Chemistry Paper 2 Questions and Answers - Mumias West Pre Mocks 2022

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QUESTIONS

  1. The table below gives information on four elements represented by letters K, L, M and N. Study it and answer the questions that follow. The letters do not represent the actual symbols of the elements.
    Element Atomic number Electron arrangement Atomic radius (nm) Ionic radius (nm)
    K   2,8,2 0.136 0.065
    L 17   0.099 0.181
    M   2,8,8,1 0.203 0.133
    N 20   0.174 0.099
    1. Complete the table by filling in the missing atomic numbers and electron arrangements (2mrks)
    2. Which two elements have similar properties? Explain (2mks)
    3. What is the formula of the oxide of M? (1mk)
    4. Which element is a non-metal? Explain (2mks)
    5. Which one of the elements is the strongest reducing agent? Explain (2mks)
    6. Explain why ionic radius of N is less than that of M (1mk)
    7. Explain why the ionic radius of L is bigger than its atomic radius (2mks)
  2.      
    1. Define the term molar enthalpy of neutralization.. (1mrk)
    2. In an experiment to determine the molar enthalpy of neutralization, 25.0cm3 of 2M sulphuric (VI) acid was added to 50cm3 of 2M sodium hydroxide in a lagged plastic beaker. The mixture was stirred with a thermometer and the final temperature attained recorded. The full results obtained in the experiment were as follows;
      volume of 2M sulphuric VI acid used 25.0cm3
      initial temperature of the acid, T1 19.0ºC
      volume of 2M sodium hydroxide used 50.0cm3
      initial temperature of the hydroxide, T2 21.0ºC
      final temperature attained T4 = 27.5ºC
      Given that the specific heat capacity of the mixture, c=4.2kJ/kg/K and that the density of the mixture is 1g/cm3, use the results above to answer the following questions.
      1. Find T3, the common initial temperature. 1mrk
      2. Calculate the heat change during the experiment. 3mks
      3. Work out the molar heat of neutralization 3mrks
      4. Write the thermochemical ionic equation for this process 1mrk
      5. Draw the energy level diagram for the process. 2mrks
      6. State any two sources of error in this experiment. 2mrks
  3. In an experiment hydrogen chloride gas was prepared and reacted with aluminium turnings to form a solid Q and gas R as shown in the diagram below.
    1
    1. Name: Liquid P (1mrk)
      Solid Q (1mk)
      Gas R (1mk)
    2. Write the chemical equation for the reaction that takes place;
      1. in the flat-bottomed flask (1mrk)
      2. in the combustion tube (1mrk)
    3. Name another substance that could serve the same purpose as the concentrated sulphuric acid. (1mk)
    4. Explain the following observation. When blue litmus paper was dipped into the water in the beaker at the end of the experiment it turned red. (1mrk)
    5. Explain why solid Q collects farther away from the heated aluminium (2mks)
    6. Given that 1.35g of aluminium reacted completely with the hydrogen chloride gas, calculate
      1. the mass of the product Q formed (Al=27, Cl= 35.5, H= 1) (2mrks)
      2. the volume of gas R formed measured at stp. (one mole of gas occupied 22.4 litres at standard temperature and pressure.) (2mks)
  4. Study the scheme given above and answer the questions that follow
    2
    1. Name the reagents used in:
      Step I ………………………………
      Step II………………………………. (1mrks)
    2. Name substances; (2mrks)
    3. Write an equation for the reaction that takes place in (3mrks
      Step iv
      Step v
      Step vi
    4. Explain one disadvantage of the continued use of items made from the compound formed in step III (1mrk)
    5. Name the type of reaction that takes place in (2mrks)
      Step i
      Step iii
      Step v
      Step vi
    6. State the conditions necessary for step ii and iii to take place (2mrks)
      Step ii
      Step iii
  5.        
    1. The flow chart below shows a sequence of reactions starting with iron metal.
      Study it and answer the questions that follow.
      3
      1. Name the reagent and state the condition necessary for the reaction in step 1.
        Reagent
        Condition
      2. Give the names of the following 3mrks
        1. Solid S
        2. Solid V
        3. Solid T
      3. Give reasons for the colour change in step 2. 1mrk
      4. Write an ionic equation for the reaction which takes place in step 3. 1mrk
      5. Name one other substance that could be used instead of sodium hydroxide in step III. 1mrk
    2. In an experiment 3.36g of iron filings were added to excess aqueous copper(II) sulphate solution. Calculate the mass of copper that was deposited. (Cu = 63.5, Fe = 56.0) 2mrks
  6.      
    1. The following diagrams show the structures of two allotropes of carbon. Study them and answer the questions that follow
      4
      1. Name allotrope 1mrk
      2. Give one use of N. 1mrk
      3. Which allotrope conducts electricity? Explain in terms of structure and bonding 2mrks
    2. In an experiment, carbon iv oxide gas was passed over heated charcoal and the gas produced collected as shown in the diagram that follows;
      5
      1. Write an equation for the reaction that takes place in the combustion tube. 1mrk
      2. State the purpose of sodium hydroxide in the set up and explain how it works using a chemical equation. 2mrks
      3. Describe a simple chemical test that can be used to distinguish between carbon iv oxide and carbon ii oxide. 2mrks
      4. Give one use of carbon ii oxide 1mrk
  7. The diagram below is set – up used by a student in an attempt to prepare hydrogen gas and react it with hot copper ii oxide in a combustion tube.
    6
    1. Name one suitable substance that can be used as solid M. 1mrk
    2. Write an equation for the reaction that would be expected to take place
      1. in the round-bottomed flask 1mrk
      2. at A 1mrk
      3. at the flame 1mrk
    3. State and explain the observations made 4mrks
      1. at point A
      2. at point B
    4. Explain why it is necessary to burn excess hydrogen. 1mrk
    5. Give two commercial uses of hydrogen. 2mrk


