# Physical Sciences Grade 12 KwaZulu Natal Mock 2016 Memo

> Grade 12 Physical Sciences memorandum from the KwaZulu Natal Mock Exam 2016, 11 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/kzn-chemistry-chemie-memo/4065397b-0892-4451-9404-c86b0adb3475
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2016
- Exam period: KwaZulu Natal Mock Exam
- Curriculum: CAPS
- Pages: 11

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Physical Sciences · Grade 12 · KwaZulu Natal Mock Exam · 2016. Memorandum, 11 pages. Read online or download the PDF.

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### Page 1

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Physical Sciences / P2                       1            Preparatory Examination September 2016
                                       NSC-Memorandum

                                  Education
                                                                     ducation
                                                                     KwaZulu-Natal
Department of Education                                                REPUBLIC OF
SOUTHAFRICA

                               PHYSICAL SCIENCES P2
                                   (CHEMISTRY)

                                    PREPARATORY

                                       NATIONAL

                                          GRAD

MARKS           : 150

TIME            : 3hours

                            This memorandum consists of 10 pages.

The marking guidelines as per 2014 Examination Guidelines, pages 34-37 must be applied when
marking this Paper.

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### Page 2

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Physical Sciences / P2                         2            Preparatory Examination September 2016
                                         NSC-Memorandum

QUESTION 1

1.1     C                                                                              (2)

1.2     B                                                                              (2)

1.3     D                                                                              (2)

1.4     B                                                                              (2)

1.5     A                                                                              (2)

1.6     C                                                                              (2)

1.7     B                                                                              (2)

1.8     A                                                                              (2)

1.9     C                                                                              (2)

1.10    D                                                                              (2)
                                                                                         [20]

QUESTION 2

2.1.    H                                                                              (2)
2.2     A                                                                               (1)
2.3     B                                                                              (2)

2.2.1 chloroethene accept 1-chloroethene                                               (2)
2.2.2 2-methylpentan-2-ol accept 2–methyl– 2-pentanol                                  (2)
2.3
                                   H   H   H H     O   H
                                   │   │   │ │     ║ │
                                H― C ― C ― C ― C ― C ― O
                                   │       │ │
                                   H       H H
                                    H―C―H 
                                       │
                                       H

                Functional group correct      rest of structure correct                (2)
                                                                                         [10]

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Physical Sciences / P2                          3              Preparatory Examination September 2016
                                          NSC-Memorandum

QUESTION 3

3.1.1 Boiling does not involve breaking the chemical (intra molecular) bonds between
      atoms.
      OR Boiling involves breaking the intermolecular forces between molecules.     (1)

3.1.2 Pentane has a longer chain of carbons than methane, therefore a greater
      surface area. Hence more sites for London force attraction and stronger
      intermolecular forces. More energy is needed to break the stronger
      intermolecular forces.                                                               (3)

3.1.3

      OPTION 1:                                    OPTION 2:
      2-methylbutane  (or methylbutane)           2,2-dimethylpropane  (or
                                                   dimethylpropane)
        H H H H
        │ │ │ │                                           H
      H─C─C─C─C─H                                       │
        │   │ │                                         H─C─H
        H   H H
                                                     H   H
         H ─C─H                                      │   │
            │                                      H─C─C─C─H                          
            H                                        │   │
                                                     H   H

                                                        H─C─H
                                                          │
                                                          H                                       (3)

3.2     The pressure exerted by a vapour at equilibrium with its liquid in a closed
        system.                                                                           (2)

3.3.1 D                                                                                   (2)

3.3.2 A                                                                                   (2)

3.4     The carboxylic acid has the strongest intermolecular forces, 
        because there are two sites for hydrogen bonding in each carboxylic acid
        molecule. The stronger the intermolecular forces the lower the vapour
        pressure (at a given temperature).  The graph indicates that D has the lowest
        vapour pressure.                                                                    (3)

                                                                                            [16]

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Physical Sciences / P2                              4          Preparatory Examination September 2016
                                              NSC-Memorandum

QUESTION 4

4.1.1 methyl ethanoate                                                                    (2)

4.1.2
        M (C3H6O2) = 74

        n = m 
            M
          = 68,88                   
             74
          = 0,93 mol

        M (CH4O) = 32
        1 : 1 (alcohol : acid)
        m = 0,93 x 32 

            = 29,78 g

        % purity : 29,78 x 100 = 59,57 % 
                   50                                                                       (5)

4.2.1 A         - hydration (or addition) 
        C       - substitution 
        D       - dehydration (or elimination) 
        E       - combustion                                                               (4)

4.2.2 (i)       Dissolve C2H5Cl in ethanol.    
      (ii)      Add concentrated NaOH solution. 
      (iii)     Heat the mixture.                                                          (3)

4.2.3 H2SO4 OR sulphuric acid                                                              (1)

4.2.4 Polythene OR Polyethene                                                              (1)

4.2.5 The chemical process in which longer chain hydrocarbon molecules are broken
      down to shorter molecules at high temperatures (and pressures).           (2)

