# Physical Sciences Grade 12 Paper 2 NSC November 2019 Memo

> Grade 12 Physical Sciences Paper 2 memorandum from the NSC November Exam 2019, 21 pages. Read online or download the free PDF.

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- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2019
- Exam period: NSC November Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 21

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Physical Sciences · Grade 12 · NSC November Exam · 2019. Memorandum, 21 pages. Read online or download the PDF.

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                                      NATIONAL
                                 SENIOR CERTIFICATE
                                     NASIONALE
                                 SENIOR SERTIFIKAAT

                                    GRADE/GRAAD 12

                      PHYSICAL SCIENCES: CHEMISTRY (P2)
                       FISIESE WETENSKAPPE: CHEMIE (V2)

                                        NOVEMBER 2019

                      MARKING GUIDELINES/NASIENRIGLYNE

MARKS/PUNTE: 150

                      These marking guidelines consist of 20 pages.
                       Hierdie nasienriglyne bestaan uit 20 bladsye.

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Physical Sciences P2/Fisiese Wetenskappe V2     2                                DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION 1/VRAAG 1

1.1      D                                                                                            (2)

1.2      C                                                                                            (2)

1.3      B                                                                                            (2)

1.4      D                                                                                            (2)

1.5      C                                                                                            (2)

1.6      B                                                                                            (2)

1.7      B                                                                                            (2)

1.8      A                                                                                            (2)

1.9      A                                                                                            (2)

1.10     C                                                                                             (2)
                                                                                                        [20]

QUESTION 2/VRAAG 2
2.1
2.1.1     C n H 2n - 2                                                                                 (1)

2.1.2                                  H
                                                          Marking criteria/Nasienriglyne
                                               
                                   H   C   H              • Functional group correct. 
                                                            Funksionele groep korrek.
                 H                         H              • 2 methyl substituents. 
                               
                                                            2 metielsubstituente.
         H       C         C   C       C   C       H     • Whole structure correct:/Hele
                                                            struktuur korrek: 3
                 H                         H                                      3

                                   H   C   H

                                       H
                                                                                                        (3)

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Physical Sciences P2/Fisiese Wetenskappe V2     3                                  DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne

2.2

2.2.1     Compounds with the same molecular formula,  but different positions of the
          side chain/substituents/functional groups on the parent chain.
          Verbindings met dieselfde molekulêre formule, maar verskillende posisies van
          die syketting/substituente/funksionele groepe op die stamketting.                              (2)

2.2.2     Pentan-3-one/3-pentanone 
          Pentan-3-oon/3-pentanoon
          Marking criteria/Nasienriglyne
          • Functional group and correct position i.e. 3 /Funksionele groep en korrekte posisie
            nl. 3. 
          • Whole name correct/Hele naam korrek. 
          Accept for ONE mark/Aanvaar vir EEN punt
          Pentanone with the 3 in incorrect place, e.g. penta-3-none.
          Pentanoon met die 3 in foutiewe plek, bv. penta-3-noon.                                        (2)

2.2.3                                                    Marking criteria/Nasienriglyne
                 H      H    H      H     O              • Whole structure correct:/Hele
                                                    
                                                           struktuur korrek: 2
           H     C      C    C      C     C     H                              2
                                                        • Only functional group correct/Slegs
                 H     H     H      H                      funksionele groep korrek Max: 1
                                                                                            2

          OR: Any correct structure of an aldehyde with five carbon atoms.
          OF:Enige korrekte struktuur van ‘n aldehied met vyf koolstofatome.
               H    H    H    O                        H     H    H    O                      

          H     C     C      C     C     H     OR/OF      H       C      C    C      C     H 
                H           H                                      H      H
                H     C     H                                             H    C     H
                                               H
                      H                                                        H
                                         H     C    H
                                         H          O          

                       OR/OF       H     C     C     C     H       
                                         H
                                          H    C     H
                                               H                                                         (2)

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Physical Sciences P2/Fisiese Wetenskappe V2     4                                DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne

2.3
2.3.1    Tertiary (alcohol)/Tersiêre (alkohol) 
         The C atom bonded to the functional group/hydroxyl (group)/-OH is bonded to
         three other C atoms. /The C-atom bonded to the hydroxyl (group) has no
         hydrogen atoms. 
         Die C-atoom gebind aan die funksionele groep/hidroksiel(groep)/-OH is
         gebind aan drie ander C-atome./ Die C-atoom gebind aan die hidroksiel
         (groep) het geen waterstofatome nie.                                                           (2)

2.3.2    2-methylbutan-2-ol/2-methyl-2-butanol/2-metielbutan-2-ol/2-metiel-2-butanol
         Marking criteria/Nasienriglyne
         • 2-methyl/2-metiel 
         • Butan-2-ol/2-butanol 
         • Any error e.g. hyphens omitted and/or incorrect sequence:
           Enige fout, bv. koppeltekens weggelaat en/of verkeerde volgorde: Max./Maks: 1                (2)
                                                                                        2

2.3.3    2-methylbut-2- ene/2-methyl-2-butene/2-metielbut-2-een/2-metiel-2-buteen
         Marking criteria/Nasienriglyne
         • 2-methyl/2-metiel 
         • But-2-ene/2-butene/But-2-een/2-buteen 
         • Any error e.g. hyphens omitted and/or incorrect sequence:
           Enige fout, bv. koppeltekens weggelaat en/of verkeerde volgorde: Max./Maks: 1                 (2)
                                                                                        2
                                                                                                        [16]

QUESTION 3/VRAAG 3

3.1       Marking guidelines/Nasienriglyne
          The underlined key phrases must be used in the CORRECT CONTEXT
          (pressure/boiling). /Die onderstreepte frases moet gebruik word in die KORREKTE
          KONTEKS (druk/kook).

