# Physical Sciences Grade 12 Limpopo November 2016 Memo

> Grade 12 Physical Sciences memorandum from the Limpopo November 2016, 11 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/lim-chemistry-chemie-memo/150a3b71-e772-47db-a2b8-0fcbccedbe4c
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2016
- Exam period: Limpopo November
- Curriculum: CAPS
- Pages: 11

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## About this document

Physical Sciences · Grade 12 · Limpopo November · 2016. Memorandum, 11 pages. Read online or download the PDF.

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### Page 1

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                          NATIONAL
                     SENIOR CERTIFICATE

                          GRADE 12

                PHYSICAL SCIENCES P2: CHEMISTRY
                FISIESE WETENSKAPPE VR 2: CHEMIE

                        SEPTEMBER 2016

                         MEMORANDUM

MARKS: 150

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                                             2
Physical Science P2 Memo                    NCS                Limpopo DoE September 2016

QUESTION?VRAAG 1
1.1       A                                                                           (2)
1.2       D                                                                           (2)
1.3       B                                                                           (2)
1.4       C                                                                           (2)
1.5       D                                                                           (2)
1.6       C                                                                           (2)
1.7       A                                                                           (2)
1.8       A                                                                           (2)
1.9       C                                                                           (2)
1.10      B                                                                           (2)
                                                                                        [20]

QUESTION/VRAAG 2
2.1
        2.1.1        Esters                                                          (1)

        2.1.2        Carboxyl group/karboksiel groep                                 (1)

        2.1.3        A                                                               (1)

        2.1.4        5-bromo-2,4-dimethylheptane //                                (3)
                     5-broom–2,4–dimetielheptaan of / 5-bromo–2,4–
                     dimetielheptaan
        2.1.5        C n H 2n                                                        (1)

        2.1.6
                          H                      Notes
                     H C      O   H   
                                                 ∙ Functional group 
                          H                      ∙ Whole structure correct           (2)

        2.1.7        propanoic acid  / propanoësuur                                  (1)

2.2
        2.2.1        Polyethene/ Polythene/ polieteen/ politeen/ polietileen
                     Accept: Polyethylene                                            (1)

        2.2.2        Addition Polymerisation  addisie polimerisasie                  (1)

        2.2.3        (a) CH 2                                                        (1)

                     (b) CH 2                                                        (1)
                                                                                      [14]

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Physical Science P2 Memo                 NCS               Limpopo DoE September 2016

QUESTION/VRAAG 3
3.1       A series of organic compounds that can be described by the same general
          formula.  / ‘n reeks organiese verbindings wat deur ‘n algemene formule
          beskryf kan word

OR:       A series of organic compounds in which one member differs from the next
          with a CH 2 group./ ‘n reeks organiese verbindings waar een van die ander met
          CH 2 groep/eenheid verskil
                                                                  NOTES: 2 or 0
3.2       Ketones / ketone                                                               (1)

3.3       Butanoic acid/ butanoöesuur                                                    (1)

3.4.1     Boiling point/ kookpunt                                                        (1)

3.4.2     Molecular mass / molekulêre massa                                              (1)

3.5       The boiling point of carboxylic acids > b.p of alcohols > b.p of aldehydes     (1)
          Kookpunt van karboksielsure > k.pt alkohole > k.pt van aldehiede
3.6       ∙ Both compounds A and B are hydrogen bonded / beide A en B het
          waterstofbinding IM kragte
          ∙ Compound A has two (more) sites for hydrogen bonding /A forms
             dimers/A is more polar/ B has less/one sites for hydrogen bonding // A het
          twee plekke vir waterstofbinding (B het slegs 1 plek vir H-bd)
          ∙ Therefore A has stronger intermolecular forces than B/B has weaker
              Intermolecular forces than A  / A het sterke IM kragte                    (3)

3.7       The stronger/weaker the intermolecular forces, the lower/higher the
          vapour pressure. // hoe sterker die IM kragte hoe swakker is die dampdruk / h (1)
          swakker die IM kragte hoe hoer is die dampdruk
                                                                                         [11]

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Physical Science P2 Memo                                  NCS                Limpopo DoE September 2016

QUESTION/VRAAG 4
4.1.1   Addition/hydrobromination / Hydrohalogeantion / Addisie/
        hidrohalogenasie aanvaar: hidrobromasie                                                             (1)

4.1.2         H       H       H       H                                           H   H   H      H

          H- C        C C         C       H       + HBr               H C C C C H
                                                                    
             H        H                                                  H H Br H                           (4)
        Notes: Reactants Product Marking Rule 6.3.10.

