# Physical Sciences Grade 12 P2 2019 Memo

> Grade 12 Physical Sciences Paper 2 memorandum from the Northern Cape November 2019, 17 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/nc-2019-paper-2-memo/0762eabb-704a-4f40-9185-846a20bedde3
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2019
- Exam period: Northern Cape November
- Paper: 2
- Curriculum: CAPS
- Pages: 17

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## About this document

Physical Sciences · Grade 12 · Northern Cape November · 2019. Memorandum, 17 pages. Read online or download the PDF.

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### Page 1

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Physical Sciences P2/Grade 12                       1
Fisiese Wetenskappe V2/Graad 12   Marking Guidelines/Nasienrigl

                                           —————————————————————
                                           DEPARTMENT OF EDUCATION
                                           —————————————————————
                                           DEPARTEMENT VAN ONDERWYS
                                           —————————————————————
                                           LEFAPHA LA THUTO
                                           —————————————————————
                                           ISEBE LEZEMFUNDO
                                           —————————————————————

                  PROVINCIAL PREPARATORY EXAMINATION

                  PROVINSIALE VOORBEREIDINGSEKSAMEN/

                                        GRADE/GRAAD 12

                       PHYSICAL SCIENCES P2: CHEMISTRY
                       FISIESE WETENSKAPPE V2: CHEMIE
                     MARKING GUIDELINES/NASIENRIGLYNE
                                        SEPTEMBER 2019

MARKS/PUNTE: 150
TIME/TYD: 3 hours/uur

                          This memorandum consists of 17 pages./
                        Hierdie memorandum bestaan uit 17 bladsye.

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Physical Sciences P2/Grade 12                       2                                 September 2019
Fisiese Wetenskappe V2/Graad 12   Marking Guidelines/Nasienriglyne

QUESTION 1/VRAAG 1

1.1      C                                                                                        (2)

1.2      B                                                                                        (2)

1.3      A                                                                                        (2)

1.4      C                                                                                        (2)

1.5      D                                                                                        (2)

1.6      D                                                                                        (2)

1.7      D                                                                                        (2)

1.8      A                                                                                        (2)

1.9      C                                                                                        (2)

1.10     B                                                                                        (2)
                                                                                                    [20]

QUESTION 2/VRAAG 2

2.1
2.1.1     E                                                                                        (1)

2.1.2     A and C 
          A en C                                                                                    (1)

2.2
2.2.1     Propanone           ACCEPT: Propan-2-one; 2-propanone
          Propanoon             AANVAAR: Propan-2-oon; 2-propanoon                                  (2)

2.2.2     Ethyl propanoate
          Etielpropanoaat                                                                           (2)

2.2.3     Methanoic acid
          Metanoësuur                                                                               (1)

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Physical Sciences P2/Grade 12                           3                                 September 2019
Fisiese Wetenskappe V2/Graad 12       Marking Guidelines/Nasienriglyne

2.3
               H             H          H     H
                                                              Marking Criteria/Nasien kriteria
                                                         
         H     C      C    C      C      C     C     H        Functional group
                                                              Funksionelegroep
               H                         H     H              Ethyl substituent on the fourth carbon 
                                                              Etielsytak op die vierde koolstof
                                                              Whole structure correct
                           H      C     H
                                                             Totale struktuur korrek
                           H      C      H

                                  H                                                                     (3)

2.4.1
          H     H

          C     C

          H     H                                                                                       (1)

2.4.2     Addition
          Addissie                                                                                      (1)

2.5                3n
          CnH2n + (   ) O2 → nCO2 + nH2O                   Marking Criteria/Nasien kriteria:
                    2                                        Balancing reactants 
                   m 192                                     Balansering reaktante
          n(O 2 ) = =      =6                                Balancing Products
                   M 32
                                                             Balansering produkte
          3n
             = 6                                            Dividing by 32
           2                                                 Deel deur 32
          n=4                                                                       3n
          ∴ CnH2n = C4H8                                    Equating/Gelykstelling    = 6
                                                                                     2
                                                             Final answer
                                                             Finale antwoord                            (5)
                                                                                                        [17]

