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

a)(i) An acid-base indicator is either a weak organic acid or base which gives different colours in an acidic or basic mediurn. (Organic dye which changes colour according to the pH/hydrogen ion concentration of the solution). (ii) Example of an acid-base indicator is: (i) phenolphthalein. (ii) methyl red. (iii) methyl orange. (iv) methyl yellow (v) alizarin yellow. (vi) litmus paper/solution. (vii) methyl violet. (viii) bromothymol blue (ix) bromocresol green. (b) Property exhibited by nitrogen (IV) oxide in each of the following equations: (i) 4Cu + 2NO\(_2\) —> 4CuO + N\(_2\), NO\(_2\) is an oxidizing agent (ii) H\(_2\)O + 2NO\(_2\) —> HNO\(_3\) + HNO\(_3\), NO\(_2\) is oxidizing agent as well as a reducing agents or NO\(_2\) serves as acid anhydride or mixed anhydride. (c)(i) Enthalpy of combustion is the amount of heat evolved or liberated when one mole of a substance is burnt mpletely in excess oxygen. (ii) Enthalpy of combustion is always negative because heat is always given out or the action is exothermic. (d) To distinguish between primary cell and secondary cell. (i) (a) Primary cells cannot be recharged while secondary cells can be recharged. (b) Primary cells do not last long while secondary cells last long. (c) Reactions in primary cells are not reversible while reactions in secondary cells are reversible. (ii) Examples of each of the cells in 1(d)(i). Primary cell: radio battery, dry cell, alkaline battery, button battery, Le'clanche cells, Daniel cell, torch battery, clock battery, bicycle lamp battery (any of these). Secondary cell: lead-acid (storage - cell), lead-acid accumulator, lithium-ion battery, battery in rechargeable lamp, car battery, nickel-metal hydride cell, phone battery, nickel-cadmium cell, (hydrogen-oxygen) fuel cell, aluminium air cell/battery (any of these). (e) Mole is defined as the amount of substance containing 6.02 x10\(^{23}\) particles of that substance as present in 12g of carbon-12 or the amount of substance containing as many elementary entities as there are atoms in 12g of carbon-12. (f) Amount of hydrochloric acid in 40.0cm\(^{3}\) of 0.40 mo/dm\(^{3}\) dilute HCl n = CV; C = concn in mo/dm\(^{3}\) n = 0.40 x \(\frac{40}{1000}\); V = volume in dm\(^{3}\) n(mol) = 0.016mol; n = amount in mol (g) Two substances used as electrodes during electrolysis of acidified water are: graphite (carbon) and platinum. (h) Two forces of attraction that can exist between covalent molecules (1) Van der waals' forces or induced dipole-dipole forces. (2) Hydrogen bonds. (3) Permanent dipole-dipole interactions or dipolar forces (any of the two): (i) Products formed when butane undergoes incomplete combustion are carbon (ii) oxide and water. (j) Electron configuration of \(_{26}\)Fe\(^{3+}\) \(_{26}\)Fe; Fe - 3\(\bar{e}\) \(\to\) Fe\(^{3+}\) lost of 3 electrons \(_{26}\)Fe 26 protons + 26 electrons \(_{26}\)Fe\(^{3+}\) 26 protons + 23 electrons \(_{26}\)Fe\(^{3+}\) \(\to\) 1s\(^2\), 2s\(^2\), 2p\(^6\), 3s\(^2\), 3P\(^2\), 3d\(^5\)

