# Physical Sciences Grade 12 Western Cape November 2016 Memo

> Grade 12 Physical Sciences memorandum from the Western Cape November 2016, 12 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/wp-metro-central-chemistry-chemie-memo/2c1a083c-28d7-46c1-9b63-32d8f46c1f40
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2016
- Exam period: Western Cape November
- Curriculum: CAPS
- Pages: 12

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## About this document

Physical Sciences · Grade 12 · Western Cape November · 2016. Memorandum, 12 pages. Read online or download the PDF.

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## Document text

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### Page 1

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                       METRO CENTRAL
                     EDUCATION DISTRICT

                            GRADE 12

          PHYSICAL SCIENCES: PAPER 2 (CHEMISTRY)
            SEPTEMBER 2016 – MARKING GUIDELINE

MARKS: 150
TIME: 3 hours

                  MARKING GUIDELINE

                  This question MEMO consists of 12 pages.

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       PHYSICAL SCIENCES/P2      2         WCED Metro Central/Sept 2016
                              MEMORANDUM

QUESTION 1

1.1      D ✓✓                                                               (2)

1.2      D ✓✓                                                               (2)

1.3      A ✓✓                                                               (2)

1.4      C ✓✓                                                               (2)

1.5      B ✓✓                                                               (2)

1.6      D ✓✓                                                               (2)

1.7      C ✓✓                                                               (2)

1.8      C ✓✓                                                               (2)

1.9      C ✓✓                                                               (2)

1.10     D ✓✓                                                               (2)

                                                                            [20]

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    PHYSICAL SCIENCES/P2                        3            WCED Metro Central/Sept 2016
                                          MEMORANDUM

QUESTION 2

2.1.1   There are single bonds between C-atoms / There are no multiple bonds
        between C atoms in their hydrocarbon chains.  / No double or triple bonds
        between C atoms                                                                      (1)

                              (substituents must be in alphabetical order)
2.1.2   3-chloro-3-ethylhexane                                                                (3)

2.1.3   Propanal ✓                                                                            (1)

2.1.4   carboxylic acids ✓                                                                    (1)

2.1.5   Methanol ✓                                                                            (1)

                 3C single bonds
2.1.6

                                    carboxylic acid functional group                         (2)

2.2.1   Propanoic acid. ✓                                                                     (1)

2.2.2   In 100 g there will be 9,81 g H    , 58,85g C   31,34g O

        number of moles:
               H :      = 9,81 mole ✓
                  C :      = 4,904 moles ✓
                  O :        = 1,959 moles ✓
            :       :
        5 :      2,5 : 1
        10 :     5 : 2 

        Molar mass/Molêre massa: 10(1) + 5(12) + 2(16) = 102 g·mol-1 ✓
        n=1
                                                                                              (6)
        C5H10O2 ✓
                                                                                              [16]

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      PHYSICAL SCIENCES/P2                       4           WCED Metro Central/Sept 2016
                                         MEMORANDUM

QUESTION 3

3.1     The temperature at which the vapour pressure of a substance✓ equals
        atmospheric pressure. ✓                                                               (2)

3.2     2-methylpropan-2-ol ✓                                                                 (1)

3.3     London force / momentarily dipole forces/ dispersion forces ✓                         (1)

3.4.1   In both pentane and 2-methylbutane there are weak London/ dispersion forces
        present. ✓ 2-methylbutane is more spherical / has a smaller surface area than
        pentane✓ and therefore there are fewer/less intermolecular forces between its
        molecules and the energy required to overcome the intermolecular forces in
        2-methylbutane is less than the energy required to overcome the
        intermolecular forces in pentane.  therefore a lower boiling point 
                                                                                              (4)

3.4.2   2-methylpropan-2-ol have stronger hydrogen bonding between molecules 
        while pentane has weaker London/dispersion forces between its molecules. 
        Therefore more energy is required to overcome the IMF in
        2-methylpropan-2-ol than in pentane. 
        And the more energy required the higher the boiling point.                           (4)

3.5      2-methylpropan-2-ol ✓                                                                (1)

3.6     n=

        M[CO2] = 12 + 2(16) = 44

             (a)   CO2:     n=    = 0,773 mol ✓

             (b)   0,773 mol CO2 is created by           3 mol C4H10✓

             (c)   Mass C4,H10: m = n x M
                                 = 0,193 x [4(12) + 10(1)]
                                 = 11,19 g✓

             (d)   Percentage purity =    x 100✓ = 43,05 % ✓        [Accept: -- 43,09%]       (5)

                                                                                              [18]

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      PHYSICAL SCIENCES/P2                   5                WCED Metro Central/Sept 2016
                                       MEMORANDUM

QUESTION 4

4.1     Cracking of alkanes                                                                     (1)

4.2.1   Addition polymerization ✓                                                                (1)

4.2.2   polyethene/ polythene/ polyethelene  (any one)                                          (1)