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Physical Sciences P2/Grade 12                       4                                 September 2019
Fisiese Wetenskappe V2/Graad 12   Marking Guidelines/Nasienriglyne

QUESTION 3/VRAAG 3

3.1       Boiling point increases with an increase in the number of carbons.
          Kookpunt neem toe met 'n toename in die aantal koolstof atome.
          OR/OF
          As the number of carbons increases, the boiling point also increases. 
          Indien die aantal koolstofatome toeneem, sal die kookpunt ook toeneem.                    (2)

3.2
3.2.1     Higher than
          Hoër as                                                                                   (1)

3.2.2     ANY TWO:
          ENIGE TWEE:
          Hydrogen bond
          Waterstofbinding
          Dipole-dipole force
          Dipool-dipool krag
          London force/induced dipole-induced dipole force/dispersion force.
          Londonkrag/geïnduseerde dipool-geïnduseerde dipool krag/dispersiekrag                     (2)

3.2.3        Butan-1-ol has strong hydrogen bonds in addition to the weak london
              forces.
              Butan-1-ol het sterk waterstofbindings saam met swak Londonkragte.

             Butane has weak London forces. 
              Butaan het swak Londonkragte

             More energy needed to overcome/break the hydrogen bonds/ IMF in
              butan-1-ol than the London forces in butane. 
              Meer energie is nodig om die waterstofbindings te oorkom/te breek/IMK in
              butan-1-ol te breek/te oorkom as die Londonkragte in butaan.
              OR/OF
             Butane has London forces/induced dipole- induced dipole forces/
              dispersion forces
              Butaan het Londonkragte/geïnduseerde dipool-geïnduseerde dipool
              krag/dispersiekrag

              Butan-1-ol has strong hydrogen bonds in addition to the weak london
              forces.
              Butan-1-ol het sterk waterstofbindings saam met swak Londonkragte.

             Less energy needed to overcome/break the London forces/ IMF in butane
              than the hydrogen bonds in butan-1-ol. 
               Minder energie is nodig om die Londonkragte te oorkom/te breek/IMK in
              butaan te oorkom/te breek as die waterstofbindings in butan-1-ol.
                                                                                                    (3)
                                                                                                    [8]

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Physical Sciences P2/Grade 12                       5                                 September 2019
Fisiese Wetenskappe V2/Graad 12   Marking Guidelines/Nasienriglyne

QUESTION 4/VRAAG 4

4.1
4.1.1     Hydrogenation
          Hidrogenasie                                                                              (1)

4.1.2     Platinum, nickel/nikkel or/of palladium/Pt, Ni ,Pd                                       (1)

4.2
                H     H       H                         Marking Criteria/Nasien kriteria:
                               H H     H               Functional group on second C-2 
         H      C     C       C   H                     Funksionele groep op die tweede C
                      O                                 Whole structure correct 
                H             H                         Totale struktuur korrek
                          
                      H

          Propan-2-ol/2- propanol                                                                  (3)

4.3       Secondary 
          Sekondêre
          The carbon atom bonded with -OH (hydroxyl) group is bonded to two other
          carbon atoms. 
          Die koolstofatoom gebind aan die -OH (hidroksiel) groep is gebind aan twee
          ander koolstofatome.                                                       (2)

4.4
4.4.1     Concentrated sulphuric acid/H2SO4/phosphoric acid/H3PO4/heat 
          Gekonsentreerde swawelsuur/H2SO4 /fosforsuur/H3PO4/hitte                                  (1)

4.4.2     Prop-1-ene/Propene
          Prop-1-een/Propeen                                                                        (1)

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Physical Sciences P2/Grade 12                        6                                 September 2019
Fisiese Wetenskappe V2/Graad 12    Marking Guidelines/Nasienriglyne