MARKING SCHEME

  1.         

    1. Element Atomic number Electron arrangement Atomic radius (nm) Ionic radius (nm)
      K 12 2,8,2 0.136 0.065
      L 17 2,8,7 0.099 0.181
      M 19 2,8,8,1 0.203 0.133
      N 20 2,8,8,2 0.174 0.099
    2. K and N √ (1 mk)
      They belong to the same chemical family √ (1mk)
    3. M2O √ (1mrk)
    4. L√ (1mrk)
      Forms ions by gaining one electron√ (1mrk)
    5. M √ (1mrk)
      Loses electrons most readily/most electropositive √ (1mrk)
    6. the ion of N has more protons √ (1/2 mrk) than that of M and hence experiences . stronger nuclear attractive force√ (1/2 mrk)
    7. L forms ions by gaining electrons √ (1mrk)
      This increases the repulsion √ (1/2 mrk) between the electrons in the outermost energy . level and reduces the effective nuclear attractive force √ (1/2mrk).
  2.      
    1. The heat change that occurs when one mole of water is formed through acid-base neutralization reaction √ (1mrk)
      Total 12
    2.      
      1. T3 = T1 + T2 = 19.0 + 21.0 = 20.0√(1/2 mrk)
                     2                  2
      2. Heat=MCΔT
        But M=DxV= 1g/cm3x (25+50)cm3 √(1/2mrk)
        =75g = 0.075kg √(1/2mrk)
        Hence Heat= 0.075kgx4.2kJ/kg/Kx(27.5-20.0)K
        =0.075X4.2kJx7.5
        = 2.3625kJ√(1mrk)
      3. Equation: H2SO4+ 2NaOH → Na2SO4 + 2H2O √(1mrk)
        Moles of water formed= moles of NaOH that reacted
        = 2.0 x 50 √(1/2 mrk)
             1000
        Hence molar heat change
        0.1 moles of water → 2.3625kJ of heat
        1 mole of water → 1.0 x 2.3625 √(1/2 mrk)
                                              0.1
        Hence ΔHn = -23.625kJ/mole √1mrk
      4. 2H+(aq) + 2OH-(aq) → 2H2O(l) ΔH=-23.625 kJ/mol √(1mrk)
        Accept H+(aq) + OH-(aq) → H2O(l) ΔH=-23.625 kJ/mol
      5. (½ mark for both axes labelled
        ½ mark for arrow in right direction
        ½ mark for reactants and products
        ½ mark for the ΔH )
      6. 1. Loss of heat to the environment
        2. Heat absorbed by the apparatus is not accounted for.
        3. Errors in volume and temperature measurements. (1mrk for any correct
        Max 2mrks)
  3.              
    1. Liquid P
      Concentrated sulphuric vi acid/ Sulphuric vi acid/H2SO4(l) / H2SO4 √(1mrk)
      Solid Q (1mk)
      Alumunium chloride / AlCl3 √(1mrk)
      Gas R (1mk)
      Hydrogen /H2 √(1mrk)
    2.      
      1.  H2SO4(l) + NaCl(s) → NaHSO4(s) + HCl(g) √(1mrk)
      2. 2 Al(s) + 6HCl(g) → 2AlCl3(s) + 3H2(g) √(1mrk)
    3. Anhydrous Calcium Chloride √(1mrk)
    4. Excess/unreacted hydrogen chloride gas dissolves in the water forming an acidic solution since it is an acidic gas. √(1mrk)
    5. At this high temperature, it forms as a gas √(1mrk). Further from the heat, on the cooler parts of the combustion tube, it sublimes √(1mrk) to solid.
    6.          
      1. From the equation,
        2moles of Al →2moles of AlCl3
        Thus 2(27)g of Al → 2(27+3(35.5))
        54g of Al → 160.5g AlCl3 √(1/2mrk)
        1.35g of Al → 1.35 x 160.5 √(1/2mrk)
                                       54
        = 4.0125g √(1mrk)
      2. From the equation;
        2moles of Al → 3moles of H2
        Thus: 2(27)g of Al → 3(22.4) litres of H2
        54g of Al → 67.2 litres of H2√(1/2mrk)
        1.35g of Al → 1.35 x 67.2 √(1/2mrk)
                                    54
        = 1.6427 litres √(1mrk)
        Total: 13
        Penalize ½ once for wrong/missing state symbols
        Some students may find moles first before converting to mass.
  4.      
    1. Step I Hydrogen / H2 √(1/2mrk)
      Step II Hydrogen chloride / HCl √(1/2mrk) (1mrks)
    2. A - Sodium Propanoate √(1/2mrk)
      B - Carbon iv Oxide √(1/2mrk)
      C - Water √(1/2mrk)
      D - Bromoethane / 1-Bromoethane √(1/2mrk)
    3. Step iv
      CH3CH2COONa + NaOH → C2H6 + Na2CO3 √(1mrk) (ignore state symbols)
      Step v
