# Physical Sciences Grade 11 Paper 2 Gauteng June 2018 Memo

> Grade 11 Physical Sciences Paper 2 memorandum from the Gauteng June 2018, 14 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/gr11-phy-p2-bilingual-june-2018-possible-answers/14d310d9-9dd9-4e24-9694-921d7e75d941
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 11
- Document type: Memorandum
- Year: 2018
- Exam period: Gauteng June
- Paper: 2
- Curriculum: CAPS
- Pages: 14

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Physical Sciences · Grade 11 · Gauteng June · 2018. Memorandum, 14 pages. Read online or download the PDF.

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               GAUTENG DEPARTMENT OF EDUCATION
           GAUTENGSE DEPARTEMENT VAN ONDERWYS
                     PROVINCIAL EXAMINATION
                      PROVINSIALE EKSAMEN
                      JUNE / JUNIE 2018
                        GRADE / GRAAD 11

                        PHYSICAL SCIENCES
                       FISIESE WETENSKAPPE

                       PAPER / VRAESTEL 2

                         MEMORANDUM

14 pages / bladsye

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                                         Physical Sciences / Fisiese Wetenskappe
                     MEMORANDUM         (Paper / Vraestel 2)   GRADE / GRAAD 11

                   GAUTENG DEPARTMENT OF EDUCATION
                GAUTENGSE DEPARTEMENT VAN ONDERWYS
                        PROVINCIAL EXAMINATION
                         PROVINSIALE EKSAMEN

                                PHYSICAL SCIENCES / FISIESE WETENSKAPPE
                                                         (Paper / Vraestel 2)

QUESTION 1: MULTIPLE-CHOICE QUESTIONS
VRAAG 1: MEERVOUDIGEKEUSE-VRAE

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

                                  2

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                                                     Physical Sciences / Fisiese Wetenskappe
                             MEMORANDUM             (Paper / Vraestel 2)   GRADE / GRAAD 11

QUESTION 2 / VRAAG 2:

2.1    Intermolecular forces are the attraction forces between molecules.
       Intermolekulêre kragte is die aantrekkingskrag tussen molekules.               (2)

2.2.1 Hydrogen bonds 
      Waterstofbindings                                                                (1)

2.2.2 Dipole-dipole forces 
      Dipool-dipool kragte                                                             (1)

2.2.3 The strength of the intermolecular forces increases as the molecular size
      increases.  HI has a bigger molecular mass than HCℓ thus HI has a
      higher melting point.  More energy is needed to break the stronger dipole-
      dipole forces in HI. 
      Die krag van die intermolekulêre kragte verhoog soos die molekulêre grootte
      verhoog.  HI het ŉ grooter molekulêre massa as die HCℓ dus het HI ŉ
      hoër kookpunt.  Meer energie is nodig om die sterker dipool –
      dipoolkragte in HI te breek.                                                    (6)

2.2.4 Hydrogen iodide. 
      Waterstofjodied.                                                                (2)

2.3.1 Boiling point is the temperature at which a liquid’s vapor pressure is equal to
      its atmospheric pressure. 
      Kookpunt is die temperatuur waarteen die dampdruk van ŉ vloeistof gelyk is
      aan die atmosferiese druk.                                                      (2)

2.3.2 There are strong Hydrogen bonds between the molecules of the methanol 
      and weak London forces (induced dipole forces) between the molecules of
      the methane.  Thus more energy is needed to break the hydrogen bonds
      between the molecules of the methanol. 
      Daar is sterk waterstofbindings tussen die molekules van die metanol  en
      swak London kragte ( geïnduseerde dipool kragte) tussen die molekules van
      die metaan.  Dus word meer energie benodig om die waterstofbindings
      tussen die metanol molekules te breek.                                           (3)
                                                                                        [17]

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                                                 Physical Sciences / Fisiese Wetenskappe
                          MEMORANDUM            (Paper / Vraestel 2)   GRADE / GRAAD 11

QUESTION 3 / VRAAG 3:

3.1      A way of representing atoms or molecules by showing valence electrons
         as dots / crosses surrounding the element symbol. 
         ŉ Manier om atome of molekules voor te stel deur die valensie-elektrone
         as kolletjies / kruisies om die simbool van die element te toon.        (2)

