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Test Observation Inference (a)(i) C + H\(_2\)O + litmus paper C dissolves, blue litmus paper turned red soluble salt solution is acidic (ii) Solution of C + NaOH\(_{(aq)}\) in drops + excess Dirty green ppt. ppt insoluble in excess (iii) Solution of C + BaCl\(_{2(aq)}\) + dil. HCl White ppt. ppt is insoluble in dil. HCl (b)(i) D + heat D changes from white to yellow and turned to white on cooling, colourless gas evolved, gas turned lime water milky (ii) Residue + dil. HCl Gas evolved. Residue dissolves from a colourless solution (iii) Solution + NaOH from b(ii) in drops + excess White gelatinous ppt. ppt soluble in excess or ppt. dissolves. (iv) Solution (b)(ii) + NH\(_{3(aq)}\) + excess White gelatinous ppt, ppt soluble in excess or ppt. dissolved

(a)(i) Covalent bond and dative or co-ordinate covalent bond. (ii) The characteristics are: —They have high melting point or boiling point. —They conduct electricity when molten or in solution —They exist as aggregates of ions —They are soluble in polar solvent e.g water and insoluble in non-polar solvents (iii) State two differences between chemical reactions and . Chemical reactions Nuclear reactions It involves (orbiting) electrons It involves nuclear particles It does not produce entities or there is no change in atomic number It results in the production of new nuclides or there is a change in atomic number Is not accompanied with great heat changes. It is accompanied with evolution of a lot of heat The rate is affected by external factors such as pressure, temperature, etc. Nuclear reactions The rate is not affected by the external factors. b)(i) 1s\(^2\)2s\(^2\)2P\(^2\) (ii) Group (II) or alkaline earth metal (iii) YX\(_2\) (iv) X is a good oxidizing agent because its outer shell contains seven electrons, it therefore accepts one electron to form an octet structure. From the efinition of an oxidizing agent, oxidizing agents are good acceptors of electrons. (v) Y (1s\(^2\)2s\(^2\)2p\(^6\)3s\(^2\)) would be expected to form a basic oxide because Y is a metal or Y is an electropositive element. c)(i) \(^{14}_6C\) \(\to\) \(^0_1X\)+ \(^{14}_7N\) X is a beta particle (ii) \(^{14}_7N\) + \(^4_{-1}Y\) \(\to\) \(^1_1H\) + \(^{17}_8O\) Y is an alpha particle d)(i) Metals are zinc, potassium and mercury (ii) Compounds that are gases at room temperature are carbon (IV) oxide and methane (iii).Non-metal that is solid at room temperature is sulphur. ,

The assumptions of kinetic theory of gases are: -The forces of attraction between gas particles are legligible -Gas molecules move randomly in straight lines -They collide with one another and with the walls of the container -The collision of the molecules is perfectly elastic. (ii) Sublimation (iii) Ammonium chloride (NH\(_4\)CI), iodine (I\(_2\), naphthalene and aluminium chloride. (iv) \(\Delta\) = \(\frac{1}{\sqrt{d}}\) or r = \(\frac{k}{\sqrt{d}}\) (b) 3Fe\(_{(s)}\) + 4H\(_2\)O \(\rightleftharpoons\) Fe\(_3\)O\(_{4(s)}\) + 4H\(_{2(g)}\); \(\Delta\)H = -150KJ mol\(^{-1}\) (i) The forward reaction is exothermic, decrease in temperature will favour the forward reaction or the position of equilibrium shifts to the right. (ii) There will be no effect on the position. of equilibrium as the number of gaseous molecules are the same on both sides. (iii) Removal of hydrogen Shifts the position of equilibrium to the right because of the decrease in concentration of H\(_2\). hence more Fe\(_{(s)}\) and water will react to produce more H\(_2\). (c)(i) Q > S > P. (ii) The larger the surface area of a solid reactant, the faster the-reaction rate. The surface area of zinc dust is the largest while that of the rod is the smallest. (iii) Zn\(_{(s)}\) + 2HCl\(_{(aq)}\) \(\to\) ZnCl\(_{2(aq)}\) + H\(_{2(g)}\) (d)(i) The pH of a solution is a measure of the degree of acidity or alkalinity of the solution. (ii) (-) The pH increases because acidity decreases (-) pH decreases because alkalinity decreases (-) pH remains constant because the salt solution is neutral (II) Non-catalysed activated complex (I). (ii) Activation energy (I) with catalyst (iii) Enthaipy change (\(\Delta\)H)

