# Physical Sciences Grade 12 Paper 2 Gauteng Mock 2019 Memo

> Grade 12 Physical Sciences Paper 2 memorandum from the Gauteng Mock Exam 2019, 12 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/2019gauteng-paper-2-memo/a25c7c32-611e-418b-a286-f1b20e9709da
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2019
- Exam period: Gauteng Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 12

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### Page 1

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       PREPARATORY EXAMINATION
                        2019
            MARKING GUIDELINES

     PHYSICAL SCIENCES: CHEMISTRY (PAPER 2) (10842)

13 pages

                           1

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QUESTION 1
1.1    A ✓✓
1.2    D ✓✓
1.3    D ✓✓
1.4    B ✓✓
1.5    D ✓✓
1.6    C ✓✓
1.7    D ✓✓
1.8    A ✓✓
1.9    C ✓✓
1.10   C ✓✓                                                                                   [20]

QUESTION 2
 2.1   2.1.1      A✓                                                                           (1)

       2.1.2      A✓                                                                           (1)

       2.1.3      C✓                                                                           (1)

       2.1.4      B✓                                                                           (1)

       2.1.5      D✓                                                                           (1)
                                               Marking guidelines:
 2.2            H           H
               HC H     H C H                  •   Correct ketone functional group ✓
              H              H           H
                                               •   Both side chains / branches correct on
        H     C C      C    C    C       C H       correct carbon atom ✓
              H H      O    H    H   H         •   Whole structure correct ✓
               2,4-dimethylhexan-3-one
                                                                                                     (3)

              ✓           ✓      ✓
 2.3    2,3-dichloro-3-fluorobutanal
                                                                                                  (3)
                                                                                                 [11]

                                                   2

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QUESTION 3

 3.1   3.1.1   Esters ✓ OR carboxylic acids                                                              (1)

       3.1.2   Ketones ✓                                                                                 (1)

 3.2   3.2.1   Chain isomers: Same molecular formula, but different types of chains ✓✓                   (2)

       3.2.2           H    H      H
                                               ✓
                                                           H H             H       ✓
               H   C C      C      C       H
                                                       H C C        C      C H
                   H        H      H                       H        H      H
                       but-1-ene                               but-2-ene
                                       ✓                                       ✓
               OR 1-butene                             OR 2-butene
                            H
                        H   C      H                               Marking guideline:
                                           ✓H
                                                                   One mark for whole structure
                   H    C   C                  C   H               One mark for correct IUPAC name
                                                                   for each of the isomers
                      H                        H
                   2-methylprop-1-ene          ✓
               OR methylpropene
               OR 2-methyl-1-propene                                                                     (6)

 3.3   3.3.1   Butan-1-ol ✓ and pentanoic acid ✓                                                         (2)

       3.3.2   Condensation ✓ OR esterification                                                          (1)

 3.4    If answered as: Butyl pentanoate has a     If answered as: Butyl butanoate has a
        higher boiling point than butyl butanoate. lower boiling point than butyl
                                                   pentanoate.
        Both esters have the same type of          Both esters have the same type of
        intermolecular forces (London forces). ✓ intermolecular forces (London forces).
        Butyl pentanoate has a longer chain        Butyl butanoate has a shorter chain
        therefore stronger forces between the      therefore weaker intermolecular forces
        molecules. ✓                               between the molecules.
        More energy is required to overcome the Less energy is required to overcome the
        intermolecular forces between butyl        intermolecular forces between butyl
        pentanoate ✓ therefore the boiling point butanoate.
        is higher.
               Marking criteria
               • same intermolecular forces must be mentioned.
                  Their names need not be mentioned. ✓
               • comparison chain length / molecular mass ✓
               • energy required ✓
                                                                                                         (3)

                                                       3

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 3.5.1   Addition polymerisation: A reaction in which small molecules join to form very large
         molecules by adding on at double bonds. ✓✓
         Condensation polymerisation: Molecules of two monomers with different functional
         groups undergo condensation reactions with the loss of small molecules, usually
         water. ✓✓                                                                                (4)

 3.6.1   Polythene or Poliethene✓                                                                 (1)

 3.6.2   Ethene ✓                                                                                 (1)
                                                                                                  [22]

