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MANGA SUB- COUNTY 233/1 CHEMISTRY PAPER 1 (THEORY) JULY/AUGUST 2015 TIME: 2 HOURS MARKING SCHEME: 1. a) Water that does not lather easily with soap1 b) Ca 2 and Mg 2 for 2 Ca 2 aq CO 3 aq CaCO 3 s c) OR Mg 2 aq CO 3 2 aq MgCO 3 s Penalise ½ mk - wrong/ missing s.s. - Penalise fully for UB equation/wrong changes. 2. 3. 4. a) b) a) b) c) Fe s S s FeS s (i) (ii) Iron II chloride/ FeCl2(aq) Hydrogen sulphide/ H2S(g) Sodium Sodium hydroxide 2Na 2 O 2 s 2H 2O l 4NaOH aqO 2 g 720 360 180 90 45 No. of half lifes 4 Age of the old wood 4 5670 yrs 22,680 yrs 5. a) b) 6. a) b) c) d) 7. a) b) (i) Group III/ 3/ Three (ii) T3+ T 2 so 3 3 Hydrogen/ H2(g) It is less dense/ lighter than air. White solid seen in the boiling tube Magnesium reacts with steam forming white magnesium oxide. Mg s H 2O g MgO s H 2 g The mass of a substance produced during electrolysis is directly proportional to the quantity of electricity passed. t 32 60 10 1930 sec dr It 0.5 1930 165c If 0.44 g 965c 88 88 g 965 0.44 193,000c 2 F ch arg e on Y 2 ©2015 Pyramid Consultants P.O BOX 67593-00200 Nairobi 0722614502/0733494581 /www.kcsetopical.co.ke 8. a) b) c) 9. a) b) c) 10. (i) Amphoteric oxides (ii) Neutral oxides Any 2 x ½ mk/ Accept correct names ZnO, PbO, Al 2 O 3 They have both acidic and basic properties. (Accept w.t.t.e) (i) Dinitrogen tetraoxide/ N2O4(l) (ii) Oxygen/ O2(g) 2PbNO3 2 s 2PbOs 4NO 2 g O 2 g - A brown gas produced. A residue that is orange when hot and yellow on cooling is formed. (Any 1x1=1mk) Mass of the hydrated salt = 28.36 – 25.14 = 3.22 g Mass of the anhydrous salt = 26.56 – 25.14 = 1.42 g Mass of water of crystallization = 3.22 – 1.42 = 1.80 g Na2SO4 H2O Mass mass 1.42 g 1.42 0.01 No. of moles = 142 0.01 1 Mole ratio = 0.01 formula of the salt is Na 2 SO 4 .10H 2O 1.8 g 1 .8 0.1 18 0.1 10 0.01 11. Add sodium hydroxide solution, to the test solutions in separate test- tubes drop by drop until in excess. The solution containing Fe2+ ions forms a dark green precipitate which is insoluble in excess sodium hydroxide solution. The solution containing Fe3+ ions forms a reddish- brown precipitate which is insoluble in excess sodium hydroxide solution. (NB: Ammonia solution can be used in place of sodium hydroxide.) 12. a) 13. 14. Hydrogen is a non- polar molecule/ compound hence insoluble in water which is a polar solvent. b) Hydrogen chloride is a polar molecule/ compound hence dissolves in water which is a polar solvent. c) All the valence electrons in diamond ore used in bounding while in graphite, three valence electrons are used in bounding leaving and delocalized electron that carries the electric charge. (Accept: Diamond does not have delocalized electrons while graphite has delocalized electrons.) Mass of gas evolved = 4.37 – 4.18 = 0.55 g If 280 cm3 weight 0.55 g 22400 0.55 g Then 22400 cm3 weighs 280 = 44 g a) C7H16 b) Butane ©2015 Pyramid Consultants P.O BOX 67593-00200 Nairobi 0722614502/0733494581 /www.kcsetopical.co.ke 15. a) b) c) 16. 2- methylpropane Water gas H 2O g C s CO g H 2 g - (penalize ½ mk for wrong/ missing s.s) Used to reduce metal oxides of some metals. Used as a fuel. Na 2 SO 4 aq BaCl 2 aq BaSO 4 s 2 NaCl aq R.M .F of BaSO 4 137 32 16 4 233 6.9 0.0296 233 moles of Na 2 SO 4 1 0.0296 No of moles of BaSO 4 0.0296 R.M .F of Na 2 SO 4 23 2 32 16 4 142 mass of Na 2 SO 4 0.0296 142 g 4.2032 g mass of KCl 10 4.2032 g 5.7968 g . 17. a) b) 18. 19. a) b) (i) Zn2+ (ii) SO422 Pb 2 aq SO 4 aq PbSO 4 s c) Bauxite Silicon IV oxide Iron III oxide (Reject formula) Carbon is below aluminium in the reactivity series therefore cannot displace aluminium. a) b) Used as fertilizer NH 3 g HNO 3 aq NH 4NO3 s 68 1000,000 17 4,000,000 moles Moles of ammonia 4,000,000 80 1,000,000 320 ions Mass of NH 4NO 3 in ions 20. a) b) c) Copper Cu s 4HNO 3 aq CuNO3 2 aq 2 NO 2 g 2H 2O l CuOH 2 s 4 NH 3 aq CuNH 3 4 2 - aq 2OH aq Penalize ½ mk for wrong/ missing s.s Penalize fully for U.B equation Penalize fully for wrong/ incorrect chemical symbols & formulae ©2015 Pyramid Consultants P.O BOX 67593-00200 Nairobi 0722614502/0733494581 /www.kcsetopical.co.ke 21. a) Al s Al 3 aq 3e Cu 2 aq 2e Cu s b) (i) 2 Al s 3Cu 2 aq 2 Al 3 aq 3Cu s (ii) = Ec 0 E E 0 0 R 0 = 0.34 1.66v 2.00v 22. b) c) 23. a) b) c) 24. a) b) 25. a) Polychloroethene (Reject polyvinyl/ chloride) 5125 No. of monomers = 62.5 = 82 monomers. Curve B. Reactants have a high concentration of particles at the beginning. As the particles collide to form products, the concentration of reactant particles decreases. Equilibrium point At the start, the concentration of products is zero. The concentration of products increases as the reaction progresses. When the reacting particles are consumed the reaction comes to an end hence no further increase in concentration of products. R. A.M of R 6.0 10 23 3.45 10 22 = 207 3.0 10 22 No. of moles of atoms = 0.05 moles 6.0 10 23 Mass of Ca 0.05 40 2 g Heat was absorbed from the surrounding to the solution in the beakers/ Heat was lost from the body (hand) to the solution. b) (Both axes labeled correctly award ½ mk) ©2015 Pyramid Consultants P.O BOX 67593-00200 Nairobi 0722614502/0733494581 /www.kcsetopical.co.ke 26. c) Endothermic reaction. a) - Atoms of the same element which have the same atomic number but different mass numbers. OR - Atoms of the same element which have the same number of protons but different number of neutrons. (Any 1x1=1mk) 9 22 1 20 R. A.M 10 198 20 218 21.8 10 10 b) 27. NB: (i) (ii) Marks are awarded only if the axes are labeled correctly. The curve s MUST become horizontal at different times but at the same maximum volume. ©2015 Pyramid Consultants P.O BOX 67593-00200 Nairobi 0722614502/0733494581 /www.kcsetopical.co.ke