         The temperature  at which the vapour pressure of a substance equals
         atmospheric/external pressure. 
         Die temperatuur waar die dampdruk van 'n stof gelyk is aan atmosferiese/
         eksterne druk.                                                                                 (2)

3.2       (Q, R and S) have same molecular mass/formulae/number of carbon and
          hydrogen atoms/are (chain) isomers. 
          (Q, R en S) het dieselfde molekulêre massa/formule/aantal koolstof en
          waterstofatome/ is (ketting)isomere.

          OR/OF
          The compounds are all alkanes /same homologous series and have the same
          number of carbon atoms.
          Die verbindings is almal alkane /dieselfde homoloë reeks en het die dieselfde
          aantal koolstofatome.                                                                         (1)

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Physical Sciences P2/Fisiese Wetenskappe V2     5                                DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne

          Marking guidelines/Nasienriglyne
          • 55 (°C) 
          • Compare all three compounds or Q and S in terms of branches/chain lengths /
             surface area. 
             Vergelyk al drie verbindings of Q en S in terme van vertakkings/kettinglengte/
             oppervlakarea.
          • Compare strengths of all three or Q and S’s IMF’s / Vergelyk sterkte van al drie of
             Q en S se IMK'e.
          • Compare energy of all three /Vergelyk energie van al drie. 
3.3       55 (°C) 
          Compare compound R with compounds Q and S:
          •    Compound R is less branched/compact/spherical/surface area than
               compound Q and more branched/compact/spherical/surface area than
               compound S. 
               OR
               Q is the most branched/compact /spherical/surface area and S is least
               branced/compact/spherical/surface area.
          •    Intermolecular forces in compound R are stronger than in compound Q
               and weaker than in compound S. 
          •    More energy needed to overcome intermolecular forces in compound R
               than in compound Q and less energy needed to overcome (break)
               intermolecular forces in compound R than in compound S. 
          OR
          •    Compound R has a longer chain length than compound Q and a shorter
               chain length than compound S. 
               OR
               S has the longest chain length and Q the shortest.
          •    Intermolecular forces increase with increase in chain length. 
          •    More energy needed to overcome intermolecular forces as chain length
               increases. 
          Vergelyk verbinding R met verbindings Q en S:.
          •   Verbinding R is minder vertak/kompak/sferieseoppervlak as verbinding Q
              en meer vertak as verbinding S.
              OF
              Q is die meeste vertak/kompak en S is die minste
              vertak/kompak/series/oppervlak.
          • Intermolekulêre kragte in verbinding R is sterker as in verbinding Q en
             swakker as in verbinding S.
          • Meer energie word benodig om intermolekulêre kragte in verbinding R te
             oorkom as in verbinding Q, en minder energie word benodig om
             intermolekulêre kragte in verbinding R te oorkom / breek as in verbinding
             S.
          OF
          • Verbinding R het 'n langer kettinglengte as verbinding Q en 'n korter
              kettinglengte as S.
              OF
              S het die langste ketting en Q die kortste.
          • Intermolekulêre kragte neem toe met toename in kettinglengte.
          • Meer energie word benodig om intermolekulêre kragte te oorkom
              wanneer kettinglengte toeneem.                                                            (4)
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Physical Sciences P2/Fisiese Wetenskappe V2     6                                DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne

3.4
3.4.1     P                                                                                           (2)

3.4.2     Marking guidelines/Nasienriglyne
          • Name type of IMFs in P/pentanal. 
             Noem tipe IMK'e in P/pentanaal.
          • Name type of IMFs in/Noem tipe IMK'e in T/pentan-1-ol.
          • Compare strength of IMFs. /Vergelyk sterkte van IMK'e. 
             OR/OF
             Compare energy needed to overcome IMFs./Vergelyk energie benodig om IMK'e
             te oorkom.
          •   In P/ pentanal/aldehydes: dipole-dipole forces (in addition to London
              forces/dispersion forces/induced dipole forces).
          •   In T/pentan-1-ol: Hydrogen bonding. (in addition to London
              forces/dispersion forces/induced dipole forces).
          •   Intermolecular forces in P/pentanal are weaker than in T/pentan-1-ol
              OR dipole-dipole forces are weaker than hydrogen bonds OR
              intermolecular forces in T/pentan-1-ol are stronger than in P/pentanal.
              OR
              More energy needed to overcome/break intermolecular forces in T.
          •   In P/pentanaal/aldehiede: dipool-dipoolkragte (tesame met Londonkragte/
              dispersiekragte/geïnduseerde dipoolkragte).
          •   In T/pentan-1-ol: Waterstofbinding. (tesame met Londonkragte/
              dispersiekragte/geïnduseerde dipoolkragte).
          •   Intermolekulêre kragte in P swakker as in T/pentan-1-ol OF
              intermolekulêre kragte in T/pentan-1-ol sterker as in P/pentanaal OF
              dipool-dipoolkragte is swakker as waterstofbindings.
              OF
              Meer energie benodig om intermolekulêre kragte te oorkom/breek in T.                       (3)
                                                                                                        [12]

