# Physical Sciences Grade 12 Paper 2 Gauteng Mock 2018 Memo

> Grade 12 Physical Sciences Paper 2 memorandum from the Gauteng Mock Exam 2018, 10 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/gp-2018-paper-2-memorandum/b76b99ee-4834-4d11-bd2c-3650944f8134
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2018
- Exam period: Gauteng Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 10

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Physical Sciences · Grade 12 · Gauteng Mock Exam · 2018. Memorandum, 10 pages. Read online or download the PDF.

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### Page 1

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           PREPARATORY EXAMINATION
                         2018
             MARKING GUIDELINES

      PHYSICAL SCIENCES: CHEMISTRY (PAPER 2) (10842)

10 pages

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                          GAUTENG DEPARTMENT OF EDUCATION
                               PREPARATORY EXAMINATION

                                                  PHYSICAL SCIENCES
                                                  (Paper 2)

                                                  MARKING GUIDELINES

QUESTION 1

1     A 

2     B 

3     C 

4     B 

5     C 

6     A 

7     A 

8     C 

9     D 

10    D                                                                              [20]

QUESTION 2

2.1   2.1.1 Compounds which have the same molecular formula but different structural
            formula. (two  or zero)                                                   (2)

                      
      2.1.2 Methyl methanoate                                                           (2)

      2.1.3       O   H               Marking guidelines:
              H O C   C    H             functional group correct 
                      H                  whole structure correct                      (2)

      2.1.4 Ethanoic acid                                                              (1)

      2.1.5 Carboxylic acids                                                           (1)

2.2   2.2.1 CnH2n                                                                      (1)
                
      2.2.2 2,5-dimethylhept-3-ene                                                     (2)

                                              2

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2.3                   H                                Marking guidelines:
           H          C   H                               OH on middle C 
          H                   H
                                                          complete structure correct 
      H   C           C       C       H
          H                   H
                      O
              H                                                                                   (2)

2.4                                                            Marking guidelines:
          H               H           H       H
                                                                functional group
      H   C       O       C           C       C   H
          H                           H       H
                          O       
                      H                                                                           (1)
                                                                                                 [14]

QUESTION 3

3.1   (Alcohols are flammable.) Do not bring it close to flames.                                 (1)

3.2   3.2.1               H       H       H                      H       OH H

                  H       C       C       C       
                                                  OH       H     C       C   C   H

                          H       H       H                      H       H   H

                  propan-1-ol                                       propan-2-ol                 (4)

      3.2.2 propan-1-ol                                                                          (1)
                          of the –OH group in a chain will influence the boiling point,
      3.2.3 The position
            because of the role of the intermolecular forces in branched structure.
            OR
            For a fair test, all alcohols need to be primary alcohols.                            (1)

3.3   3.3.1 The melting point of butan-1-ol will be higher. 

                  Reasons:
                      The hydrogen bond between butan-1-ol molecules are stronger than
                       between the butan-2-ol molecules, because the hydroxyl group is
                       more exposed in the butan-1-ol than in the butan-2-ol, which causes
                       the hydrogen bond to have a stronger influence. 
                      More energy is required to weaken IM Forces between the butan-1-ol
                       molecules, causing a higher melting point. 
                      Straight chains have a higher melting point than branched chains.
                  OR
                      Butan-1-ol has a larger surface area (chain length)
                      Therefore stronger IMF
                      More energy needed to overcome IMF
                  Any TWO reasons                                                                 (3)

                                                                     3

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      3.3.2 Butan-2-ol 

            Reasons:

                If the hydroxyl group is on a terminal carbon atom, the intermolecular
                 force (hydrogen bond) is stronger than when the –OH group is on the
                 second carbon atom, where it is screened and has a smaller
                 influence on the energy needed to cause a phase change (boil) in the
                 second isomer, the butan-2-ol. 
                The boiling point of butan-2-ol will be lower and therefore the vapour
                 pressure of butan-2-ol will be higher than the butan-1-ol. Less
                 energy required to overcome the hydrogen bonds and to weaken the
                 bonds. 
            Any TWO reasons                                                                 (3)

      3.3.3 Increase                                                                       (1)

      3.3.4       Van der Waals forces (London forces) increase with an increase in
                   molecular mass / chain length / size of molecule. 
                  The longer the chain, the higher the boiling point, the stronger the
                   London forces.                                                          (2)

3.4   3.4.1 Polymerisation                                                                 (1)

