P2=1.013×10⁵Nm⁻² (constant pressure), so Charles' law applies. T1=300K, V1=150cm³, V2=300cm³ (doubled). T2 = (V2/V1)×T1 = (300/150)×300 = 600K = 327°C. (v) Increasing kinetic energy: Solid<Liquid<Gas. Increasing forces of cohesion: Gas<Liquid<Solid. (b)(i) Le Chatelier's principle states that when a physical constraint is imposed on a chemical system at equilibrium, the equilibrium position adjusts so as to neutralize the effect. (ii) Let 100% = 100g. For the first oxide: mass of oxygen = 11.1g, so mass of metal = 100−11.1 = 88.9g. Fixing a mass of oxygen (say 50g): (88.9/11.1)×50 = 400g of the metal. For the second oxide: mass of oxygen = 20g, mass of metal = 100−20 = 80g. (80/20)×50 = 200g of the metal. Ratio of the metal in the two oxides = 400:200 = 2:1, a simple whole-number ratio, confirming the law of multiple proportion. (c)(i) Water should be added to the mixture. Since C is soluble, it dissolves. B and A are then filtered off as residue, since they are not soluble. The dry form of C can be obtained by evaporating the filtrate. The residue, containing A and B, should be heated until it melts. A and B are then separated using simple distillation, since their boiling points are far apart (A boils at 117°C while B boils at 160°C). (ii) Mass of mixture = (mass of A)+(mass of B)+(mass of C). 25.25g = 7.52g+8.48g+(mass of C). Mass of C = 25.25−16 = 9.25g. % of C in the mixture = (9.25/25.25)×100% = 36.63%.

Model answer

(a)(i) Nuclear fission is a process by which a heavy nucleus splits into lighter nuclei of comparable masses, with the release of energy. (ii) Let a represent the number of beta particles and b represent the number of alpha particles. Since beta particle does not have mass, the difference between the mass of ₂₉³⁸U and ₉₀²³⁰Th will be the mass of alpha particles: mass of alpha particle = 238−230 = 8. Since one alpha particle has a mass of 4, there are two alpha particles. b=2. Substituting b=2 into the nuclear equation and balancing: ₂₉³⁸U → ₉₀²³⁰Th + a(−1⁰β) + 2(₂⁴α). 92=90+a(−1)+2(2); 92=90−a+4; a=94−92=2. So two beta particles are also involved. (b)(i) The equilibrium position shifts to the right, favouring production of more AB3. (ii) The equilibrium position shifts to the left, favouring production of more A2 and B2 (since there are more moles of gas on the reactant side). (iii) No change; a catalyst alters both the forward and backward reaction rate alike, so equilibrium position is unaffected. (c) Of the three halogens, fluorine is the most electronegative. This is why NaF has the highest lattice energy. Generally speaking, for a given cation, the lattice energy increases as the electronegativity increases and vice versa. (F<Br<I decreasing electronegativity corresponds to NaF>NaBr>NaI decreasing lattice energy.) (d)(i) Oxidizing property (NO2 oxidizes Cu to CuO while itself being reduced to N2). (ii) Acid anhydride (NO2 disproportionates in water to form nitric and nitrous acids). (e)(i) Top of the furnace: Fe2O3+3CO→2Fe+3CO2. (ii) Middle of the furnace: CaCO3 →(Δ) CaO+CO2. (iii) Bottom of the furnace: C+O2→CO2; then CO2+C→2CO. (f)(i) The products of destructive distillation of coal are coke, coal tar, coal gas and ammoniacal liquor. (ii) Coke is used in the extraction of metals; coal tar is used for the synthesis of many chemical substances; coal gas is utilized as fuel; ammoniacal liquor is used for making fertilizers.

