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C2H4(g) + 3O2(g) \(\to\) 2CO2(g) + 2H2O(l) 1 mole of ethene required 3 moles of O2 28g of ethene required 3 x 22.4dm3 of O2 14g will require \(\frac{14 \times 3 \times 22.4dm^3}{28}\) = 33.6dm3

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Global warming is a gradual increase in the earth’s temperature generally due to the greenhouse effect caused by increased levels of carbon dioxide, CFCs, and other pollutants., consequences of global warming include: Seasonal/climate change, Rising oceans, Increase in flooding occurrences. Melting ice caps or polar ice. Increase in the average temperature of the world.

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(a) Nucleons refer to the protons and neutrons that are found in the nucleus OR Nucleons are nuclear particles which are protons and neutrons. (b) Graham's law of diffusion states that, at constant temperature and pressure, the rate of diffussion of a gas is inversely proportional to the square root of its density or molar mass or vapour density (c) Aluminum oxide layer forms on the surface of the metal. The oxide layer forms a potective coating layer which does not flake off. (d) — atomic radius, Ionic radius, ionization energy. —redox potentials, electronegativity, electron affinity. — melting point, boiling point, electrical conductivity. (e) — Molecules of real gases have significant attractive forces between them. — The volume of the molecules of real gases is not negligible. --- Collision between the molecules are not perfectly elastic. (f) — Separation of crude oil into its components / petroleurn refining. —Purification of ethanol. —Separation of liquefied air (g) —Position of the ions in the electrochemical series —Concentration of the ions in the electrolyte —Nature of the electrodes used. (h)(i) Substitution reaction. (ii) Addition reaction. (i) — Carbon or soot, Carbon two oxide (carbon II oxide), Water. (j) Filtration, Crystallisation, Evaporation, decantation, centrifugation

(a)(i) Isotopy (ii) They have different mass numbers or number of neutrons. (iii) —Hydrogen, —Chlorine. —Carbon, —Oxygen. (iv) 1 S\(^2\)2S\(^2\)2P\(^62\)3S\(^2\)3P\(^5\) (b)(i) 1. DIFFERENCES IN PHYSICAL PROPERTIES METALS NON — METALS Good conductors of heat and electricity Poor conductors of heat and electricity Malleable - can be beaten into thin sheets Not malleable / brittle Ductile — can be stretched into wire Non ductile / biittle Possess metallic luster Do not possess metallic luster Are sonorous Not sonorous Solid at room temperature except mercury / high melting point Solids, liquids or gases at room temperature / low melting point 2. DIFFERENCES IN CHEMICAL PROPERTIES METALS NON METALS Lose their valence electrons easily / ionize by electron loss Gain or share valence electrons easily / ionize by electron gain Form oxides that are basic / amphoteric Form oxides that are acidic / neutral Are good reducing agents Are good oxidizing agents Have lower electro negativities Have higher electro negativities Have low ionization energy Have high ionization energy (Some reactive metals) react with dilute acids to give off hydrogen Do not react with dilute acids to give off hydrogen. (ii) (l) BeO, Al\(_2\)O\(_3\), ZnO, SnO\(_2\), PbO, PbO\(_2\) (II) LiH, NaH, KH, CaCO\(_3\) MgH\(_2\) (III) Li\(_2\)CO\(_3\), BeCO\(_3\), MgCO\(_3\), CaCO\(_3\), ZnCO\(_3\), PbCO\(_3\) (IV) AlCl\(_3\), MgCl\(_2\), PCl\(_3\), SiCl\(_4\), NH\(_4\)Cl (c)(i) - form coloured compounds / ions, - exhibit variable oxidation states. - form complex compounds / ion. - have magnetic properties (paramagnetism). - have catalytic properties. (ii) —> 1s\(^2\)2s\(^2\)2p\(^6\)3s\(^2\)3p\(^6\)3d\(^{10}\)4s\(^2\) are fully filled (d) Mr(MgCO\(_3\)) = 24 + 12 + (16 x 3) 84 Mr(Na\(_2\)CO\(_3\)) = (23 x 2) +12 + (3 x 16) = 106 moles of MgCO\(_3\) = \(\frac{3.36}{84}\) = 0.04 moles of Na\(_2\)CO\(_3\) = moles of MgCO\(_3\) = 0.04 mass of NaCO\(_3\) = 0.04 x106 = 4.24g OR Mr(MgCO\(_3\)) = 24 + 12 + (1 6 x 3) 84 Mr(Na\(_2\)CO\(_3\)) = (23 x 2) +12 + (3 x 16) = 106 84g of MgCO\(_3\) produced from 10\(^6\)g Na\(_2\)CO\(_3\) 3.36g of MgCO\(_3\) will be produced From 106 x \(\frac{3.36}{84}\) = 4.24g

