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(a) Solubility is the maximum amount of solute that would dissolve / be contained in 1 dm\(^3\) of solution at a given temperature. (b)(i) (ii) At 35^\circ solubility = 0.50 to 0.55 moldm\(^{-3}\) range (iii) Solubility @ 55^\circC = 0.95 to 0.98 moldm\(^{-3}\) 1 moldm\(^{-3}\) of salt Z = 0.97; 100 = 97g 100cm\(^{-3}\) will contain = \(\frac{97 \times 100}{100}\) = 9.7g Solubility @ 35^\circC = 0.54moldm\(^{-3}\) 1 drn\(^{-3}\) contain = 0.54 x 100 = 54g 100cm\(^{-3}\) will contain 54 x100 = 5.4g 10 Mass of salt that would crystallize out = 9.7 - 5.4 = 4.3g (c)(i) (I) Al\(_2\)O\(_{3(s)}\) + 6HCl\(_{(aq)}\) \(\to\) 2AlCl\(_{3(s)}\) + 3H\(_2\)O\(_{(l)}\) (II) Al\(_2\)O\(_{3(g)}\) 2NaOH\(_{(aq)}\) + 3H\(_2\)O\(_{(l)}\) \(\to\) 2NaAl(OH)\(_4\) (ii) Amphoteric (iii) ZnO / PbO / SnO /Al\(_2\)O\(_3\) d)(i) I [Al(H\(_2\)O)\(_6\)]\(^{3+}\) = x + (0) = +3 x = +3 Al = +3 (II) NaH = 1+ x = 0 x = —1 H = — 1 (ii)(I) Copper (II) tetraoxosulphate VI pentahydrate (II) Calcium tripxocarbonate IV (III) Potassium tetraoxomanganate VII.

(a)(i) It is an atom or group of bonded atoms that is responsible for the chemical behaviour of an organ compound (ii)(I) — OH or hydroxyl group or alkanol (C) C\(_4\) H\(_8\) + H\(_2\)SO\(_4\) \(\to\) C\(_4\)H\(_8\)HSO\(_4\) C\(_4\)H\(_9\)HSO\(_4\) + H\(_2\)O \(\to\) C\(_4\)H\(_8\)OH + H\(_2\)SO\(_4\) OR CH\(_2\) = CHCH\(_2\)CH + H\(_2\)O \(\to\) CH\(_3\)OHCHCH\(_2\)CH\(_3\) OR C\(_4\)H\(_8\) + H\(_2\)O \(\to\) C\(_4\)H\(_8\)OH (d)(i) (I) Conc. H\(_2\)SO\(_4\) / H\(_3\)PO\(_4\) (II) Acidified KMnO\(_4\) / K\(_2\)Cr\(_2\)O\(_7\) / Na\(_2\)Cr\(_2\)O\(_7\) (Ill) Ethanol and Conc. H\(_2\)SO\(_4\) (e)(i) (I) Effervescence occurs (II) Yellow precipitate (III) Bromine water decolourized (ii) (I) or — COOH or alkanoic group (II) CH\(_3\)CHOH / CH\(_3\) adjacent to CHOH (III) C = C or carbon — carbon double bond

(a) X — CaCO\(_3\) Y — Ca(NO\(_3\))\(_2\) (b)i) 2NH\(_{3(g)}\) + 3CuO\(_{(s)}\) \(\to\) 3Cu\(_{(s)}\) + 3H\(_2\)O\(_{(g)}\) + N\(_{2(g)}\) (ii) *Explosion may occur. *Because ammonium dioxonitrate (Ill) is unstable and decomposes exothermically. (iii) *Manufacture of NH\(_3\) etc *liquid nitrogen as cooling agent * gas carrier in gas chromatography * as preventive against oxidation. (c)(i) Dilute H\(_2\)SO\(_4\) reacts with CaCO\(_{3(s)}\) to form our insoluble salt CaSO\(_4\), which covers the CaCO\(_3\), thus preventing further reaction or production of CO\(_{2(g)}\) , after a short time. (ii) Conc. H\(_2\)SO\(_4\) reacts with ammonia to form a salt (NHq)\(_2\)SO\(_4\) (d)i) Physical Properties of Metals * / ductility *Good conductor of heat / electricity *Sonorous *relatively high density * high melting / boiling points (ii) Chemical Properties of Metals * Ionize by electron loss / reducing agent *Form basic oxides *(some) displace hydrogen from dilute oxide *A more reactive metal displaces or loss reactive metal from a solution cf its salt. (e)(i) (I) Oxidation no of CI = 0 (II) Oxidation no of CI = + 5 (ii) Oxidation no of V = + 5 (f)(i) Half - life is the time taken for half of the total number of atoms in a given sample of the element to decay. (ii)(I) Q is more stable (II) Stability is dependent on half life, the longer it takes substance to decay the more stable it is.