3.2.1    NaCℓ – Ionic bond  / Ioniese binding
         H2O & NH3 – polar covalent bond / polêre kovalente binding
         H2 – non-polar covalent bond  / nie-polêre kovalente binding             (3)

3.2.2    NaCℓ – ionic / not a molecule/ ionies / nie ŉ molekule
         H2O – Angular / bent / hoekig / gebuig 
         NH3 – Pyramidal / piramidaal 
         H2 – linear / linieêr                                                    (3)

3.2.3    H2O / water to form H3O+ / water om H3O+ te vorm 
         and / en
         NH3 / ammonium / amoniak to form NH4+ / NH3 / ammonium /
         ammoniak om NH4+ te vorm                                                 (2)

3.2.4

                                                                                (3)

         OR / OF

                                                         
                                                                                   [13]

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                                                     Physical Sciences / Fisiese Wetenskappe
                             MEMORANDUM             (Paper / Vraestel 2)   GRADE / GRAAD 11

QUESTION 4 / VRAAG 4:

4.1   Strong Hydrogen bonds  /Sterk waterstofbindings                                (2)

4.2   Heat of vaporisation is the amount of heat required to make water
      evaporate. 
      Verdampingshitte is die hoeveelheid hitte benodig om water te laat
      verdamp.                                                                       (2)

4.3   POLAR  Because of
        1. The EN difference, H2O EN = 3,5 – 2,1 = 1,4  polar covalent 
           O attract shared electron pair more than H 
        2. Because of the non-symmetrical shape and two lone pairs on the
           oxygen.                                                                    (4)
      POLÊR  omdat
        1. H2O EN = 3,5 – 2,1 = 1.4  polêr kovalent  O trek die
           gedeelde elektronpaar meer as die waterstof. 
        2. As gevolg van die nie-simmetriese vorm en die twee paar
           ongepaarde elektrone op die suurstof. 

4.4   Molecular dipoles occur due to the unequal sharing of electrons  between
      atoms in a molecule. Those atoms that are more electronegative pull the
      bonded electrons closer to themselves creating a positive and a negative
      side to the opposite sides of the molecule.                                     (2)
      ŉ Molekulêre dipool ontstaan wanneer daar ŉ oneweredige verdeling van
      elektrone tussen die atome  in die molekule ontstaan a.g.v. ŉ groot verskil
      in elektronegatiwiteit. Dit gee dan aanleiding tot die vorming van ŉ positiewe
      en negatiewe lading aan die teenoorgestelde kante van die molekule. 

4.5   KCℓ                                                                             (1)

4.6   KCℓ – forms an ionic bond and will be able to dissociate in water to form K +
      and Cℓ - ions,  where I2 forms a non-polar bond and will not mix with the
      polar water molecules.                                                          (2)
      KCℓ – vorm ŉ ioniese binding wat in staat is om te dissosieer in water om K+
      en Cℓ - ione te vorm,  terwyl I2 ŉ nie-polêre molekuul is en dus nie met die
      polêre watermolekules sal meng nie. 

4.7   Capillary action  – the adhesion forces between the molecules of
      different origin e.g. water and glass are bigger than the cohesion forces of
      the water molecules.                                                            (3)
      Kapillêre werking  – die adhesiekragte tussen die molekules van
      verskillende stowwe is groter as die adhesiekragte tussen die
      watermolekules. 
                                                                                       [16]

                                                5

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                                                       Physical Sciences / Fisiese Wetenskappe
                               MEMORANDUM             (Paper / Vraestel 2)   GRADE / GRAAD 11

QUESTION 5 / VRAAG 5:

5.1     Ideal gas: It is a hypothetical gas that will obey all the gas laws under all
        conditions of pressure and temperature.  (two marks or none)                  (2)
        Ideale gas: ŉ Hipotetiese gas wat al die gaswette nakom onder alle
        omstandighede van temperatuur en druk.  (twee of geen punte)

5.2     Hydrogen gas (H2) OR Helium Gas (He)                                          (2)
        Waterstofgas (H2) OF Heliumgas (He)

5.3     Temperature                                                                   (2)
        Temperatuur 

5.4.1 Dependent variable: Volume of the gas 
      Independent variable: Different temperatures of the water bath.                  (2)
      Afhanklike veranderlike: Volume van die gas 
      Onafhanklike veranderlike: Verskillende temperature van die
      waterbaddens. 