      2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l) √(1mrk) (ignore state symbols)
      Accept correct structural formulae
      Step vi
      C2H6 + Br2 → C2H5Br + HBr √(1mrk) (ignore state symbols)
      Accept correct structural formulae
    4. It is nonbiodegradable and hence a pollutant of the environment/Produces poisonous gases when burnt. √(1mrk)
    5. Step i
      Addition / Hydrogenation √(1/2mrk)
      Step iii
      Polymerization √(1/2mrk)
      Step v
      Combustion/ Burning √(1/2mrk)
      Step vi
      Substitution √(1/2mrk)
    6. Step ii
      Presence of nickel catalyst √ (1/2 mark) and Heat √ (1/2 mark)
      Step iii
      High temperature √(1/2 mark) and high Pressure √ (1/2 mark)
      Total: 11
  5.      
    1.        
      1. Reagent
        Chlorine gas √ (1mark)
        Condition
        Heat √(1mark)
      2.        
        1. Solid S
          Iron iii oxide √(1mark)
        2. Solid V
          Iron ii sulphide √(1mark)
        3. Solid T
          Mg/Zn/Ca √(1mark)
      3. Hydrogen peroxide is an oxidizing agent √(1/2mark) and oxidizes iron (ii) ions
        to iron (iii) ions √(1/2mark)
      4. Fe3+(aq) + 3OH (aq) → Fe(OH)3(s) √(1mark)
      5. Potassium Hydroxide solution √(1mark)
    2. Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s) √(1mark)
      1mole(56g) Fe → 1 mole(63.5g) Cu
      ⁖ 3.65g Fe → 3.65 x 63.5 √(1mark)
                                   56
  6.      
    1.      
      1. M
        Graphite √(1mark)
        N
        Diamond √(1mark)
      2. -In jewelry
        - Making glass cutters and drilling bits
        = 4,1388g Cu
        Total: 10
        (1 mark for any one, max 1 mark)
      3. M/Graphite √(1/2mark). In graphite, every carbon atom is bonded to three other carbon atoms √(1/2marks) in a layer of hexagonal rings using three √(1/2mark)of the four valence electrons. The fourth valence electron remains delocalized √(1/2mark) making graphite a good electrical conductor.
    2.      
      1. CO2(g) + C(s) → 2CO(g) √(1mark) (penalize ½ for wrong/no state symbols)
      2. To absorb the unreacted/excess carbon jv oxide. √(1mark)
        NaOH(aq) + CO2(g) → NaHCO3(aq) √(1mark)
      3.      
        1. Pass the gases separately through fresh lime water √(1mark). Carbon iv oxide forms a white ppt √(1/2mark) but carbon (ii) oxide does not √(1/2mark).
        2. Try to separately ignite √(1mark) them. Carbon (ii) oxide burns √(1/2mark) with a blue flame while CO2 does not burn √(1/2mark)
          (any one or any other that is workable, max 2 marks)
      4. Give one use of carbon ii oxide 1mrk
        1. As a fuel
        2. As a reducing agent in extraction of metals
          (1mark for any correct, max 1 mark)
  7.        
    1. Zinc/Zn, Iron/Fe, Magnesium/Mg (any one for I mark)
    2.      
      1. in the round-bottomed flask 1mrk
        1mark for correct equation based the correct answer in (a) above
      2. at A 1mrk
        CuO(s) + H2(g) → Cu(s) + H2O(g) √(1mark)
      3. at the flame 1mrk
        2H2(g) + O2(g) → 2H2O(g) √(1mark)
    3.      
      1. Black copper ii oxide turns brown √(1mark). Hydrogen gas reduces black copper ii oxide to brown copper metal √(1mark).
      2. The anhydrous copper ii sulphate changes from white powder to blue crystals √(1mark). White anhydrous copper ii sulphate combines with the steam formed and changes to hydrated copper ii sulphate √(1mark).
    4. Because a mixture of hydrogen with air is explosive. √(1mark)
    5.      
      • As a rocket fuel
      • In hardening of oils to form fats during the manufacture of margarine.
      • In the manufacture of ammonia.
      • In welding
      • In weather balloons
      • Manufacture of hydrochloric acid
      • As a fuel in fuel cells
        (1 mark for any correct max 2 marks )
        Total: 10
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