(a)(i) (ii) Rate = change in concentration = \(\frac{0.002 — 0.000}{80}\) = \(\frac{0.002}{80}\) = 2.5 x 10\(^{-5}\) moldm\(^{-3}\) sec\(^{-1}\) (b) (i) (I) Manufactise of baking powder (II) Manufacture of tetraoxosulphate (VI) plaster of paris (P.O.P.) (III) Cement making, extraction of iron, solvay process, manufacture of glass, manufacture of putty (ii) (I) Conc. H\(_2\)SO\(_4\) (II) Conc. H\(_2\)SO\(_4\) (III) CaO / Silica gel. (c)(i) C\(_{(s)}\) + O\(_{2(g)}\) \(\to\) CO\(_{2(g)}\) (II) 12g C gives 1 mol CO\(_2\) 2. 5g C gives \(\frac{1 \times 2.5}{12}\) = 0.208 mol. (iii)(I) Extraction of metals, manufacturer of fuel, as gaseous fuel. (II) * Manufacture of fizzy drinks * Fire extinguisher * Cooling nuclear reactors *Refrigeration. (d) Mole: Fe = \(\frac{28.1}{56}\); Cl = \(\frac{35.7}{35.5}\); H\(_2\)O = \(\frac{36.2}{18}\) Mole ratio; Fe = \(\frac{0.5}{0.5}\); Cl = \(\frac{1}{0.5}\); H\(_2\)O = \(\frac{2}{0.5}\) Fe =1; Cl = 2; H\(_2\)O = 4 (e)(i) Mercury (H\(_g\)) (ii) Bromine (Br\(_2\))

(a) = CO\(_2\) or Carbon IV oxide gas Q = ZnCO\(_3\) or Zinc trioxocarbonate IV gas R = CaCO\(_3\) or calcium trioxocarbonate IV gas (b)(i) Dark brown/amber coloured bottle (ii) Bunsen burner (iii) Separating funnel

TEST OBSERVATION INFERENCE (a) C + water, mixture stirred and filtered partly dissolved, colorless filtrate. White residue C contains soluble and insoluble salts (b) Filterate + Pb(NO\(_3\))\(_2\) Boiled then cooled White precipitate formed Precipitate dissolved Precipitate reappears Cl\(^-\), or SO\(_4^{2-}\) or CO\(_3^{2-}\) Cl\(^-\) confirmed (c)(i) Residue + dil. HNO\(_3\) Effervescence or bubbles or gas evolved. Colourless, odourless gas which turns lime water milky CO\(_2\) gas evolved from CO\(_3^{2-}\) (ii) Solution from (i)+ NaOH\(_{(aq)}\) in drops then in excess White gelatinous precipitate formed, precipitate dissolves Zn\(^{2+}\) or Al\(^{3+}\) Zn\(^{2+}\) or Al\(^{3+}\) (iii) Solution from (i)+ NH\(_{3(aq)}\) in drops then in excess White gelatinous precipitate formed, precipitate dissolves Zn\(^{2+}\) or Al\(^{3+}\) Zn\(^{2+}\) confirmed

Rough cm(\(^3\)) 1st Titration (cm\(^3\)) 2nd Titration (cm\(^3\)) Final Burette Readings 24.10 47.00 23.10 Initial Burette Readings 01.10 24.10 00.00 Volume of Acid used 23.00 22.90 23.10 Average Titre = \(\frac{23.00 + 22.90 + 23.100}{3}\) = 23.0 cm\(^3\) (b)(i) No of moles of acid = \(\frac{0.100 \times 23.00}{1000}\) = 0.00230 moles (ii) No of moles of KOH in B 500cm\(^3\) of B contains 2.8g of KOH 1000cm\(^3\) of B will contain \(\frac{2.8 \times 1000}{500}\) = 5.60g KOH Molar mass of KOH = 39 + 16 + 1 = 56 g mcl\(^{-1}\) Conc. of B = \(\frac{5.6}{56} = 0.100 mol dm^{-3}\) Number of moles of KOH in B = \(\frac{0.100 \times 25}{1000}\) = 0.00250 moles (ii) Mole ratio of acid to base = 0.0023 : 0.0025 = 1 : 1

The Blood is a mixture which consists of water, proteins, fats, oil, sugar, vitamins, hormones, enzymes,e\ etc

When a cold milk bottle is taken out from the refrigerator, the water vapor in the air around the bottle comes in contact with the cold surface of the bottle and gets condensed.

Calcium trioxonitrate(V) is (CaNO\(_3\))\(_2\) The molecular mass of CaNO\(_3\))\(_2\) = 40 + (14 + (16 X 3)) x 2 = 164 Relative molecular mass of nitrogen(N\(_2\)) = 14 x 2 = 28 % by mass of nitrogen in CaNO\(_3\))\(_2\) = \(\frac{28}{164}\) x \(\frac{100}{1}\) = 17.073 ≈ 17.1%

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

Nitrogen has 7 electrons in all. Its electronic configuration is = \(1s^22s^22p^3\) So, it has three unpaired electrons in its p-orbitals. Note that the p orbital can accommodate a maximum of 6 electrons

The maximum number of electrons present in a particular shell is calculated by the formula 2n2, where “n” represents the shell number. For instance, K shell is the first shell and it can hold up to 2(1)2 = 2 electrons. Similarly, L shell is the second shell and it can hold up to 2(2)2 = 8 electrons.

Given: V\(_1\) = V cm\(^3\) P\(_1\) = P mmHg V\(_2\) = ? P\(_2\) = \(\frac{P}{2}\) At constant temperature ⇒ Boyle's Law P\(_1\)V\(_1\) = P\(_2\)V\(_2\) V\(_2\) = \(\frac{{P_1}{V_1}}{P_2}\) = \(\frac{{P}\times{V}\times {2}}{P}\) V\(_2\) = 2V

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