Chemistry Paper 1 Questions and Answers - Pavement Mock Exams 2021/2022

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INSTRUCTIONS TO CANDIDATES

  • Answer all the questions.
  • Mathematical tables and electronic calculators may be used
  • All working must be clearly shown where necessary.
  1. State and explain the change in mass that occur when the following substances are separately heated in open crucibles.
    1. Copper metal (1½ marks)
    2. Copper (II) nitrate (1½ marks)
  2.  
    1. State Graham’s law of diffusion (1 mark)
    2. A volume of 120cm3 of nitrogen gas diffused through a membrane in 40 seconds, how long will 240cm3 of carbon (IV) oxide defuse through the same membrane? (2 marks)
  3. A reaction of Propane with chlorine gas gave a compound of formula C3H7Cl.
    1. What condition is necessary for the above reaction to take place. (1 mark)
    2. Draw the structural formula of the compound C3H7Cl (2 marks)
  4. Name a gas which is used together with Oxygen in welding. (1 mark)
  5. Study the table below and answer the questions that follow.
    (The letters are not the actual symbols of the elements)
     Element   B   C   D   E   F 
     Atomic number   18   5   3   5   20 
     Mass number   40   10   7   11   40 
    1. Which two letters represent the same elements? Give reason. (2 marks)
    2. Give the number of neutrons in an atom of element D. (Show your working) (1 mark)
  6. A hydrated salt of copper has the formula CuSO4.nH2O. About 25g of the salt was heated until all the water evaporated. If the mass of the anhydrous salt is 16.0g, find the value of n. (Cu = 64.0, S = 32.0, O = 16.0, H = 1)
    (3 marks)
  7. The table below shows the pH values of the solutions I, II, III and IV
     Solution   I   II   III   IV 
     pH  2  7   11   14 
    1. Which solution is likely to be that of calcium hydroxide? (1 mark)
    2. Select the solution in which a sample of aluminum oxide is likely to dissolve. Give a reason for your answer.
      (2 marks)
    3. Select a pair of solutions that would likely give a pH of 7 when equal volumes are reacted with each other.
      (1 mark)
  8. Sodium chloride has a higher melting point than hydrogen chloride, explain. (2 marks)
  9. Study the table below and answer the questions that follow
     Substance   M.pt °C   B.pt °C   Electrical conductivity in solid state   Electrical conductivity in molten state 
         J     365     463       Nil      Nil 
         K     1323    2773       Good     Good 
         L     1046    1680       Nil      Good 
         M     2156    2776      Nil      Nil
    Place J, K, L and M in the appropriate categories from the following:
    1. Metallic solid ______________________________ (1 mark)
    2. Covalent network solid ________________________ (1 mark)
    3. Ionic solid __________________________________ (1 mark)
    4. Covalent molecular solid_______________________ (1 mark)
  10. The diagram below shows how two gases, P and Q were collected.
         Chem PVM Q10 2122 PP1
    1. Name the two methods used. (2 marks)
    2. State properties of P and Q that enable them to be collected through the methods shown.   (2 marks)
  11. Study the information in the table below and answer the questions in the table below and answer the questions below the table
     Bond   Bond Energy (KJmol-1
     C-H       414 
     Cl-Cl       244
     C-Cl       326
     H-Cl       431
    Calculate the enthalpy change of the following reaction
                                    U.V. light
    CH4(g) + Cl2(g) arrowCH3Cl(g) + HCl(g)              (3 marks )
  12. Study the diagram below used to investigate the property of steam on aluminium
            Chem PVM Q12 2122 PP1
    1. Explain why no gas was collected in the set up above. (1 mark)
    2. Explain why the reaction between aluminium and steam stops after a short time.   (2 marks)