(a) (b)(i) Hydrogen is lighter or-4ess dense than air. The hydrogen particles therefore diffuse faster than air. (ii) The MnO\(_2\) catalyses the decomposition of the hydrogen peroxide into-water and oxygen (iii) HCl is covalent, it produces no ions in a non-polar solvent like methylbenzene. In water it ionizes to give H\(^+\) or \(H_3\)O\)(^+\) as the only positive charge which turns blue litmus paper red. (c)(i) MnO\(^{-4}\). Since oxidation number of Mn decreases from +7 to +2. (ii) C\(_2\)O\(_4\)\(^{-2}\). Since the oxidation number of carbon increases from +3 to +4 (iii) MnO\(^{-4}\) + 8H\(^+\)+ 5e\(^-\) \(\to\) Mn\(^{2+}\) + 4H\(_2\)O (d)(i) An electrochemical cell is a device which converts chemical energy into electrical energy. ii) Electrochemical cell Electrolytic cell 1. Chemical reaction produces electricity Electric current is required for the reaction to occur 2. Electrons are produced as a result of oxidation at an electrode Electrons are pushed by an outside source such as a battery. 3: Cathode is the positive electrode Cathode is the negative electrode 4. Anode is negative electrode Anode is positive electrode

(a) The condition of the reaction is the presence of ultra-violet light or sunlight. (ii) Substitution reaction. (iii) Uses of methane are: (1) It is used as a fuel (2) It is used for making hydrogen, carbon black, carbon (iv) oxide; etc. (iv) Methane occur as natural gas or refinery gas (b)(i) (i) Hydrogen is evolved. (ii) Resulting liquid is both colourless (iii) Both reactions are exothermic (ii) (iii) Heavy-chemicals are chemicals that pre made in large quantities while fine chemicals are produced in small quantities. (iv) Examples of heavy chemicals are: H\(_2\)SO\(_4\), NaOH, fertilizer, alum. Examples of fine chemicals are dyes, photographic chemicals, analytical chemicals, drugs. (c)(i) Empirical formula of X and Y; No of atoms of C = \(\frac{92.3}{12}\) = 7.69 No. of atoms of H = \(\frac{7.7}{1}\) = 7.7 Ratio of atoms atoms of C:H = \(\frac{7.69}{7.69}\) : \(\frac{7.7}{7.69}\) ; 1 : 1 Empirical formula = CH (ii) Let the molecular formula of X be (CH)n ; (12 + 1)n = 13n ; 13n = 26 n = \(\frac{26}{13}\) = 2 molecular formula of X =C \(_2\)H \(_2\); Let the molecular formula of Y be (CH)Y (12 + 1) Y = 13Y ; 13Y = 78 ; Y = \(\frac{78}{13}\) = 6 ; Molecular formula of Y = C \(_6\)H \(_6\). (d)(i) Hydrolysis (ii) Amino acids (iii) Functional groups or amino acids are: amino group (--NH \(_2\)) ai carboxylic group (--COOH).