(a)(i) The periodic law states that the properties of the elements are a periodic function of their atomic numbers. (ii) Periodic property of element(s) is any physical or chemical property of an element which changes regularly with increasing atomic number. (ii) Three properties of an element which show periodicity are: melting/boiling point, ionization energy, electron affinity, electronegativity/electropositivity, atomic/ionic radius, eletrical and thermal - conductivities/metallic and non metallic properties. (iii) Brief explanation of properties of an element which shows periodicity across the period: (1) Melting/boiling points of metallic elements increase due to increase in strength of the metallic bond while melting/boiling point for non-metallic elements decreases due to decrease in Van der waals' forces of attraction. (2) lonizatioin-energy increases due to increase in proton number or increase nuclear charge or decreased atomic radius. (3) Eleetron affinity increases due to decrease of atomic radius or increase in nuclear charge. (4) Eleclronegativity of elements increases due to increased nuclear charged or decreased atomic radius. (5) Electropositivity of elements decreases due to increased nuclear charge (6) Atomic/ionic radius decreases due to increased nuclear charge or increase in proton number or increase in effective nuclear charge. (7) Metallic character (electrical/thermal) decreases because of increase in covalent character (b) Relative atomic mass defined as the number of times the average mass of one atom of an element is heavier than one twelfth the mass of one atom of carbon-l2. OR the ratio of the average mass of one atom of an element to one twelfth the mass of an atom of carbon-12 OR the average mass of one atom of an element on a scale in which one atom of carbon 12 weighs 12 a.m.u. (c) (i) Phenomenon exhibited by elements Z is isotopy. (ii) Difference in the number of neutrons accounts for the difference in mass number. (iii) 75% ( \(^{35}_{17} Z\) )+ \(^{37}_{17} Z\); 100 - 75 = 25% 75% \(^{35}_{17} Z\) + 25% \(^{37}_{17} Z\) (\(\frac{75}{100} \times 35\)) + (\(\frac{25}{100} \times 37\)) = 26.25 + 9.25 = 35.5 (d)(i) I. CO\(_{2(g)}\) collected by downward delivery or upward displacement of air. II. HC\(I_g\) collected by downward delivery or upward displacement of air. III. H\(_{2(g)}\) collected by upward delivery or downward displacement of air. (ii) Reason for answer stated in d(i)I & d(i)II. I. CO\(_{2(g)}\) is heavier / denser than air. II. HC\(l_g\) is heavier / denser than air.

(a)(i) Structure of two isomers of X (ii) Catalytic cracking (iii) C\(_4\)H\(_8\) (12 - 8) = 4, (26 - 18) = 8 (iv) Name of Isomers drawn in a(iii) are: (A) But - 1- ene (B) But - 2 - ene (C) 2-methyl propene (v) Balanced equation for the reaction between X and hydrogen. C\(_4\)H\(_8\) + H\(_2\) ---> C\(_4\)H\(_{10}\) (b) Test for fats: A small amount of the substance placed on a filter paper then rub or warm or melt on the filter paper. Formation of a translucent spot on the filter paper shows or indicates the presence of fats presence of fat or a small amount of the substance treated with Sudan III reagent/or Add Sudan III to the sample. Observation of a red colour shows the presence of fat. (c) Molar mass of H\(_2\)SO\(_4\): [(1 x 2) + 32 + (4 x16)] = 2 + 32 + 64 = 98 — 98g mol\(^{-2}\) 1 mole of O\(_{2(g)}\) --> 2 moles of H\(_2\)SO\(_2\) 1.5 mole of O\(_{2(g)}\) --> (2 x 1.5) moles of H\(_2\)SO\(_4\) 3 moles of H\(_2\)SO\(_4\) Mass of 3 moles of H\(_2\)SO\(_4\) = 3 x 98 = 294g. (d) (i) Enthalpy changes for reactions in their increasing magnitude. \(\bigtriangleup\)H\(_2\) < \(\bigtriangleup\)H\(_1\) < \(\bigtriangleup\)\(_3\) or \(\bigtriangleup\)H\(_2\) \(\bigtriangleup\)H\(_1\) \(\bigtriangleup\)H\(_3\), or \(\bigtriangleup\)H\(_3\) > \(\bigtriangleup\)H\(_1\) > \(\bigtriangleup\)H\(_2\) < lesser than > greater than (ii) The explanation in the order of arrangement in (d) (i): Neutralization involves the combination of H\(^+\) and OH\(^-\) ions to give a specific amount of energy. The greater the ionization, the greater the energy given off \(\bigtriangleup\)H\(_2\) involves a weak acid and a weak base/weak electrolyte. \(\bigtriangleup\)H\(_1\), involves a stronge base and a weak acid. \(\bigtriangleup\)H\(_3\) involves a strong acid and a strong base. (e) (i) Cu\(_{(s)}\) can conduct electricity, (ii) NaH\(_{(g)}\) , BeCl\(_{2(s)}\), HF\(_{(l)}\) are soluble in water.