4.2.3
                                                                   (any one)
                    CH2   CH2
                                n
                                                                                                 (1)
                                                                                             ✓
4.3
                                                   ✓

                                                  Mild heat                 ✓

                                                  Dilute H2SO4

                                                                                                 (3)

                                         +
4.4     Hydration ✓                                                                              (1)

4.5     Ethanol ✓
                                                                                                 (1)

4.6     Excess of a conc. strong acid (H2SO4) ✓
        Mild heat ✓                                                                              (2)

4.7.1   Substitution     NOT: Halogenation/Bromination                                          (1)

4.7.2   Add NaBr in presence of dilute H2SO4  Mild heat ✓           add                       (2)
                                                                   Add HBr and mild heat

4.7.3   Sodium hydroxide                                            water                      (1)

                                                                                                 [15]

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      PHYSICAL SCIENCES/P2                       6             WCED Metro Central/Sept 2016
                                            MEMORANDUM

QUESTION 5

5.1     Amount of mole Zn available at 0 s:

                                                     = 0,00025 mol ✓   (2,462 x 10-4 mol)

        Amount of mole Zn available at 12 s:

                                                     = 0,00014 mol ✓   (1,385 x 10-4 mol)
        Ave Rate = -      (no mark for formula)
                                        ✓
                 = - 9,167 x 10-6 mol∙s-1 ✓                      (= - 8,975 x 10-6 mol·s-1)
        Ave Rate = 9,167 x 10-6 mol·s-1                          (= 8,975 x 10-6 mol·s-1)

         n = m/MM
            = (0,009 – 0,016)/ 65   
            = - 1,077 x 10-4 mol

          Average Rate = - n/t 
                       = - [-1,077 x 10-4 / (12 – 0)] 
                       = 8,974 x 10-6 mol·s-1.                                                 (4)

5.2     The HCℓ is used up/depleted / HCℓ is the limiting reactant ✓                            (1)

5.3     The rate decreases as time passes. ✓                                                    (1)

5.4.1                                       ✓✓
5.4.2
                                                         ✓✓

                                                                                                (4)

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      PHYSICAL SCIENCES/P2                     7            WCED Metro Central/Sept 2016
                                         MEMORANDUM

5.5

                                                       ✓✓

                                                                                              (2)

5.6     An increase in temperature increases the number of particles having minimum
        kinetic energy. 
        Therefore there are more collisions per second/ unit time/ frequency of
        collision increases
        More effective collisions per second/ unit time / frequency of effective collision
        increases 
                                                                                              (2)
        Which increases rate of reaction
                                                                                              [14]

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      PHYSICAL SCIENCES/P2                        8           WCED Metro Central/Sept 2016
                                           MEMORANDUM

QUESTION 6

6.1     It is a dynamic equilibrium when the rate of the forward reaction equals the
        rate of the reverse reaction and the reactions occur simultaneously.
                                                                               [2 or 0]         (2)

6.2                                                N2              3 H2          2 NH3
                                           33,6               24
         Initial moles                         /28 = 1,2        /2 = 12          0
         Change in moles                       - 1               - 3          + 2
                                           5,6
         Equilibrium moles                    /28 = 0,2            9             2
                                             0,2                 9             2
         Equilibrium conc (= n/V)               /5 = 0,04         /5 = 1,8      /5 = 0,4

        [N2] = 0,04 mol·dm-3        AND         [H2] = 1,8 mol·dm-3 

        mass of N2 used during the reaction?
        33,6 - 5,6 = 28 g
        Therefore 1 mol N2 was used during the reaction. ✓
        1 mol N2 react with three mole H2
        Therefore 6 g H2 was used during the reaction✓

        concentration N2 at equilibrium:
                                                      ✓
                                              = 0,04 mol∙dm-3✓

        Amount H2 at equilibrium:
                                       =          ✓
                                       =1,8 mol∙dm-3✓                                           (6)

6.3     Kc =           ✓
           = (0,4)2/(0,04)(1,8)3 ✓
           = 0,686 ✓                                                                            (3)

6.4     Increases ✓                                                                             (1)

6.5     Exothermic ✓                                                                            (1)

6.6     When the temperature increases, the Kc value decreases, which means the
        concentration of the reactants increased and the concentration of the products
        decreased. ✓ Therefore the reverse reaction was favoured. ✓ An increase of
        temperature favours the endothermic reaction, ✓ therefore the forward
        reaction must be exothermic.                                                            (3)
                                                                                                [16]

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      PHYSICAL SCIENCES/P2                      9            WCED Metro Central/Sept 2016
                                           MEMORANDUM

QUESTION 7

7.1     It dissociates completely in water  to produce a high concentration of
        OH- ions.                                                                            (2)

7.2     (a)     n=
                  =
                  = 0,158 mol    ✓
                              HO
        (b)     Ba(OH)2       2
                                       Ba2+ (aq) + 2 OH- (aq)
                Therefore 0,158 mol Ba(OH)2 produces 2 x 0,158 = 0,316 mol OH- ✓

