# Physical Sciences Grade 12 P2 Eastern Cape Prelim 2024 Memo

> Grade 12 Physical Sciences Paper 2 memorandum from the Eastern Cape Prelim Exam 2024, 21 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/xhosa-versioned-memo-ec-phys-sciences-p2-memo-sept2024-final/f9c71b2f-e2e2-42d9-8911-d0cf64643370
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2024
- Exam period: Eastern Cape Prelim Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 21

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Physical Sciences · Grade 12 · Eastern Cape Prelim Exam · 2024. Memorandum, 21 pages. Read online or download the PDF.

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### Page 1

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                          NATIONAL
                     SENIOR CERTIFICATE

                        GRADE/IBANGA 12

                       SEPTEMBA 2024

 PHYSICAL SCIENCES P2 IFIZIKHALI SAYENSIZI P2

  MARKING GUIDELINE/UMGAQO WOKUMAKISHA

AMANQAKU:   150

                  This marking guideline consists of 21 pages./Lo
                  mgaqowokumakisha unamaphepha angama 21

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2             PHYSICAL SCIENCES P2/IFIZIKHALI SAYENSIZ P2   (EC/SEPTEMBA 2024)

QUESTION 1 /UMBUZO 1

1.1   C                                                                   (2)

1.2   C                                                                   (2)

1.3   B                                                                   (2)

1.4   A                                                                   (2)

1.5   B                                                                   (2)

1.6   B                                                                   (2)

1.7   C                                                                   (2)

1.8   A                                                                   (2)

1.9   B                                                                   (2)

1.10 B                                                                   (2)
                                                                          [20]

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(EC/SEPTEMBA 2024)              PHYSICAL SCIENCES P1/IFIZIKHALI SAYENSIZI P2                  3

 QUESTION 2 / UMBUZO 2

 2.1     Marking criteria/ Umgaqo woniko manqaku
         If any of the underlined key words/phrases in the correct context are omitted:
         - 1 mark per word/phrase.
         Ukuba esinye seziqwengana ezingu ndoqo esinomgca ngaphantsi ngokwe
         khonteksti eyiyo sishiyiwe, nyiba inqaku libe-1.

         A bond or an atom or a group of atoms that determine(s) the physical and
         chemical properties of a group of organic compounds. 

         Yibhondi okanye i-athom okanye igruphu yee-athomu edithemina iifizikhali nee
         khemikhali propatizi ze gruphu yee oganikhi khompawundi.                               (2)

 2.2.1 E                                                                                         (1)

 2.2.2 A                                                                                         (1)

 2.2.3 C                                                                                         (1)

 2.3     EQUAL TO. 

         Compound E and ethyl methanoate are functional isomers / structural isomers
         / have the same molecular formula / same number and type of atoms / same
         number of C, H and O atoms 

         ILINGANA NO. 

         Ukhompawundi E no ethyl methanoate baziifunctional isomers/ structural
         ispmers/ bane molecularfomyula efanayo/bane nani elilingayo netayiphu yee
         athom/ banenani lee athom zika C,H no O elilinganayo.                                  (3)

 2.4.1                 H
                        │                              Marking criteria/ Umgaqo woniko
              H      H― C ― H O           H            manqaku
                                                         • Functional group correct 
              │         │     II          │
                                                              iFunctional gruphu
          H ― C      ― C ― C ―            C ―     H    .      Whole structure correct 
              │         │                 │                   Isitraktsha sonke sichanekile
              H         H                 H
                                                                                                  (2)

 2.4.2 Propanoic acid / Propanoic asidi                                                         (2)

 2.4.3 4-methylhex-2-yne / 4-methyl-2-hexyne Marking criteria/ Umgaqo woniko
                                                 manqaku
                                                    • Hexyne 
                                                    • Methyl                                     (3)
                                                    • Whole name correct / igama lonke
                                                       lichanekile

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4                PHYSICAL SCIENCES P2/IFIZIKHALI SAYENSIZ P2                   (EC/SEPTEMBA 2024)

                                                  Marking criteria/ Umgaqo woniko
                                                  manqaku

2.5           H   H   H                               •    Functional group on 2nd carbon 
              │   │   │                                    iFunctional gruphu ikwikhabhoni
                                                           yesibini
          H ― C ― C ― C ― H
                                                       • Whole structure correct 
              │   │  │                           Isitraktsha sonke sichanekile
              H   Cℓ  H                                                                              (2)

2.6    C7H12 + 10 O2 → 7 CO2 + 6 H2O             Bal 

       NOTES / AMANQAKU
         • Reactant / Irhiecktenti   Products/ iProwduakti  Balancing / ukuBalansisa 
                                                                                                     (3)
                                                                                                    [20]