5.4.2 Charles’s law                                                                   (2)
      Charles se wet 

5.4.3 Type of gas
      Mass of gas
      Pressure of the gas (any two)                                                   (2)

        Tipe gas
        Massa van die gas
        Druk van die gas (enige twee) 

5.4.4 58 – 60 cm3                                                                     (2)

5.4.5     pV = nRT 
                                                       p=?
           p x 7,5 x 10 -4  = 12 x 8,31 x 298        V = 750 ÷ (100)3
                                                         = 7,5 x 10 -4 m3
              p=    12 x 8,31 x 298                    T = 298 K
                       7,5 x 10 -4                     R = 8,31
                                                       n = 12 mol
              p   = 39622080 Pa       / 3,96 x 107 Pa                                  (4)

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                                                 Physical Sciences / Fisiese Wetenskappe
                          MEMORANDUM            (Paper / Vraestel 2)   GRADE / GRAAD 11

5.5.1 LOWER THAN 
      MINDER AS                                                                (1)

5.5.2 1 mole of gas at the same temperature and pressure has the same volume
      (pV =nRT  p  n)  if the pressure is ½ then the molar mass is less    (2)
      1 mol van enige gas teen dieselfde temperatuur en druk het dieselfde
      volume (pV =nRT  p  n)   dus as die druk ½ is, is die molêre
      massa minder
                                                                                [21]

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                                                      Physical Sciences / Fisiese Wetenskappe
                             MEMORANDUM              (Paper / Vraestel 2)   GRADE / GRAAD 11

QUESTION 6 / VRAAG 6:
6.1.1 Zn(s) + 2 HCℓ (aq)            ZnCℓ2 (s) + H2 (g)                          (2)
6.1.2                  Zn(s) +           2 HCℓ (aq)         ZnCℓ2(s)   + H2(g)
        Mole ratio:          1                  2               1          1
        Mol
        verhouding:
        Initial mole   n =m/M            = 50 / 36,5
        Aanvangs       = 45,5g / 65      = 1,37 mol 
        mol            = 0,7 mol 
        Change in                   Mol ratio /
        mol            Have enough. verhouding:
        Verandering    Het genoeg   1,37 ÷2 =0,685
        in mol         Zn           mol 
                                    limiting
                                    reactant 
                                                                                   (4)
                                    Beperkende
                                    reagens
6.1.3                       Zn(s) +   2 HCℓ (aq)            ZnCℓ2 (s) + H2 (g)
        Change in
        mol          Used 0,685    If 1,37 mol used
        Verandering mol (mole      gebruik
        in mol       ratio used /
                     mol
                     verhouding
                     gebruik) 
        Mol left or  0,7 – 0,685
        formed at    = 0,015 mol 
        end          m =n x M
        Mol           = 0,015 x 65
        oorgelaat of  = 0,975 g                                                   (3)
        aan die          Zn left /
        einde        Zn oor
        gevorm

                                                 8

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                                                    Physical Sciences / Fisiese Wetenskappe
                            MEMORANDUM             (Paper / Vraestel 2)   GRADE / GRAAD 11

6.1.4                  Zn(s) + 2 HCℓ (aq)         ZnCℓ2 (s) + H2 (g)
        Change in
        mol                    0,137                         0,685 mol
        Verandering                                          (mole ratio used /
        in mol                                               gebruik) 
        Mol left or                                           n = V/Vm 
        formed at                                             = 0,685 ÷ 22,4
        end                                                  = 15,34 dm3 
        Mol oor /
        gevorm aan                                                                (3)
        die einde
6.2.1                    CH4      +    2 O2           CO2      +    2 H2O
        Mole ratio:         1           2                1               2
        Mol
        verhouding:
        Initial mole        3      excess /              0               0
        Aanvangs                   oortollig
        mol
        Change in
        mol          If all used /                 Ratio 1:1  /
        Verandering Indien alles                   verhouding 1:1
        in mol       gebruik                        3 mole
                           3
        Mol left or                                m =n xM
        formed at                                  = 3 x (44) 
        end                                        =132g 
        Mol oor /                                  CO2 that can be
        gevorm aan                                 produced
        die einde                                  CO2 wat
                                                   geproduseer
                                                   kan word                       (4)

