# Physical Sciences Grade 12 P2 KwaZulu Natal Mock 2018 Memo

> Grade 12 Physical Sciences Paper 2 memorandum from the KwaZulu Natal Mock Exam 2018, 9 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/kzn-2018-paper-2-memorandum/d081943f-947a-4487-96ff-7126fe94a219
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2018
- Exam period: KwaZulu Natal Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 9

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Physical Sciences · Grade 12 · KwaZulu Natal Mock Exam · 2018. Memorandum, 9 pages. Read online or download the PDF.

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### Page 1

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Physical Sciences / P2                        1
                                        NSC-Memorandum

                                   Education
        ducation                  KwaZulu-Natal Department of Education

                         PHYSICAL SCIENCES P2 (CHEMISTRY)

                            PREPARATORY EXAMINATION

                                   SEPTEMBER 2018

                                    MEMORANDUM

                                      NATIONAL
                                 SENIOR CERTIFICATE

                                        GRADE 12

MARKS           : 150

                             This memorandum consists of 9 pages.

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

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Physical Sciences / P2                      2          Preparatory Examination September 2018
                                      NSC-Memorandum

QUESTION 1

1.1     B                                                                         (2)

1.2     C                                                                         (2)

1.3     C                                                                         (2)

1.4     A                                                                         (2)

1.5     B                                                                         (2)

1.6     B                                                                         (2)

1.7     D                                                                         (2)

1.8     A                                                                         (2)

1.9     A                                                                         (2)

1.10    C                                                                         (2)
                                                                                    [20]

QUESTION 2

2.1.1   E                                                                          (1)
2.1.2   B                                                                          (1)
2.1.3   D                                                                          (1)
2.1.4   F                                                                          (1)
2.1.5   G                                                                          (1)

2.2.1   2,4,4-trimethylpent-2-ene                                                  (2)
2.2.2   CnH2n                                                                      (1)
2.3.1   ethanol                                                                    (1)
2.3.2   sulphuric acid                                                             (1)
                                                                                   [10]

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Physical Sciences / P2                           3               Preparatory Examination September 2018
                                           NSC-Memorandum

QUESTION 3

3.1.1   a series of organic compounds that can be described by the same general formula
        in which one member differs from the next with a CH2 group.                             (2)
3.1.2   the temperature at which the vapour pressure equals atmospheric/external
        pressure.  (2 or 0)                                                                    (2)
3.2     C
        As the boiling point increases the vapour pressure decreases. 
        C has the highest boiling point.                                                        (3)
3.3     B                                                                                       (1)
3.4.1   118,50 ºC                                                                               (1)
3.4.2In addition to London forces and dipole-dipole forces, C has two sites for hydrogen
     bonding between the molecules resulting in the strongest intermolecular forces occurring
     between molecules of C.
     The intermolecular forces between molecules of C require the most amount of energy
     to overcome. 
     C will therefore have the highest boiling point.                                       (4)
                                                                                             [13]
QUESTION 4
4.1.1   Addition/hydrohalogenation                                                              (1)

4.1.2   Substitution/hydrolysis                                                                 (1)

4.2                                                             Whole structure correct: 2/2
                                                                Only functional group correct
                 H        H        H       H                     1/2
                 │        │        │       │                    More than one functional group
        H    ―   C   ―    C    ―   C   ―   C   ―   H            0/2
                 │        │        │       │
                 H        Br       H       H
                               
        2-bromobutane                                                                           (3)

4.3     Secondary
        The carbon to which the ―O―H is bonded to, is bonded to TWO other
        carbon atoms.                                                                           (3)
4.4     Dehydration                                                                            (1)
4.5     (Gentle) heat
        Aqueous/dilute strong base (accept NaOH(dilute) or KOH(dilute)                          (2)
4.6.1   Compounds with the same molecular formula,  but different positions of the side
        chain/substituents/functional groups on the parent chain.                               (2)
4.6.2   Elimination                                                                             (1)
4.6.3
               H H H H                                 H H H H
               │ │ │ │                                 │ │ │ │
             H─C─C─C─C─H                          H ─ C ─ C ═ C ─ C ─ H  + H─Br
               │ │ │ │                                 │           │
               H Br H H                                H           H
             1 mark for each reactant and product                                                (3)
                                                                                                 [17]

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Physical Sciences / P2                                  4                 Preparatory Examination September 2018
                                                  NSC-Memorandum
QUESTION 5
5.1.1   ANY ONE
                The change in concentration of reactants/products per unit time. 
                Rate of change in concentration of reactants or products. 
                Change in amount/number of moles/volume/mass of reactants/products 
                 per (unit) time. 
                Amount/number of moles/volume/mass of products formed OR reactants used 
                 per (unit) time.                                                         (2)
5.1.2   60 - 61(s)                                                                                      (1)
5.1.3   n(CO2)           = n(CaCO3)                 Marking criteria
                                  m                      Use mol ratio: n(CO2) = n(CaCO3) = 1:1 
                         = (n      )                                     86  40        m
                                  M                         Substitute           in n    
                                                                           100           M
                               86  40                      Substitute 0,46 mols and
                         = (           )
                                100                                                        V
                                                            Substitute 22,4 dm3 in n 
                         = 0,46 mols                                                       Vm
                          V                                 Final answer: V = 10,304 dm3
                       n
                          Vm