(a)(i) Structural Isomers of C\(_4\)H\(_4\) (ii) (I)Tertiary(3^\circ)alkanol. (II) Primary (1^\circ)alkanol. (III) Secondary (2^\circ) alkanol (b)(i) (I) CH\(_3\)CH\(_2\)COOK, H\(_2\) (II) CH\(_3\)CH\(_2\)COOC\(_4\)H\(_9\) H\(_2\)O (III) CH\(_3\)CH\(_2\)CH\(_2\)COOH (ii) Major products in (b)(i) (1) — Potassuim propanoate (II) — Butyl propanoate (III) — Butanoic acid (C)(i) % carbon by mass = 100 – 14.3 (l) = 85.7% Ratio of atoms = \(\frac{C}{\frac{85.7}{12.0}}\) = 7.1415; \(\frac{H}{\frac{14.3}{1.0}}\) = 14.3 Divide by smaller number = \(\frac{7.1415}{7.1415}\) = 1; \(\frac{14.3015}{7.1415}\) = 2 Empirical Formular = CH\(_2\) (i) 2.24dm\(^3\) = 7.0g of R 22.4dm\(^3\) = \(\frac{7.0}{2.24}\) x 22.4 = 70gmol\(^1\) OR Molar volume of gas = 22.4 dm\(^3\) at s.t.p 22.4 22.4 dm\(^3\) gas = 1 mole 22.4 dm\(^3\) = \(\frac{2.24}{22.4} \times 1\) = 0.1 mol But numbers of moles = 0.1 Mr = \(\frac{7}{0.1}\) = 70gmol\(^1\) Molecular formular = C\(_5\)H\(_{10}\) (CH_2)n = 70 14n = 70 n = \(\frac{70}{4}\) = 5 (d) The occurence of / existence of / condition whereby compounds with the same molecular formula have different structural formulae.

(a) (i) Concentrated tetraoxosulphate (vi) acid fused calcuim chloride. (ii) Pass or bubble each gas into lime water, if lime water turns milky then CO, is present if no reaction then CO is present. (b) (i) –Atmospheric air is compressed and cooled to liquefy it. – The liquefied air undergoes fractional distillation. – Nitrogen boils out first at – 196^\circC followed by oxygen at – 183^\circC. (ii) – Splint burns brighter / continues to burn in oxygen. Splint is extinguished in nitrogen (c) (i) – Bauxite is more abundant than any other ore. – Bauxite has less percentage of silicates as impurity. (ii) . – iron (III) oxide. – Trioxosilicates(IV) / silica / sand. (iii) To dissolve the aluminium oxide so that the impurities can be filtered off (iv) Aluminium is very reactive to be reduced by the eommon reducing agenfleoaction. (v) Al\(_2\)O\(_3\) + 2NaOH \(\to\) 2NaAlO\(_2\) +H\(_2\)O or Al\(_2\)O\(_2\) + 2NaOH + 3H\(_2\)O --> 2NaAl(OH)\(_4\) (d) (i) The ions are not free at 25^\circC since sodium chloride is in the solid state but between 801^\circC and 1413^\circC the ions are free / mobile since it is now in the molten / liquid state. (ii) Sodium metal is very reactive with air/ it is easily oxidised in air / tarnishes easily. (e) (i) Brown precipitate / reddish brown / orange solid formed (ii) 3Na\(_2\)CO\(_{3(aq)}\) + 2Fe\(^{3+}\) --> F\(e_2\) (CO\(_3\))\(_{3(s)}\) + 6Na\(^{+}\)

(a) (i) let x = oxidation number of sulphur 2(1) + 2 + 3(-2) = 0 x = +2 (ii) Allotropes of sulphur are: – rhombic / alpha sulphur. – monoclinic / Beta / prismatic sulphur. – amorphous / delta sulphur. – plastic sulphur (iii) – Macromolecule / giant molecule /3 dimensional. – carbon atoms are covalently bonded. – crystalline in nature. (b) (i) Carbon(IV)oxide. – methane. – nitrogen(I)oxide. –water vapour – ozone – chlorofluorocarbons. (ii) – Global warming / climate change. (iii) – Ammonium trioxonitrate (V) /aerosol (accept ammonium nitrate). – Combustion of fuel at high temperature. (iv) (I) 2KNO\(_3\) --> 2KNO\(_2\) + O\(_2\) (II) 2AgNO\(_{_3(s)}\) ---> 2Ag\(_s\) + 2NO\(_{2(g)}\) + O\(_2\) (c) (i) – Heat the solution to concentrate it or leave in a warm place. - partially evaporate solution and allow to cool filter off crystals. - pick out crystals - Dry crystals with filter paper. (ii)(1) Ammonia readily associates with water molecules through hydrogen bonding. (II) Chlorine molecules are smaller or chlorine molecules have fewer number of electrons and the strength of van der waals forces between molecules are weaker as compared to that of iodine OR iodine molecules are higher and the wander waal's forces between molecules are stronger as compared to that of chlorine. (d) Mr(NaCl) = 23.0 + 35.5 = 58.5g/mol\(^1\) Mass of NaCl = 5.85g Molar mass of NaCl= 58.5g/mol\(^1\) Amount of NaCl in 85g = \(\frac{5.85g}{58.5gmol^{-1}}\) = 0.1mol 1 mol of NaCI = 22.4dm\(^3\) of HCI :. 0.1 mol = \(\frac{0.1}{1}\) x 22.4 = 2.24dm\(^3\) OR Molar mass NaCl = 23.0 + 35.5 = 58.5gg/mol\(^1\) 58.5g of NaCl = 22.4dm\(^3\) HCl 58.5g = \(\frac{22.4}{58.5}\) x 5.85 = 2.24dm\(^3\)

