# Physical Sciences Grade 12 Paper 2 Mpumalanga Mock 2018 Memo

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

- Page: https://www.hlayiso.com/library/mp-2018-paper-2-memorandum/625a278e-4cfe-4d04-b275-814147323e10
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2018
- Exam period: Mpumalanga Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 9

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## About this document

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

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### Page 1

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                              NATIONAL
                         SENIOR CERTIFICATE

                               GRADE 12

                 PHYSICAL SCIENCES: CHEMISTRY (P2)
                       MARKING GUIDELINES

                              SEPTEMBER 2018

MARKS: 150

TIME: 3 hours

                  This marking guidelines consists of 9 pages

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                                           2
Physical Sciences P2 Memorandum           NSC                         MDE/September 2018

 QUESTION 1

 1.1     B 

 1.2     D 

 1.3     C 

 1.4     D 

 1.5     D 

 1.6     C 

 1.7     C 

 1.8     C 

 1.9     A 

 1.10    B 
                                                                                   [20]

 QUESTION 2

 2.1     A series of organic compounds that can be described by the same general
         formula OR in which one member differs from the next with a CH2 group.
         
         ‘n Reeks organiese verindings met dieselfde algmene formule OF waarvan
         die lede verskilmet ‘n CH2 groep.                                         (2)

 2.2.1   U                                                                        (1)

 2.2.2   S                                                                        (1)

 2.2.3   P                                                                        (1)

 2.2.4   Q and S                                                                 (2)

              
 2.3.1   3,4-dibromo-2,2-dimethylpentane / 3,4-dibromo-2,2-dimetielpentaan         (3)

          
 2.3.2   Heksan-3-one / Heksan-3-oon                                               (2)

 2.4.1                                 1 mark for functional group

                                       1 mark for whole structure
                              
                                                                                   (2)

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                                                3
Physical Sciences P2 Memorandum                NSC                         MDE/September 2018
 2.4.2
                                           1 mark for functional group

                                           1 mark for whole structure

                                                                                      (2)
                                                                                      [16]

 QUESTION 3

 3.1     The temperature where the vapour pressure is equal to the atmospheric
         pressure.
         Die temperatuur waar die dampdruk gelyk is aan atmosferiese druk.             (2)

 3.2.1   Homologous series/ Functional group / Homoloë reeks/ Funksionele
         groep                                                                         (1)

 3.2.2   The number of C-atoms/ The length of the carbon chain 
         Die aantal C-atome / Die lengte van die C-ketting                             (1)

 3.3     A                                                                            (1)

 3.4.1   London forces/ Londonkragte                                                 (1)

 3.4.2   Dipole-dipole forces  / Dipool-dipoolkragte                                  (1)

 3.5     A
         Lower boiling point / Laer kookpunt                                         (2)

 3.6     Compound C: Alcohol-One site for hydrogen bonding 
         Compound D: Carboxylic acid- Two sites for hydrogen bonding 
         More energy needed to overcome the intermolecular forces of D.
         Verbinding C:Alkohol-Een plek vir waterstofbinding
         Verbinding D: Karboksielsuur – Twee plekke vir waterstofbinding
         Meer energie word benodig om die intermolekulêre kragte van D te oorkom.      (3)

 3.7     LOWER THAN. / LAER AS                                                        (1)
                                                                                       [13]
 QUESTION 4

 4.1.1   Substitution.  / Substitusie                                                 (1)

 4.1.2   Elimination / Dehydrohalogenation /Eliminasie / Dehidrohalogenering         (1)

 4.1.3   Addition / Hydrogenation / Addisie / Hidrogenering                           (1)

 4.2.1   2-chloropropane  / 2-chloropropaan                                          (2)

 4.2.2
                              + KOH        →                        + KCℓ + H2O
                                                                                (4)

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Physical Sciences P2 Memorandum          NSC                            MDE/September 2018
 4.3.1   An alcohol where the C-atom bonded to the –OH group, is bonded to two
         other C-atoms. 
         ‘n Alkohol waar die C-atoom wat gebind is aan die –OH groep, aan twee
         ander C-atome gebind is.                                                   (2)
           