QUESTION 3 / UMBUZO 3

3.1.1 Marking criteria/ Umgaqo woniko manqaku

       If any of the underlined key words/phrases in the correct context are omitted:
       - 1 mark per word/phrase.
       Ukuba esinye seziqwengana ezingu ndoqo esinomgca ngaphantsi ngokwe
       khonteksti eyiyo sishiyiwe, nyiba inqaku libe-1

       The temperature at which the vapour pressure of a substance / liquid equals
       the atmospheric pressure 
       Itempritsha apho ivapour pressure yesabsitensi/ ilikhwidi ilingana ne
       atmospheric presha.                                                                           (2)

3.1.2 All Primary alcohols / Zonke iiprayimari alkhoholi                                            (1)

3.1.3 London forces / dispersion forces/ induced-dipole forces 
      iiLondon fosi/ iidispersion fosi/ ii- induced-dipole fosi                                      (1)

3.1.4 Pentan-1-ol / 1-pentanol / i-Pentan-1-ol / i-1-pentanol /                                    (2)

3.2.1 Compound C / propanoic acid 
      Ukhomawundi C / ipropanoikhi aside                                                             (1)

3.2.2 Marking criteria/ Umgaqo woniko manqaku
      • Type of intermolecular forces in compound A 
      • Compound B and C have hydrogen bonds 
      • Compare the number of sites for hydrogen bonding in B and C 
      • Compare the strength of intermolecular forces 
      • Compare energy required to overcome the intermolecular forces 

      •   Itayiphu ye intamoletyhula fosi ku khompawundi A
      •   Ukhompawundi B no C bane hydrogen bhondi
      •   Thelekisa inani lee site ze hydrogen bhodi ku B naku C
      •   Thelekisa isitrhithi se-itamoletyhula fosi

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(EC/SEPTEMBA 2024)               PHYSICAL SCIENCES P1/IFIZIKHALI SAYENSIZI P2             5
        •    Thelekisa i=eneji efunekayo ukoyisa ii-intamoletyhula fosi

        •    Compound A/butanone has dipole-dipole forces  (and London forces /
             dispersion forces/ induced-dipole forces)
        •    Compound B/butan-1-ol and C/propanoic acid has hydrogen bonds (and
             London forces / dispersion forces/ induced-dipole forces)
        •    Compound B/butan-1-ol has one site for hydrogen bonding and C/propanoic
             acid has two sites for hydrogen bonding 
        •    Strength of the intermolecular forces increases from compound A /
             butanone to compound B / butan-1-ol to compound C / propanoic acid 
        •    More energy is needed to overcome the intermolecular forces in compound
             C / propanoic acid than compounds A/ butanone and B/butan-1-ol 

         •   Ukhompawundi A/ ibutanone unee dipole-dipole fosi (neeLondon forces /
             needispersion fosi/nee-induced-dipole fosi)
         •    Ukhompawundi B/ ibutan-1-ol no C/ ipropanoic asidi inee hydrogen
             bhond (neeLondon fosi / iidispersion fosi/ ii-induced-dipole fosi)
             • UkhompawundiB/ ibutan-1-ol ine site enye yehydrogen bhondingi noC
                /ipropanoic asidi inee site ezimbini zehydrogen bhondingi
             • iiStrength se-intermolecular fosi siyanda ukusuka kukhompawundi A /
                ibutanone ukuya kukhompawundi B / ibutan-1-ol ukuya
                kukhompawundi C / ipropanoic asidi
         •   i-eneji eninzi iyafuneka ukoyisa ii- intermolecular fosi kukhompawundi C /
             kwipropanoic asidi kuno khompawundi A/ ibutanone noB/ ibutan-1-ol            (5)
                                                                                          [12]

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6                  PHYSICAL SCIENCES P2/IFIZIKHALI SAYENSIZ P2            (EC/SEPTEMBA 2024)

QUESTION 4 / UMBUZO 4

4.1.1   UNSATURATED  It contains a double bond between its carbon-carbon
        atoms in the hydrocarbon chain 
        I-ANSATSHUREYITHEDI  Inee double bhondi phakathi kwee carbon-
        carbon athom kwihydrocarbon chain.
                                                                                               (2)

4.1.2   Substitution / Hydrolysis of a haloalkane 
        yiSabsitityushini /y iHydrolysis yeehaloalkane                                         (1)

4.1.3   Elimination / dehydrohalogenation of haloalkanes 
        yiElimination / yidehydrohalogenation /yidehydrohalogenation
        yeehaloalkane                                                                          (1)
                                                         Marking criteria/ Umgaqo
                                                         woniko manqaku
4.1.4         H       H     H    H     H            •    Functional group correct 
                                                         iFunctional gruphu
         H     C      C     C    C     C    H           .        Whole structure
                                                         correct 
                                                        Isitraktsha sonke sichanekile
               H      Cℓ    H    H     H                                                       (2)