4.1.3   2-bromobutane / 2-broombutaan/2-bromobutaan                                 Notes
                                                                                      ∙ 2-bromo 
                                                                                      ∙ butane 
                                                                                                            (2)

4.2.1   Substituition  /Hydrolysis                   / substitusie/ hidroliese                             (1)
4.2.2                                                            Notes:

                      H       H       H           H              ∙ Functional group correct /
                                                                  Funksionele groep korrek
              H       C       C       C           C      H
                      H       H        OH          H            ∙ Whole structure correct /
                                                                  Hele struktuur korrek
                                                                                                            (2)

4.2.3     Butan-1-ol / 1- butanol                                                                         (2)
                                                                  Marking Rules:
4.3               H       H       H           H                   Functional group correct / korrekte
         H       C        C       C       C        H              funksionele groep
                                                                 Whole structure correct  / hele
              H                               H                  struktuur reg
                                                                                                            (2)
4.4
4.4.1     Alkene  / alkene                                                                                 (1)
4.4.2     Potassim bromide/KBr / kaliumbromied
          Water /H 2 O                                                                                     (2)
                                                                                                           [17]

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Physical Science P2 Memo                                                         NCS              Limpopo DoE September 2016

QUESTION/VRAAG 5
5.1               Carbon dioxide/CO 2 / koolstofdioksied                                                                    (2)
5.2.1             The decrease in concentration of hydrochloric acid per unit time. /
                  die afname in soutsuur konsentrasie per eenheidstyd                                                       (2)
                  (2 or 0)
5.2.2             Concentration(of the acid) / konsentrasie                                                                (1)
5.2.3             For a fair test/comparison / vir ‘n regverdige toets
                                                                                                                            (1)
5.2.4             Higher acid concentration in experiment 2 means:
                     • More particles/molecules per unit volume
                     • More particles have kinetic energy equal to or greater than
                        activation energy/More particles have enough kintetic energy 
                     • More effective collisions per unit time/frequency of effective
                        collisions increases/Rate of effective collisions increases.                                       (3)
                  Hoër konsentrasie suur vir eksperiment beteken:
                     • Meer deeltjies/ eenheidsvolume
                     • Meer deeltjies het genoeg E K gelyk of groter as aktiverings
                         energie/ genoeg E K
                     • Meer effektiewe botsings/ eenheidstyd
                                                               m
5.2.5             n(CaCO 3 ) =                                 𝑀𝑀
                                                    4
                                                  = 100                               Marking Guidelines:
                                                                                                                 𝑚𝑚
                                                     = 0,04 mol                        ∙ Substitution into n =
                                                                                                                 𝑀𝑀
                                                                                                         1
                  n(HCℓ):n(CaCO 3 ) = 2 : 1                                            ∙ Ratio n(HCℓ) = n(CaCO3)
                                                                                                         2
                            ∴ n(HCℓ) = 2(0,04) = 0,08 mol                             ∙ Substitution in c = n/V
                                                                                       ∙ Final answer
                                                          𝑛𝑛
                              c(HCℓ) = 𝑉𝑉
                                                        0,08
                                   0,4 = 𝑉𝑉 
                                   ∴ V = 0,2 dm3 = 200 cm3
                                                                                                                            (4)
5.3               POTENTIAL ENERGY VERSUS REACTION COORDINATE GRAPH
                                                                    Activated Complex / geaktiveerde kompleks
                                                        Ea

        Potential energy/    Potensiële energie
                                                                                         Shape /

                                                                    ∆H                   vorm

                                                                                                                            (4)
                                                          Reaction Coordinate / reaksie ko-ordinate                         [17]

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                                               6
Physical Science P2 Memo                     NCS                Limpopo DoE September 2016

QUESTION/VRAAG 6
6.1         3 to 4 minutes (Any value from 3 min to 4 min can be accepted) / 3 - 4          (1)
            minute
6.2.1       Greater than/ groter as                                                        (1)