  13. A pupil analyzed a commercial vinegar solution by titration and found that 24.5cm3 of 0.09 M sodium hydroxide solution was required for titration of 10cm3 of vinegar. Calculate the molarity of ethanoic acid CH3COOH in vinegar. (3 marks)
  14. The diagram below shows a ‘jiko’ when in use. Study it and answer the questions that follow
            Chem PVM Q14 2122 PP1
    1. Identify the gas formed at region B (1 mark)
    2. Using an equation, explain what happens at region A (2 marks)
  15. Sodium chloride is contaminated with copper (II) oxide. Explain how pure sodium chloride can be obtained from the mixture. (3 marks)
  16. The table below gives three experiments on the reaction of excess sulphuric (VI) acid and 0.5g of zinc done under different conditions. In each the volume of gas was recorded at different time intervals.
     Experiment   Form of zinc   Sulphuric (VI) acid solution 
         I   Powder    0.8M 
        II   Powder    1.0M
        III   Granules    0.8M
    On the axis below, draw and label the three curves that could be obtained from such results.    (3 marks)
    Chem PVM Q16 2122 PP1
  17. The set-up of apparatus below used to prepare sulphur (VI) oxide:
          Chem PVM Q17 2122 PP1
    1. Name Gas N………………………… and Gas M………………………….. (1 mark)
    2. Catalyst X …………………………………………………………………... (1 mark)
    3. Why is it necessary to use drying agent Y? (1 mark)
  18. State and explain the observation made when chlorine gas is bubbled into potassium iodide solution. (2 marks)
  19. The diagram below shows an arrangement of electrons in Aluminium chloride dimer.
            Chem PVM Q19 2122 PP1
    1. Write down the formula of the above molecule. (1 mark)
    2. On the diagram, indicate using an arrow the dative bond. (1 mark)
  20. When Magnesium metal is burnt in air, it reacts with both oxygen and nitrogen gases giving a white solid with black specs. Write two equations for the reactions that take place. (2 marks)
  21. Below is a representation of an electrochemical cell.
    Pb(s)|Pb2+(aq)||Ag+(aq)|Ag(s)
    1. What does || represent? (1 mark)
    2. Given the following:
                                                              Eᶿ (volts)
      Pb2+(aq) + 2e-  →  Pb(s)                  -0.13
      Ag+(aq) + e-  →  Ag(s)                      +0.80
      Calculate the E.M.F of the electrochemical cell (2 marks)
  22. When hot concentrated nitric (V) acid is added to sulphur, a red – brown gas and a colourless liquid are formed.
    1. Write an equation for the reaction. (1 mark)
    2. Identify the oxidizing agent in the reaction above. (1 mark)
    3. State one environmental hazard of the nitrogen compounds. (1 mark)
  23. Draw the dot (.) and cross (x) structure of:
    1. Carbon (II) oxide - CO (2 marks)
    2. Ammonium ion - NH4+ (2 marks)
  24. Using sodium hydroxide solution, describe a chemical test that can be used to distinguish between copper (II) ions and iron (II) ions. (3 marks)
  25. State and explain what would be observed if concentrated sulphuric (VI) acid is added to:
    1. Sugar crystals (1 ½ marks)
    2. Copper (II) sulphate crystals (1 ½ marks)
  26. When 100cm3 of 0.5 M sulphuric acid solution, H2SO4, react with 100cm3 of 1 M sodium hydroxide solution, NaOH, the temperature rises by 6.85 Kelvins.
    (Density = 1.0g/cm3, specific heat capacity = 4.2kJkg-1K-1)
    Calculate the molar heat of neutralization described by the equation: (3 marks)
    H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
  27. Filtration is carried out in the apparatus shown
            Chem PVM Q27 2122 PP1
    1. Name X ……………………………………………………………………… (12 mark)
    2. State one property that makes it possible to separate mixtures using filtration. (12 mark)
  28. Calculate the oxidation numbers of sulphur in the following species: (3 marks)
    1. SO32-
    2. SO3
    3. S2O32-   