 4.3.2   Propan-2-ol                                                                (2)

                                                                                    [13]

 QUESTION 5

 5.1     Rate of reaction / Time Reaksietempo / Tyd                               (1)

 5.2     Temperature / Temperatuur                                                (1)

 5.3               ΔV
         Rate =       
                   Δt
                   0,06−0
               =                               60−0                        2
                   30−0                     =          = 2 cm3∙s-1   MAX:
                                                30−0                        3
               = 2 X 10-3 dm3∙s-1                                                  (3)

 5.4     LOWER THAN / LAER AS                                                     (1)

 5.5     The small pieces of magnesium has a greater surface area than the one
         piece. 
         More particles are exposed. 
         More effective collisions per second. 
         Higher reaction rate.
         Die klein stukkies magnesium het ‘n groterkontakoppervlakte.
         Meer deeltjies is blootgestel.
         Meer effektiewebotsings per sekonde.
         Hoër reaksietempo.                                                         (3)

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Physical Sciences P2 Memorandum                     NSC                            MDE/September 2018

 5.6     Mg              +            2HCℓ          →          MgCℓ2       +       H2
         25 g                        n=cXV
                                                                Marking Rule
                                      = 0,1 X 0,1
                                                                   Mol HCℓ
               m                                                   Use of ratio 2:1
         n=                           = 0,01 mole                 Substitute 24 g.mol-1
               M
                                                                   1,04 mol
                                                                   Subtraction:
                25
           =                                                        initial mol - mol used
               24                                                 Final answer
          = 1,04 mole 
                                                                    Range (24,8 -25)

         0,005 mole                           0,01 mole (ratio)

         1,04 – 0,005 = 1,035 mole 
         m = n X M = 1,035 X 24 = 24,84 g                                                    (6)

                                                                                               [15]

 QUESTION 6

 6.1     HOMOGENEOUS / HOMOGEEN                                                              (1)

 6.2     The reactants and products are in the same phase.                                   (2)
         Die reaktante en produkte is in dieselfde fase.
 6.3               m       138,6
             n=      =         = 0,9 
                   M        154
             Ratio : 1:3:1:1 
             Calculation of mole at equilibrium 
             Divide by 2 
             Correct Kc equation
             Correct substitution in Kc equation 
             Answer = 4,92                                                                   (8)

                                       CS2              3Cℓ2           S2Cℓ2        CCℓ4
          Mole initial                  1,2               4              0            0
          Mole react/form              -0,9             -2,7           +0,9         +0,9
          Mole equillibrium             0,3              1,3            0,9          0,9
          []                           0,15             0,65           0,45         0,45

                     [S2Cℓ2][CCℓ4]
          Kc =
                      [CS2][Cℓ2]3

               =      (0,45)(0,45)
                       (0,15)(0,65)3

               =       4,92

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Physical Sciences P2 Memorandum             NSC                          MDE/September 2018

 6.4.1   DECREASES / NEEM AF                                                         (1)

 6.4.2   INCREASES / NEEM TOE                                                        (1)

 6.4.3   REMAINS THE SAME / BLY DIESELFDE                                            (1)

 6.5     If the concentration of S2Cℓ2 is increased, the system will according to Le
         Chatelier act in such a way to decrease the concentration of S2Cℓ2. 
         The reverse reaction will be favoured. 
         The amount of reactants will increase. 
         As die konsentrasie an die S2Cℓ2 verhoog word, sal die sisteem volgens Le
         Chatelier, sodanig reageer dat die konsentrasie weer sal verlaag.
         Die terugwaartse reaksie word bevoordeel.
         Die aantal mol reaktante sal toeneem.                                         (3)

 6.6     The pressure is decreased.  / Die druk word verlaag.                         (1)

 6.7     INCREASE  / TOENEEM                                                          (1)

 6.8     The rate of both forward and reverse reactions has increased.                 (1)
         Die tempo van beide voorwaartse en terugwaartse reaskies het verhoog.         [20]

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Physical Sciences P2 Memorandum             NSC                               MDE/September 2018