      3.4.2 H C   CH2    CH2   CH2   CH2         CH2       CH2   CH2   CH2   CH3
             3
                                             n
            Ethene.                                                                        (1)

      3.4.3 Any ONE use:

            Packaging material 
            Squeeze bottles
            Electrical insulation
            Industrial protective clothing
            Toys, etc.                                                                      (1)
                                                                                           [19]

QUESTION 4

4.1   4.1.1 Compound X = C 3H5Br                                                           (1)

      4.1.2 Addition / hydrohalogenation                                                   (1)

      4.1.3 Propan-2-ol                                                                    (1)

      4.1.4 An alcohol; where the hydroxyl group is attached to a carbon atom which is
            bonded to two other carbon atoms.                                              (1)

      4.1.5 Prop-1-ene OR propene                                                          (1)

      4.1.6 Dehydrohalogenation                                                            (1)

                                                       4

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4.2   4.2.1 Elimination                                                                    (1)

      4.2.2 Hydration                                                                      (1)

      4.2.3 Hydrohalogenation                                                              (1)

      4.2.4 H2SO4                                                                          (1)
                   
      4.2.5 2-bromo-2-methylpentane                                                        (2)
                                                                                           [12]

QUESTION 5

5.1   5.1.1 Use powdered CaCO3 instead of lumps 
            Use of a more concentrated HCℓ solution 
            Heat the reaction mixture 
            Add a suitable catalyst
            Any THREE ways                                                                  (3)

      5.1.2 Measure rate of CO2 formed by measuring the volume produced at certain
            time intervals. 
            Measure the rate at which the mass decreases, by placing the reaction
            container on a sensitive mass-meter and record the decrease in the mass
            per unit time.                                                                (4)

      5.1.3 It is not a closed system. 
            The CO2 gas escapes from the reaction.                                         (2)

5.2   5.2.1 5 cm3 / min OR 5 cm3 / min–1                                                   (1)

      5.2.2 Steeper gradient                                                               (1)

      5.2.3 The rate of production of hydrogen gas will be faster per unit time 
            because the reaction proceeds at a higher rate at a higher temperature.        (2)

      5.2.4 The rate of production of hydrogen gas will increase as the length / surface
            area of the magnesium ribbon increases. 
            DO NOT ACCEPT: Graph is directly proportional; does not go through the
            origin.                                                                         (2)
                                                                                           [15]

                                               5

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QUESTION 6

6.1   If a stress is applied to a system in equilibrium, the system will respond in such a
      way as to relieve the stress and restore the equilibrium under a new set of
      conditions.                                                                             (2)

6.2   6.2.1 A                                                                                  (1)
                                            
      6.2.2 The forward reaction is endothermic. Decreasing the temperature from the
            equilibrium system favours the reverse reaction.  The reverse reaction is
            an exothermic reaction. Therefore more iodine and hydrogen molecules are
            formed and less HI will be in the reaction mixture.                                (3)

6.3   6.3.1 Kc = [H3O+][CH3COO-] 
               = (1,34 x 10-3)( 1,34 x 10-3) 
               = 1,80 x 10-6                                                                   (3)

      6.3.2 INCREASES.                                                                         (1)

      6.3.3 When sodium ethanoate, CH3COONa, dissolves in aqueous ethanoic
            (acetic) acid, CH3COOH, it dissociates into Na + ions and acetate ions,
            CH3COO-.  The acetate ion increases the total acetate ion concentration in
            the solution.  The equilibrium re-establishes to the left and more ethanoic
            (acetic) acid forms,  decreasing the [H3O+] concentration, therefore the pH
            increases.                                                                          (3)

6.4   6.4.1 0,5 mol of gas X2Y3                                                                (1)

      6.4.2
                                           2X(g)        3Y(g)         X2Y3(g)
              Initial no. of moles               4           4               0
              No of moles formed                 0           0              0,5
                                                                                  ratio
              No of moles used                   1          1,5              0
                                                                                   used
              No of moles at equilibrium         3          2.5             0,5 
              Equilibrium concentration       1,5           1,25            0,25  ÷ 2
              0,25

                     X 2 Y3 
                   X2 Y3 
              Kc =

                    0,25
              =                
                1,52 1,253
              = 0,057                                                                          (6)
                                                                                               [20]

                                                   6

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QUESTION 7

7.1   7.1.1 HSO 4 
            DO NOT ACCEPT: Hydrogen sulphate ion                                            (1)

      7.1.2 CN-                                                                            (1)
                                         
7.2   7.2.1 Na2CO3.10H2O + 2HCℓ  2NaCℓ + 11H2O +CO2                                        (2)