4.5
                       H                                      H       H                       
                H            H                                              
                                  HH   H 
          H      C     C      C    H + NaOH              H     C      C     C     H + NaCℓ + H2O
                                                                                             
                 H     Cℓ                                      H           H
                              H

                           Marking Criteria/Nasien kriteria:
                           Reactants/Reaktante: 2-chloropropane 
                                                    2-chloropropaan
                                                    NaOH
                           Products/Produkte:
                                     Double bond on prop-1-ene
                                     Dubbelbinding op prop-1-een
                                     Whole structure of prop-1-ene correct 
                                     Totale struktuur van prop-1-een korrek
                                     NaCℓ + H2O
                           Any additional reactants and/or products/Addisionele
                           reaktante en/produkte: Max. 4/5
                           Molecular/condensed formulae/
                           Molekulêre/gekondenseerde formules: Max. 4/5
                           Missing arrow/Sonder pyl: Max. 4/5
                           Reactants only /Slegs reaktante:Max. 2/5
                           Products only/Slegs produkte: Max. 3/5
                                                                                                     (5)

4.6
4.6.1     Substitution/hydrolysis
          Substitusie/hidrolise                                                                      (1)

4.6.2     Dilute strong base
          Verdunde sterk basis
          Mild heat                                                                                 (2)
          Matige hitte                                                                               [17]

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Physical Sciences P2/Grade 12                       7                                 September 2019
Fisiese Wetenskappe V2/Graad 12   Marking Guidelines/Nasienriglyne

QUESTION 5/VRAAG 5

5.1       ANY ONE/
          ENIGE EEN

             Change in concentration of reactant/product per (unit) time.
              Verandering in konsentrasie van reaktant/produk per (eenheid) tyd.

             Rate of change in concentration.
              Tempo van verandering in konsentrasie

             Change in amount/number of moles/volume/mass of products or
              reactants per (unit) time
              Verandering in hoeveelheid/aantal mol/volume/massa van produkte of
              reaktante per (eenheid) tyd

             Amount/number of moles/volume/mass of products formed or reactants
              used per (unit) time.
              Hoeveelheid/aantal mol/volume/massa van produkte gevorm of reaktante
              opgebruik per (eenheid) tyd.                                         (2)

5.2
5.2.1     Surface area/state of division
          Oppervlak area/toestand van verdeling                                                     (1)

5.2.2     ANY ONE/
          ENIGE EEN
          Concentration of the acid
          Konsentrasie van die suur
          Temperature of the acid
          Temperatuur van die suur
          Mass of the sodium carbonate
          Massa van die natriumkarbonaat                                                            (1)

5.3       Carbon dioxide/CO2 
          Koolstofdioksied/CO2                                                                      (1)

5.4       90 seconds 
          90 sekondes                                                                               (1)

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Physical Sciences P2/Grade 12                       8                                 September 2019
Fisiese Wetenskappe V2/Graad 12   Marking Guidelines/Nasienriglyne

5.5       Greater than 
          Groter as

                 In experiment 2, the sodium carbonate has a greater surface area
                  which means more reaction sites. 
                  In eksperiment 2 het die natriumkarbonaat 'n groter oppervlak wat
                  meer plekke waar reaksies kan plaasvind beteken.

                  OR/OF
                  More particles with correct orientation/more contact points
                  Meer deeltjies met die regte oriëntering/meer kontakpunte.

