# Physical Sciences Grade 11 P2 North West November 2024 Memo

> Grade 11 Physical Sciences Paper 2 memorandum from the North West November Exam 2024 (English), 10 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-11-physical-sciences-2024-north-west-november-exam-question-paper-paper-2-english/b8a6be24-d3ec-4af3-a410-887c621b997b
- Updated: 2026-10-29
- Subject: Physical Sciences
- Grade: Grade 11
- Language: English
- Document type: Memorandum
- Year: 2024
- Exam period: North West November Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 10

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## About this document

Physical Sciences · Grade 11 · North West November Exam · 2024 · English. Memorandum. Read online or download the PDF.

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### Page 1

PROVINCIAL ASSESSMENT
                           PROVINSIALE ASSESSERING

                                GRADE/GRAAD 11

                         PHYSICAL SCIENCES P2
                        FISIESE WETENSKAPPE V2
                   MARKING GUIDELINES/NASIENRIGLYNE
                             NOVEMBER 2024

 MARKS/PUNTE: 150

                  These marking guidelines consists of 10 pages.
                   Hierdie nasienriglyne bestaan uit 10 bladsye.

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### Page 2

Physical Sciences P2/Fisiese Wetenskappe V2      2                            NW/November 2024
                          Grade/Graad 11 – Marking Guidelines/Nasienriglyne

 QUESTION 1/VRAAG 1

 1.1    D                                                                                         (2)

 1.2    D                                                                                         (2)

 1.3    B                                                                                         (2)

 1.4    B                                                                                         (2)

 1.5    A                                                                                         (2)

 1.6    C                                                                                         (2)

 1.7    B                                                                                         (2)

 1.8    A                                                                                         (2)

 1.9    A                                                                                         (2)

 1.10 D                                                                                           (2)
                                                                                                   [20]

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### Page 3

Physical Sciences P2/Fisiese Wetenskappe V2      3                            NW/November 2024
                          Grade/Graad 11 – Marking Guidelines/Nasienriglyne

 QUESTION 2/VRAAG 2

 2.1 Electronegativity as a measure of the tendency of an atom in a molecule to
     attract bonding electrons. 
     Elektronegatiwiteit is ‘n maatstaf van die neiging van ‘n atoom in ‘n molekuul om
     bindingselektrone aan te trek.                                                  (2)

       2.2.1 ∆E = 4 – 1 = 3; Ionic bond/ioniese binding.                                         (2)

       2.2.2 ∆E = 3 – 3 = 0; non polar/nie polêr.                                                (2)

       2.3.1 Valence electrons are the electrons in the highest energy level of an
             atom in which there are electrons. 
             Valenselektrone is die elektrone in die hoogste energievlak van ‘n atoom
             waarin daar elektrone is.                                                           (2)

       2.3.2 4                                                                                    (1)

       2.3.3                        
                                                                                                   (2)

       2.3.4
                            

                                                                                                   (2)

       2.4.1 Linear/Liniêr                                                                        (1)

       2.4.2 Tetrahedral/Tetrahedries                                                             (1)

       2.5.1 Bond length is the average distance between the nuclei of two bonded
             atoms.
             Bindingslengte is die gemiddelde afstand tussen die kerne van twee
             gebonde atome.                                                                      (2)

       2.5.2 C2H4 have double bonds between carbon atoms  while C2H6 have
             single bonds between carbon atoms.
             Bond length decreases with the number of bonds between the
             atoms.
             C2H4 het dubbelbindings tussen koolstofatome terwyl C2H6
             enkelbindings tussen koolstofatome het.
             Bindingslengte neem af met die aantal bindings tussen die atome.                    (3)

       2.5.3 The shorter the bond length the higher the bond energy. OR The longer
             the bond length the smaller the bond energy.
             Hoe korter die bindingslengte hoe hoër is die bindingsenergie. OF Hoe
             langer die bindingslengte hoe kleiner die bindingsenergie.                            (2)