                                V
                0,46  =
                               22,4 
                 V       =     10,304 dm3                                                             (5)
5.1.4 40 g                                                                                            (1)
5.1.5 INCREASES                                                                                       (1)
5.1.6 See attached graph.
                                                                                                       (3)
                                               Curve starts at 86 g and ends at 40g
                                               The completion time is above 60 or 61s
                                               The curve above the original

5.2.1 Collision theory                                                                                (1)

5.2.2The shaded areas in the distribution curves represent the number of molecules with
     sufficient kinetic energy to overcome the activation energy . An increase in the
     temperature of the system results in a greater number of particles with sufficient kinetic
     energy to overcome the activation energy of the reaction . This results in more effective
     collisions per unit time OR a higher chance of an effective collision occurring, resulting in
     a higher reaction rate.                                                                      (3)
                                                                                                  [17]
QUESTION 6
6.1     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)

6.2.1   REMAINS THE SAME                                                                              (1)
6.2.2   INCREASES                                                                                     (1)
6.2.3   REMAINS THE SAME                                                                              (1)
6.2.4   INCREASES                                                                                     (1)

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Physical Sciences / P2                                  5          Preparatory Examination September 2018
                                                  NSC-Memorandum

Apply negative marking from 6.2.4

6.3     According to Le Chatelier’s Principle a decrease in temperature favours the exothermic reaction
        A decrease in temperature increases the equilibrium constant.Therefore the forward reaction is
        favoured                                                                                  (3)

6.4
          Marking criteria:
           Indicating that the number of mols of H2 decreases by an unknown amout
           Correct mol ratio
           Calculating in terms of x the quantity(mol) at equilibrium of all three substances 
                                         n
           Substitute V = 4 dm3 in c      to determine concentration at equilibrium of H2/I2
                                         V
             and HI.
           Kc expression
           Substitution of concentrations in Kc expression 
           Substitution of 49 for Kc
                                 m
             Equation: n         
                                 M
             Substituting in the above equation
             Final answer: 399,36 g

      No Kc expression, correct substitution: Max. 9
                                                        10

      Wrong Kc expression :        Max. 6
                                            10
                                                  H2          I2           HI
         Initial quantity(mol)                    2           2            0
         Change(mol)                             -x          -x          +2x        Ratio 

         Quantity at                             2-x         2-x           2x        
         equilibrium(mol)
         Equilibrium                             2x         2x            x        Divide by 4
         concentration(mol.dm-3)                  4           4             2

                                                  x
                                                 ( )2
                      [HI ] 2                     2
                Kc =              =                          =   49 
                     [H 2 ][I 2 ]             2x 2x
                                            (     )(    )
                                               4      4

                         x         =    1,56 mol

                m(HI)              =     nM
                                   =    (2)(1,56)(128) 
                                   =    399,36 g                                                   (10)
                                                                                                    [19]

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Physical Sciences / P2                                    6              Preparatory Examination September 2018
                                                    NSC-Memorandum

QUESTION 7
7.1     It dissociates/ionises completely in water to form a high concentration of OH- ions.                    (1)
7.2     It contains a small amount (number of moles) of base in proportion to the volume
        of water                                                                                               (2)
7.3

             Formula pH = - log [H3O+] / pOH              = - log [OH-] 
             Substitute 13,45 for pH/0,55 for pOH
             c(OH-) =        0,282 mol.dm-3
             Using ratio of 1: 2 to calculate c((Ba(OH)2) 
             Formula m       =       cVM
             Substituting into the above formula
             Answer

      Option 1:                                               Option 2:
           pH               = - log [H3O+] 
                                                                      pOH      = - log [OH-] 
                                        +
             13,45         = - log [H3O ]
                                                                 0,55         = - log [OH-]
      [H3O ]    +
                            = 3,54 x 10-14 mol.dm-3
                                                               [ OH-]         = 0,282 mol.dm-3
       [H3O+][OH-]          =      1 x 10-14
                                                              c((Ba(OH)2) = 0,141 mol.dm-3
                        -                      -3
             c(OH ) =     0,282 mol.dm                              m     = cVM
      c((Ba(OH)2) = 0,141 mol.dm-3                                        = (0,141)(0,25)(171) 
             m      =     cVM                                             = 6,03 g
                    =     (0,141)(0,25)(171) 
                    =     6,03 g

                                                                                                                (7)
7.4 Positive marking from question 7.3: concentration of Ba(OH)2

         Marking guidelines
                                    n         c  Va n a
                Formulae: c         /n = cV/ a         
                                    V         c b  Vb nb
          Substitution of: 0,141 x 60 /0,141 x 0,06 
          Use mol ratio: na : nb = 2 : 1 
          Final answer: 33,84 cm3 / 0,03384 dm3 