4.1.4   Pentan-2-ol / 2-pentanol /iPentan-2-ol / i-2-pentanol                                (2)

4.1.6   Dilute strong base / NaOH  and mild heat 
        iDilute strong base / iNaOH  nemild heat                                              (2)

4.1.7   Positional isomers 
        iPositional isomers                                                                    (2)

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(EC/SEPTEMBA 2024)              PHYSICAL SCIENCES P1/IFIZIKHALI SAYENSIZI P2       7

 4.2      Marking criteria / Umgaqo woniko manqaku Reaction 1:
          Dehydration reaction and correct chemicals. 
          Reactants with correct condensed structural formula 
          Products with correct condensed structural formula 

          uRhiektshini 1:
          iDehydration rhiekshini neekhemikhali ezichanekileyo.
          iiRhiekhenti eziziKhocondenisidi straktshaliral fomyula echanekileyo
          iiProwdacti eziziKhondenisidi straktshaliral fomyula echanekileyo.

          Reaction 2:
          Addition reaction and correct chemicals used. 
          Reactants with correct condensed structural formula 
          Products with correct condensed structural formula 
          uRhiekshini 2:
          i-Addition rhiekshini nokusetyenziswa kweekhemikhali ezichanekileyo
          iiRhiekithenti ezizikhondensinisidi straktsharali fomyula
          iiProwudakti ezizikhondensinisidi straktsharali fomyula

          Reaction 1
                                  Conc H2SO4
          CH3CH2CH2CH2OH  Heat / Hitte CH3CH2CHCH2 + H2O  (correct
          reaction / korrekte reaksie)
          Urhiekshini 1:
                                 Conc H2SO4
          CH3CH2CH2CH2OH           Heat    CH3CH2CHCH2 + H2O  (irhiekshini
          echanekinikileyo)

          Reaction 2
                                    Pt
          CH3CH2CHCH2 + H2                   CH3CH2CH2CH3  (correct reaction)
                                   Heat

          Urhiekshini 2
                                    Pt
          CH3CH2CHCH2 + H2                   CH3CH2CH2CH3  ( irhiekshini
                                   Heat
          echanekinikileyo)

                                                                                    (6)
                                                                                   [18]

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8               PHYSICAL SCIENCES P2/IFIZIKHALI SAYENSIZ P2            (EC/SEPTEMBA 2024)

QUESTION 5 / UMBUZO 4

5.1   Marking criteria/ Umgaqo woniko manqaku
      If any of the underlined key words/phrases in the correct context are
      omitted: - 1 mark per word/phrase.
      Ukuba esinye seziqwengana ezingu ndoqo esinomgca ngaphantsi ngokwe
      khonteksti eyiyo sishiyiwe, nyiba inqaku libe-1.

      ANY ONE
      Change in concentration of reactant or product per (unit) time. 

      Change in amount/number of moles/volume/mass of products or reactants
      per (unit) time. 

      Change in amount/number of moles/volume/mass of products formed or
      reactants used reactants per (unit) time. 

      NAYIPHI ENYE
      Uthintsho kwi khonsentreyishini yee prowudakthi/rhiekthenti per (unit) time.
      Utshintsho kwi amawunti/inumber of molesi/ivoluyumu/imass yee
      prowudakthi okanye yee rhiekthenti per (unit) time.

       Utshintsho kwi amawunti/inumber of molesi/ivoluyumu/imass yee
      prowudakthi ezifonyishiweyo okanye yee rhiekthenti ezisetyenzisiweyo per
      (unit) time.

                                        OR/OF

      The rate of change in concentration / amount of moles / number of moles /
      volume / mass.  (2 or 0).
      Irheyithi yotshintsho kwi khonsentreyishini/ I-amawunti yee mowuli/inumber
      of molesi/ivoluyumu/imass (2 of 0)
                                                                                       (2)

5.2   Marking Criteria for investigative question / Umgaqo woniko manqaku
      kwi investigative question

      The independent and dependent variables are stated 

      I-independent nedependent variables zichaziwe
      Ask a question about the relationship between the independent and
      dependent variables 
      Buza umbuzo ngereleyinshiphu phakathi kwe-independent ne dependent
      variables

      What is the effect of the increase / decrease / change of concentration on
      the reaction rate? 

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(EC/SEPTEMBA 2024)                              PHYSICAL SCIENCES P1/IFIZIKHALI SAYENSIZI P2                    9

         Ithini impembelelo yokonyusa/nokunciphisa/utshintsho kwikhonsentreyishini
         kwi reaction rate?