(a) Industrial preparation of ammonia (NH\(_{3(g)}\)) (i) Dry nitrogen (N\(_{2(g)}\)) mixed with dry hydrogen (H\(_{2(g)}\)) in the volume proportion of 1 : 3 over a pressure of 200 — 500 atm at temperature of 450^\circC — 500^\circC. Finely divided iron used as catalyst. (ii) N\(_{7(g)}\) 3H\(_{2(g)}\) 2NH\(_{3(g)}\) Balanced equation of the reaction written above. (iii) Way of increasing the yield of ammonia (NH\(_{3(g)}\)) is continuous removal of ammonia as it is formed or liquefying of the ammonia gas at a low temperature or high pressure. (iv) Two uses of ammonia: (1) used in manufacture of fertilizers. (2) used as a precipitating agent. Other uses: Manufacture of trioxonitrate (v) acid; manufacture of sodium trioxocarbonate. (iv) textiles, pesticides, used in detergent/cleasing agent, manufacture of plastics, laundry works, pharmaceuticals as a refrigerant, softening temperarily hard water, removal of insect stings. (b)(i) Chemical test for hydrogen H\(_{2(g)}\), a lighted wooden splint is brought to the tip of a test tube containing the gas. The gas gives a pop sound shows the presence of Hydrogen H\(_{2(g)}\). (ii) CO\(_{2(g)}\), The gas bubble through limewater Ca(OH)\(_{2}\) solution. The lime water turns milky shows the presence of CO\(_{2(g)}\) (iii) Oxygen O\(_{2(g)}\), a glowing (wooden) splint dipped into a test tube of the gas. The (wooden) splint rekindles or relights shows that O\(_{2(g)}\) is present. (c)(i) Composition of water gas: Carbon (II) oxide (C\(O_g\)) and Hydrogen gas H\(_{2(g)}\) (ii) 2 Uses of water gas: (1) industrial manufacture of hydrogen. (2) industrial manufacture of methanol butanol. (3) used as fuel. (d) A simple experiment to determine type of hardness in a sample of water. Sample of water boiled, soap added and shaken, if it forms lather easily it is temporary hardness, if it forms lather with difficultly or does not form lather easily it is permanent hardness.

(a)(i) Test Observation i. Sample of concentrated H\(_2\)SO\(_4\) + heat Brown fumes of NO\(_2\) evolved ii. Sample + Copper + cone H\(_2\)SO\(_4\) and warm Reddish brown fume of NO\(_2\) iii. Cold Sample solution + freshly prepared FeSO\(_4\) solution and followed by conc H\(_2\)SO\(_4\) (few drops) to the sample solution + freshly prepared FeSO\(_4\) in slant form A brown ring observed between the junction of the acid and sample solution. (ii) Two uses of each of the following compounds: (1) Sodium Chloride - food seasoning, food preservative, glazing pottery/earthen ware. soap production, crude oil refilling, water softener, production of chlorine/hydrogen/ sodium metal, NaOH/NaClO\(_3\)/NaOCl/Na\(_2\)CO\(_3\) lowering melting point of ice preparation of freezing mixture. Used in tannery industry/tanning. (2) Uses of sodium trioxocarbonate (IV) - Manufacture of glass/borax in detergents/ soaps, manufacture of paper softening hardwater, manufacture of NaOH treating (concentrated) acid splash, standardization of acids/analytical reagent. (b)(i) Balanced equations for the reactions involved in the extraction of iron in the blast furnace. C\(_{(g)}\) + O\(_{2(g)}\) ----> CO\(_{2(g)}\) CO\(_{2(g)}\) + C\(_{(g)}\) --> 2CO\(_{(g)}\) F\(e_2\)\(O_3\) + 3CO\(_{(g)}\) \(\to\) 2Fe + 3CO\(_{2(g)}\) CaO + Si\(O_2\) -> CaSi\(O_3\) CaC\(O_3\) -> CaO + C\(O_2\) (ii) Faraday's first law of electrolysis states that the mass of an element discharged at the electrode is directly proportional to the quantity of electricity passed through the electrolyte. (iii) Two applications of electrolysis are: Electroplating of metals, Extraction of metals, Purification of metals, Preparation of compounds/elements, Production of gases \(H_2\), \(O_2\), C\(l_2\) (any 2) (c) Sugar crystals + concentrated \(H_2\)S\(O_4\) in a beaker. Observations are as follows: (1) Concentrated acid dehydrates the sugar. (2) A mass of carbon/black soot/ sugar charcoal is formed. (3) Reaction is exothermic (heat evolved or released). (d) Equations for the reaction of Zinc powder with dilute tetraoxosulphate (VI) acid: (i) Zn\(_{(g)}\) + \(H_2\)SO\(_{4(aq)}\) -> ZnSO\(_{4(aq)}\) + H\(_{2(g)}\) (ii) Reaction with dilute concentrated H\(_2\)SO\(_4\) (iii) Zn\(_{(g)}\) + 2H\(_2\)SO\(_{4(aq)}\) -> ZnSO\(_{4(aq)}\) + SO\(_{2(g)}\) + 2H\(_2\)O\(_{(l)}\) (e) Property of concentrated tetraoxosulphate (VI) acid shown in (d)(1) is oxidizing property.