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Physical Sciences P2/Fisiese Wetenskappe V2     7                                      DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION 4/VRAAG 4

4.1      Haloalkane/alkyl halide 
         Haloalkaan/alkielhalied                                                                              (1)

4.2
4.2.1    Elimination/dehydrohalogenation 
         Eliminasie/dehidrohalogenering                                                                       (1)

4.2.2    Substitution/hydrolysis 
         Substitusie/hidrolise                                                                                (1)

4.2.3    Esterification/condensation 
         Esterifikasie/kondensasie/verestering                                                                (1)

4.3
4.3.1     • (Mild) heat/Heating/(matige) hitte/ verhitting 
          • Dilute (strong base)/Verdunde (sterk basis)/(NaOH/KOH/LiOH) 
            OR/OR
            Add water/H 2 O/Voeg water/H 2 O by                                                               (2)

4.3.2     Propan-1-ol/1-propanol 
          Marking criteria/Nasienriglyne:
          • Correct stem and functional group i.e. propanol/Korrekte stam en funksionele
            groep, d.i. propanol. 
          • Whole name correct:/Hele naam korrek: propan-1-ol                                                (2)

4.4                 H     H     H                         Marking criteria/Nasienriglyne
                                                          • Whole structure correct:/Hele struktuur
              H     C     C     C   H
                                                            korrek: 2
                                                                      2
                    H     O     H                         • Only functional group correct/Slegs
                          H                                funksionele groep korrek: 1
                                                                                        2

          Notes/Aantekeninge
          • Accept –OH as condensed. /Aanvaar –OH as gekondenseerd.
          • Condensed or semi-structural formula:
             Gekondenseerde of semi-struktuurformule: Max./Maks. 1
                                                                          2
          • Molecular formula/Molekulêre formule:        0
                                                           2
          • If functional group is incorrect/Indien funksionele groep verkeerd is: 0
                                                                                          2
          •       If more than one functional group:
                  Indien meer as een funksionele groep:     0
                                                                2                                             (2)

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Physical Sciences P2/Fisiese Wetenskappe V2     8                                    DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne

4.5
            POSITIVE MARKING FROM Q4.3.2 ONLY IF THE COMPOUND IN Q4.3.2
            IS AN ALCOHOL. /POSITIEWE NASIEN VANAF V4.3.2 SLEGS INDIEN DIE
            VERBINDING IN Q4.3.2 'N ALKOHOL IS.

4.5.1                                                   Marking criteria/Nasienriglyne
                 H    H     H         O      H
                                                         • Whole structure correct:/Hele struktuur
            H    C    C     C    O    C      C   H        korrek: 2
                                                                     2
                                                         • Only functional group correct/Slegs
                H     H     H                H
                                                           funksionele groep korrek: 1
                                                                                       2

            Notes/Aantekeninge
            • Condensed or semi-structural formula:
               Gekondenseerde of semistruktuurformule:          Max./Maks. 1
                                                                            2
            •   Molecular formula/Molekulêre formule:       0
                                                               2
            •   If functional group is incorrect/Indien funksionele groep verkeerd is: 0                    (2)
                                                                                           2

4.5.2       (Concentrated) sulphuric acid/(Gekonsentreerde) swawelsuur/H 2 SO 4                             (1)
                                                                                                            [13]

QUESTION 5/VRAAG 5
5.1         Exothermic/Eksotermies 
        -
            H < 0/Energy is released/Energie word vrygestel                                               (2)

5.2                             ∆m                OR/OF
            rate/tempo = −
                                ∆t                                ∆m
                                                  rate/tempo = −
                               0,25 − 2                          ∆t
                          = −
                                  30                            − 1,75 
                                                             = −
                          = 0,06 (g·s-1)                          30 
                            (0,0583 g·s-1)                   = 0,06 (g·s-1) 
                                                               (0,0583 g·s-1)
                                                                                                            (3)
            Notes/Aantekeninge
                                                                                  -1
            Accept negative answer i.e./Aanvaar negatiewe antwoord d.i. - 0,06 g·s .

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Physical Sciences P2/Fisiese Wetenskappe V2     9                                   DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne

5.3       Marking guidelines
          • Calculate/Bereken: m(CaCO 3 ) reacted/reageer or / of V(CO 2 ) produced/gevorm.
            
          • Substitute/Vervang: 100 g∙mol-1. 
          • USE mol ratio/GEBRUIK molverhouding: n(CO 2 ) : n(CaCO 3 ) = 1 : 1 
          • Use of/ /Gebruik van 22,4 dm-3∙mol-1.
          • Final answer/Finale antwoord: 0,18 dm3 (0,1792 dm3) 

