HIGHER CHEMISTRY REVISION

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					               HIGHER CHEMISTRY REVISION.
               Unit 1:- The Mole
                1.    The balanced equation for the decomposition of hydrogen peroxide into
                      water and oxygen is:
                             2H2O2 (l)       2H2O(l) + O2 (g)




24000 cm3 of gas = 1 mol of O2
                                    Using information from the above graph, calculate the mass of hydrogen
So 40 cm3 = 0.00167 mol
                                    peroxide used in the reaction, assuming all the hydrogen peroxide
                                    decomposed.
1 mole of O2 weighs 32g
So 0.00167 mol weighs 0.0534g       (Take the molar volume of oxygen to be 24 litres mol-1)
                              2. . A student electrolysed dilute sulphuric acid using the apparatus shown in
                                    order to estimate the volume of one mole of hydrogen gas.




(a) Q= I x t
     = 0.5 x 14 x 60
     = 420C
2H+ + 2e   -            H2
   2 x 96500C           1 mol              Current = 0.5 A
     193000C  1 mol                        Time = 14 minutes.
                                            Volume of hydrogen collected = 52 cm3.
        420C gives 52 cm3
So 193000C gives 23 895 cm3              (a) Calculate the molar volume of hydrogen gas.
               or 23.895 litres          (b) What change could be made to the apparatus to reduce a possible
(b) Introduce a variable resistor            significant source of error?
   so that current can be kept
    constant.
3. The symbol for the Avogadro Constant is L.
      Identify the two true statements.


  A          64.2 g of sulphur contains approximately L atoms.
  B          16.0 g of oxygen contains approximately L molecules.
  C          6.0 g of water contains approximately L atoms.
  D          1.0g of hydrogen contains approximately L protons.
  E          2.0 litres of 0.50 mol l-1 sulphuric acid contains approximately
             L hydrogen atoms.
  F          1.0 litres of 1.0 mol l     -1   barium hydroxide solution contains
            approximately L hydroxide atoms.

  C and D

      4. A student heated a compound which gave off carbon dioxide
          gas and water vapour.
         The volume of carbon dioxide collected was 240 cm3.
          Calculate the number of molecules in this volume.
      (Take the molar volume of carbon dioxide to be 24 litres mol-1.)

       24 litres = 24000 cm3.
       24000 cm3 contains 6.02 x 1023 CO2 molecules.
       So 240 cm3 contains   240/
                                 24000   x 6.02 x 1023 = 6.02 x 1021 CO2 molecules.
5.   A student added 0.20 g of silver nitrate, AgNO3, to 25 cm3 of
     water. This solution was then added to 20 cm3 of 0.0010 mol l-1
     hydrochloric acid as shown in the diagram




     The equation for the reaction which occurs is:
       AgNO3 (aq)    + HCl(aq)        AgCl(s)       + HNO3 (aq)
     (a) Name the type of reaction which takes place.
     (b) Show by calculation which reactant is in excess.


       (a)   Precipitation.
       (b)   No of moles of HCl = C x V(litres)          =   20/
                                                                    1000    x 0.001
                                                        = 2 x 10-5 mol
             No of moles of AgNO3 =    mass/
                                               gfm      =    0.2/
                                                                    169.9

                                                        = 1.2 x 10-3 mol
       So the silver nitrate is in excess.
                               6. When sodium hydrogencarbonate is heated to 112oC it decomposes and the gas
                                     carbon dioxide is given off:
                                              2NaHCO3(s)           Na2CO3(s)   + CO2(g)   + H2O(g)
                                     The following apparatus can be used to measure the volume of carbon dioxide
                                      produced in the reaction.




(a) Water boils at 100oC and           (a) Why is an oil bath used and not a water bath?

   so could not raise the              (b) (i) Calculate the theoretical volume of carbon dioxide produced by the complete

   temperature to 112oC.                       decomposition of 1.68 g of sodium hydrogencarbonate.

(b) (i) From equation                          Take the molar volume of carbon dioxide to be 23 litres mol-1)

      2mol of NaHCO3                       (ii) Assuming that all of the sodium hydrogen carbonate is decomposed, suggest

      gives 1 mol of CO2.                      why the volume of carbon dioxide collected in the measuring cylinder would

168 g NaHCO3  23 litresCO2.                   be less than the theoretical value.

So 1.68 g          0.23 litres
b) (ii) Some of the carbon dioxide
      dissolves in water.
7.   The concentration of a solution of sodium thiosulphate can be found by reaction with iodine.
     The iodine is produced by electrolysis of an iodide solution using the apparatus shown.




         The current is noted and the time when the indicator detects the end-point of the
         reaction is recorded.
         (a) Iodine is produced from the iodide solution according to the following equation:
                          2I-(aq)                          I2(aq)      + 2e-
              Calculate the number of moles of iodine generated during the electrolysis given
              the following results.
                  Current = 0.010 A                         Time        = 1 min 37 s

                Q=I xt                           = 0.01 x 97                       = 0.97C
                2I-(aq)                         I2(aq)   + 2e-
                                                 1 mol     2F
                                                 1 mol    2 x 96500 C
                                                 1 mol    193000C
                    So 0.97 C    0.97
                                      /193000   x1                = 5 x 10-6 mol
7. (b) The iodine produced reacts with the thiosulphate ions according to the equation:
             I2(aq)   + 2S2O32-(aq)        2I-(aq)       + S4O62-(aq)
            iodine      thiosulphate ions
     At the end-point of the reaction, excess iodine is detected by the indicator.
   (i) Name the indicator which could be used to detect the excess iodine present
       at the end-point.
  (ii) In a second experiment it was found that 1.2 x 10-5 mol of iodine reacted with
      3.0 cm3 of the sodium thiosulphate solution.
      Use this information to calculate the concentration of the sodium thiosulphate
       solution in mol l-1.