QUESTION 4

 4.1     Elimination / dehydrohalogenation / dehydrobromination ✓                                 (1)

 4.2     •     heat ✓                             ✓
         •     Concentrated sodium hydroxide (NaOH) / Concentrated potassium hydroxide
               (KOH) / Concentrated strong base
         OR
                Hot ethanolic concentrated sodium hydroxide / potassium hydroxide / KOH /
                NaOH                                                                              (2)

 4.3                       H
                                          Marking guidelines:
              H H    H H   C   H          • Correct hydroxyl functional group on C-2 ✓
         H    C C    C     C   O H        • Side chains / branch correct on correct
                                             carbon atom ✓
              H H    H H   C   H          • Whole structure correct ✓
                           H              • If condensed structural formula given -1              (3)

 4.4     H2O / water ✓                                                                            (1)

 4.5     Addition / Hydration ✓                                                                   (1)
                                                                                                  [8]

                                               4

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QUESTION 5
 5.1      To prevent loss of any solution / acid from the flask ✓
          OR
          To allow gas to escape
          OR
          To prevent any solids / liquids getting in / out
          OR
          To prevent spurting
          NOTE:     Answer must not be given in terms of preventing evaporation or
                    condensation.                                                                     (1)

 5.2                       change in mass 164,84 − 165,00 ✓
         Reaction rate =                   =                = –0,016 g·s-1 ✓
                            change in time   10,0     0 ✓
         Accept positive sign or no sign in the answer.
         (Negative value indicates loss in mass during reaction.)                                     (3)

 5.3.1

                                                             MARKING CRITERIA:
                                                             • Plotting of all points correctly ✓✓
                                                               (Allow 1 plotting error.)
                                                             • Draw best fit line through the plotted
                                                               points ✓
                                                             • Dotted line represents new line.

                                                                                                      (3)

                                                  5

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5.3.3   The reaction will be faster. ✓
        Increasing the temperature, increases the kinetic energy so that more
        particles have kinetic energy that is higher than the activation energy. ✓
        More collisions with correct orientation ✓
        More successful or effective collisions per unit time take place. ✓           (4)

5.4.1   2✓                                                                            (1)

5.4.2   State of division changed – Size of the zinc particles increased / surface
        area increased – originally lumps were used. ✓
        Exp 1 powder and exp 2 lumps
        Accept: removed catalyst.                                                     (1)

5.4.3   0,5 g / half of the amount as in experiment 1. ✓                              (1)
                                                                                      [14]

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QUESTION 6
 6.1   A catalyst was used in the experiment that produced the graph for experiment C. ✓       (1)

 6.2   The concentration of the products in both graphs at equilibrium are the same and
       the concentration of the reactants at equilibrium are the same, ✓ but equilibrium was
       reached faster in Experiment C than in Experiment A. ✓
       OR
       Equilibrium concentration is the same, equilibrium was not disturbed.
       OR
       Accept:
       A catalyst was added and the rate of both forward and reverse reactions were
       increased.                                                                              (2)

 6.3   EXOTHERMIC ✓                                                                            (1)

 6.4   – negative marking from QUESTION 6.3
       In experiment C the concentration of the reactants is higher than the concentration
       of the reactants in Experiment A when equilibrium is established. ✓
        The reverse reaction is favoured, ✓
       Endothermic reaction is favoured when the temperature is increased. ✓
       Therefore reverse reaction is endothermic OR the forward reaction is exothermic.        (3)

 6.5   Use the data given to calculate the equilibrium constant at 500 o C.

                                     H2(g)                I2(g)              2HI(g)
        Initial mol                   0,5                  0,5                  0
        Change in mol                -0,2                 -0,2                0,4 ✓
        Mol at equilibrium            0,3          ✓       0,3          0,4 ratio used ✓
        Concentration at              0,3
                                                           0,3                 0,4
        equilibrium
                                             MARKING CRITERIA:
       Kc =
               HI2 ✓                       • Change in mol of HI ✓
              H 2 I 2                    • Using ratio 1:1:2 ✓
           (0,4)2 ✓ (positive marking)       • Addition and subtraction to get the correct
       =
         (0,3)(0,3)                            mol at equilibrium / concentration at
                                               equilibrium ✓
       = 1,78 ✓
                                             • Correct Kc expression ✓
                                             • Correct substitution of concentration at
                                               equilibrium values from this table✓
                                             • Correct final answer ✓