                                                      OR / OKANYE

         What is the relationship between the concentration and reaction rate? 
         Ithini irhileyishinishiphu phakathi kwekhonsentreyishini ne reaction rate?                                 (2)

 5.3.1   HIGHER THAN / NGAPHEZULU KUNO                                                                             (1)

            Volume of hydrogen gas (cm3) /
                                                                         Marking criteria / Umgaqo woniko
 5.3.2                                                                   manqaku
                                                                         •   Gradient B of is higher than A 
                                                B                    A       IGradient kaB ingaphezulu kune
                                                                             ka A.

            iVoluyumu ka hydrogen gas
                                                                         •   Final volume of curve A and B
                                                                             is the same and is horizontal /
                                                                             iFinal volumyu kakhevu A noB
                                                                             iyafana kwihorizontal 

            (cm3)
                                                                         NOTE: A or B must be indicated
                                                                         Ignore the labels of the axes.
                                                                         QAPHELA: uA noB
                                                                         mababonakaliswe. Zinanze
                                                        Time / ixesha
                                                                         iileyibheli ze-axes
                                                                                                                    (2)

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10              PHYSICAL SCIENCES P2/IFIZIKHALI SAYENSIZ P2       (EC/SEPTEMBA 2024)

5.4   Marking criteria
          a) Subst. into rate equation 
                                                    V
          b) Subst. 486,62 and 25 000 into in n =     
                                                   Vm
          c) Use of mol ratio n (Zn) : n (H2) 
          d) Subst. 0,02 and 65 into m = nM 
          e) Final answer 
      Umgaqo woniko manqaku
          a) Sabst. Kwi-ikhweyzhini ye rheyithi n 
          b) Sabst. u486,62 no 25 000 ku "n = " "V" /Vm 
          c) ukusetyenziswa okwe mol ratio ku n (Zn) : n (H2) 
          d) Subst. 0,02 and 65 into m = nM 
          e) Impendulo yokugqibela
              ΔV
      rate =
              Δt
               V-0
      8,39 =       (𝑎𝑎) 
                58

      V = 486,62 cm3

            V
      n=
           Vm

           486,62
      n=          (b) 
           24000

      n(H2) = 0,020 mol

      n (Zn) = n (H2) = 0,020 (c) 

      m (Zn) = nM

      m (Zn) = (0,020)(65) (d) 

      m (Zn) = 1,30 g (e)                                                             (5)

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(EC/SEPTEMBA 2024)              PHYSICAL SCIENCES P1/IFIZIKHALI SAYENSIZI P2               11

 5.5      INCREASES / IYNYUKA                                                              (1)

 5.6     •   At higher temperature the average kinetic energy of the particles is
             higher 
         •   More particles have enough / sufficient kinetic energy or more particles
             have kinetic energy equal or higher than the activation energy 
         •   More effective collisions per unit time/ Higher frequency of effective
             collisions 

         •   Kwitthempritsha ephezulu i-average kinetic energy yeeparticles
             iyenyuka 
              • Iiparticles ezinininzi zine-enough / isufficient kinetic energy okanye
                   iiparticles ezinizi zikinetic energy elingana okanye engaphezulu kune
                   activation energy 
         •    ii-effective collisions ezininzi per unit time/ ifrequency ephezulu yee
             effective collisions.                                                          (3)
                                                                                           [16]

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12               PHYSICAL SCIENCES P2/IFIZIKHALI SAYENSIZ P2             (EC/SEPTEMBA 2024)

QUESTION 6 / UMBUZO 6

6.1.1 Marking criteria/ Umgaqo woniko manqaku
      If any of the underlined key words/phrases in the correct context are
      omitted: - 1 mark per word/phrase.
       Ukuba esinye seziqwengana ezingu ndoqo esinomgca ngaphantsi
       ngokwe khonteksti eyiyo sishiyiwe, nyiba inqaku libe-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/cancel the
       disturbance. 

       Xa i-ekhwilibriyam kwiclosed system is disturbediphazanyisiwe, isystem iyal
       ure-instata i-ekhwilibriyam entsha ngoku feyiva irhiekshinieza ku-opposa/
       icancel isiphazamiso.                                                                  (2)

6.1.2 Forward (reaction).  The equilibrium concentration of B is higher than A. 
      Fowadi (rhiekishini). i-ekhwelilbriyam khonsentreyishini kaB iphezulu kune
      ka-A.                                                                                   (2)