          OPTION 1/OPSIE 1                     OPTION 2/OPSIE 2
                            40                 For 2 g antacid/teensuurtablet:
          m(CaCO 3 ) =          x2            100 g CaCO 3 ......22,4 dm3  CO 2
                           100
                       = 0,8 g                 2 g CaCO 3 .....0,448 dm3 
                               m
          n(CaCO 3 ) reacted =                 100% CO 2 .......... 0,448 dm3 
                               M
                                               40% CO 2 ............. 0,18 dm3 
                               0,8
                             =                 OPTION 3/OPSIE 3
                               100 
                             = 8 x 10-3 mol    100% CaCO 3 .........2 g
                                               40% .......................0,8 g 
          n(CO 2 ) = n(CaCO 3 )               100 g ……. 1 mol
                   = 8 x 10-3 mol              0,8 g .......... 8 x 10-3 mol 
          V(CO 2 ) = 8 x 10-3 x 22,4          1 mol ...............22,4 dm3 
                   = 0,18 dm3                 8 x 10-3 mol ......0,18 dm3                              (5)

5.4       ANY ONE/ENIGE EEN:
          •   Concentration (of acid)/Konsenterasie (van suur) 
          •   Size/mass of tablet/Identical tablet /Type of tablet.
              Grootte/massa van tablet/Identiese tablet./Tipe tablet.
          •   State of division / Surface area / Toestand van verdeeldheid /
              reaksieoppervlak.                                                                          (1)

5.5       Criteria for conclusion/Riglyne vir gevolgtrekking:
          Dependent [(reaction) rate/time] and independent (temperature)
          variables correctly identified.
                                                                                                
          Afhanklike    [(reaksie)tempo/tyd]   en   onafhanklike    (temperatuur)
          veranderlikes korrek geïdentifiseer.
          Relationship between the independent and dependent variables correctly
          stated./Verwantskap tussen die afhanklike en onafhanklike veranderlikes               
          korrek genoem.

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Physical Sciences P2/Fisiese Wetenskappe V2     10                                DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne

          Examples/Voorbeelde:
                                1
          • Reaction rate (        ) increases with increase in temperature.
                             time
                                 1
             Reaksietempo (         ) neem toe met toename in temperatuur.
                               time
                                1
          • Reaction rate (        ) decreases with decrease in temperature.
                             time
                                 1
             Reaksietempo (         ) neem af met afname in temperatuur.
                               time
          • Time taken for reaction decreases when temperature                         increases.
            Tyd vir die reaksie neem af wanneer temperatuur toeneem.
          • Time taken for reaction increases when temperature decreases.
            Tyd vir die reaksie neem toe as temperatuur afneem.

          IF/INDIEN
          Reaction rate is DIRECTLY proportional to temperature:     Max. 1
                                                                           2
          Reaksietempo is DIREK eweredig aan temperatuur:            Maks. 1
                                                                             2                          (2)

5.6       • Increase in temperature increases the average kinetic energy/molecules
            move faster. /Toename in temperatuur verhoog die gemiddelde kinetiese
            energie/molekule beweeg vinniger. 
          • More molecules have enough/sufficient kinetic energy/More molecules have
            E k > E a. 
            Meer molekule het genoeg/voldoende kinetiese energie/Meer molekule het
            E k > E a.
          • More effective collisions per unit time/second. /Frequency of effective
            collisions increases. 
            Meer effektiewe botsings per eenheidtyd/sekonde./Frekwensie van
            effektiewe botsings neem toe.                                                               (3)

5.7       Marking guidelines/Nasienriglyne
          • For each value of temperature, the CURVE Y must be above the given CURVE. /
            Vir elke waarde van temperatuur, moet kurwe Y bo die gegewe kurwe wees. 
          • CURVE Y must have an increasing rate with an increase in temperature. /
            KURWE Y moet ‘n toenemende tempo het soos die temperatuur toeneem. 

           1                                                                  Y
                  (s-1)
       time / tyd

                                        Temperature/Temperatuur (°C)
                                                                                                         (2)
                                                                                                        [18]

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Physical Sciences P2/Fisiese Wetenskappe V2     11                               DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION 6/VRAAG 6

6.1      (The stage in a chemical reaction when the) rate of forward reaction equals
         the rate of reverse reaction. 
         (Die stadium in 'n chemiese reaksie wanneer die) tempo van die voorwaartse
         reaksie gelyk is aan die tempo van die terugwaartse reaksie.         (2 or/of 0)
         OR/OF
         (The stage in a chemical reaction when the) concentrations of reactants and
         products remain constant.
         (Die stadium in 'n chemiese reaksie wanneer die) konsentrasies van
         reaktanse en produkte konstant bly.                                (2 or/of 0)                 (2)

6.2      CALCULATIONS USING NUMBER OF MOLES
         BEREKENINGE WAT AANTAL MOL GEBRUIK
6.2.1    Marking guidelines/Nasienriglyne
         •   Substitute/Vervang: 44 g∙mol-1. 
         •   Equilibrium concentration of CO 2 multiply by 3 dm3
             Ewewigskonsentrasie van CO 2 vermenigvuldig met 3 dm3 
             AND/EN n(CO) eq divide by /deel deur 3 dm3
         •   Use mole ratio/Gebruik molverhouding: 1:2 / n(CO) = 2n(CO 2 ). 
         •   n(CO 2 ) change = n(CO 2 ) initial - n(CO 2 ) final
                                                                 
             n(CO) eq/ewe = n(CO) initial/begin + ∆n(CO)
         •   Correct K c expression (formulae in square brackets). 
             Korrekte K c -uitdrukking (formules in vierkanthakies).
         •   Substitution of concentrations into K c expression. 
             Vervanging van konsentrasies in K c -uitdrukking.
         •   Final answer/Finale antwoord: 12,24 (range/gebied: 11,85 – 12,66) 