        (c)     Concentration of hydroxide ions:
                c =
                  = 0,316/2
                  = 0,158 mol∙dm-3

        (d)       Kw = [OH-][H+] ✓                   OR:
               10-14 = [0,158][H+] ✓                 Calc: pOH = -log [OH-] 
                [H+] = 6,329 x 10-14 mol·dm-3                  = - log (0,158) 
                                                               = 0,801 
         (e) pH = -log [H+] ✓
                = - log[6,329 x 10-14] ✓              pH = 14 - 0,801     
                = 13,19 ✓                                 = 13,20 
                                                                                              (7)

7.3     Burette ✓                                                                             (1)

7.4     An acid is a proton (H+-ion) donor. ✓✓ (2 or 0)                                       (2)

7.5       Ba(OH)2   +        2 HCℓ                  BaCℓ2 + 2H2O
          nb = 1             na = 2
         cb = 0,079          ca = 2,5
         Vb = 2 dm3          Va = ?

        0,158 mol Ba(OH)2 will be neutralized by 0,316 mol HCℓ✓
        c= ✓
        2,5 =        ✓
        V = 0,126 dm-3 or 0,13 dm-3 ✓.

        CaVa
              /CbVb = na/nb       
                   
        (2,5)Vb
               /(0,079)(2) = 2/1 
               Vb = 0,126 dm3                                                                (4)

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         PHYSICAL SCIENCES/P2                               10       WCED Metro Central/Sept 2016
                                                       MEMORANDUM

 7.6       bromothymol blue changes colours when the pH is around 7. This is also
           the end point for a reaction between a strong acid and a strong base / ✓
           Phenolphthalein is an effective indicator for a reaction between a strong base              (2)
           and a weak acid.

 7.7       REMAINS YELLOW ✓                                                                            (1)
                                                                                                       [19]

 QUESTION 8

 8.1       Aℓ ✓                                                                                        (1)

 8.2.1     Aℓ → Aℓ3+ + 3 e- ✓✓                                                                         (2)

 8.2.2      Co3+ ✓                                                                                     (1)

 8.3       Decreases ✓                                                                                 (1)

 8.4                                              
                                                                        Phase must be indicated.
            Aℓ(s)     Aℓ3+ (aq)      Co   3+          2+
                                               (aq) Co (aq) Pt(s)✓
                     (1 mol·dm-3)    (1 mol·dm-3)                       1 mol·dm-3 not necessary

accept     Aℓ(s)    Aℓ3+ (aq)       Co3+ (aq) Co2+(aq) , Pt(s)✓                                        (3)
                             -3
                    (1 mol·dm )     (1 mol·dm-3)

 8.5       Eϴ = Ereduction - Eoxidation ✓

              = 1,81 - (-0,76) ✓
                                                                                                       (3)
              = 2,57 V ✓

                                                                                                       [11]

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      PHYSICAL SCIENCES/P2                    11            WCED Metro Central/Sept 2016
                                         MEMORANDUM

QUESTION 9

9.1     The chemical process in which electrical energy  is converted to
        chemical energy 

        OR

         The use of electrical energy  to produce a chemical change .                      (2)

9.2.1   Chlorine gas/ Cℓ2                                                                   (1)

9.2.2   Hydrogen / H2                                                                       (1)

9.3     H2O has a stronger oxidizing ability than Na+ /
        Na+ is a weaker oxidizing agent than H2O 
                                                                                             (2)

9.4     (a)   Amount of mole Cℓ2 that formed:

              n=     =

                = 0,1 mol Cℓ2 formed.

        (b)   0,1 mol Cℓ2 is formed from 0,2 mol NaCℓ✓

               Initial amount of NaCℓ available
                 n = cV = 2,5 x 0,5 ✓ = 1,25 mol ✓

        (c)   Amount NaCℓ left in solution after electrolysis: 1,25 - 0,2 = 1,05 mol ✓       (4)
                                                                                             [10]

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       PHYSICAL SCIENCES/P2                    12         WCED Metro Central/Sept 2016
                                          MEMORANDUM

QUESTION 10

10.1

10.1.1    Oxygen / O2 ✓                                                                    (1)

10.1.2    Haber process ✓                                                                  (1)

10.1.3    H2SO4 ✓                                                                          (1)

10.1.4    The temperature at which the reaction takes place is approx. 450 oC and water is
          a vapour. Also the H2SO4 that will be formed is a vapour/mist and cannot be
                                                                                           (2)
          collected easily. ✓

10.2.1    Nitrogen ✓and phosphorous ✓                                                      (2)

10.2.2    Mass of nutrient = 35/100 x 40 = 14 g 

          50% of the fertilizer consist of phosphorous:
          Mass of phosphorus = 0,5 x 14 = 7 g ✓

          n=

           = 7/31 ✓

           = 0,226 mol ✓                                                                   (4)

                                                                                           [10]

                                                                               TOTAL       150

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