(a)(i) Solution of lime juice turns to pink or red, (ii) The FeCl\(_3\) solution changes colour from brown to green and a yellow deposit. (iii) The purple colour of KmnO\(_4\) solution turns colourless or decolourised. (iv) The solution of K\(_2\)CO\(_3\) reacts with the addition of ethanoic acid to evolve a colourless, odourless gas with bubbling or effervescence

TEST OBSERVATION INTERFERENCE (a) F + distil water + stir + filter Effervescence (bubbling) occurs. Colourless and odourless gas evolved after filtration pale blue/colourless filtrate and green residue (b)(i) Filtrate + NaOH\(_{(aq)}\) in drops Then in excess White precipitate formed Precipitate insoluble in excess NaOH\(_{(aq)}\) Ca\(^{2+}\)/pb\(^{2+}\) Ca\(^{2+}\) present (b)(ii) Filtrate + NH\(_3\) \(_{(aq)}\) in drops Then in excess No visible reaction No visible reaction Ca\(^{2+}\) present Ca\(^{2+}\) confirmed (b)(i) Filtrate + NaOH\(_{(aq)}\) in drops Then in excess Blue gelatinous precipitate formed Precipitate insoluble in excess Cu\(^{2+}\) is present (b)(ii) Filtrate + NH\(_3\) \(_{(aq)}\) in drops Then in excess Pale (light) blue gelatinous precipitate formed Precipitate dissolves to form a deep blue solution Cu\(^{2+}\) Cu\(^{2+}\) confirmed (c) Filtrate + NHO\(_3\) \(_{(aq)}\) + AgNO\(_3\)\(_{(aq)}\) No visible reaction No gas involved White precipitate formed Cl\(^{-}\) present (d)(i) Residue + HNO\(_{3(aq)}\) Effervescence occurs, colourless and odourless gas evolves that turn blue litmus paper to red and turns lime water milky CO\(_{3(g)}\) CO\(_{3^{2-}}\) present (d)(ii) Solution from (d)(i) + NaOH\(_{(aq)}\) in drops Then in excess Blue precipitate formed present Precipitate insoluble in excess NaOH\(_{(aq)}\) Cu\(^{2+}\) Cu\(^{2+}\) present

Indicator = Methyl Orange Volume of the base used = 25.00cm\(^3\) Titration Rough Titre 1st Titre 2nd Titre 3rd Titre Final burette readings cm\(^3\) 24.70 24.80 24.70 24.90 Initial burette reading cm\(^3\) 0.00 0.00 0.00 0.00 Volume of acid used cm\(^3\) 24.70 24.80 24.70 24.90 Average Titre = \(\frac{1st + 2nd + 3rd}{3}\) = \(\frac{24.80 + 24.70 + 24.90}{3}\) = 24.80cm\(^3\) Alternatively 2 concordant titres can be used to calculate average titre Equation of thereaction; Na\(_2\)CO\(_3\).XH\(_2\)O + 2HCl\(_{(aq)}\) \(\to\) 2Nacl\(_{(aq)}\) \(\to\) 2Nacl\(_{(aq)}\) + CO\(_{2(aq)}\) + (X + 1)H\(_2\)O\(_{(l)}\) \(\frac{C_AV_A}{C_BV_B} = \frac{n_A}{n_B}\) C\(_A\) = Molar concentration of HCl\(_{(aq)}\) in moldm\(^{3}\) V\(_A\) = Volume of acid used in cm\(^3\) = 24.80cm\(^3\) C\(_B\) = Molar concentration of Na\(_2\)Cu\(_3\) . xH\(_2\)O in molddm\(^3\) n\(_A\) = 2 n\(_B\) = 1 (b)(i) Concentration of C in moldm\(^{-3}\) = ? From the equation reaction; \(\frac{C_AV_A}{C_BV_B} = \frac{n_A}{n_B}\) C\(_A\) = 0.200 molddm\(^{-3}\) C\(_B\) = ? V\(_A\) = 24.80 cm V\(_B\) = 25.00 cm\(^3\) Substitution of known values \(\frac{0.200 \times 24.80}{C_B \times 25.00} = \frac{2}{1}\) C\(_B\) = \(\frac{1 \times 0.200 \times 24.80}{2 \times 25.00}\) C\(_B\) = 0.0992 moldm\(^{-3}\) (ii) Concentration of C in gdm\(^3\) = ? 500 cm\(^2\) \(\to\) 14.3 g 1000cm\(^3\) \(\to\) \(\frac{14.3}{500}\) x 1000g = 28.6 dm\(^{-3}\) (iii) Molar mass of Na\(_2\)CO\(_3\) . XH\(_2\)O Molar conc in moldm\(^{-3}\) = \(\frac{\text{conc. in gdm}^{-3}} {\text{molar mass} }\) Molar mass gmol\(^{-1}\) = \(\frac{\text{conc in g dm}^{-3}}{\text{molar conc.in moldm}^{-3}}\) \(\frac{28.6 gdm^{-3}}{0.0992 moldm^{-3}}\) = 288.3065 = 288 gmol\(^{-1}\) (iv) Value of x in Na\(_2\)CO\(_3\) . xH\(_2\)O? [H = 1.0, C = 12.0, O = 16.0, Na = 23.0] N\(_2\)CO\(_3\) . xH\(_2\)O = 288 2(23) + 12 + 3 (16) + x(2(1) + 16) = 288 46 + 12 +48 +18x = 288 106 + 18x = 288 18x = 288 - 106 18x = 182 x = \(\frac{182}{18}\) x = 10.11 x = 10