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Physical Sciences P2/Fisiese Wetenskappe V2     12                                             DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne

         OPTION 1/OPSIE 1
                    m
         n(CO 2 ) =
                    M
                        60,8
                    =
                         44 
                    = 1,382 mol

                                                  CO 2        CO
         Initial quantity (mol)
                                                 1,382         0
         Aanvangshoeveelheid (mol)
         Change (mol)                          1,22                   Use ratio/Gebruik
                                                              2,44
         Verandering (mol)                                             verhouding 
         Quantity at equilibrium (mol)/
                                                 0,162        2,44     Divide/multiply by
         Hoeveelheid by ewewig (mol)
                                           -3
         Equilibrium concentration (mol∙dm )                           3/Deel/ vermenig-
                                                 0,054       0,813     vuldig met 3 
         Ewewigskonsentrasie (mol∙dm-3)
               [CO] 2
         Kc =                          No Kc expression, correct substitution/Geen Kc-
               [CO 2 ]                                                                6
                                                      uitdrukking, korrekte substitusie: Max./Maks.
                (0,813 ) 2                                                                                  7
              =            
                  0,054                               Wrong Kc expression/Verkeerde Kc- uitdrukking:
              = 12,24                                Max./Maks. 4
                                                                   7

         OPTION 2/OPSIE 2
                    m
         n(CO 2 ) =
                    M
                    60,8
                  =
                     44 
                  = 1,382 mol
         n(CO 2 ) change = n(CO 2 ) initial/begin - n(CO 2 ) final/finaal
                         = 1,382 – 0,162
                         = 1,22 mol

         n(CO) change = 2(CO 2 )         
                      = 2(1,22) 
                      = 2,44 mol
         n(CO) eq = n(CO) change = 2,44 mol
                   n
         c(CO) =
                   V
                   2.44
                =
                     3
                 = 0,813 mol∙dm-3
              [CO] 2
         Kc =          
              [CO 2 ]
               (0,813)2
            =             
                 0,054
            = 12,24  (Accept range/Aanvaar gebied: 11,85 – 12,66).
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Physical Sciences P2/Fisiese Wetenskappe V2     13                                DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne

         CALCULATIONS USING CONCENTRATION
         BEREKENINGE WAT KONSENTRASIE GEBRUIK
         Marking guidelines/Nasienriglyne
         • Substitute 44 g∙mol-1. 
         • Initial n(CO 2 )divide by 3 dm3. 
            Aanvanklike n(CO 2 ) gedeel deur 3 dm3.
         • USE ratio/GEBRUIK verhouding: c(CO 2 ) : c(CO) = 1 : 2 
         • ∆c(CO 2 ) = c(CO 2 ) initial/begin - c(CO 2 ) eq/ewe .
                                                                  
             c(CO) eq/ewe = c(CO) initial/begin + ∆c(CO).
         • Correct K c expression (formulae in square brackets). 
             Korrekte K c uitdrukking (formules in vierkanthakies).
         • Substitution of concentrations into K c expression. 
             Vervanging van konsentrasies in K c -uitdrukking.
         •   Final answer/Finale antwoord: 12,15 (range/gebied: 11,85 – 12,66) 

         OPTION 3/OPSIE 3
                    m
         n(CO 2 ) =
                    M
                       60,8
                   =
                        44 
                   = 1,382 mol

                                                          CO 2            CO
          Initial concentration (mol∙dm-3)                                         Divide by /Deel
                                                         0,4607            0
          Aanvanklike konsentrasie (mol∙dm-3)                      
                                                                                   deur 3 dm3 
          Change (mol∙dm-3)                                                         ratio 
                                                         0,4067           0,813
          Verandering (mol∙dm-3)                                                    verhouding
          Equilibrium concentration (mol∙dm-3)
                                                          0,054           0,813
          Ewewigskonsentrasie (mol∙dm-3)

                 [CO] 2                      No Kc expression, correct substitution/Geen Kc-
         Kc =            
                 [CO 2 ]                     uitdrukking, korrekte substitusie: Max./Maks. 6
                                                                                             7
               (0,813) 2
             =                              Wrong Kc expression/Verkeerde Kc-uitdrukking:
                 0,054
             = 12,15                        Max./Maks. 4
                                                          7
                                                                                                        (7)

6.2.2     POSITIVE MARKING FROM Q6.2.1/POSITIEWE NASIEN VANAF V6.2.1
          n(C) reacted/reageer =            Marking guidelines
          n(CO 2 ) reacted/reageer          • USE mol ratio/ GEBRUIK molverhouding:
                               = 1,22 mol    n(C) = n(CO 2 ). 
                                                                       -1
                                                   • Substitute/Vervang: 12 g∙mol . 
          m(C) = nM                                • Final answer/Finale antwoord: 14,64 g. 
                        