(a) Removal of permament hardness in water (i) Physical method - by distillation. Chemical method - addition of washing soda and the use of ion-exchange resin. (ii) Removal of CaCO\(_3\) Physical method ---Filtration. Chemical method - Bubbling CO\(_2\) into the mixture for a long time or addition of an acid. (ii) Disadvantages of hard water - Waste soap - causes furring in pans and kettles - results in burst pipe narrow tubes in boilers. (b)(i) The crystals lose their water of crystallization to the atmosphere. (ii) Each mole of CaCl\(_2\) ionizes to give two moles of Cl\(^-\) - whereas one mole of NaCl produces one mole of CL\(^-\) on ionization. (iii) With the iron filings, a greater surface area is exposed than in the nails and since raa increases with increase in surface area. (c)(i) (i) Amphoteric oxides - ZnO (ii) Neutral oxide - CO (iii) Adidic oxide - NO\(_2\) (ii) (i) Acid Anhydride (H\(_2\)CO\(_3\)) is CO\(_2\) (ii) Acid Anhydride (H\(_2\)SO\(_4\)) is SO\(_3\) (iii) COCl\(_2\) is cobalt (II) chloride Mg(NO\(_3\))\(_2\) is magnesium trioxonitrate (V) (d)(i) Chlorofluorocarbon are aerosols etc. (ii) 1 mole of O\(_2\) occupies 22.4 dm\(^3\) at s.t.p; 0.125 mole O\(_2\) occupies (22.4 x 0.125) dm\(^3\) at s.t.p. from \(\frac{P_1V_1}{T_1}\) = \(\frac{P_2V_2}{T_2}\) = 28 dm\(^3\) = \(\frac{28 \times 1.01 \times 10^5}{273}\) = \(\frac{v^2 \times 2.02 \times 10^5}{300}\) \(v^2 = \frac{2.8 \times 1.01 \times 300}{273 + 2.02} = 154dm^3\) (iii) Fractional distillation of liquid air or electrolysis of water

(a)(i) Copper is more reactive than silver or copper is higher in the electrochemical series than silver. Hence it will displace silver ions from solution. Copper, being less reactive than lead cannot displace lead, henc it remains insoluble in Pb(NO\(_3\))\(_{2(aq)}\). (ii) No. of mole of Ag\(^+\) in the solution. \(\frac{250cm^3 \times 0.2}{1000}\) mole Ag\(^+\) = 0.05mole; Cu + 2Ag+ 2Ag\(^+\) \(\to\) 2Ag\(^+\) + Cu\(^{2+}\) 63.5g Cu dissolves in 2 mole Ag\(^+\) 1.06g Cu will dissolve in \(\frac{2 \times 1.06}{63.5}\) mol Ag\(^+\) = 0.033 (iii) Amount of (Ag\(^+\)) silver ion present = 0.05. Since this is more than the amount required (0.033) i.e. the AgNO\(_{3(aq)}\) is in excess, all the copper turnings will dissolve (b)(i) Acids react with electropositive metals to liberate hydrogen (ii) Acids neutralize bases to form salts. (iii) Acids react with trioxocarbonate (IV) salts to liberate CO\(_{2(g)}\) (iv) Acid react with trioxosulphate (IV) salts to liberate SO\(_2\) (ii) Large scale uses of HNO\(_3\) (1) For manufacturing nitrocompounds such as nitrocellulose which is used for making artificial silk, explosives or compounds for manufacturing dyes. (2) For manufacturing fertilizer. (3) As an oxidizing agent. (4) rocket fuel. (iii) 2Pb(NO\(_3\))\(_{2(s)}\) \(\to\) 2PbO\(_{(s)}\) + 4NO\(_{2(g)}\) + \(O_2\) - 2AgNO\(_{3(s)}\) -> 2Ag\(_{(s)}\) + 2NO\(_{2(g)}\) + O\(_{2(g)}\) (c)(i) The litmus paper turns pink or red and then white or bleached. (ii) Redox reaction (iii) Bleachinc agent (d)(i) Coke, coal tar, coal gas and ammoniacal liquor (ii) uses Coke - a reducing agent or used as fuel. Coal tar -in manufacture of drugs, dyes, explosives. Coal gas - as fuel. Ammoniacal liquor -manufacture of fertilizer.