see diagram in textbook

Burette Reading (cm\(^3\)) Rough 1st Titration 2nd Titration 3rd Titration Final Burette Readings (cm\(^3\)) 36.50 30.40 41.70 26.80 Initial Burette readings (cm\(^3\)) 12.30 05.90 17.10 01.90 Vol. of acid used (cm\(^3\)) 24.20 24.50 24.60 24.90 Titre value = \(\frac{24.50 + 24.60}{2}\) = 24.55cm\(^3\) (b)(i) Molar concentration of KMNO\(_4\) \(\frac{C(MnO_4^-) \times V(MnO_4^-)}{C(C_2O_4^{2-}) \times V(C_2O_4^{2-})} = \frac{n(MnO_4^-)}{n(C_2O_4^{2-})}\) \(\frac{C(MnO_4^-) \times 24.55}{0.05 \times 25} = \frac{2}{5}\) C(MnO\(_4^-\)) = \(\frac{0.05 \times 25 \times 2}{24.55 \times 5}\) = 0.0204 mol/dm (ii) Mass concentration of KmnO\(_4\) Molar mass of KmnO\(_4\) = 39 + 55 + (16 x 4) = 158g/mol Mass conc. KmnO\(_4\) = molar conc. x molar mass = 0.0204 x 158 = 3.22g/dm\(^3\) (iii) Volume of CO\(_2\) (gas evolved) \(\frac{n(CO_2)}{n (H_2C_2O_4)}\) = \(\frac{10}{5}\) n(CO\(_2\)) = \(\frac{10}{5}\) x 0.05 = 2 x 0.05 = 0.100 mol V(CO\(_2\)) at s.t.p = n x Vm 25cm\(^3\) of H\(_2\)C\(_2\)O\(_4\) = 0.100 x 22.4dm\(^3\)mol\(^{-1}\) x \(\frac{25}{1000}\) = 0.056dm\(^3\) or 56.0 cm\(^3\) Alternative method Volume of CO\(_2\) gas evolved Amount of H\(_2\)C\(_2\)O\(_4\) used = 0.050 x \(\frac{25}{1000}\) = 0.00125 mol From the equation of the reaction 5 mol of C\(_2\)O\(_4^{2-}\) = 10 mol of CO\(_2\) 0.00125 mol C\(_2\)O\(_4^{2-}\) = 0.00125 mol x \(\frac{10}{5}\) = 0.0025 mol CO\(_2\) @S.T.P 1 mol of CO\(_2\) = 22.4dm\(^3\) 0.00250 mol = 22.4 x 0.0025 = 0.056 dm\(_3\) or 56.0 cm\(^3\)

(a)(i) Wide difference in boiling points of components (for miscible liquids)/ difference in physical state (for liquid - solid solution) (ii) Difference in absorption/retention rate of components (ii) Close difference in boiling points of components. (b)(i) To make the endpoint colour change colour easily visible/ in order not to overshoot the endpoint. (ii) To ensure (high) accuracy of the titre value (iii) Using Dilution formula M\(_1\)V\(_1\) = M\(_2\)V\(_2\) M\(_1\) = 2.5, V\(_1\) = ?, M\(_2\) = 0.20, V\(_2\) = 500 \(\frac{M_1V_1}{M_1}\) = \(\frac{M_2V_2}{M_1}\) Make V\(_1\) the subject V\(_1\) = \(\frac{M_2V_2}{m_1}\) = \(\frac{0.20 \times 500}{2.5}\) = 40.0 cm\(^3\)

Test Observation Inference (a) C\(_{(s)}\) + distilled water + filter Party dissolved/soluble colourless filtrate obtained, water residue C contains soluble and insoluble salts (b) Filtrate C + AgNO\(_{3(aq)}\) + HNO\(_{3(aq)}\) + NH\(_{3(aq)}\) White precipitate Precipitate insoluble Precipitate soluble/dissolved Cl\(^{-}\) or CO\(_3^{2-}\) or SO\(_3^{2-}\) Cl\(^-\) Cl\(^-\) present (c)(i) Residue + HNO\(_{3(aq)}\) Effervescence, colourless, Odourless gas evolved Gas turned lime water milky CO\(_2\) gas from CO\(_3^{2-}\) evolved (ii) Solution from C(i) + NaOH\(_{3(aq)}\) drops Then in excess White gelatinous precipitate Precipitate dissolved/soluble Zn\(^{2+}\) or Al\(^{3+}\) present (iii) Solution from C(i) + NH\(_{3(aq)}\) drops Then in excess White gelatinous precipitate Precipitate dissolved/soluble Zn\(^{2+}\) or Al\(^{3+}\) present Zn\(^{2+}\) present