6.1.3 INCREASES / IYENYUKA                                                                   (1)

6.1.4 NO EFFECT/ AKUKHO MPEMBELELO                                                           (1)

6.1.5 EXOTHERMIC / I-EKZOTHEMIKHI                                                            (1)

6.1.6 •    The concentration of A increased / [B] decreased 
      •    (According to Le Chatelier’s principle) an increase in temperature favours
           the endothermic reaction. 
       •   The reverse reaction was favoured / The equilibrium position shifted
           towards the left 

       •   I-khonsentreyishini ka A yonyukile / [B] yehlile
       •   (ngoku kaLe Chatelier’s principle) ukonyuka kwethempritsha kufeyiva i-
           endothemikhi rhiekshini.
       •   irivesi rhiekshini feyiviwe / i- equilibrium position ishiftele ngase kunxele      (3)

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(EC/SEPTEMBA 2024)               PHYSICAL SCIENCES P1/IFIZIKHALI SAYENSIZI P2           13

 6.2     OPTION 1: CONCENTRATION : Marking criteria
         a) Correct Kc expression with square brackets 
         b) Substitution of 6,34 x 10-4 into correct kc expression 
         c) Correct substitution of [Br2] into correct kc expression 
         d) Use of the correct ratio [Br2] : [Br] 
         e) Determine initial [Br2] 
         f) Substitution into c = n/V 
         g) Final answer 

         INDLELA 1: IKHONSENTREYISHINI/Umgaqo woniko manqaku
            a) I-Kc expression echanekileyo enesquare brackets.
            b) iSabstityushini ka 6,34 x 10-4 kwi kc expression echanekileyo.
            c) iSabstityushini ka[Br2] ka kc expression echanekileyo.
            d) Ukusetyenziswa kweratio [Br2] : [Br] echanekileyo.
            e) Fumana [Br2] yokuqala
            f) iSabstityushini ku c = n/V
            g) Impendulo yokugqibela                  • No Kc expression, correct
                                                         substitution / akho Kc expression,
                  [Br] 2                                 iSabstityushini echanekileyo.
          Kc =           (a)                            Max / Makz 6/7
                  [Br2]
                                                      • Wrong Kc expression / Kc
                                   [Br]2                 expression engachanekanga. Max.
          6,34 x 10-4(b)      =           (c)          Makz 3/7
                                 (2,074)

          [Br] = 0,03626 mol·dm-3

                                                Br2               2 Br
          Initial conc
                                          2,09213  (e)             -
          Change in conc                                     0,03626  (d)
                                             -0,01813
          Utshintsho kwi khons
          Equilibrium conc.
                                               2,074            0,03626

                     n
          V     =
                     c

                       1,05
          V     =            (f) 
                     2,09213

          V     = 0,50 dm3 (g) 

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14               PHYSICAL SCIENCES P2/IFIZIKHALI SAYENSIZ P2                (EC/SEPTEMBA 2024)

     OPTION 2: CONCENTRATION : Marking criteria
     a) Determine Δ [Br2] 
     b) Use of the correct ratio [Br2] : [Br] 
     c) Correct Kc expression with square brackets 
     d) Substitution of 6,34 x 10-4 into correct Kc expression 
     e) Correct substitution of [Br2] and [Br] into correct Kc expression 
     f) Substitution into c = nV 
     g) Final answer 

     INDLELA 2: IKHONSENTREYISHINI/Umgaqo woniko manqaku

          a) Fumana i-Δ [Br2]
          b) sebenzisa iratio ka [Br2] : [Br] echanekileyo
          c) iKc expression enesquare brackets echanekileyo
          d) iSabstityushini ka6,34 x 10-4 kwiKc expression echanekileyo
          e) Isabstityushini ka [Br2] echanekileyo no [Br] kwiKc expression
             echanekileyo
          f) iSabstityushini kuc = nV
          g) impendulo yokuggqibela

     Δ[Br2] = x(0,865/100) = 0,0865x  (a)

                                               Br2                  2 Br
      Initial conc
                                                  x                   -
      Change in conc
                                          -0,00865x            0,0173x  (b)
      Utshintsho kwi khons
      Equilibrium conc.
                                              2,074               0,0173x

                [Br]2
      Kc =               (c) 
                [Br2]

                                 (0,0173x)2
      6,34 x 10-4 (d)  =                         (e) 
                                   (2,074)

      x / [Br2]o = 2,096 mol·dm-3

                 n                            •       No Kc expression, correct
      V     =
                 c                                    substitution / akho Kc expression,
                                                      iSabstityushini echanekileyo.
                 1,05                                 Max / Makz 6/7
      V     =                (f) 
                 2,096                        •       Wrong Kc expression / Kc
                                                      expression engachanekanga. Max.
      V     = 0,50 dm3 (g)                           Makz 3/7

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(EC/SEPTEMBA 2024)                    PHYSICAL SCIENCES P1/IFIZIKHALI SAYENSIZI P2               15