                 There will be more effective collisions per unit time in experiment 2
                  than in experiment 1. 
                  Daar sal meer effektiewe botsings per tydseenheid wees in
                  eksperiment 2 as in eksperiment 1.                                    (3)

5.6
5.6.1     Mass of CO2 formed/Massa CO2 gevorm = 150 -145,60 
                                              = 4,4 g
             m
          n=
             M

                 4,4
          n=
                 44 

          n = 0,1moles/mol

          n (CO2) : n (Na2CO3) = 1:1

          m (Na2CO3) = n x M
                     = (0,1)(106)
                     = 10,6 g
                                                                                                    (5)

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Physical Sciences P2/Grade 12                        9                                 September 2019
Fisiese Wetenskappe V2/Graad 12    Marking Guidelines/Nasienriglyne

5.6.2
                      150,00

                                                    Experiment 2

                          mass/g
                                                    Eksperiment 2

                      145,60

                                             time/s

          Marking Criteria/Nasien kriteria:
             Graph for experiment 2 should have a steeper gradient than
                experiment 1. 
                 Grafiek vir eksperiment 2 moet 'n steiler helling hê as eksperiment 1.

                Time for reaction to come to completion must be shorter. 
                 Tyd vir reaksie om te voltooi moet korter wees.                                      (2)
                                                                                                      [16]

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Physical Sciences P2/Grade 12                       10                                   September 2019
Fisiese Wetenskappe V2/Graad 12   Marking Guidelines/Nasienriglyne

QUESTION 6/VRAAG 6

6.1       A stage in a reversible reaction where the rate of the forward reaction equals
          the rate of the reverse reaction. 
          'n Tydstip tydens 'n omkeerbare reaksie waar die tempo van die voorwaartse
          reaksie gelyk is aan die tempo van die terugwaartse reaksie.
                                              OR/OF
          The stage in a chemical reaction when the concentration/amounts of
          reactants and products remain constant.
          Die tydstip in 'n chemiese reaksie wanneer die konsentrase/hoeveelheid van
          die reaktante en produkte onveranderd bly.                                                  (2)

6.2       CALCULATIONS USING NUMBER OF MOLES
          BEREKENINGE WAT AANTAL MOL GEBRUIK
          Mark allocation/Punte toekenning:
           Change n(H2) 
             Verandering n(H2)
           Ratio to determine n(CO) reacted and n(CH3OH) formed = 1 : 1 
             Verhouding om te bepaal n(CO) gereageer en n (CH3OH) gevorm=1:1
           n(CO) & n(CH3OH) at equilibrium 
             n(CO) & n(CH3OH) tydens ewewig
           Divide three equilibrium amounts by 2 (calculation of concentration) 
             Verdeel drie ewewighoeveelhede deur 2 (berekening van konsentrasie)
           Kc expression
             KC uitdrukking
           Substitution into Kc expression
             Vervanging in KC uitdrukking
           Answer : X = 0,8 mol 
             Antwoord: X= 0,8 mol
           Substitution into m = n x M
             Vervanging in m= n x M
           Final answer m = 22,4 g
          Finale antwoord m = 22,4 g

          OPTION 1/OPSIE 1:

                                    CO (g)          +     2 H2(g)          →       CH3OH(g)
                                    1 mole/mol          2 moles/mol          1 mole/mol
          Start/begin (mol)             X               1,5                  0
          Reacted (mol)             0,375               0,75                  0,375 ratio

          Equilibrium (mol)                 X-0,375            0,75                 0,375

          Equilibrium                   X - 0,375       0,75                     0,375
                            n                                                              
          Concentration c =                 2             2                        2
                            V
          (mol∙dm )
                 -3
                                                        =0,375               =0,1875

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Physical Sciences P2/Grade 12                       11                                  September 2019
Fisiese Wetenskappe V2/Graad 12   Marking Guidelines/Nasienriglyne

                 [CH3 OH]
         Kc =                
                [H2 ] 2 [CO]

                       0,1875
           6,27 =                     
                        2 X - 0,375
                 [0,375] [          ]
                               2
         x = 0,8 mol
             m
         n=
             M
         m = (0,8)(28)
         m = 22,4 g 