         No Kc expression, correct substitution Max ⅚

         Wrong Kc expression      Max 3/6

                                                                                               (6)

                                               7

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      OPTION 2
      Concentration of I2 at equilibrium is 0,3 mol.dm-3.
                        n                                         n
      c (I2 ) initial =                         c (H2 ) initial =
                        V                                         V
                        0,5                                      0,5
                      =                 and                   =
                         1                                        1
                     = 0,5 mol  dm -3
                                                              = 0,5 mol  dm-3

                                     H2(g)                 I2(g)                 2HI(g)
       Initial                        0,5                   0,5                    0
       concentration
       Change in                      -0,2                 -0,2                  0,4 ✓
       concentration
       concentration at                 0,3   ✓            0,3            0,4 ratio used ✓
       equilibrium

      Kc =
              HI2 ✓
            H 2 I 2 
      =
          (0,4)2 ✓ (positive marking)
        (0,3)(0,3)
      = 1,78 ✓

          No Kc expression, correct substitution Max ⅚

          Wrong Kc expression     Max 3/6
                                                                                                   (6)

6.6   Lower than ✓ positive marking from QUESTION 6.3.
                                           answer would then be: higher than                           (1)

6.7   +
      Kc =
              HI2
             H 2 I 2 
      =
         (0,38)
                2
                   (re-calculate equilibrium concentrations)
                                                                   Note: may use calculation
        (0,3)(0,3)                                                 or use explanation.
      = 1,6
      The [product] is smaller and the [reactant] is higher at the higher temperature. ✓
      This results in a lower Kc value at a higher temperature. ✓
      Or
      Equilibrium is to the left and reverse reaction is favoured.                                      (2)
                                                                                                       [16]

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QUESTION 7

 7.1   Temperature = 25o C / 298 K ✓
       Pressure = 101,3 kPa / 1,013 x 10 5 / 1 atm) ✓
       [Cl-] = 1 moldm-3 ✓                                                                              (3)

 7.2
       7.2.1 Mn → Mn2+ + 2e- ✓✓                                                                          (2)
                     Marking criteria:

                     Mn  Mn2 + + 2e-                              (2 2)
                     Mn  Mn2+ + 2e -           ⇌   Mn             (12)
                     Mn2 + + 2e −  Mn                             (0 2)
                     Mn      ⇌       Mn2+ + 2e −                   (0 2)
                     Ignore if charge is omitted on electron.

                     If charge (+) is omitted on Mn2+               Max. ½

       7.2.2 Mn + Cℓ2 → 2Cℓ- + Mn2+
                                                           Marking criteria:
                                                           ✓ Reactants ✓ Products ✓ Balancing            (3)

            ✓                         ✓           ✓
 7.3   Mn(s) / Mn 2+(aq)(1 moldm-3) // Cℓ2(g)(1 atm) / Cℓ-(aq)(1 moldm-3) / Pt

       NOTE:
       Do not penalise if phases and conditions are not included.                                        (3)

 7.4   MnCℓ2 / Mn(NO3)2 / MnSO4✓                                                                         (1)

 7.5   Chlorine gas ✓ OR Cℓ2                                                                             (1)

 7.6   E θcell = E reduction
                   θ
                             − E θoxidation ✓       Note:
               = +1,36 – (-1,18) ✓                  Accept any other correct formula from the
               = +2,54 V ✓                          data sheet.
                                                    Any other formula using unconventional
                                                    abbreviations, e.g. Ecell = EOA – ERA
                                                    Followed by correct substitutions: Max 2
                                                                                             3

                                                                                                          (3)
                                                                                                         [16]

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QUESTION 8
 8.1   8.1.1   P✓                                                                               (1)

       8.1.2   P is connected to the positive terminal of the battery. ✓                        (1)

 8.2   8.2.1   Ni2+(aq) ✓ Note: (aq) may be omitted.                                            (1)

       8.2.2   Q✓                                                                               (1)

       8.2.3   Q ✓                                                                              (1)

       8.2.4   Ni2+ + 2e- → Ni ✓✓

                 Marking criteria:

                  Ni  Ni2+ + 2e-                           (2 2)
                  Ni2+ + 2e -     ⇌    Ni                   (12)
                  Ni2+ + 2e −  Ni                          (0 2)
                  Ni ⇌          Ni2+ + 2e −                 (0 2)
                 Ignore if charge is omitted on electron.