       2.5.4 SMALLER THAN/KLEINER AS                                                              (1)

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### Page 4

Physical Sciences P2/Fisiese Wetenskappe V2      4                            NW/November 2024
                          Grade/Graad 11 – Marking Guidelines/Nasienriglyne

 2.6 Both CHCℓ3 and CCℓ4 molecules have polar bonds and both have a
     tetrahedral shape. CHCℓ3 molecule is asymmetrical/have uneven distribution of
     electrons while CCℓ4 molecule is symmetrical/ have even distribution of
     electrons. This difference in the charge distribution makes CHCℓ3 to be a polar
     molecule and CCℓ4 to be a non-polar molecule.
     Beide CHCℓ3 en CCℓ4 molekules het polêre bindings en beide het ‘n
     tetraëdriese vorm. CHCℓ3 molekule is asimmetries/het oneweredige
     verspreiding van elektrone terwyl CCℓ4 molekule simmetries is/ het eweredige
     verspreiding van elektrone. Die verskil in die lading verspreiding veroorsaak
     dat CHCℓ3 ‘n meer polêre molekule en CCℓ4 ‘n nie-polêre molekule is.                          (4)
                                                                                                   [27]

 QUESTION 3/VRAAG 3

 3.1      Melting point is temperature at which the solid and liquid phases of a
          substance are at equilibrium.
          Smeltpunt is die temperatuur waarby die vaste- en vloeistoffases van ‘n stof
          in ewewig is.                                                                             (2)

          3.2.1 The stronger the intermolecular forces the higher the melting point.  (2)
                Hoe sterker die intermolekulêre kragte, hoe hoër die smeltpunt.

          3.2.2 The molecular mass of HBr is larger than that of HCℓ.
                The intermolecular (London) forces between molecules of HBr are
                stronger than the intermolecular force between the molecules of
                HCℓ.
                More energy is required to overcome the intermolecular forces
                between the molecules of HBr than of HCℓ.
                Die molekulêre massa van HBr is groter as die van HCℓ.
                Die intermolekulêre (London) kragte tussen molekules van HBr is
                sterker as die intermolekulêre kragte tussen die molekules van
                HCℓ.
                Meer energie benodig om die intermolekulêre kragte tussen
                molekules van HBr te oorkom as van HCℓ.                                            (3)

 3.3      HI                                                                                       (1)

 3.4      CCℓ4 molecules have London forces only while H2O has (London forces)
          and hydrogen bond.
          Intermolecular forces are not of comparable strength.                                   (2)
          CCℓ4 molekules het slegs Londonkragte terwyl H2O waterstofbindings het.
          Intermolekulêre kragte is nie van vergelykbare sterkte.
                                                                                                   [10]

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### Page 5

Physical Sciences P2/Fisiese Wetenskappe V2      5                            NW/November 2024
                          Grade/Graad 11 – Marking Guidelines/Nasienriglyne

 QUESTION 4/VRAAG 4

 4.1   Boyle’s law: the pressure of an enclosed gas is inversely proportional to the
       volume it occupies at constant temperature.
       Boyle se wet: die druk van ‘n ingeslote gas is omgekeerd eweredig aan die
       volume wat dit by konstante temperatuur beslaan.                                           (2)

       4.2.1     Pressure/Druk                                                                     (1)

       4.2.2     Mass (of gas) and temperature. 
                 Massa (gas)  en temperatuur.                                                     (2)

 4.2           p1V1 = p2V2
       128,5 × 55 = 201,93×V2
                V2 = 35 cm3                                                                        (3)