(a) The dry gas does not contain H\(_3\)O\(^+\) or H\(^-\) or OH\(^+\) which changes the colour of litmus or the ions H\(_2\)O or H\(^+\) or OH\(^-\) are produced only in the presence of water. (i) The reagent barium chloride can form a white precipitate with other ions if the solution is neutral. Addition of a mineral acid removes the interfering (CO\(_3^{2-}\), SO\(^{2-}_3\)) ions if they are present: (b)(i) Chlorine, carbon (II) oxide, hydrogen sulphide, hydrogen chloride, sulphur (IV) oxide, nitrogen (V) oxide and ammonia. (ii) They should be prepared in the fume chamber. (I) For keeping solvents such as distilled water (II) it is used as a support for flasks or beakers during heating (C)(i) Pb(NO\(_3\))\(_2\) decomposes, brown fumes are given off, yellow residue or reddish-brown residue left. (i) Greenish-yellow gas evolved.

Titration number Rough 1 2 3 Final reading 17.90 18.00 18.00 28.00 Initial reading 0.00 0.00 0.00 0.00 Volume of solution A used/cm\(^3\) 17.90 18.00 18.00 18.00 HCl\(_{(aq)}\) + KOH\(_{(aq)}\) + H\(_2\)O\(_{(l)}\) Average volume of acid used (cm\(^{3}\) = \(\frac{18.00 + 18.00 + 18.00}{3} = \frac{54.00}{3}\) = 18.00 cm\(^3\) (b)(i) To calculate the concentration of solution A molar mass of HCl = 1 + 35.5 = 36.5 mol \(^{-1}\) concentration in g/dm\(^3\) = 5 concentration A (C\(_A\)) = \(\frac{5.0}{36.5}\) = 0.137 mol/dm\(^3\) (ii) To calculate the concentration of solution B \(\frac{C_AV_A}{C_BV_B}\) = \(\frac{N_A}{N_B}\) HCl\(_{(aq)}\) + KOH\(_{(aq)}\) \(\to\) KCl\(_{(aq)}\) + H\(_2\)O\(_{(l)}\) N\(_A\) = mole ratio of acid = 1 N\(_B\) = mole ration of base = 1 C\(_A\) = concentration acid = 0.137 mol/dm\(^3\) C\(_B\) = concentration of base = ? V\(_A\) = volume of acid used in cm\(^3\) = 18.00 cm\(^3\) V\(_B\) = volume of base used = 25.00 cm\(^{3}\) C\(_B\) = \(\frac{C_AV_A \times N_B}{V_B \times N_A} = \frac{0.137 \times 18 \times 1}{25 \times 1}\) = \(\frac{2.466}{25}\) = 0.099 mol/dm\(^{3}\) (iii) To calculate percentage purity of KOH in B; Molar mass of KOH = 56g/dm\(^3\) Concentration in mol/dm\(^3\) = \(\frac{\text{concentration in g/dm}^3}{\text{molar mass}}\) Molar mass = 0.099 x 56 = 5.54 g/dm\(^3\) Percentage purity = \(\frac{\text{mass concentrate}}{\text{mass of impure KOH}}\) x \(\frac{100}{1}\) = \(\frac{5.54}{6.50} \times \frac{100}{1}\) = 85.23%

Test Observation Inference (a)(i) C + H\(_2\)O + litmus paper C dissolves, blue litmus paper turned red soluble salt solution is acidic (ii) Solution of C + NaOH\(_{(aq)}\) in drops + excess Dirty green ppt. ppt insoluble in excess (iii) Solution of C + BaCl\(_{2(aq)}\) + dil. HCl White ppt. ppt is insoluble in dil. HCl (b)(i) D + heat D changes from white to yellow and turned to white on cooling, colourless gas evolved, gas turned lime water milky (ii) Residue + dil. HCl Gas evolved. Residue dissolves from a colourless solution (iii) Solution + NaOH from b(ii) in drops + excess White gelatinous ppt. ppt soluble in excess or ppt. dissolves. (iv) Solution (b)(ii) + NH\(_{3(aq)}\) + excess White gelatinous ppt, ppt soluble in excess or ppt. dissolved

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