6.2.2       Equal to / gelyk aan                                                           (1)

6.3         CALCULATION USING NUMBER OF MOLES / bereken met aantal
            mol

        Marking Criteria:
        • Correct Kc expression/ korrekte Kc uitdrukking
        • Substituting the Kc value in the expression/ vervanging van Kc waarde
        • Substituting concentration value of PCℓ5(g) in Kc/ vervanging van [PCℓ5]] in Kc
        • Equilibrium concentration PCℓ3 and Cℓ2 = 0,91 mol·dm-3/ [ewewig] PCℓ3 = Cℓ2 =
          0,91
        • Using n = cV to find the n of PCℓ5 (multiplying c by 2)/ x 2 om n van PCℓ5 te kry
        • Ratio of n(PCℓ5): n(PCℓ3): n(Cℓ2) = 1:1:1/ verhouding [PCℓ5]: [PCℓ3]: [Cℓ2] =
          1:1:1
        • Calculating n (PCℓ5) initial = 2,12 mol/ berekening van aanvanklik PCℓ5 = 2,12

            Assume that the equilibrium concentration of PCℓ 3 (g) is x
                   [PCl3][Cl2]
            Kc =
                     [PCl5] 
                   ( x )( x )
            5.55 =
                     0,15 
            ∴x =0.91 mol·dm-3 

                                                PCℓ 5       PCℓ 3          Cℓ 2
             Initial Quantity (mol)            2,12          0             0
             Change in quantity (mol)           -1,82       +1,82         +1,82       R

             Equilibrium quantity (mol)         0,30        1,82         1,82        X2
             Concentration at equilibrium          0,15      0,91         0,91
             (mol·dm-3) V = 2 dm3
                 𝑚𝑚
             n = 𝑀𝑀

                      m
            2,12 =
                      102   
            m = 216,24 g 

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    Physical Science P2 Memo                      NCS                 Limpopo DoE September 2016

           CALCULATION USING CONCENTRATION/ berekening met [ ]
•    Correct Kc expression/ korrekte Kc uitdrukking
•    Substituting the Kc value in the expression/ vervanging van Kc waarde
•    Substituting concentration value of PCℓ5(g) in Kc / vervanging van [PCℓ5]] in Kc
•    Equilibrium concentration for PCℓ3 and Cℓ2 = 0,91 mol·dm-3 / [ewewig] PCℓ3 = Cℓ2 = 0,91
•    Ratio of [PCℓ5]: [PCℓ3]: [Cℓ2] = 1:1:1 / verhouding [PCℓ5]: [PCℓ3]: [Cℓ2] = 1:1:1
•    Initial concentration of PCℓ5(g) / aanvanklike [PCℓ5]
•    Substituting 2 in c = n/V / deel deur 2 vir [ ]
•    Substituting 102 in n = m/M / vervanging van 102 in n = m/M
•    Final answer / finale antwoord

           Assume that the equilibrium concentration of PCℓ 3 (g) is x

                  [PCl3][Cl2]
           Kc =
                    [PCl5] 
                     ( x )( x )
           5.55 =
                       0,15       

         ∴x =0.91 mol·dm-3 
                                                   PCℓ 5      PCℓ 3         Cℓ 2
    Initial concentration (mol·dm-3 )             1,06        0             0
    aanvanklik
    Change in concentration (mol·dm-3 )            -0,91      +0,91        +0,91      Ratio/
    Verandering                                                                       verhouding
    Equilibrium concentration (mol·dm-3 )          0,15       +0,91        +0,91
    Ewewig

                                                        𝑛𝑛               OR/ of
                                                 𝑐𝑐 =
                                                        𝑉𝑉               m = cMV
                                                         n
                                               1.06 = 2                 =(1,06)(102)(2)
                                               n = 2,12 mol              =216,24 g

                                      m = n M = (2,12)(31+5 x 31,5) 
                                             = 443,02 g                                       (9)

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Physical Science P2 Memo                 NCS               Limpopo DoE September 2016

