Ii the atom of d has a bigger stronger nuclear charge

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ii) The atom of D has a bigger / stronger nuclear charge than that of C. The number of energy level pulled is the same. iii) A is more reactive than B. This is because A loses only one electron while B looses two electrons to obtain an octet configuration. More ionisation energy is requird for B to react than is required for A. d) i) Noble gases ii) Helium is used in weather balloons. Neon is used in electric lamps. e) The bond between B and chlorine is ionic formed by transfer of electrons from B to chlorine. On the other hand, the bond between C and chlorine is covalent, formed by equal sharing of electrons, hence a molecular compound. 2. A mixture canbe separated by physical means (eg filtration, distillation etc) a compound cannot. - The components of air are not in definite whole number ratio in a compound, the elements are combined in definite whole number ratios, - A compound is a pure substance whilst air is a mixture of several substances. (any (2 x 1) = 2mks b) i) A : It separates the dye to the greatest number of components ie separates the dye the most. ii) The dye is insoluble in the solvent C c) i) Baseline / Origin ii) The solvent front iii) Yellow - It moves the shortest distance from the origin. d) i) Fractional distillation ii) The gases are used as a fuel iii) Gases, Petrol, Lubricating oil, Bitumen 3. a) MgCO 3 (s) + 2HNO 3(aq) Mg (NO 3 ) 2(aq) + CO 2(g) + H 2 O (l) b) - Scale (at least 1 / 2 page used) Label the axis (Both) - Plotting (1mk) Smooth curve (1mk) ii) 900 - 650 = 250 = 12.5cm 3 /sec 40 - 20 20 Working on the graph must be shown II 1070 - 900 = 170 = 8.5 cm 3 /sec 60 - 40 20 Working on the graph must be shown c) The rate of reaction decreases as time progresses, since the concentration of reactants reduces as the reaction progresses. d) The reactions has reached completion//the reactants are used up. e) 1 mole of CO 2 occupies 22400cm 3 ? ,, ,, ,, 1120 cm 3 ? = 1120cm 3 x 1 mole = 0.05 mole of 22400cm 3 CO 2 f) Mole ratio of MgCO 3 : CO 2 1 : 1 ? : 0.05 Top grade predictor publishers Page | 298
Chemistry paper 1, 2&3 .: Moles of MgCO 3 = 0.5 x 1 = 0.5 mole of MgCO 3 1 Mass of MgCO 3 used = R.F.M of x Number of moles MgCO 3 = [24 + 12 = (16 x 3)] x 0.05 = 84 x 0.05 = 4.2g = 4.2g of MgCO 3 g) % purity of MgCO 3 = 4.2 x 100 9.5 = 44.21% h) Mole ratio of MgCO3 : HNO3 1 : 2 0.05 : ? .: Moles of HNO 3 used = 0.05 x 2 = 0.1 mole 1 ==> 0.1 mole of HNO 3 is in 50cm 3 ? 1000cm 3 ? = 0.1 x 1000 = 2.0M 50 4. a) i) Heating ii) Oxygen gas II S : Nitric (III) acid R : Nitric (V) acid b) PbO (s) + H 2(g) Pb (s) + H 2 O (l) II) 2HNO 2 (aq) + O 2(g) 2HNO 3(aq) c) i) The reaction produces Lead (II) sulphate which is not soluble in water. Thus, the lead (II) sulphate forms an insoluble layer which prevents further contact between lead (II) nitrate and sulphuric (VI) acid. Hence, the reaction stops. ii) Potassium Nitrate d) a) Coke reduces lead (II) oxide to lead metal b) Limestone produces calcium oxide which combines with silica to form calcium siliate and hence removes silica impurities. c) Scrap Iron reduces any remaining lead (II) sulphite to lead metal.

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