         OPTION 2 (USING CONCENTRATIONS)/OPSIE 2 (GEBRUIK VAN
         KONSENTRASIES)
                                                     CO (g)       +          2 H2(g)          →
                                                     CH3OH(g)
          Ratio/Verhouding                           1                2                1
          Initial concentration/Aanvanklike          X                1,5
          konsentrasie                                 = 0,5 X            = 0,75       0    
                                                     2                 2
          (mol∙dm-3)
          Change in
                                                                                           0,1875
          concentration/Verandering in                   0,1875
          konsentrasie                                                    0,375           ratio
          (mol∙dm-3)
          Equilibrium                                    0,5X -       0,75
          Concentration/Ewewigskonsentrasie                                = 0,375         0,1875
                                                         0,1875         2
          (mol∙dm-3)

                 [CH3 OH]
         Kc =                
                [H2 ] 2 [CO]

                        [0,1875]
           Kc =          2
                                        
                 [0,375] [0,5 X 0,1875]
         x = 0,8 mol
         Initial mass of CO injected in the container/Aanvanklike massa CO toegevoeg
         tot houer (m) = nxM
                      m = (0,8)(28)
                      m = 22,4g 
                                                                                                      (9)

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Physical Sciences P2/Grade 12                       12                                 September 2019
Fisiese Wetenskappe V2/Graad 12   Marking Guidelines/Nasienriglyne

                                                                         1
6.3       Kc( reverse reaction/terugwaartse reaksie)         =
                                                               Kc ( forward reaction)
                                                                 1                             
          Kc( reverse reaction/terugwaartse reaksie)         =
                                                               6,27
          Kc(reverse reaction/terugwaartse reaksie )         = 0,16

          NOTE Give 2 marks if the learners write only the correct answer.
          NB Gee 2 punte as die leerders slegs die korrekte antwoord gegee het.                     (2)

6.4
6.4.1     Remains the same 
          Bly dieselfde
          The temperature is constant, Kc value is only affected by a change in
          temperature.
          Die temperatuur is konstant, KC se waarde word slegs beïnvloed deur ’n
          verandering in temperatuur.                                            (2)

6.4.2     Decreases 
          Neem af.
          A larger volume (results in a lower pressure which) will favour the reaction
          producing a larger number of moles/reverse reaction is favoured.
          'n Groter volume (het tot gevolg 'n verlaagde druk) sal die reaksie bevoordeel
          wat 'n groter aantal mol sal produseer/terugwaartse reaksie word bevoordeel. (2)

6.5       Decreases 
          Neem af
          A decrease in temperature favours the exothermic reaction. The reverse
          reaction will be favoured.
          'n Afname in temperatuur bevoordeel die eksotermiese reaksie. Die
          terugwaartse reaksie word bevoordeel.                                   (3)
                                                                                  [20]

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Physical Sciences P2/Grade 12                           13                                 September 2019
Fisiese Wetenskappe V2/Graad 12       Marking Guidelines/Nasienriglyne

QUESTION 7/VRAAG 7

7.1       An acid is a proton donor. 
          'n Suur is 'n protonskenker.                                                                     (2)

7.2       NaOH → Na+ +OH-
          0,75 mol∙dm-3 (NaOH) gives/gee 0,75 mol∙dm-3 OH- ( Ratio/Verhouding 1:1) 

          OPTION 1/OPSIE 1                                OPTION 2/OPSIE 2
          pOH = - log [OH-]                              [H+][OH-] =10-14

          pOH = - log[0,75 ]                             [H+][0,75] =10-14 

          pOH = 0,1249                                    [H+] = 1,3333x10-14 mol∙dm-3              

          pH + pOH = 14                                   pH = - log [H+] 

          pH + 0,1249 = 14                               pH = - log [1,3333x10-14]

          pH = 13,88                                     pH = 13,88                                      (5)