      Option 1:                                             Option 2:

      n(HCℓ) = 2n((Ba(OH)2)
                                                                cA      VA           nA
                     = 2cV
                                                                cB      VB     =     nB          
                     = 2(0,141)(0,06) 
                                                              0,5       VA          2
                     = 0.01692 mols
                                                              0,141     0,06   =   1            
      c(HCℓ)         = n/V
                                                                     VA    =     0,03384 dm3
        0,5 = 0,01692/V                                                      3
                                                            Accept VB = 60 cm
             V       = 0,03384 dm3/33,84cm3                VA      =      33,84       cm3
                                                                                                                 (4)
                                                                                                               [14]

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Physical Sciences / P2                                7              Preparatory Examination September 2018
                                                NSC-Memorandum
QUESTION 8
8.1     GALVANIC,  converts chemical energy to electrical energy or no dc power supply.             (2)
8.2     Temperature of 25 ºC/298K
        Pressure 101,3 kPa
        Concentration of electrolyte of 1mol.dm-3                                                    (3)
8.3     Chlorine (molecule)                                                                         (2)

8.4     OPTION 1                                    Notes
                                                     Accept any other correct formula from the
        EΘcell = EΘcathode - EΘanode 
                                                      data sheet.
                  = 1,36  - (-1,66)                Any other formula using unconventional
                                                      abbreviations, e.g. E°cell = E°OA - E°RA
                  = 3,02 V 
                                                      followed by correct substitutions:
                                                      E°sel = E°OM - E°RM Max/: 3
                                                                                    4
        OPTION 2
                       Cℓ2(g) + 2e-  2Cℓ-(aq)                 1,36 (V) 
                    Aℓ(s) Aℓ 3+(aq) + 3e-                   -1,66 (V) 
                    Aℓ(s) + Cℓ2(g) Aℓ 3+(aq) + 2Cℓ- (aq)      3,02 (V)                               (4)

8.5     3Cℓ2(g) + 2Aℓ(s)  6Cℓ-(aq) + 2Aℓ 3+(aq)

          Notes
             Reactants          Products          Balancing 
           Ignore phases.
             Marking rule 6.3.10
             Marking rule 3.9.
             Marking rule 3.4: One mark is forfeited when the charge of an ion is omitted per
              equation (not for the charge on the electron)

                                                                                                      (3)
8.6.1   REMAINS THE SAME                                                                             (1)
8.6.2   DECREASES                                                                                    (1)
                                                                                                      [16]
QUESTION 9
9.1     A solution that conducts electricity through the movement of ions.                          (1)
9.2     Cu2+                                                                                        (1)
9.3.1   Decreases                                                                                   (1)
                                 -
9.3.2   Cu → Cu2+ + 2 e

         Notes
          Cu2+ + 2e- ← Cu               (2 )                     Cu2+ + 2e- ⇌ Cu     (0 )
                                             2                                           2
                            2+
                  Cu ⇌ Cu + 2e     -
                                         ( 1   )                     2+
                                                                  Cu + 2e → Cu        ( 0   )
                                            2                                             2
                 Ignore if charge on electron is omitted.
                 If a charge of an ion is omitted e.g. Cu → Cu2 + 2e- is Cu → Cu2 + 2e- Max.: 1
                                                                                                2       (2)

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Physical Sciences / P2                                8            Preparatory Examination September 2018
                                                NSC-Memorandum

9.3.4
          Marking criteria
           Calculate number of mols of cations: 2,259 x 1024 = n(6,023 x 1023)
                               m
             Formula:    n     
                               M
                                                                                 m
             Substitute calculated number of moles of cations and 63,5 in n      
                                                                                 M
             Final answer 238,125 g

                         ne = nNA
                2,259 x 1024 = n(6,023 x 1023) 
                          n       =       3,75 mols
                          m       =       nM
                                  =       (3,75)(63,5) 
                                  =       238,125 g.                                                   (4)
                                                                                                        [9]
QUESTION 10
10.1.1          Haber                                                                                 (1)
10.1.2          Catalytic oxidation of ammonia                                                        (1)
10.1.3          Nitrogen dioxide                                                                       (1)
10.1.4          Ammonium nitrate                                                                      (1)
10.2.1          Sulphuric acid/H2SO4                                                                   (1)
10.2.2          H2SO4 + 2NH3               (NH4)2SO4

                 Notes:
                    Reactants         Products           Balancing 
                    Marking rule 6.3.10.
                                                                                                         (3)

10.3     %N     =        14/20 x 36
                =        25,2%
         Mass of N       = 25,2/100 x m
         12,60=         25,2/100 x m
         m      =        50 kg                                                                        (4)
10.4     Fertiliser A
         Fertilizer A has a high percentage of Phosphorus compared to fertilizer B.                 (3)
                                                                                                  [15]
                                                                                       TOTAL MARKS: 150

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Physical Sciences / P2                               9                  Preparatory Examination September 2018
                                               NSC-Memorandum

                                               QUESTION 5.1.6

                        100

                              ●
                        80
                                  ●
                                            lumps

    Mass of CaCO3 (g)
                                       ●
                        60
                                                    ●
                                                                    ●
                                                                           ●           ●           ●              ●
                        40

                        20

                         0
                             0    10   20           30          40        50          60          70
                                                         Time (s)

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