               = 1,22(12)
               = 14,64 g                                                                               (3)

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Physical Sciences P2/Fisiese Wetenskappe V2     14                               DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne

6.3
6.3.1       Remains the same/Bly dieselfde                                                             (1)
6.3.2       Decreases/Afneem 
        -
            •   (When pressure is increased) the reaction that leads to the smaller
                amount/number of moles/volume of gas is favoured. 
                (Wanneer die druk verhoog word,) word die reaksie wat tot die kleiner
                hoeveelheid/aantal mol/volume gas lei, bevoordeel.
            •   The reverse reaction is favoured. / More CO 2 is formed. 
                Die terugwaartse reaksie word bevoordeel./ meer CO 2 word gevorm.                       (3)
6.4
6.4.1       Endothermic/Endotermies 
        -
            •   When the temperature increases the mol/percentage CO(g)/product
                increases/forward reaction is favoured./Wanneer die temperatuur
                toeneem, neem die mol/persentasie CO(g)/produk toe/voorwaartse
                reaksie word bevoordeel. 
            •   An increase in temperature favours the endothermic reaction/Toename in
                temperatuur bevoordeel die endotermiese reaksie.                                       (3)

6.4.2       POSITIVE MARKING FROM Q6.2.1./POSITIEWE NASIEN VANAF V6.2.1.
            Marking guidelines/Nasienriglyne
            •   Calculate total volume/mol of gas at equilibrium/Bereken totale
                volume/mol gas by ewewig: 0,162 + 2,44 = 2,606 dm3 /mol 
                OR/OF
                Calculate the total concentration at equilibrium/Bereken die totale
                konsentrasie by ewewig: 0,054 + 0,813 = 0,867 mol∙dm-3
            •   Calculate percentage of ANY one gas/Bereken persentasie van ENIGE
                een gas (CO 2 or/of CO). 
            •   Final answer/Finale antwoord: T = 827 °C 
            OPTION 1/OPSIE 1
            V total eq = 0,162 + 2,44 
                       = 2,606 dm3
                       0,162
            % CO 2 =          x 100         OR/OF
                       2,606                                       2,44
                                                       % CO =           x 100 
                     = 6,225 %                                    2,606
                                                                = 93,63 %
            OPTION 2/OPSIE 2
            c total eq = 0,054 + 0,813
                       = 0,867 mol∙dm-3
                       0,054                                       0,813
            % CO 2 =          x 100         OR/OF % CO =                x 100 
                       0,867                                       0,867
                     = 6,228 %                                  = 93,77 %

            ∴ T = 827 °C                                                                                (3)
                                                                                                        [22]

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Physical Sciences P2/Fisiese Wetenskappe V2     15                               DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION 7/VRAAG 7
7.1
        - Strong (acid)/Sterk (suur) 
          Large/Groot K a value/waarde/ K a > 1 / (HBr) ionises completely/ioniseer
          volledig                                                                                     (2)
7.2       H2O 
            -
          Br                                                                                           (2)
7.3
7.3.1     Marking guidelines/Nasienriglyne
                                n          c × Va n a
          • Formula/Formule: c = /n = cV / a       =   
                                V          c b × Vb nb
          • Substitution of/Vervanging van: (0,5)(0,0165)/(0,5)(16,5) 
          • Use mol ratio/Gebruik molverhouding: 1:1/n(HBr) = n(NaOH) 
          • Substitute/Vervang: V = 0,09 dm3 /90 cm3 
          • Formula/Formule: pH = -log[H 3 O+] 
          • Substitute [H 3 O+] in pH formula. 
          • Final answer/Finale antwoord: pH = 1,04 (range/gebied: 1,036 – 1,05) 

          OPTION 1/OPSIE 1                                       OPTION 2/OPSIE 2
          n(NaOH) reacted/reageer = cV                          c a Va na
                                  = 0,5(0,0165)                         =        
                                                                 c b Vb nb
                                  = 0,00825 mol
                                                                   c a (90)  1
                                                                               = 
          n(HBr) excess/oormaat = n(NaOH) = 0,00825 mol         (0,5)(16,5) 1
                       n                                                 c a = 0,092 mol·dm-3
          c(H 3 O+) =
                       V
                      0,00825                                    pH = -log[H 3 O+] 
                   =                                                = -log(0,092) 
                         0,09 
                                                                    = 1,04 
                   = 0,092 mol·dm-3

          pH = -log[H 3 O+] 
             = -log(0,092) 
             = 1,04                                                                                    (7)

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Physical Sciences P2/Fisiese Wetenskappe V2     16                                        DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne

7.3.2    Marking guidelines/Nasienriglyne
         • Calculate/Bereken n(HBr) initial/aanvanklik : substitute/vervang (0,45)(0,09) in n = cV 
         • Subtraction/Aftrekking:
            n(HBr) reacted/reageer = n(HBr) initial/aanvanklik – n(HBr) reacted with/reageer met NaOH . 
           OR/OF: c(HBr) reacted/reageer = c(HBr) initial/aanvanklik – c(H 3 O+) excess/oormaat
         • Use mol ratio/Gebruik molverhouding: n(Zn(OH) 2 ) : n(HBr) = 1 : 2 
         • Substitution of/Vervanging van: 99 g∙mol-1 
         • Final answer/Finale antwoord: 1,5964 g (range/gebied: 1,58 – 1,68) 