       4.3.1 • Particles are in continual motion in all directions.
             • Particles do not contribute to the volume of the gas.
             • There are no forces between the particles, or the particles
               and the wall of the container, except during collisions.
             • Collisions are perfectly elastic with no loss of total energy
               of the molecules.
             • All molecules are identical.
             • Collisions of particles on the surface cause pressure.
               (Any two)
             • Deeltjies in voortdurende beweging in alle rigtings.
             • Deeltjies dra nie by tot die volume van die gas nie.
             • Daar is geen kragte tussen die deeltjies, of die deeltjies en die kant van
               die houer, behalwe tydens botsings.
             • Botsings tussen die molekules is volkome elasties met geen verlies van
               kinetiese energie van die molekules.
             • Al die molekules is identies.
             • Botsings van deeltjies teen die oppervlak veroorsaak druk.
             (Enige twee)                                                                           (2)

       4.3.2 High temperature and low pressure.
             Hoë temperatuur  en lae druk.                                                        (2)

 4.4   As the temperature increases the average kinetic energy of the molecules
       increases. pressure of the molecules on each other and on the sides of
       the container increases as there will be more collisions.
       As die temperatuur verhoog, verhoog die gemiddelde kinetiese energie van die
       molekules. druk uitgeoefen deur die molekules op mekaar en die kante van
       die houer verhoog en daar is meer botsings.                                                 (2)
                                                                                                   [14]

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### Page 6

Physical Sciences P2/Fisiese Wetenskappe V2      6                            NW/November 2024
                          Grade/Graad 11 – Marking Guidelines/Nasienriglyne

 QUESTION 5/VRAAG 5

 5.1       EXOTHERMIC. Energy is released OR ∆H < 0 OR more energy is released
           than absorbed.
           EKSOTERMIES. Energie word vrygestel OF ∆H < 0 OF meer energie word
           vrygestel as geabsorbeer.                                                              (2)

 5.2       ∆H = HP – HR
              = -86-25
              = -111 k∙J∙mol-1                                                                     (2)

           5.3.1       Decreases/Afneem                                                            (1)

           5.3.2       Remain the same/Bly dieselfde.                                              (1)

           5.4.1       Activated complex is the unstable transition state from reactants to
                       products.                                                                  (2)
                       Geaktiveerde kompleks is die onstabiele oorgangstoestand van
                       reaktanse na produkte.

           5.4.2       154 k∙J∙mol-1                                                               (2)

 5.5       V SO3 : V SO2
               2 : 2
           V SO3 = V SO2 = 50 cm3                                                                  (2)
                                                                                                   [12]

 QUESTION 6/VRAAG 6

 6.1 Empirical formula is the simplest whole-number ratio of atoms in a
     compound.
     Empiriese formule is die eenvoudigste heelgetal verhouding in ‘n verbinding.                   (2)

 6.2 nC:nH:nN
         m
     n=
         M
     74.02    8.71    17.27
           :      :       
      12        1      14
       6.17 : 8.71 : 1.23
          5 : 7 : 1
       Empirical formula/Empiriese formule: C5H7N
                                                                                                    (5)

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### Page 7

Physical Sciences P2/Fisiese Wetenskappe V2      7                            NW/November 2024
                          Grade/Graad 11 – Marking Guidelines/Nasienriglyne

 6.3            m
          n=
                M
                40,57
       0,25 =           
                    M
          M = 162,28 g 
       Empirical formula mass/Empiriese formule massa = 5 ×12 + 7 × 1 + 14 = 81
       Ratio of the empirical formula mass: molecular formula mass
       Verhouding van empiriese formule massa : molekulêre formule massa
       81 : 162,28
        1 : 2
       ∴ Molecular formula/molekulêre formule: C10H14N2                                            (5)
                                                                                                   [12]

 QUESTION 7/VRAAG 7

 7.1 Limiting reagent is a substance that is completely used up in a chemical
     reaction.
     Beperkte reaktans is ‘n stof wat volledig opgebruik word in ‘n chemiese reaksie.
                                                                                                  (2)

 7.2 Calcium carbonate/Kalsiumkarbonaat/CaCO3                                                      (1)

       7.3.1                m
                nCO2 =
                            M
                            4,4
                        =         
                            44
                        = 0,1 mol             Any one/Enige een
                            V
                nCO2 =
                            Vm
                         VCO
                                  2
                0,1 =                 
                            22,4
                VCO2 = 2,24 dm3                                                                   (4)