6.4      Endothermic.  endotermies

         Decrease in temperature increases [PCℓ 5 ] and decreases [Cℓ 2 ] and
          [PCℓ 3 ], which implies that the reverse reaction is favoured/equilibrium
         position shifts to the left. According to Le Chatelier ‘s Principle, the
         decrease in temperature favours the exothermic reaction.
         Therefore, the forward reaction is endothermic. /
         Afname in temperatuur verhoog [PCℓ 5 ] en verlaag [Cℓ 2 ] en [PCℓ 3 ], dus word
         terugwaartse reaksie bevoordeel/ ewewig skuif na links
         Volgens Le Chatelier se beginsel sal ‘n afname in temperatuur die
         eksotermiese reaksie bevoordeel (d.w.s terugwaartse reaksie is endotermies )
         Dus is die voorwaartse reaksie endotermies                    (3)
                                                                                    [15]

QUESTION/VRAAG 7
7.1
7.1.1   Concentrated acid- contains a large amount (number of moles) of
        acid in proportion to the volume of water.
        Dilute acid – contains a small amount (number of moles) of acid in
        proportion to the volume of water./                                       (2)
        Gekonsentreede suur: bevat ‘n groot hoeveelheid /mol suur in
        verhouding to die volume water
        Verdunde suur bevat ‘n klein hoeveelheid / aantal mol suur in
        verhouding tot die volume water
7.1.2   It ionises completely in water to form a high concentration of H 3 O+
        ions. // dit ioniseer volledig in water en vorm ‘n hoë konsentrasie
        H 3 O+ ione                                                                (2)
                                                               Note: 2 or 0
7.1.3   pH = -log[H 3 O+]
           = - log(0,20)
           = 0,7                                                                  (3)
7.2.1   Basic / basies
                                                                                   (1)
7.2.2      CO 3 (aq) + H 2 O (ℓ)  HCO 3 (aq) + OH (aq)
                2-                       -           -

        (reactants  products )
        Excess OH- (aq) ions are produced/it is a salt of strong base and
        weak acid (and the resultant solution is basic.) 
        ‘n Oormaat OH- ione word geproduseer/ dit is die sout van ‘n sterk
        basis en ‘n swak suur                                                      (3)

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Physical Science P2 Memo                 NCS                Limpopo DoE September 2016

                                          Marking Criteria / nasienkriteria:
7.3.1                 V
         n(CO2) =                         •    Substitution into/vervanging in
                     Vm                               𝑉𝑉
                     4,48                      n =
                   =                                𝑉𝑉𝑚𝑚
                     22,4                 •    Using mole ratio/ gebruik molverhouding
                   = 0,2 mol              •    Multiplying by/ vermenigvuldig met 106
                                          •    Percentage purity / pesentasie
        n(Na 2 CO 3 ): n(CO 2 ) = 1:1          suiwerheid
        ∴ n(Na 2 CO 3 ) = 0,2 mol 
        m(Na 2 CO 3 ) = n·M = (0,2)(106)  = 21,2 g
                          mass of pure Na2 CO3
        ∴ % purity =                           × 100%
                        mass of impure Na2 CO3
                        21,2
                     =       × 100 
                         25
                                                                                   (5)
                     = 84,8%
7.3.2   Mass of impurity/ massa onsuiwerheid = 25 -21,2 = 3,8 g
        OR % impurity/ onsuiwerheid = 100-84,8 = 15,2%
        15,2
               x 25 = 3,8 g
         100
                                                                                   (2)
                                                                                   [18]

QUESTION/VRAAG 8
8.1.1   Reduction is a gain in electron(s) (by a chemical substance)// die       (2)
        bykry van elektrone
            Note: 2 or 0
8.1.2   Aluminium is a stronger reducing agent than Sn/ Ag/ Hg and
        therefore it is easily oxidised and a current flows, simulating the
        pain sensors./                                                             (3)
        Al is ‘n sterker reduseermiddel as sn/Ag/Hg en word daarom maklike
        geoksideer. ‘n Stroom vloei wat pyn veroorsaak
8.2.1   2Cr2+(aq) + Cu2+(aq) 2Cr3+(aq) + Cu(s)
        (reactants, products, balancing)                                        (3)
8.2.2   Eθcell = Eθcathode − Eθanode 
               =+0,34-(-0,41)
               = 0,75 V 
        OR
        Eθcell = Eθcathode − Eθanode 
               = +0,34+0,41                                                      (4)
               = 0,75 V
8.2.3   Increases  / verhoog
        Increase in temperature increases the rate of electron transfer as well
        as the movement of ions in the electrolyte.
        (This increases the rate of flow of charge in the cell.)/                  (2)