         POSITIVE MARKING FROM Q7.3.1/POSITIEWE NASIEN VANAF V7.3.1
         OPTION 1/OPSIE 1
         n(HBr) initial/begin = cV
                              = (0,45)(0,09) 
                              = 0,0405 mol

         n(HBr reacted with/reageer met Zn(OH) 2 ) = 0,0405 – 0,00825 
                                                   = 0,03224 mol

         n(Zn(OH) 2 ) = ½n(HBr) = ½(0,03224) = 0,016125 mol

         m(Zn(OH) 2 ) = nM
                      = (0,016125)(99) 
                      = 1,596 g 
         OPTION 2/OPSIE 2
         c(HBr) = 0,45 – 0,092 
                = 0,358 mol∙dm-3
         n(HBr reacted/reageer) = cV
                                 = 0,358 x 0,09 
                                 = 0,0322 mol
         n(Zn(OH) 2 ) = ½n(HBr) = ½(0,0322) = 0,01611 mol

         m(Zn(OH) 2 ) = nM
                      = 0,01611 x 99 
                      = 1,595 g  (1,60 g)                                                                      (6)
                                                                                                               [17]

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Physical Sciences P2/Fisiese Wetenskappe V2     17                                  DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION 8/VRAAG 8

8.1         Chemical to electrical/Chemies na elektries                                                   (1)

8.2         Provides path for movement of ions./ Completes the circuit./Ensures electrical
            neutrality in the cell./Restore charge balance. 
            Verskaf pad vir beweging van ione./Voltooi die stroombaan./Verseker
            elektriese neutraliteit in die sel./Herstel balans van lading.                                 (1)

8.3         OPTION 1/OPTION 1
              θ
            E cell     θ
                   = E cathode     θ
                               − E anode       Notes/Aantekeninge
                              θ
                                                • Accept any other correct formula from the data
             1,49 = 1,36 − E anode                 sheet. /Aanvaar enige ander korrekte formule
              θ
            E anode = 1,36 − 1,49                  vanaf gegewensblad.
                                                • Any other formula using unconventional
                  = − 0,13 ( V )                  abbreviations, e.g. E° cell = E° OA - E° RA
            X is Pb/Lead/Lood                    followed by correct substitutions:/Enige ander
                                                   formule wat onkonvensionele afkortings gebruik,
                                                   bv. E° sel = E° OM - E° RM gevolg deur korrekte
                                                   vervangings : 4
                                                                     5
            OPTION 2/OPSIE 2
            Cℓ 2 + 2e- → 2Cℓ-                E θ = 1,36 V 
        
            X → X2+ + 2e-                    E θ = 0,13 V 
            Cℓ 2 + X → X2+ + 2Cℓ-             E θ = 1,49 V 

            X is Pb/Lead/Lood                                                                             (5)

            POSITIVE MARKING FROM Q8.3/POSITIEWE NASIEN VANAF V8.3
8.4         X/Pb/Lead/Lood                                                                                (1)

8.5
8.5.1       Reaction reached equilibrium./(In each half cell) the rate of oxidation is equal
            to rate of reduction./Rate of the forward reaction is equal to the rate of the
            reverse reaction. 
            Reaksie bereik ewewig./(In elke halfsel) die tempo van oksidasie is gelyk aan
            tempo van reduksie./Tempo van die voorwaartse reaksie is gelyk aan die
            tempo van die terugwaartse reaksie.                                                            (1)

8.5.2       Increases/Toeneem                                                                             (1)

8.5.3       • [Cℓ-] decreases/neem af. 
            • Forward reaction is favoured./Voorwaartse reaksie word bevoordeel.                           (2)
                                                                                                           [12]

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Physical Sciences P2/Fisiese Wetenskappe V2     18                               DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION 9/VRAAG 9

9.1       Marking guidelines/Nasienriglyne
          If any one of the underlined key phrases in the correct context is omitted, deduct 1
          mark./Indien enige van die onderstreepte frases in die korrekte konteks uitgelaat is,
          trek 1 punt af.
          The chemical process in which electrical energy is converted to
          chemical energy. 
          Die chemiese proses waarin elektriese energie omgeskakel word na
          chemiese energie.
          OR/OF
          The use of electrical energy to produce a chemical change.
          Die gebruik van elektriese energie om 'n chemiese verandering teweeg te
          bring.
          OR/OF
          The process during which an electrical current passes through a
          solution/molten ionic compound.
          Die proses waar 'n elektriese stroom deur 'n oplossing/gesmelte ioniese
          verbinding gestuur word.                                                                      (2)
9.2
9.2.1    2H 2 O(ℓ) + 2e-  H 2 (g) + 2OH-(aq) 
         Ignore phases/Ignoreer fases
         Marking guidelines/Nasienriglyne
          • H 2 (g) + 2OH-(aq) ← 2H 2 O(ℓ) + 2e- ( 2 ) 2H 2 O(ℓ) + 2e- ⇌ H 2 (g) + 2OH-(aq)
                                                       2
              (  1  )
                  2
             H 2 (g) + 2OH-(aq) ⇌ 2H 2 O(ℓ) + 2e- ( 0 ) 2H 2 O(ℓ) + 2e- ← H 2 (g) + 2OH-(aq)
                                                     2
         ( 0    )
             2
         • Ignore if charge omitted on electron./Ignoreer indien lading weggelaat op
              elektron.
         • If charge (-) omitted on OH- /Indien lading (-) weggelaat op OH-:
             Example/Voorbeeld: 2H 2 O(ℓ) + 2e-  H 2 (g) + 2OH(aq)             Max./Maks: 1
                                                                                              2         (2)