7.3       ninitial/aanvanklik (HCℓ)
                                   n
                             c=      
                                  V
          n initial/aanvanklik = cV
                                = (0,5) (0,075)
                                = 0,0375 moles
          For the reaction between HCℓ (excess) and NaOH/Vir die reaksie tussen HCℓ
          (oormaat) en NaOH
          n(NaOH) reacted/gereageer n = cV
                                              n = 0,75X 0,022
                                              n = 0,0165 moles
             n(NaOH) : n(HCℓ) = 1:1
          nexcess/oormaat (HCℓ) = 0,0165 moles(ratio/verhouding)
          nreacted/gereageer (HCℓ) with/met CaCO3
                                  n= 0,0375-0,0165
                                  n=0,021moles/mol
                                            1
          nreacted/gereageer(CaCO3) = × 0,021 = 0,0105 moles/mol
                                            2
          m=nxM
          m = (0,0105)(100)
          m =1,05 g
                          1,05
          % CaCO3 =            × 100 
                            5
                                                                                                            (8)
                       = 21 %                                                                             [15]

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Physical Sciences P2/Grade 12                       14                                 September 2019
Fisiese Wetenskappe V2/Graad 12   Marking Guidelines/Nasienriglyne

QUESTION 8/VRAAG 8

8.1       Salt bridge.
          Soutbrug.                                                                                    (1)

8.2       Standard hydrogen electrode.                                                                (1)
          Standaard waterstofelektrode.

8.3       Provides a surface for the transfer of electrons/Allows conduction.                         (1)
          Voorsien 'n oppervlakte vir die oordrag van elektrone/Laat geleiding toe.

8.4       Mg (s) | Mg2+ (aq)  || I2 (g) | I- (aq) |Pt                                               (3)

8.5       Eθcell  Eθcathode  Eθanode 
          E0sel = E0katode- E0anode
          Ecell/sel = 0,54- (-2,36)
          Ecel/sel = 2,90 V
          Ecell is less than 3 V so the bulb will not glow to its maximum brightness. 
          Esel is minder as 3 V en die gloeilamp sal nie gloei teen sy maksimum
          helderheid nie.
                                                                                                     (5)

8.6       The bulb will switch off/it will no longer glow at all.
          Die gloeilamp sal afskakel/dit sal nie langer brand nie.
          The cell potential will be 0 V.
          Die selpotensiaal sal 0 V wees.                                                           (2)
                                                                                                    [13]

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Physical Sciences P2/Grade 12                       15                                 September 2019
Fisiese Wetenskappe V2/Graad 12   Marking Guidelines/Nasienriglyne

QUESTION 9/VRAAG 9

9.1       Electrical energy to chemical energy.                                                       (1)
          Elektriese energie na chemiese energie.

9.2.      2O2-(ℓ)→O2 (g) + 4e- OR/OF 2O2- (ℓ )+ C(s)→O2(g) + 4e                                      (2)

9.3       They undergo a combustion reaction and get used up.
          Hulle ondergaan 'n verbrandingsreaksie en word opgebruik.
          ACCEPT/AANVAAR:
          They get burnt. ( since they react with oxygen at a high temperature)
          Hulle verbrand. (aangesien hulle met suurstof reageer by hoë temperatuur)                    (2)

9.4       To reduce the working temperature  and hence save on extraction cost.
          Om die werkende temperatuur te verlaag en om sodoende ekstraksiekoste te
          bespaar.                                                                                     (1)

9.5       ANY TWO/ENIGE TWEE (Consider any other reasonable answers from
          learners/Oorweeg enige ander redelike antwoord van die leerders.)

                 Excessive CO2 (which is a green house gas) is formed during the
                  extraction process and this causes global warming.
                  Oormaat CO2 (wat 'n kweekhuisgas is) word gevorm gedurende die
                  ekstraksieproses en dit veroorsaak Aardverwarming.

                 The CO2 formed can also react with water when it rains and this
                  causes acid rain which corrodes buildings and damages plants. 
                  Die gevormde CO2 kan ook met water reageer wanneer dit reën en dit
                  veroorsaak suurreën wat geboue verweer en plante beskadig.