         OPTION 3: MOLE CALCULATION : Marking criteria
         a) Determine Δ mol Br2 
         b) Use of the correct ratio Br2 : Br 
         c) Correct Kc expression with square brackets 
         d) Substitution of 6,34 x 10-4 into correct Kc expression 
         e) Correct substitution of [Br] and [Br] into correct Kc expression 
         f) Substitution into c = nV 
         g) Final answer 

         INDLELA 3: I-MOLE KHALTYHULEYISHINI: Umgaqo woniko manqaku
            a) Fumana i-Δmol Br2
            b) sebenzisa iratio kaBr2 : Br echanekileyo
            c) iKc expression enesquare brackets echanekileyo
            d) iSabstityushini ka6,34 x 10-4 kwiKc expression echanekileyo
            e) Isabstityushini ka [Br2] echanekileyo no [Br] kwiKc expression
               echanekileyo
            f) iSabstityushini kuc = nV
            g) impendulo yokuggqibela

         Δn(Br2) = 1,05 x 0,865 / 100 = 0,0090825 mol  (a)

                                                         Br2                  2 Br
          Initial mol
                                                         1,05                   -
          Change in mol
                                                   -0,0090825            0,0818165  (b)
          Utshintsho kwi mol
          Equilibrium mol.
                                                        2,074V             0,0818165
          Equilibrium conc.
          Equilibrium khons.                            2,074             0,018165 / V

                     [Br]2
          Kc =                (c) 
                     [Br2]                                       •   No Kc expression, correct
                                                                     substitution / akho Kc expression,
                                        [Br]2                        iSabstityushini echanekileyo.
          6,34 x 10-4 (d)  =                   (e) 
                                      (2,074)                        Max / Makz 6/7
                                                                 •   Wrong Kc expression / Kc
          [Br]o = 0,03626 mol·dm-3                                   expression engachanekanga. Max.
                                                                     Makz 3/7
                     n
          c    =                                                 •
                     V

                             0,018165
          0,03626 =                   (f) 
                                 V

          V              = 0,50 dm3 (g) 

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16              PHYSICAL SCIENCES P2/IFIZIKHALI SAYENSIZ P2             (EC/SEPTEMBA 2024)

     OPTION 4: MOLE CALCULATION: Marking criteria
     a) Correct Kc expression with square brackets 
     b) Substitution of 6,34 x 10-4 into correct Kc expression 
     c) Correct substitution of [Br] and [Br] into correct Kc expression 
     d) Determine change in mol Br 
     e) Use of the correct ratio [Br2] : [Br] 
     f) Substitution into ne = ni – Δn for Br2 
     g) Final answer 

     INDLELA 4: IMOLE KHALTYHULEYISHINI /Umgaqo woniko manqaku
        a) I-Kc expression echanekileyo enesquare brackets.
        b) iSabstityushini ka 6,34 x 10-4 kwi kc expression echanekileyo.
        c) iSabstityushini ka[Br2] ka kc expression echanekileyo.
        d) Fumana utshintsho kwimoli zika Br
        e) Ukusetyenziswa kweratio [Br2] : [Br] echanekileyo
        f) iSabstityushini ku ne = ni – Δn ku Br2
        g) Impendulo yokugqibela
                                                    • No Kc expression, correct
                                                       substitution / akho Kc
                                                       expression, iSabstityushini
              [Br]2                                    echanekileyo. Max / Makz
      Kc =             (a)                            6/7
              [Br2]
                                                    • Wrong Kc expression / Kc
                               [Br] 2                  expression engachanekanga.
      6,34 x 10-4 (b)  =             (c)             Max. Makz 3/7
                             (2,074)

      [Br] = 0,03626 mol·dm-3

                                               Br2                2 Br
      Initial mol
                                              1,05                  -
      Change in mol
                                        -0,01813V  (e)     0,03626V  (d)
      utshintsho kwimol
      Equilibrium mol.
                                             2,074V            0,03626V
      Equilibrium conc.
      Equilibrium khons.                     2,074              0,03626

     ne = ni – Δn

     2,074V = 1,05 - 0,01813V  (f)

     V = 0,5 dm3  (g)                                                                   (7)
                                                                                        [17]

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(EC/SEPTEMBA 2024)                   PHYSICAL SCIENCES P1/IFIZIKHALI SAYENSIZI P2   17

 QUESTION 7 / UMBUZO 7

 7.1.1 Strong acids ionise completely in water  to produce a high concentration
       of the hydronium ions / (H3O+) 

         iiStrong acid ziayonayiza ngokupheleleyo emanzini completely 
         ukuprodyusa ikhonsentreyishini ephezulu yeehydronium ions / (H3O+) .        (2)