9.2.2     Water/ H 2 O                         -                                                       (1)
9.3       H2O is a stronger oxidising agent  than Na+  and will be reduced  (to H2).
          H2O is 'n sterker oksideermiddel as Na+ en sal gereduseer word (na H2).
          OR/OF                                -
             +
          Na is a weaker oxidizing agent than H 2 O and therefore H 2 O will be
          reduced  (to H 2 )
          Na+ is ‘n swakker oksideermiddel as H 2 O en daarom sal H 2 O gereduseer
          word (na H 2 )

          OR/OF
          The half-reaction that produces H2(g) has a more positive reduction potential
          (-0,83 V)  than the half-reaction that produces Na (-2,71 V). 
          Therefore water/H2O will be reduced  to H2./Na+ will not be reduced to Na.
          Die halfreaksie wat H2(g) vorm, het 'n meer positiewe reduksiepotensiaal
          (-0,83 V) as die halfreaksie wat Na vorm (-2,71 V).
          Daarom word water/H2O na H2 gereduseer./Na+ sal nie gereduseer word na                        (3)
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Physical Sciences P2/Fisiese Wetenskappe V2     19                               DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne
          Na nie.
                                                                                                        [8]

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Physical Sciences P2/Fisiese Wetenskappe V2     20                               DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION 10/VRAAG 10

10.1
10.1.1    Hydrogen/Waterstof/H 2                                                                       (1)

10.1.2    Nitrogen monoxide/Stikstofmonoksied/NO                                                       (1)

10.1.3    Nitric acid/Salpetersuur/HNO 3                                                               (1)

10.2
10.2.1    (Catalytic) oxidation/Redox/(Katalitiese) oksidasie/Redoks                                   (1)

10.2.2    NH 3   + HNO 3  → NH 4 NO 3          Bal 
          Notes/Aantekeninge
          • Reactants         Products            Balancing 
            Reaktanse          Produkte             Balansering
          • Ignore double arrows (⇌) and phases./Ignoreer dubbelpyle (⇌)en fases.
          • Marking rule 6.3.10./Nasienreël 6.3.10.                                                     (3)

10.3
10.3.1    (Total) percentage of nutrients/fertiliser/N,P,K. 
          (Totale) persentasie nutriente/ kunsmis/N,P, K.                                               (1)

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Physical Sciences P2/Fisiese Wetenskappe V2     21                           DBE/November 2019
                             NSC/NSS – Marking Guidelines/Nasienriglyne

10.3.2    Marking guidelines/Nasienriglyne
          • Calculate mass fertiliser in A./Bereken massa kunsmis in A 
          • Calculate mass fertiliser in B./ Bereken massa kunsmis in B 
          • Calculate mass P in A and B ./Bereken massa P in A en B
          • Final answer/Finale antwoord:
             B has more phosphorous than/het meer fosfor as A. 

          OPTION 1/OPSIE 1                          OPTION 3/OPSIE 3
          Mass fertiliser in A:                     Mass phosphorous in A/
          Massa kunsmis in A:                       Massa fosfor in A:
                21                                        3
          m=       × 50  = 10,5 kg                 %P = × 21= 7,88%
               100                                        8
          Mass fertiliser in B:                             7,88
          /Massa kunsmis in B:                      m(P) =       × 50  = 3,94 kg
                                                            100
                27                                                                  
          m=       × 40  = 10,8 kg                 Mass(P) in B
               100                                  Massa (P) in B:
          Mass phosphorous in A/                            3
          Massa fosfor in A:                        %(P) = × 27 = 10,13%
                                                            8
           3                                            10,13
             × 10,5 = 3,94 kg                                  × 40  = 4,05 kg
           8                                        m=
                                                         100
          Mass phosphorous in B/ 
          Massa fosfor in B:                        Fertiliser B has more phosphorous
           3                                        than fertiliser A. /Kunsmis B het meer
             × 10,8 = 4,05 kg                       fosfor as kunsmis A 
           8
          Fertiliser B has more phosphorous         OPTION 4/OPSIE 4
          than fertiliser A.                       Mass fertiliser in A:
                                                    Massa kunsmis in A:
          OPTION 2/OPSIE 2                                 21
                                                    m=         × 50  = 10,5 kg
          Mass phosphorous in A/                          100
          Massa fosfor in A:                        Mass fertiliser in B:
              3 21                                  /Massa kunsmis in B:
          m= ×        × 50  = 3,94 kg
              8 100                                        27
                                                    m=         × 40  = 10,8 kg
          Mass(P) in B                                   100
          Massa (P) in B:                           For the same NPK ratio 
              3 27                                  the bag with more fertiliser will have
          m= ×        × 40  = 4,05 kg
              8 100                                 more phosphorous ∴ bag B
                                                    Vir dieselfde NPK verhouding, die sake
          Fertiliser B has more phosphorous         met meer kunsmis sal meer fosfor het
          than fertiliser A. /Kunsmis B het
                                                    ∴ sak B
          meer fosfor as kunsmis A.                                                              (4)
                                                                                                 [12]
                                                                          TOTAL/TOTAAL:          150

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## Collection

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

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