MARKING SCHEME

  1.  
    1. Increase in mass. Copper metal combines with oxygen in air to form copper (II) oxide
    2. Decrease in mass. Copper (II) nitrate decomposes to form copper (II) oxide, nitrogen (IV) oxide and oxygen gas. The gaseous products escape to the atmosphere.
  2.  
    1. Graham’s law states that the rate of diffusion of a gas is inversely proportional to the square root of its density given that the temperature and pressure are kept constant.
    2.  R.M.M N2 = 14 × 2 = 28
      R.M.M CO2 = 12 + ( 16 × 2) = 44
      Rate of diffusion of N2 
      120cm3 = 3cm3/s
        40s
      Chem PVM Ans 2 2122 PP1
      Rco2 = 3 × √28  = 2.393cm3/s
                    √44
      1s → 2.393cm3
       ?  → 240cm3  
      240 × 1 = 100.3s
        2.393
  3.  
    1. UV light or sunlight
    2.  
      Chem PVM Ans 3b 2122 PP1
  4.  Acetylene or ethyne
  5.  
    1. C and E. They have the same atomic number.
    2. 7 – 4 = 3 neutrons
  6. Mass of water 25 − 16 = 9g
     Compound   CuSO4   H2
     Mass
    R.F.M/R.M.M 
    Moles
    Mole ratio
     16.0
      160
     16/160 = 0.1 
     0.1/0.1 = 1
     9.0
      18
     9/18 = 0.5 
     0.5/0.1 = 5
    CuSO4. 5H2O
       n = 5
  7.  
    1. III
    2. I and IV. Aluminium oxide is amphoteric.
    3. I and IV
  8. Sodium chloride has strong ionic bonds in a giant ionic structure that require more energy to break compared to the weak Van der Waals forces of attraction between hydrogen chloride molecules.
  9.  
    1. K
    2. M
    3. L
    4. J
  10.  
    1.  
      1. Downward delivery or upward displacement method
      2. Upward delivery or downward displacement method
    2.  
      1. the gas is denser/heavier than air
      2. the gas is less dense/lighter than air.
  11.  
    Chem PVM Ans 11 2122 PP1
    [4(414) + 244] − [3(414) + 326 + 431]
    = 1900 − 1999
    = − 99 kJmol
  12.  
    1. There is no heating in the set up.
    2. The formation of aluminium oxide coating that prevents further reaction.
  13.  
    0.09 mol → 1000cm3
       ?          →  24.5cm3   
    24.5 × 0.09 = 0.002205 moles
        1000
    M:R   1:1
    ∴ 0.002205 moles of CH3COOH
    0.002205mol  →  1cm3
          ?               →   1000cm3 
    1000 × 0.002205 = 0.2205M
                10
  14.  
    1. carbon (IV) oxide / CO2
    2. C(s) + O2(g) → CO2(g) Carbon (charcoal) reacts with sufficient air to form carbon (IV) oxide.
  15. Add water to the mixture and stir. Sodium chloride dissolves but not copper (II) oxide. Filter the mixture to obtain sodium chloride solution as the filtrate. Heat the filtrate to evaporate the water and remain with sodium chloride.
  16.  
    Chem LJM PP1 Ans16 2122
  17.  
    1. Gas N – sulphur (IV) oxide ; gas M – oxygen OR gas N – oxygen ; gas M – sulphur (IV) oxide.
    2. Platinum
    3. Sulphur (VI) oxide crystals are readily hydrolyzed by water.
  18. A brown solution is formed. Chlorine displaces iodide ions from the solution to form iodine.
  19.  
    1. Al2Cl3
    2.  
      Chem LJM PP1 Ans19ii 2122
  20. 2Mg(s) + O2(g) → 2MgO(s) and 3Mg(s) + N2(g) → Mg3N2(s)   
  21. Ecell =  Ered − Eoxy 
            = +0.80 − (−0.13)
            = + 0.80 + 0.13
            = + 0.93V
  22.  
    1. 6HNO3 (aq) + S (s)  H2SO4 (aq) + 6NO2 (g) + 2H2O (l)
    2. concentrated nitric (V) acid
    3.  
      • acid rain
      • Eutrophication
  23.  
    1.  
      Chem PVM Ans 23a 2122 PP1
    2.  
      Chem PVM Ans 23b 2122 PP1
  24. Add sodium hydroxide drop wise until in excess to separate test tubes with solutions containing copper (II) ions and iron (II) ions. A green precipitate is formed in the test tube with iron (II) ions while a pale blue precipitate is formed in test tube containing copper (II) ions.
  25.  
    1. White sugar crystals are charred into a black mass. Concentrated sulphuric (VI) acid is a strong dehydrating agent and removed elements of water from sugar.
    2. Blue crystals turn white. . Concentrated sulphuric (VI) acid is a strong dehydrating agent and removed the water of crystalisation from the copper (II) sulphate crystals.
  26. mass = 200cm3 × 1g/cm3 = 200g
    ΔH = mcΔT
          = 200/1000 × 4.2 × 6.85 = 5.754kJ
    Moles of NaOH used
     1 mole → 1000cm3 
        ?      → 100cm3  
    100 × 1 = 0.1moles
      1000
    0.1 moles → 5.754kJ
        1 mole  →   ?
    1 × 5.754 = 57.54kJ/mol
         0.1     
    = − 57.54kJ/mol
  27.  
    1. residue
    2. Immiscibility
  28.  
    1. S + 3(−2) = −2
      S – 6 = −2
      S = + 4
    2. S + 3(−2) = 0
      S – 6 = 0
      S = + 6
    3. 2S + 3(−2) = −2
      2S – 6 = −2
      2S = +4
      S = + 2

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