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(a)(i) Structural isomers are compounds with the same molecular formula but different structures. (ii) Structural isomers of C4H10O (alkanols): CH3CH2CH2CH2OH (butan-1-ol); CH3CH2CH(OH)CH3 (butan-2-ol); (CH3)2CHCH2OH (2-methylpropan-1-ol); (CH3)3COH (2-methylpropan-2-ol). (iii) 2-methylpropan-2-ol does not react easily with K2Cr2O7, since it is a tertiary alkanol. (b)(i) C5H12+Cl2→C5H11Cl+HCl. (ii) The major product is chloropentane, e.g. CH3CH2CH2CH2CH2Cl (1-chloropentane). (iii) Light serves as a catalyst for the reaction. (iv) Pentane would distil off first. This is because chloropentane has a higher relative molecular mass than pentane (C5H11Cl ≈ 106.5, pentane C5H12 ≈ 72), and generally the lower-boiling component distils off first. (v) If but-1-ene were used instead of pentane, the main product formed would be C4H8Cl2 (1,2-dichlorobutane). (c)(i) Hydration of CH3CH(CH3)CH=CH2 gives 3-methylbutan-2-ol: CH3CH(CH3)CH(OH)CH3. (ii) Hydration of CH2=CHCOOH gives 2-hydroxypropanoic acid: CH3CH(OH)COOH. (d)(i) In acidic medium: CH3CH(NH3⁺)COOH. In alkaline medium: CH3CH(NH2)COO⁻. (ii) C=60.5%, H=6.5%. N = (0.0313/0.309)×100% = 10.129%. C+H+N+O=100%, so 60.5+6.5+10.129+O=100%, O = 22.871%. Mole ratio: C=60.5/12=5.042, H=6.5/1=6.5, N=10.129/14=0.7235, O=22.871/16=1.4292. Dividing by smallest (0.7235): C=7, H=9, N=1, O=2. Empirical formula = C7H9NO2.

Model answer

7(a)(i) The standard electrode potential is the potential difference between the electrode and the hydrogen electrode under standard conditions of 298K temperature, 1moldm⁻³ concentration and 1 atm pressure. (ii) Temperature and concentration. (iii) To calculate the emf of an electrochemical cell; to determine the feasibility of a reaction. (iv) The diagram shows a Daniell-type cell: a zinc electrode in ZnSO4 solution connected via a salt bridge to a copper electrode in CuSO4 solution, with a voltmeter (V) connected across the two metal electrodes. (v) The shorthand notation for the cell is Zn/Zn²⁺//Cu²⁺/Cu. E°cell = E°red(Right) − E°red(Left), where E°red(Right) is the standard reduction potential of the compartment on the right and E°red(Left) is that of the compartment on the left. E°cell = 0.34V − (−0.76V) = +1.10V. (b)(i) I. Oxidation is defined as loss of electron(s). II. An oxidizing agent is the substance that gains electron(s). (ii) Reduction half equation: MnO4⁻ → Mn²⁺. Adding 4 moles of H2O to the right and balancing with hydrogen ions: MnO4⁻+8H⁺→Mn²⁺+4H2O. Oxidation half equation: I⁻→I2 i.e. 2I⁻→I2+2e⁻. To make electrons equal, the oxidation half equation is multiplied by 5 while the reduction half equation is multiplied by 2: 10I⁻→5I2+10e⁻; 2MnO4⁻+16H⁺+10e⁻→2Mn²⁺+8H2O. Adding the two and eliminating electrons: 2MnO4⁻+16H⁺+10I⁻→2Mn²⁺+5I2+8H2O. (c)(i) Carbon (II) oxide is neutral. (ii) Sulphur (IV) oxide is acidic. (iii) Aluminium oxide is amphoteric. (iv) Lithium oxide is basic. (d) Hydrogen bonding is an intermolecular force present in molecules where hydrogen is bonded to very small electronegative elements such as nitrogen, oxygen and fluorine.

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8(a)(i) I. Biotechnology refers to the use of biological organisms for developing products suitable for human use. II. Biogas refers to a fuel obtained from organic waste using some bacteria as agents through the process of anaerobic decomposition. (ii) Biotechnology plays a key role in waste management; it is harnessed in making substances such as drugs and fertilizers. (b)(i) Yeast should be added to cane sugar juice and then allowed to be for a while. The enzyme sucrose hydrolyses the sucrose in the cane sugar to glucose. The enzyme zymase then ferments the glucose into ethanol. (ii) The by-product is carbon dioxide. (iii) Uses: it is used for preserving soft drinks; it is used as a fire extinguisher; it is used in the Solvay process for the manufacture of NaHCO3 and Na2CO3. (iv) The ethanol from petroleum is said to be non-renewable because the petroleum used for making the ethanol cannot be retrieved once depleted. But sugarcane can be re-planted, from which cane sugar can then be obtained. Thus, the ethanol form cane sugar is said to be renewable. (c) A heavy chemical is produced in large quantity while a fine chemical is produced in small quantity. H2SO4 is an example of a heavy chemical while dyes are examples of fine chemicals. (d) Increasing order of deviation from ideal gas behaviour: O2 < Cl2 < HCl. Deviation from ideal gas behaviour increases as the intermolecular forces increase. The strongest intermolecular forces are found in HCl since it is a polar molecule. Hence, HCl shows the highest deviation from ideal gas behaviour. Both O2 and Cl2 are non-polar; however, the Van der Waals' forces in Cl2 are greater than that of O2 because Cl2 has a greater relative molecular mass.