                                               9

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                                                        Physical Sciences / Fisiese Wetenskappe
                               MEMORANDUM              (Paper / Vraestel 2)   GRADE / GRAAD 11

6.2.2    % Yield / opbrengs =     actual yield / ware opbrengs    
                                Max possible yield / maks moontlike opbrengs
                            = 87    x 100 
                              132

                           = 65,91%                                                 (3)

6.3                     C                   H            O
                     39,9 g               6,7 g         53,4 g   = 100g
        m =           39,9                6,7           53,4
                                                                  row / ry
        M             12                   1            16

        ÷ smallest    3,3                 6,7            3,3
                                                                  row / ry
          kleinste    3,3                  3,3           3,3

        Mol ratio:     1         :         2      :       1       row / ry

         IPF = CH2O = M = 12 + 2 + 16
                       = 30 g.mol-1 
        Molecular Formula / Molekulêre formule = 60 ÷ 30 = 2 x CH2O

                           C2H4O2                                                   (6)
                                                                                     [25]

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                                                  Physical Sciences / Fisiese Wetenskappe
                           MEMORANDUM            (Paper / Vraestel 2)   GRADE / GRAAD 11

QUESTION 7 / VRAAG 7:
7.1      n = m/M
           = 12
             65
           = 0,1846 mol 
      But in each mole of NaN3 there are 3 moles of N
      Maar in elke mol NaN3 is daar 3 mol N
           nN = 0,1846 x 3 = 0,554 mol 

         n = No
             NA 

        0,554 = No ÷ 6,02 x 10 23 
       No N atoms / atome = 3,45 x 10 23 atoms / atome                          (5)

7.2    2NaN3 (s)  → 2Na(s) + 3N2(g)                                            (3)

7.3                       2NaN3 (s)   → 2Na(s) + 3N2(g)
                                              n = V
                                                    VM 
                                                 = 85
                                                  22,4
            Use mole ratio                     = 3,795 mol 
           Gebruik mol vehouding
               2,53 mol 
       Mass NaN3 needed to fill bag:
       Massa NaN3 benodig om sak te vul:
                  m=nxM
                    = 2,53 x 65 
                    = 164,45 g                                                   (5)

7.4   The chemical formula that shows the simplest ratio between the atoms of a
      compound.                                                                 (2)
      Die chemiese formule wat die eenvoudigste verhouding tussen die atome in
      ŉ verbinding aantoon.

                                            11

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                                                      Physical Sciences / Fisiese Wetenskappe
                            MEMORANDUM               (Paper / Vraestel 2)   GRADE / GRAAD 11

7.5                          Na                         N
             %             35,39                     64,61       = 100

      m =                   35,39                     64,61
      M                      23                        14

                            1,549                      4,615

      ÷ smallest            1,549                      4,615
        kleinste            1,549                      1,549

      Mol ratio:              1             :           3       row / ry

               IPF = NaN3                                                               9
      Empiriese formule = NaN3                                                     (3)

7.6.1 DECREASE                                                                   (2)
      VERMINDER

7.6.2 p  T Pressure of a gas is directly proportional to temperature /
      Druk van ŉ gas is direk verwant aan die temperatuur daarvan
      If the temperature decreases, the average kinetic energy of the gas
      molecules decreases  and the pressure decreases.                           (3)
      As die temperatuur van die gas afneem sal die gemiddelde kinetiese
      energie van al die deeltjies ook afneem en dus sal die druk verlaag.
                                                                                   [23]

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                                                    Physical Sciences / Fisiese Wetenskappe
                            MEMORANDUM             (Paper / Vraestel 2)   GRADE / GRAAD 11

QUESTION 8 / VRAAG 8:
8.1. Oxygen                                                                    (2)
     Suurstof