 7.1.2 It donates two protons/ H+ / Idoneytha iiproton ezimbini / H+ / H+           (1)

 7.1.3 H2O                                                                          (1)

 7.1.4 HSO4 –                                                                      (2)

 7.1.5 Marking criteria/ Umgaqo woniko manqaku

         a) Ratio / iRatio [H3O+] : [H2SO4] 
         b) Formula / iFomyula pH = - log [H3O+] 
         c) pH value substituted into formula / i-pH value isabstitythwe
            kwifomyula
         d) Final answer / impendulo yokugqibela 

         [H3O+] = 2(0,1)  (a) = 0,2 mol·dm-3

         pH = - log[H3O+]  (b)

         pH = - log (0,2)  (c)

         pH = 0,70  (d)                                                             (4)

 7.2.1 OPTION 1/ INDLELA 1                       OPTION 2 / INDLELA2

                     m                                    m
          c =                                    n   =
                     MV                                   M

                          1,2                             1,2
          c =                                    n   =
                     (90)(50x10-3)                        90

          c = 0,27 mol· dm-3                     n   = 0,013 mol· dm-3
                                                         n
                                                  c   =      
                                                         V

                                                           0,013
                                                  c   =             
                                                          50x10-3

                                                  c   = 0,26 mol· dm-3              (3)

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18              PHYSICAL SCIENCES P2/IFIZIKHALI SAYENSIZ P2                   (EC/SEPTEMBA 2024)

7.2.2 Positive marking from / IPozitivi            Positive marking from /
      makhingi ukusuka ku 7.2.1                    IPozitivi makhingi ukusuka ku
      Marking criteria / Umgaqo woniko             7.2.1
      manqaku                                      Marking criteria / Umgaqo
                                                   woniko manqaku
      a) Subst. values of H2C2O4 into / n =
         cV Sabst. iiveliyu zikaH2C2O4 ku n =      a) Subst. values of/ Sabst
         cV                                                                c V
                                                      iveliyu ka na / nb ku ca Va =
      b) Using ratio / sebenzisa iratio                 na
                                                                                    b b

         H2C2O4: NaOH 1:2                                   
                                                        nb
      c) Subst of values of / sabst iveliyu ka     b) Subst. values of/ Sabst
         NaOH ku c = n/V                             iveliyu ka caVa into/ ku
      d) Subst into values / Vervang waardes          ca Va   na
                                                            =    
         van NaOH into/ in                c1V1 =      c   b Vb       nb
         c2V2                                      c)   Subst. values of/ Sabst
      e) Formula / iFomyula m = cMV                    iveliyu in/ka Vb ku
                                                        ca Va   n
      f) Subst into / sabst m = cMV                          = na 
                                                        c b Vb
      g) Final answer / impendulo                                     b
                                                   d) Subst into values/ Sabst
         yokugqibela 
                                                      iiveliyu zika NaOH into/ ku
                                                      c1V1 = c2V2 
       n (H2C2O4)    = cV
                                                   e) Formula / iFomyula
                                                      m = cMV 
                     = 0,27 x 25 x 10-3 (a) 
                                                   f) Subst into/ Sabst         m=
                                                      cMV 
                     = 6,75 x 10-3 mol
                                                   g) Final answer/ Impendulo
                                                      okugqibela 
       n (NaOH)      = 2 x 6,75 x 10-3 (b) 
                                                   caVa              na
       n (NaOH)      = 0,0135 mol                                =
                                                   cbVb              nb
                      n
       c(NaOH) =                                   (0,27)(25) (b)              1
                      V                                                   =          (a) 
                                                   cb(43,8) (c)                2
                       0,0135
       c(NaOH) =                    (c)           cb = 0,308 mol·dm-3
                      43,8 x 10-3

       c(NaOH) = 0,308 mol·dm-3

       c1V1 = c2V2

       c1(25) = (0,308)(100)   (d) 

       c1 = 1,232 mol· dm-3

       m = cMV (e) 

       m = (1,232)(40)(2) (f) 

       m = 98,56 g (g)                                                                         (7)
                                                                                               [20]

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(EC/SEPTEMBA 2024)                     PHYSICAL SCIENCES P1/IFIZIKHALI SAYENSIZI P2                      19

 QUESTION 8 / UMBUZO 8

 8.1     Chemical energy to electrical energy 
         iChemical energy kwielectrical energy                                                            (2)

 8.2     Provide path for movement of ions / Bied pad vir beweging van ione
         Ensures electrical neutrality in the cell /
         Iprovayida indlela yokuhambahamba kwee-ayoni / iqinisekisa i- electrical
         neutrality kwiseli (Any one/ Nayiphi enye)                                                      (1)

 8.3     Fe3+ + e- → Fe2+                                                                               (2)

         Marking criteria / Umgaqo woniko manqaku

         •    Fe3+ + e - ⇌ Fe2+                 ½
         •    Fe2+ ← Fe3+ + e-                  2/2
         •    Fe2+ ⇌ Fe3+ + e -                 0/2
         •    Fe2+ → Fe3+ + e -                 0/2

         Ignore if the charge omitted on electron / INanza ukuba itshaji kwignoreer
         kwi elektroni ishiyiwe.