4.2.6 C4H10                                                                               (2)
                                                                                            [20]

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Physical Sciences / P2                         5              Preparatory Examination September 2016
                                         NSC-Memorandum

QUESTION 5

5.1     The amount of product formed/reactant used up per unit time (per second).        (2)

5.2     concentration of HCl                                                              (1)

5.3     Product CO2 is an insoluble gas and escapes from the beaker.                     (2)

5.4     Calcium carbonate is used up and the reaction stops, no more CO2 formed.  (2)

5.5.1 graph B                                                                             (1)

5.5.2 graph D                                                                             (1)

5.5.3 graph C                                                                             (1)

5.6     graph C.
        Catalyst speeds up the rate of reaction, therefore gradient of graph will be
        steepest for same initial mass.                                                   (4)

5.7     Remains the same.                                                                 (1)
                                                                                           [15]

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Physical Sciences / P2                            6                  Preparatory Examination September 2016
                                            NSC-Memorandum

QUESTION 6

6.1.1 Increases                                                                                  (1)

6.1.2 When the equilibrium in a closed system is disturbed, the system will
      re-instate a new equilibrium by favouring the reaction that will oppose the
      disturbance.                                                                              (2)

6.1.3 Increase in temperature increases Kc 
      Increase in Kc indicates that the forward reaction has been favoured ,
      Increase in temperature favours the endothermic reaction ,
      Therefore the forward reaction is endothermic .                                            (4)

6.1.4 Add a catalyst 
      Increase pressure OR decrease volume of container                                          (2)

6.2
                                                A2      +       2B         →          2AB
                 Initial (mol)                  x                2                0    

                  Used / formed                -0,2            -0,4             0,4            RATIO
                 Equilibrium (mol)           x – 0,2            1,6               0,4 
                 [equilibrium]               x – 0,2            1,6               0,4           Addition for
                                                                                                 all three
                                                                 For conc. only
                                                                 
                                                A2      +       2B         →          2AB
                 Initial conc.(mol.dm-3)        x                2                0    
                                                                                                RATIO
                  Used / formed                -0,2            -0,4             0,4  
                 [equilibrium]               x – 0,2            1,6              0,4           Addition for
                                                                                                all three
         Kc = [AB]2 
                                           Marking criteria:
             [A2][B]2
                                              • Initial mol/conc correctly indicated. 
                                              • Mol/conc of AB produced = 0,4.
             =      (0,4)2                   • Ratio applied correctly. 
              (x – 0,2) (1,6)2                • Equilibrium mol:       initial – used
                                                                        Initial + produced
             = 0,5                               OR
                                                 Equilibrium conc:  initial – used
      ∴ 0,5 [((x – 0,2) (1,6)2] = (0,4)2                                Initial + produced
                                              • Equilibrium mol ÷ 1 = eq conc. 
           ∴ 1,28x - 0,256 = 0,16             • Correct Kc expression. 
                                              • Substitution of eq conc to Kc expression. 
          ∴ X = 0,325 mol                    • Final answer. 

                                             Wrong Kc expression – (Max.: 4/8)
                                             Wrong values on the table but used
                                             in the calculation (positive marking)
Copyright Reserved                           - (Max.: 3/8) no mark for answer.              Please Turn Over

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Physical Sciences / P2                  7          Preparatory Examination September 2016
                                  NSC-Memorandum

                                                                                (8)
                                                                                 [17]

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Physical Sciences / P2                            8             Preparatory Examination September 2016
                                            NSC-Memorandum

QUESTION 7
7.1.1
        Option 1:

        pH = - log [H3O+] 
           =1
        ∴ [H3O+] = 10-1 = 0,1 
         ∴ [HCl] = 0,1 mol.dm-3 

        Option 2:

        pH = - log [H3O+] 
           =1
         ∴ [HCl] = 0,1 mol.dm-3 
                                                                                             (3)

7.1.2 increase                                                                              (1)
7.1.3 Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l)                                            (3)
                Reactants                         Products           Balancing 

7.2      When an acid reacts with a base to produce a salt and water.  OR
         Chemically equivalent quantities of acid and base are reacted.                    (2)

7.3

Option 1:
2NaOH + H2SO4 → Na2SO4 + 2H2O 
n(NaOH)       = c x V  = 0,1 x 0,012 = 0,0012 mol
n(H2SO4)             = ½ x n(NaOH)  = 0,0006 mol
V(H2SO4)             = n/C = 0,0006 / 0,05 = 0,012 dm3

V(H2SO4) that reacts with Ba(OH)2 = 54 – 12  = 42 cm3 (OR 0,054 - 0,012 = 0,042 dm3)