                 The electricity used in the extraction process is generated from the
                  burning of coal which pollutes the environment and depletes a non
                  renewable energy source.
                  Die elektrisiteit wat gebruik word gedurende die ekstraksieproses word
                  opgewek deur die verbranding van steenkool wat besoedeling van die
                  omgewing veroorsaak en 'n nie-hernubare energiebron uitput.

                 The plant requires a very large landscape so people might be moved
                  from their places of origin OR massive deforestation can occur in order
                  to accommodate the plant.
                  Die aanleg benodig 'n baie groot gebied en mense word sodoende
                  verskuif van hulle oorspronklike blyplekke OF grootskaalse ontbossing
                  kan gebeur om die aanleg te akkomodeer.

                 Red mud used in the extraction process can contaminate underground
                  water reserves.
                  Rooi modder wat gebruik word gedurende die ekstraksie kan die                        (2)
                  ondergrondse waterreserwes besoedel.                                                 [8]
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Physical Sciences P2/Grade 12                       16                                 September 2019
Fisiese Wetenskappe V2/Graad 12   Marking Guidelines/Nasienriglyne

QUESTION 10/VRAAG 10
10.1.1   Nitrogen (gas) 
         Stikstof (gas)                                                                               (1)
10.1.2    Haber process
          Haberproses                                                                                 (1)

10.1.3    Ammonium sulphate/(NH4)2SO4             
          Ammoniumsulfaat/(NH4)2SO4                                                                   (1)

10.1.4    Nitrogen dioxide/NO2
          Stikstofdioksied/NO2                                                                        (1)

10.2      3NO2 + H2O           → 2HNO3 + NO                   (balancing/balansering)              (3)

10.3
10.3.1    OPTION 1/OPSIE 1                              OPTION 2/OPSIE 2
          Using NPK ratios: P:K = 2:1                   Using NPK ratios
          Gebruik NPK verhoudings                       Gebruik NPK verhoudings
                  2                                     NP: K = 9:1
          m (P) = × 1,2                                             9
                  1                                     m ( NP)= × 1,2 
          m(P) =2,4 kg                                              1
                                Any One                m ( NP) =10,8 kg
                 7
          m(N) = × 1,2                                  m (NPK) =10,8 + 1,2
                 1                                      m (fertiliser/kunsmis) = 12 kg
          m (N) =8,4 kg

          m (fertiliser/kunsmis) = 1,2 + 2,4                                       12
                                                        % fertiliser/kunsmis =        ×100 
          +8,4                                                                     50
                        = 12 kg
                                  12                     % fertiliser/kunsmis = 24%
          % fertiliser/kunsmis =      ×100 
                                  50
                                                        X = 24
          % fertiliser = 24%: X = 24

          OPTION 3/OPSIE 3

          If 100% fertiliser/Indien 100% kunsmis
                  1
          m (K)=     × 50 
                 10

          m (K) =5 kg

                  1,2
           m (K)=     × 100 
                   5
                X = 24%                                                                            (4)

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Physical Sciences P2/Grade 12                       17                                 September 2019
Fisiese Wetenskappe V2/Graad 12   Marking Guidelines/Nasienriglyne

10.3.2    No.
          Nee
          The chosen fertilizer contains too much nitrogen.
          Die gekose kunsmis bevat te veel stikstof.
          This will make the leaves of the maize plant grow big and green and not
          promote the actual growth of the maize crop.
          Dit sal veroorsaak dat die blare van die mielieplant groot en groen groei en
          sal nie die eintlike groei van die mielie-oes bevoordeel nie.                                (3)

10.4      Use organic fertilisers like manure in place of inorganic fertilisers. 
          Gebruik organiese misstowwe soos beesmis in plaas van kunsmis.
          Avoid putting too much or excessive inorganic fertilisers. 
          Moet nie te veel anorganiese kunsmisstowwe gebruik nie.                                      (2)
                                                                                                      [16]

                                                        GRAND TOTAL/GROOTTOTAAL:                      150

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