CHEMISTRY PAPER 2 Marking Scheme - 2017 KITUI MOCK EXAMINATION

Marking Scheme

1.  
   1. Carbon = 6 (½ mk), Hydrogen = 1 (½ mk), Sodium = 11 (½ mk), Neon = 10 (½ mk)
   2. Ca²⁺ 2.8.8 (1 mk)                 P^3- 2.8.8 (1 mk)
   3. –259 + 273 = 14K (1 mk)
   4. Red phosphorous (1 mk) it has a higher melting point (1 mk)
   5. The one with atomic number 24 (1 mk) because the mass number is closer to the relative atomic mass(R.A.M) showing that it contributes to R.A.M more than the other two isotopes (1 mk)
   6. Al₄C₃ (1 mk)
   7. Melting point of magnesium is higher than that of sodium (1 mk) because magnesium has more (2) delocalized electrons compared to sodium (1) hence it has strong metallic bonds (1 mk)
2.  
   1.  
      1. Hot compressed air (1 mk)
      2. In order to melt suplhur in the deposits (1 mk)
      3. It is insoluble in water (1 mk)
         Its melting point ids lower than that of super-heated water (1 mk)
   2.  
      1. S_(s) + O_2(g) → SO_2(g) (1 mk)
      2. To dry sulphur (IV) oxide and air
      3. Platinum (1 mk), Vanadium (V) oxide (1 mk)
      4. It cools the products from the catalytic chamber (1 mk)
         It prevents the reactants getting into the catalytic chamber (1 mk)
   3. It dissolves in water vapour in the air leading to formation of acid rain (1 mk)
   4. A black mas is formed, concentrated H2SO4 is a dehydrating agent. It removes elements of water (hydrogen and oxygen) leaving behind carbon (1 mk)
3.  
   1. Type of flame produced by the fuel
      Amount of heat produced during combustion
      Effects of products formed on the environment (Any two)
   2.  
      1. Heat produced = MCΔT
         ΔT = 46.5 – 25 = 21.5^oC (1 mk)
         ΔH = 450 x 4.2 x 21.5 (1 mk)
         = – 40635J (1 mk)
      2. Moles of ethanol =^1.5/₄₆ (¹/₂ mark)
         =0.0326 moles
         Molar heat = ⁴⁰⁶⁵³/_0.0326 (¹/₂ mark)
         =-1246472.392 J/mole (¹/₂ mark)
         NB:- penalize ½ mk for wrong units
         Penalize ½ mk for missing or wrong sign
   3. C₂H₅OH_(aq) + 3O_2(g) → 2CO_2(g) + 3H₂OH_(l) (1 mk)
      NB: Penalize ½ for wrong/missing state symbol
   4. Heat loss by radiation / conduction / convectional current (1 mk)
      
      
4.  
   1.  
      1. Fractional distillation (1 mk)
      2. Boiling point (1 mk)
         Molecular mass / density (1 mk)
   2.  
      1. C₄H₈ (1 mk)
      2. Use acidified potassium manganate (VII), C₃H₈ does not decolourise it while C₄H₈ decolourises it from purple OR Use acidified potassium dichromate (VI), C₃H₈ does not change it from orange to green will C₄H₈ does (Any one 2 mks)
   3.  
      
   4.  
      1. Ethanol (1 mk)
      2. It is slightly soluble in water (1 mk)
   5. Name: polythene (1 mk)
      Disadvantage: it is non-biodegradable hence causes pollution (1 mk)
5.  
   1.  
      1. Cl₂ (1 mk) It has a positive standard electrode potential hence a higher tendency to gain electrons (1 mk)
      2.  
         
   2.  
      1. Metal S: it has the highest negative e.m.f hence a high oxidizing power therefore low tendency to gain electrons and be displaced (1 mk)
         
         
      2.  
         1.  
            
         2. P:     P²⁺ _(aq) + 2e^- → P_(s)
            Q:    Q_(s) → Q²⁺ _(aq) + 2e^-
         3. It completes the circuit (1 mk)
            It ensures electrical neutrality and balance of ions between the two cells (1 mk)
      3. The salt in the bridge should not react with ions in the solution
         1. E^θ cell = E^θ reduction + E^θ oxidation
            +1.71 = Z – – 079
            Z = +1.71 – 0.76
            = + 0.95V
         2. Q, R, Z, P (2 marks)
6.  
   1. Solubility is the maximum mass for solute that dissolves in 100g of a solvent (water) at a particular temperature to make a saturated solution.  
   2.  
      
   3.  
      1. X: 31.5 (± 0.2)g / 100g of water (½ mk)
         Y: 31.5 (± 0.2)g / 100g of water (½ mk)
      2. 60.5±1^oC
   4. 60 – 20.7 = 39.3g (1 mk)
   5. 13 g (½ mk) of X will crystallize, salt Y will be in solution ü½ mk the solubility of X is lower than the mass dissolved while that of Y is higher than the mass dissolved   (1 mk)
   6. Fractional crystallization (1 mk)
7.  
   1. To remove any oxide film on it (1 mk)
   2. A white solid is formed (1 mk)
   3. Due to the oxygen which combines with magnesium (1 mk)
   4. 2Mg_(s) + O_2(g) → 2MgO_(s) (1 mk)
   5. The filtrate is magnesium hydroxide ü1 mk which is an alkaline, there was no change in the blue litmus paper (½ mk) but the red litmus paper turned blue (½ mk)
   6. 2(24g) → 2400cm³
      2.4g →?             (1 mk)
      2.4g x 2400cm³/_48g (1 mk)     
      = 120cm³ (1 mk)