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Physical Science P2 Memo                 NCS                Limpopo DoE September 2016

        ‘n verhoging in temperatuur verhoog die tempo van electron oordrag en
        die beweging van ione in die elektroliet
8.2.4   As the cell operates, [cation]/[Cr 3+ ]increases in the anode half cell and
        [cation]/[Cu 2+]decreases in the cathode half cell.
        Anions from the salt bridge move into the anode half-cell to balance out
        excess cations.
        Cations from the salt bridge move into the cathode half-cell to balance
        out shortage of cations./                                                  (3)
                                                     3+
        Soos die sel werk, verhoog [katione] / [Cr ] in die anode/Cr halfsel en
        [katione] / [Cu 2+] verlaag in die katode/Cu halfsel. Anione /
        negatiewe ione va die soutbrug beweeg na die anode/ Cr halfsel om
        die oormaat katione/ Cr 3+ uit te balanseer
        Katione / negatiewe ione van die soutbrug beweeg na die katode/ Cu
        halfsel om die tekort van katione/ Cu2+ te balanseer
                                                                                    [17]

QUESTION/VRAAG 9
9.1     Bauxite / Bauksiet                                                        (2)

9.2      Carbon / Koolstof                                                         (1)

9.3     The anode reacts with the oxygen produced to form CO 2. // Die             (1)
        anode reageer met die suurstof wat geproduseer word en vorm CO 2.
9.4     Cryolite / Krioliet                                                        (1)

9.5     Aℓ + 3e- Aℓ                                                              (2)

9.6     1890 C~ 0,176 g Aℓ
        96500 C ~ x g Aℓ
             (96500 )(0,176 )
        ∴x =                  
                 (1890 )
           = 8,98 g

        1 mol of metal ~ 27 g
        Since 8,98 g of metal are discharged by 1 mol of electrons,
                                          27g
        27 g of metal are discharged by          =3 moles of electrons.
                                         8,98g
        If 1 mole of metal needs 3 moles of electrons, the charge on the
        cations is +3/
        1 mol metaal ~ 27 g
        Omdat 8,98 g metaal ontlaai word deur 1 mol elektrone, sal
                                        27g
        27 g metaal ontlaai word deur          =3 mol elektrone
                                       8,98g
        Indien 1 mol metal 3 mol elektrone benodig, is die lading van die
        katioon +3                                                                   (3)
                                                                                    [10]

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Physical Science P2 Memo                    NCS                  Limpopo DoE September 2016

QUESTION/VRAAG 10
10.1
10.1.1 Increase in population requires an increase in the production of food.
       OR Removing plant material from the soil depletes the nutrients so
       more fertilizer must be added to the soil. /
       Toename in bevolking vereis ‘n toename in voedselproduksie OF                      (1)
       verwydering van plantmateriaal vanuit die grond veroorsaak uitputting
       van voedingstowwe so meer kunsmis moet tot die grond toegevoeg
       word
10.1.2 Allows nutrients to be absorbed through roots with water./ laat toe               (1)
       dat voedingstowwe deur die wortels opgeneem word
10.1.3 Run off/ leaching results in the cost of lost nutrients/ eutrophication/          (1)
       eutrifikasie/ afloop /loging veroorsaak verhoogte koste as gevolg van
       verlore voedingstowwe
10.2
10.2.1 Contact process/ kontakproses                                                     (1)

10.2.2 Oleum/ pyrosulphuric acid/ fuming sulphuric acid/oleum/rokende                    (1)
       swael
10.2.3 H 2 S 2 O 7 + H 2 O  2H 2 SO 4
       (reactants, products, balancing) Note: Marking rule 6.3.10                      (3)

10.2.4    Ammonium sulpahte /(NH 4 ) 2 SO 4 / ammoniumsulfaat                            (1)

10.2.5    (a)    For stems and leaves development / vir ontwikkeling van                 (1)
                blare en stamme
                                                                                          (1)
          (b)    Fractional distillation of liquid air / fraksionele distillasie van
                vloeibare lug
                                                                                  [11]
                                                                   GRAND TOTAL : 150

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## Collection

- [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers)

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