            (b) (i) Starch solution.
               (ii)     I2(aq)   + 2S2O32-(aq)           2I-(aq)      + S4O62-(aq)
                       1 mol       2 mol


                 So 1.2 x 10-5 mol of I2(aq) reacts with 2.4 x 10-5 mol of S2O32-(aq


                Concentration of thiosulphate         =   no of moles/
                                                                      volume (litres)

                                                           2.4 x 10-5
                                                      =
                                                           0.003

                                                      = 0.008 mol l-1.
8. Aluminium is manufactured in cells by the electrolysis of aluminium oxide
   dissolved in molten cryolite.
   What mass of aluminium is produced each hour, if the current passing
    through the liquid is 180 000 A?

          Al3+(l) + 3e-    Al(l)
                  3F         1 mol
            3 x 96500C  27 g
               289500C  27 g


          Q=Ixt
            = 180 000 x 60 x 60
           = 648 000 000 C


            289500C  27 g
          So 648 000 000 C           648 000 000/
                                                     289500   x 27
                               = 60 435 g
                              = 60.435 kg of Al
9. Calcite is a very pure form of calcium carbonate which reacts with nitric
   acid as follows:

        CaCO3(s) + 2HNO3 (aq)                    Ca(NO3) 2(aq) +             H2O(l)   + CO2(g)

A 2.14 g piece of calcite was added to 50.0 cm3 of 0.200 mol l-1 nitric acid in a
beaker.
(a) Calculate the mass of calcite, in grams, left unreacted.
(b) Describe what could be done to check the result obtained in (a)

  (a)     No. of moles of calcite =    mass/gfm     =   2.14/
                                                             100     = 0.0214 mol.
          No. of moles of acid = C x V(litres) = 0.2 x             50/
                                                                      1000   = 0.01
          From equation 0.0214 mol of calcite reacts with 0.0418 mol of acid.
          So all the acid is used up
          0.01 mol of acid reacts with 0.005 mol of calcite = 0.005 x 100 = 0.5 g
          Mass of calcite left over = 2.14 – 0.5g = 1.64 g.
  (b) Filter off any unreacted calcite, dry and then weigh it.
10. Diphosphine, P2H4, is a hydride of phosphorus. All of the covalent bonds
      in diphosphine molecules are non-polar because the elements present have
      the same electronegativity.
(a)     What is meant by the term”electronegativity”?
(b) The balanced equation for the complete combustion of diphosphine is:
                2P2H4(g)     + 7O2 (g)              P4O10(s)       +    4H2O(l)
      What volume of oxygen would be required for the complete combustion o
      10 cm3 of diphosphine?
(c) Calculate the volume occupied by 0.330 g of diphosphine.
      (Take the molar volume to be 24.0 litres mol-1.)


        (a) Electronegativity is a measure of the attraction of an atom
              for electrons it shares with other atoms.
        (b) 2P2H4(g)       + 7O2 (g)              P4O10(s)     +       4H2O(l)
              2 mol         7 mol
              2 vol         7 vol
              10cm3        35cm3
        (c)     1 mole of P2H4 weighs 66g
               66g occupies 24.0 litres
        So 0.330g occupies     0.33/
                                       66   x 24   = 0.12 litres
11. In 1996, the scientists Robert Curl, Harold Kroto and Richard Smalley won the
   Nobel Prize in Chemistry for their contribution to the discovery of new forms of
    carbon called fullerines.
   (a) In what way does the structure of fullerines differ from the other forms of
         carbon, diamond and graphite?
   (b) One form of fullerine, C60, forms a superconducting crystalline compound with
          potassium.
          Its formula can be represented as K3C60.
   A sample of this compound was found to contain 2.88 g of carbon.
     (i) Calculate the number of moles of fullerine used to make this compound.
     (ii) Calculate the mass of potassium, in grams, in the sample.



   (a) Fullerines are covalent molecular solids – diamond and graphite are covalen
         network solids.
   (b)     1 mole of K3C60 contains 60 x 12 = 720g of carbon.
              So 2.88g of carbon =   2.88/
                                          720   =   0.004 mol of fullerine.
   (c)     1 mole of K3C60 contains 3 x 39 g of potassium.
          So 0.004 moles contains 3 x 39 x 0.004 g
                                  = 0.468 g of potassium
12. Ionisation energies provide information about the structure of atoms.
   (a) Write the equation, showing state symbols, for the first ionisation
      energy of sodium.
   (b) Calculate the number of electrons lost when one mole of boron atoms
      is converted into one mole of boron atoms with a charge of 3+.


      (a)   Na(g)     Na+(g) + e-
      (b)   B(g)     B3+(g) + 3e-
            1 mol             3 mol
                              3 x 6.02 x 1023
                             1.806 x 1024 electrons

				
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