                 If charge (+) is omitted on Ni2+           Max. ½

                                                                                                (2)

       8.2.5   Remains the same ✓                                                               (1)

       8.2.6   For each mole or atom of nickel oxidised at the anode, a mole or atom of
               nickel is reduced at the cathode. ✓✓
               Or
               Rate of oxidation equals the rate of reduction.                               (2)
                                                                                            [10]

                                              10

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QUESTION 9
 9.1.1     A substance that can act as either an acid or a base ✓✓                     (2)

 9.1.2     H 2O ✓                                                                      (1)

 9.1.3     n = cV ✓
            = (0,05)(0,036) ✓
            = 1,8 x 10-3 mol H2SO4 ✓
           1,8 x 10-3 mol H2SO4 neutralised 1,8 x 10-3 mol Na2CO3 in 25 cm3
           + positive marking from QUESTION 9.1.3                                      (3)

 9.1.4     m = nM ✓                            Marking criteria:
             = 1,8 x 10-3 ✓ x 106 ✓            ✓ formula
             = 0,1908 g ✓                      ✓ ratio of mol
                                               ✓ 106
                                               ✓ correct answer
           Also accept:
           nb    c V
               = b b
           na    c a Va
               1   c b 25
                 =
               1 0,05 x 36
           c b = 0,072 mol  dm-3
                       m       ✓
                  c=
                       MV
           0,072 ==     mm       ✓
         ✓ 0,072    106x x0,025
                            0,25
                   106
               m = 1,908 g
              m = 0,1908 g ✓                                                           (4)

 9.1.5     Positive marking from Question 9.1.2
                                                  Marking criteria:
           % Na2CO3 =
                      Actual mass
                                    x 100         ✓ correct calculation of mass
                      Original mass               ✓ substitution of 5,13
           = (10)(0,0018)(106) = 1,908 g Na2CO3 ✓ ✓ correct answer
               1,908
           =         ✓ x 100
                5,13
           = 37,19 % ✓

           Or

           25 cm3 has 0,1908 g✓
           250 cm3 has 10✓ x 0,1908 g = 1,908 g ✓
                            1,908
           % Na2CO3 =             x 100 ✓ = 37,19 %   ✓
                             5,13                                                      (3)

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 9.2     CO32- + H2O ⇌ HCO-3 + OH- ✓reactants; ✓products
         OR
         CO2- + 2H O ⇌ H CO + 2OH-
            3        2         2     3                                                 (2)
                                                                                      [15]

QUESTION 10

 10.1   10.1.1   Total percentage mass fertiliser in the bag ✓                               (1)

        10.1.2   Nitrogen, ✓ phosphorous ✓ OR potassium (Any two)                            (2)

        10.1.3            2✓
                 %N=        x 40 ✓
                         11
                  = 7,27 % ✓                                                                 (3)

 10.2   10.2.1   Ammonium nitrate or NH4NO3 ✓                                                (1)

        10.2.2   Haber process                                                               (1)

        10.2.3   (NH4)2SO4 ✓                                                                 (1)

        10.2.4   Ostwald process ✓ OR catalytic oxidation of ammonia                         (1)

        10.2.5   So that plants can absorb them from the soil ✓                              (1)

        10.2.6   Eutrophication ✓
                 Any given examples that apply to rivers and dams.
                 Note: do not accept red tide.                                               (1)

        10.2.7   Nitrogen ✓ and hydrogen ✓                                                   (2)

        10.2.8   Sulphur dioxide ✓                                                           (1)

        10.2.9   Enhance growth of crops / plants to produce more food for humans ✓
                 Production of fertiliser results in job creation. ✓
                 Selling of fertilisers stimulates the economy. ✓
                 (Any relevant positive impact)                                               (3)
                                                                                             [18]

                                                                               TOTAL:        150

                                              12

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