 8.4     Eθcell = Eθcathode/reduction/oxidising agent –Eθanode/oxidation/reducing agent 
                                                Marking criteria/ Umgaqo woniko manqaku
         Eθcell = (0,77)  – ( - 0,13)            • Any other formula using unconventional
                                                       abbreviation , e.g. /
                                                       Eθcell = EθOA –Eθ RA followed by the correct
         Eθcell = 0,90 V                              substitution Max ¾
                                                   • Nayiphi ifomyula ekusetyenziswe i-
                                                       unconventional abbreviation , umzl.. /
                                                       Eθcell = EθOA –Eθ RA ilandelwe                     (4)
                                                       yisabstityushininechanekileyo followed by the c
                                                       Makz ¾
         8.5.1 Increases / Iyanda                                                                        (1)

         8.5.2 Remains the same / ihlala injalo                                                          (1)

         8.5.3Increases / iyanda                                                                         (1)

 8.6     Zn is a stronger reducing agent than Pb 
         More energy per unit charge is released for the reaction between Zn and
         Fe3+ / The reaction between Zn and Fe3+ is more strongly product-favoured /
         equilibrium position lies further to the right than Pb and Fe3+ 

         i-Zn yireducing agent enamandla kunoPb 
         I-ieneji eninzi iyakhutshwa per unit charge kwirhiekshini ephakathi koZn no
         Fe3+ / Irhiekshini ephakathi koZn no Fe3+ iproduct-favoured kakhulu
         ngamandla / i-equilibrium position ikekelele kakhulu ngase kunene kune
         kaPb no Fe3+                                                                                     (2)
                                                                                                         [14]

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20               PHYSICAL SCIENCES P2/IFIZIKHALI SAYENSIZ P2            (EC/SEPTEMBA 2024)

QUESTION 9 / UMBUZO 9

9.1     The chemical process in which electrical energy is converted to chemical
        energy (2 or 0) 
        yichemical process apho i-electrical energy iguqulelwa kwichemical energy
        (2 of 0)
                                       OR / OKANYE
        The use of electrical energy to produce a chemical change (2 or 0) 
        Kuksetyenziswa kwe-electrical energy ukuprodyusa ichemical change (2 of
        0)                                                                                   (2)

9.2     Battery/ ibhetri                                                                    (1)

9.3     Hydrogen gas / H2                                                                   (1)

9.4     2 Cℓ− (aq)+ 2e− → Cℓ2(g)       Ignore phases / Nanza iphase

        Marking criteria / Umgaqo woniko
        manqaku
                                                     1/2
           •   2 Cℓ− (aq) + 2e−⇌ Cℓ2(g)              2/2
           •   Cℓ2(g) ← 2 Cℓ− (aq)+ 2e−              0/2
           •   Cℓ2(g) ⇌ 2 Cℓ− (aq)+ 2e−              0/2
           •   Cℓ2(g) → 2 Cℓ− (aq)+ 2e−

        Ignore if the charge omitted on electron/
        Nanza ukuba itshaji kwi-elektroni ishiyiwe                                           (2)

9.5     Pink / iPinki  . OH−                                                               (2)

9.6.1   Zinc/ iZinc                                                                         (1)

9.6 2   Marking criteria / Umgaqo woniko manqaku

        a) Using ratio / Sebenzisa iratio Cu: e- = 1:2 
        b) Subst. of values of / Sabst iveliyu ka Cu into / ku m = nM 
        c) Adding the mass of Cu reduced to initial mass /Ukwengeza imass ka
           Cu emva kwephuyurifikheyishini.phuryifawa. 
        d) Final answer / Finale antwoord 

        n (Cu) = ½ x 1,38 x 10-2 (a) 

        n (Cu) = 6,9 x 10-3 mol

        m = nM

        m = (6,9 x 10-3)(63,5) (b) 

        m = 0,43815 g

        m (cathode /katode) = 2 + 0,43815 (c)                                               (4)

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(EC/SEPTEMBA 2024)            PHYSICAL SCIENCES P1/IFIZIKHALI SAYENSIZI P2       21

          m (cathode/katode) = 2,43815 g (d) 
                                                                                 [13]

                                                              TOTAL / ITOTALI:   150

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