CH2CH2 - CCH3CH2 Start the numbering from the terminal carbon.

Indicator = Methyl Orange Volume of the base used = 25.00cm\(^3\) Titration Rough Titre 1st Titre 2nd Titre 3rd Titre Final burette readings cm\(^3\) 24.70 24.80 24.70 24.90 Initial burette reading cm\(^3\) 0.00 0.00 0.00 0.00 Volume of acid used cm\(^3\) 24.70 24.80 24.70 24.90 Average Titre = \(\frac{1st + 2nd + 3rd}{3}\) = \(\frac{24.80 + 24.70 + 24.90}{3}\) = 24.80cm\(^3\) Alternatively 2 concordant titres can be used to calculate average titre Equation of thereaction; Na\(_2\)CO\(_3\).XH\(_2\)O + 2HCl\(_{(aq)}\) \(\to\) 2Nacl\(_{(aq)}\) \(\to\) 2Nacl\(_{(aq)}\) + CO\(_{2(aq)}\) + (X + 1)H\(_2\)O\(_{(l)}\) \(\frac{C_AV_A}{C_BV_B} = \frac{n_A}{n_B}\) C\(_A\) = Molar concentration of HCl\(_{(aq)}\) in moldm\(^{3}\) V\(_A\) = Volume of acid used in cm\(^3\) = 24.80cm\(^3\) C\(_B\) = Molar concentration of Na\(_2\)Cu\(_3\) . xH\(_2\)O in molddm\(^3\) n\(_A\) = 2 n\(_B\) = 1 (b)(i) Concentration of C in moldm\(^{-3}\) = ? From the equation reaction; \(\frac{C_AV_A}{C_BV_B} = \frac{n_A}{n_B}\) C\(_A\) = 0.200 molddm\(^{-3}\) C\(_B\) = ? V\(_A\) = 24.80 cm V\(_B\) = 25.00 cm\(^3\) Substitution of known values \(\frac{0.200 \times 24.80}{C_B \times 25.00} = \frac{2}{1}\) C\(_B\) = \(\frac{1 \times 0.200 \times 24.80}{2 \times 25.00}\) C\(_B\) = 0.0992 moldm\(^{-3}\) (ii) Concentration of C in gdm\(^3\) = ? 500 cm\(^2\) \(\to\) 14.3 g 1000cm\(^3\) \(\to\) \(\frac{14.3}{500}\) x 1000g = 28.6 dm\(^{-3}\) (iii) Molar mass of Na\(_2\)CO\(_3\) . XH\(_2\)O Molar conc in moldm\(^{-3}\) = \(\frac{\text{conc. in gdm}^{-3}} {\text{molar mass} }\) Molar mass gmol\(^{-1}\) = \(\frac{\text{conc in g dm}^{-3}}{\text{molar conc.in moldm}^{-3}}\) \(\frac{28.6 gdm^{-3}}{0.0992 moldm^{-3}}\) = 288.3065 = 288 gmol\(^{-1}\) (iv) Value of x in Na\(_2\)CO\(_3\) . xH\(_2\)O? [H = 1.0, C = 12.0, O = 16.0, Na = 23.0] N\(_2\)CO\(_3\) . xH\(_2\)O = 288 2(23) + 12 + 3 (16) + x(2(1) + 16) = 288 46 + 12 +48 +18x = 288 106 + 18x = 288 18x = 288 - 106 18x = 182 x = \(\frac{182}{18}\) x = 10.11 x = 10

The reactivity of halogens decreases as you move down Group 7. Halogens become less reactive as you move down the group as the atomic radius of the element increases which means that the distance between the outer shell of electrons and the nucleus increases.

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

the conditions necessary for corrosion to take place are; 1 presence of oxygen 2 moisture(water)

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