Ba(OH)2 + H2SO4 → BaSO4 + 2H2O 
n(H2SO4) that reacts with Ba(OH)2 = c x V = 0,05 x (42 x 10-3 mol) = 0,0021 mol
         n(Ba(OH)2) = 1 x 0,0021 = 0,0021 mol

         [Ba(OH)2] = n/v = 0,0021 x 0,048       = 0,04375 mol∙dm-3 

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Physical Sciences / P2                       9                Preparatory Examination September 2016
                                       NSC-Memorandum

7.3
Option 2:                                         Option 3:
2NaOH + H2SO4 → Na2SO4 + 2H2O                2NaOH + H2SO4 → Na2SO4 + 2H2O
n(NaOH)       = c x V  = 0,1 x 0,012        na     = Va x ca 
                                              nb       Vb x cb
= 0,0012 mol
n(H2SO4) = ½ x n(NaOH)                       1  = Va x 0,05 
                                              2     0,1 x 12
= 0,0006 mol
n(H2SO4) tot = c x V                               Va = 12 cm3
= 0,05 x (54 x10-3)
                                              V(H2SO4) with Ba(OH)2 = 54 – 12 
= 0,0027 mol                                                    = 42 cm3

                                              Ba(OH)2 + H2SO4 → BaSO4 + 2H2O 
n(H2SO4) with Ba(OH)2
                                              na     = Va x ca
= 0,0027 - 0,0006  = 0,0021 mol              nb       Vb x cb
Ba(OH)2 + H2SO4 → BaSO4 + 2H2O
                                              1     = 0,05 x 42 
n(Ba(OH)2) = 1 x 0,0021                       1       48 x cb
                      = 0,0021 mol
                                              [Ba(OH)2] = 0,04375 mol∙dm-3 
[Ba(OH)2] = n / v = 0,0021 x 0,048      =
0,04375 mol∙dm-3 
                                                                                           (8)
                                                                                           [17]
QUESTION 8

8.1.1 Pt or the electrode where oxygen reacts or where peroxide is formed                 (1)

8.1.2 Cu → Cu2+ + 2e-                  Cu ⇌ Cu2+ + 2e- (1)                               (2)
                                         Cu ← Cu2+ + 2e- (0)

8.1.3 O2 + 2H+ + 2e-  → H2O2                                                             (2)

8.1.4 EΘcell = EΘcathode - EΘanode 
               = 0,68  - 0,34 
               = 0,34 V                                                                   (4)

8.2     2Ce + 3Pd2+  → 2Ce3+ + 3Pd  (balance )                                          (3)
                                                                                           [12]

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Physical Sciences / P2                           10             Preparatory Examination September 2016
                                           NSC-Memorandum

QUESTION 9
  9.1          Electrical energy to chemical energy. 
                                                                                                   (1)
  9.2          Decrease in mass/ decrease in size/ gets used up/ errosion
                                                                                                   (2)

  9.3          Oxidation                                                                          (1)
  9.4                                               Note:
                     +   -
  9.4.1        Ag + e → Ag                        Marking rule 3.3
                                                                                                   (2)
  9.4.2        Silver nitrate OR AgNO3
               OR
               Silver ethanoate/silver acetate 
               CH3COOAg/ AgC2H3O2/ AgCH3CO2                                                        (1)
               OR
               Ag+ ions

  9.5          Rate of oxidation is equal to the rate of reduction. 

               ACCEPT: Reduction and oxidation take place simultaneously

                                                                                                   (2)

                                                                                                   [9]
QUESTION 10

10.1.1 2SO2 + O2 ⇌ 2SO3                  (ACCEPT:      2SO2 + O2 → 2SO3 )                      (2)

10.1.2 Step 2                                                                                     (1)

10.1.3 Vanadium pentoxide                                                                         (1)

10.2.1 Fertilisers must replenish nutrients used by growing of crops.                             (1)

10.2.2 Any ONE:

         Excessive fertiliser seeps/washes into groundwater and contaminates
         Drinking Water OR
         Excessive fertiliser run-off can lead to eutrophication which depletes aquatic
         life which serves as a food source OR
         Excessive fertilisation can damage a crop which leads to a smaller harvest.              (1)

10.2.3          Eutrophication                                                                    (1)

10.2.4          (NH4)2SO4                                                                        (2)

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Physical Sciences / P2                         11             Preparatory Examination September 2016
                                         NSC-Memorandum

10.3.1 Total percentage of active fertiliser. OR Combined percentage of N, P and K.
                           (1)

10.3.2 OPTION 1

        13 = 5  x 26 
             x

        x = 10 

  thus N = 3

   N:P:K = 3: 2 : 5                                                                    (4)

        OPTION 2

(N + P + K) in the bag      = 26/100 x 20 = 5,2 kg

K in the bag                = 13/100 x 5,2 = 2,6 kg

        2,6/5,2 x 100  = 50% = 5 parts 

P = 2 parts = 20 % 

Therefore, N = 30% = 3 parts

Unknown component = 3.                                                                       (4)
                                                                                              [14]

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