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### Page 8

Physical Sciences P2/Fisiese Wetenskappe V2      8                              NW/November 2024
                          Grade/Graad 11 – Marking Guidelines/Nasienriglyne

       7.3.2 n(CaCO3) : n(CO2)
                             1:1
               n(CaCO3) reacted/gereageer = n(CO2) formed/gevorm = 0,1 mol
                           mCaCO
                                 3
               n(CaCO3 ) =         
                             M
                           mCaCO
                                 3
                    0,1 =          
                             100
               m(CaCO3 ) = 10 g 
                                 mass of CaCO3 reacted
               % purity =                                  x 100%
                                mass of IMPURE CaCO3

                                10
                            =        x 100% 
                                12

                            = 83 ,33%                                                                (6)
       7.3.3        n
               c=       
                    V
                        n
               2=           
                    0,15
               n = 0,15 mol
               n(HCℓ) reacted/gereageer : n(CO2) formed/gevorm
                                                2:1
                                                  1
               n(HCℓ) reacted/gereageer =             x 0,1 = 0.05 mol
                                                  2
               n(HCℓ) in excess/oormaat = n(HCℓ) initial/begin - n(HCℓ)
               reacted/gereageer
                                                 = 0,15 – 0,05 
                                                 = 0,1 mol                                            (5)
                                                                                                      [18]

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### Page 9

Physical Sciences P2/Fisiese Wetenskappe V2      9                            NW/November 2024
                          Grade/Graad 11 – Marking Guidelines/Nasienriglyne

 QUESTION 8/VRAAG 8
 8.1   Base is a proton/H+ ion acceptor. 
       Basis is ‘n proton/ H+- ioon ontvanger.                                                    (2)

 8.2 pH = - log[H3O+] 
       = - log(0,1)
       = 1                                                                                         (3)

 8.3 Cℓ-.                                                                                          (1)
 8.4 It is a substance that can act as either acid or base.                                      (2)
     Dit is ‘n stof wat as ‘n suur of ‘n basis kan optree. 

 8.5 HSO-4 + H2O ⇌ H3O+ + SO42-                                                                  (3)

       8.6.1         m
               c=       
                     MV
                       7,9
                 =          
                   56(0,25)
                 = 0,56 mol∙dm-3                                                                   (3)

       8.6.2    ca Va        na
                         =   
                 cb Vb nb
                 ca x40       1
                          = 
               0,56 x 25      2
                       ca = 0,175 mol∙dm-3                                                         (4)

       8.6.3 c1V1 = c2V2 
               c1 × 10 = 0,175 × 500
               c1 = 8,75 mol∙dm-3                                                                  (4)
                                                                                                   [22]

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### Page 10

Physical Sciences P2/Fisiese Wetenskappe V2      10                           NW/November 2024
                          Grade/Graad 11 – Marking Guidelines/Nasienriglyne

 QUESTION 9/VRAAG 9

 9.1 Reduction is a decrease in oxidation number.
     Reduksie is ‘n afname in oksidasiegetal.                                                   (2)

       9.2.1 +2                                                                                  (1)

       9.2.2 +3                                                                                  (1)

 9.3 Mn2+/ MnO .                                                                                 (3)
     Oxidation number of Mn2+ decreases.
     Oksidasiegetal van Mn2+ neem af. 

       9.4.1 Cu ⟶ Cu2+ + 2e-                                                                    (2)

       9.4.2 NO3- + 4H+ + 3e- ⟶ NO + 2H2O                                                       (2)

       9.4.3 Cu ⟶ Cu2+ + 2e- × 3 
             NO3- + 4H+ + 3e- ⟶ NO + 2H2O × 2 
             3Cu + 2NO3- + 8H+ ⟶ 3 Cu2+ + 2NO + 4H2O                                            (4)
                                                                                                 [15]

                                                                          TOTAL/TOTAAL: 150

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