      7.2.2 Methyl orange                                                                  (1)
                                                  / The equivalence point is in the pH
      7.2.3 Strong acid is titrated with a weak base
            range (3 – 4,4) / Low pH / Acidic solution after titration end point. Any ONE
            answer will be correct                                                          (2)

      7.2.4 Red to yellow.                                                                 (1)

      7.2.5 Option 1:

                   n
             cHCℓ =    
                   V
              nHCℓ = CV
                              24,8                       Marking guidelines:
                      = 0,1 x
                             1000                           Calculate n(HCℓ) 
                   = 2,48 x 10-3 mol HCℓ                   Use formula
                                                                 n
                                                             C=       
             1 mol Na2CO3 reacts with 2 mol HCℓ                 V
             n (Na2CO3) in 500 cm3:                         Use ratio, 1 : 2 
                 2,48 x 10  x 500
                           3                               Calculate n(Na2CO3) 
             =                                              Calculate m of Na2CO3 
                      2         25
                          -3
             = 2,48 x 10 mol Na2CO3 

                  m
             n=
                  M
             m = nM
             = 2,48 x 10-3 x 286 gmol-1
             = 7,092 g 

                                               7

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      Option 2:

      na Ca Va
                       
      nb   Cb Vb

            0,1 24.5 
      2           1000 
     1
           Cb  25 
                  1000 
      Cb = 0,0496 mol·dm-3 
            n
      C=
            V
       n = (0,0496)(0,5)
         = 0,0248 mol Na2CO3 
           m
      n=
           M
      m = nM = (0,0248) (286) = 7,093 g Na2CO3                        (5)

7.2.6 Positive marking from question 7.2.6
      % Na2CO3 in commercial washing soda:

            actual mass
      =                    x 100 
          theoretical mass
          7,092
      =          x 100
           7,6
      = 93,32 % 

      If answer of 7,093 g is used then the answer is 93,33 %          (3)
                                                                      [16]

                                       8

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QUESTION 8
                    
8.1   8.1.1 Mg|Mg2+||Ag+|Ag                                                                        (3)

      8.1.2 Silver                                                                                (1)

                                                   agent
      8.1.3 E  = E  oxidising agent – E  reducing
              cell

              = 0,8 V –(-2,36 V) 
              = 3,16 V                                                                            (3)

      8.1.4 25°C or 298 K                                                                         (1)

      8.1.5 1 moldm-3                                                                            (1)

      8.1.6 Option 1
                                                               
              The reaction rate is too low to supply enough charges to maintain a large
              enough current. 

              Option 2

                   P   6W
              I=     =    =2A
                   V   3V

              The light bulb is manufactured to operate effectively if it is connected to a
              source of 3 V which delivers a current of 2 A. Although the potential
              difference is sufficiently large, the current is probably too small (due to a too
              large internal cell resistance).                                                     (3)
                                                                                                  [12]

QUESTION 9

9.1   9.1.1 To the cathode                                                                        (1)

      9.1.2 2H3O+ + 2e- → H2 + 2H2O                                                              (2)

      9.1.3 Chlorine (C2) gas                                                                    (1)

9.2   9.2.1 The covering of the surface or an object with a thin layer of a metal by
            means of electrolysis.                                                               (2)

      9.2.2 Any ONE use of electroplating.
             Protection of the surface of a base metal. 
             Decorative purposes.                                                                 (1)

      9.2.3 Negative electrode / cathode.                                                         (1)
                                                                                                   [8]

                                                      9

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QUESTION 10

10.1 10.1.1 Wheat                                                                        (1)

     10.1.2 Ammonium nitrate / NH4NO3                                                    (1)

     10.1.3 Wheat needs a soil low in potassium, medium in nitrogen and high in
            phosphorous.  Nitrogen needs to be increased in the soil.  More nitrogen
            needs to be added and ammonium nitrate has the most nitrogen.                (3)

     10.1.4 %nitrogen in NH NO =            mass nitrogen
                           4  3                                  x 100 
                                     molar mass ammonium nitrate
                 28
             =       x 100 = 35% 
                 80                                                                      (4)

     10.1.5 A        Nitrogen / N2 
            B        Hydrogen / H2 
            C        Nitrogen dioxide / NO2 
            D        Ammonium nitrate / NH4NO3 
            E        Ammonium sulphate / (NH4)2SO4                                       (5)
                                                                                         [14]

                                                                             TOTAL:      150

                                            10

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