 QUESTION 7

 7.1     It dissociates completely in water and produce a high concentration of
         OH- ions. 
         DIt dissosieer volledig in water en vorm ‘n hoë konsentrasie OH- ione.           (2)

 7.2     [H3O+] = 0,4 mol∙dm-3

         pH = -log [H3O+] 
            = -log (0,4)
            = 0,4                                                                        (3)

 7.3     pH = 13
         [H3O+] = 10-13 
         [H3O+][OH-] = 1 x 10-14
         [OH-] = 10-1 mol∙dm-3
         [NaOH] = 0,1 mol∙dm-3                                                           (3)

 7.4.1   BASIC                                                                           (1)
             
 7.4.2   CO3 + H2O ⇌ HCO3- + OH-
            2-                                                                            (2)

 7.5     Na2CO3                    →                   CO2
          Marking Rule
                         V
           Formula 𝑛 = V
                              m
             Substitute 22,4 dm3.mol-1
             Ratio: 1:1
             Subtitute 106 g.mol-1
             Final answer

         OPTION 1                                     4,48 dm3

         m=nXM                                               V         4,48
          = 0,2 X 106                                n =                   
                                                             Vm        22,4
          = 21,2 g
          0,2 mole                                       = 0,2 mole (ratio)

              21,2
         %=          X 100 = 84,8%                                                       (5)
               25

         OPTION 2

              25                           0,2
         n=         = 0,236 mole          0,236
                                                  X 100 = 84,75%
              106

                                                                                          [16]

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Physical Sciences P2 Memorandum              NSC                          MDE/September 2018

 QUESTION 8

 8.1     Oxidation is an increase in oxidation number. 
         Oksidasie is ‘n toename in oksidasiegetal.                                   (2)

 8.2     Complete the circuit. / Voltooi die stroombaan,
         Maintain electrical neutrality./ Handhaaf elektriese eutraliteit.         (2)

 8.3.1   Fe3+ + 3 e- → Fe                                                           (2)
            
 8.3.2   Ti(s)│Ti3+ (aq)║ Fe3+ (aq)│Fe(s)
                 OR
         Ti │ Ti3+ ║ Fe3+ │ Fe                                                        (3)

 8.4     E°cell = E°reduction - E°oxidation 
         1,57  = -0,06 - E°oxidation
         E°oxidation = -1,63 V                                                       (4)

 8.5     0 (V)                                                                       (1)

                                                                                      [14]

 QUESTION 9

 9.1     The process where electrical energy is converted to chemical energy.      (2)
         Die proses waar elektriese energie omgeskakel word in chemiese
         energie.
 9.2.1   Ag → Ag+ + e-                                                              (2)

 9.2.2   Ag + Ag+ → Ag + Ag+                                                        (2)

 9.3     Directly proportional / Direk eweredig                                     (1)

 9.4.1   Q                                                                           (1)

 9.4.2   Equal to 17 g                                                               (1)

 9.4.3        m        17
         n=        =          = 0,157 mole                                         (2)
              M        108

 9.4.4   Number of e- = 0,157 X 6,02 X 1023  = 9,45 X 1022                          (2)

                                                                                      [13]

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Physical Sciences P2 Memorandum            NSC                           MDE/September 2018

 QUESTION 10

 10.1.1 Fe / FeO                                                                     (1)

 10.1.2 NH4NO3                                                                       (1)
              
 10.1.3 4NH3 + 5O2 → 4NO + 6H2O (bal)                                               (3)

 10.2.1 X                                                                            (1)

 10.2.2 X has the highest percentage of potassium that is needed for the growth
        of fruit. 
        X het die hoogste persentasie kalium wat nodig is vir die groei van vrugte.   (1)

 10.2.3 OPTION1

         15+3+3 = 21

          3
              X 51 = 7,29% K
         21

         7,29
                X 20 = 1,45 kg                                                      (3)
         100

         OPTION 2

          𝟓𝟏
                X 20 = 10,2 kg
         𝟏𝟎𝟎

          3
               X 10,2 = 1,46 kg                                                     [10]
         21
                                                                           TOTAL      150

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