8.2   2KCℓO3   2 KCℓ  + 3 O2  balancing / balansering
      One for each correct formula / Een punt vir elke korrekte formule.         (4)

8.3   n = m/M
            = 16,2
              122,5
            = 0,1322 mol 
      In each mole of KCℓO3 there is 1 mol of K 
      In elke mol KCℓO3 is daar 1 mol K
          n = No
                NA
          0,1322 = No ÷ 6,02 x 10 23 
       No K ions/ ione = 7,97 x 10 22 ions/ ione                               (4)

8.4   2KCℓO3                             2 KCℓ     +       3 O2

      If all off 16,2 g KCℓO3 decomposed
      As 16,2 g KCℓO3 alles ontbind

      n = m/M
          = 16,2                                         use ratio 
            122,5                                      gebruiksverhouding
          = 0,132 mol    KCℓO3                     0,132 KCℓ
                                                     m=nxM
                                                        = 0,132 x 74,5
                                                        = 9,83g 

      % Purity = Actual yield x 100 %Suiwerheid = ware opbrengs x100
                  Max yield                maks moontlike opbrengs

                           = 7,2         x 100
                             9,83 

                           = 73,25 %                                            (5)
                                                                                 [15]

                                              13

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                                                                          Physical Sciences / Fisiese Wetenskappe
                                      MEMORANDUM                         (Paper / Vraestel 2)   GRADE / GRAAD 11

Taxonomy Grid / Taksonomietabel
                       Recall / Onthou            Comprehension /               Analysis /                  Evaluation /
                                                     Verstaan                   Analiseer                    Evalueer
                      Q no: /       Mark /        Q no: /  Mark /           Q no: /   Mark /             Q no: /    Mark /
                      Vr nr:        Punt          Vr nr:    Punt            Vr nr:     Punt              Vr nr:      Punt
                       1.1            2            1.3        2              1.7         2                8.1          4

                         1.2             2           1.4             2        1.10            2            9.2             7
                         2.1             2           1.5             2         3.3            4            9.3             3
                         3.1             4           1.6             2         3.5            2
                         4.1             1           1.8             2         4.4            2
                       5.3.1             2           1.9             2         4.5            2
                       5.3.2             2           2.2             4         5.1            5
                       5.3.3             2           2.3             6         5.2            2
                         5.4             2           3.2             2         6.4            2
                         6.1             3           3.4             4         6.5            2
                       7.4.2             2           4.2             2         6.7            5
                                                     4.3             2         7.2            3
                                                     6.2             6        7.4.3           6
                                                     6.3             2         8.3            3
                                                     6.6             2         8.4            4
                                                     7.1             2         9.1            6
                                                     7.3             6
                                                    7.4.1            2
                                                    7.4.4            4
                                                     8.2             3

 Total mark /
                        16%         24            39,33%        59          34,67%       52              9,3 %            14
 Totale punte
Total / Totaal %
                          P1&2: 15%               P1:35%/P2:40%              P1:40%/P2:35%                   P1&2: 10%
    / 100%

Correct application of Bloom’s / Barrett’s Taxonomy: /
Korrekte toepassing van Bloom / Barrett se taksonomie:

Level 1:   Recall of information (what? which? when? list; label; name; define; give; describe)
Vlak 1:    Oproep van inligting (wat? watter? wanneer? lys; benoem; definieer; voorsien; beskryf)
Level 2:   Understanding and using information (summarise; classify; apply rules; discuss)
           Applying information (distinguish; specify; compare; design; explain; investigate; interpret; calculate; give your input)
Vlak 2:    Verstaan en gebruik inligting (som op klassifiseer); pas reëls toe; bespreek)
Level 3:   Analysis of information (classify; explain; identify; interpret; compare; give reasons; prove; give causes and effects)
Vlak 3:    Toepassing van inligting (onderskei; spesifiseer; vergelyk; ontwerp; verduidelik; ondersoek; interreteer; bereken; gee jou
           opinie)
Level 4:   Synthesize information (summarize; construct; argue; create; relate; design; formulate)
           Evaluate information (judge; assess; evaluate; choose; support; compare; estimate)
Vlak 4:    Sintetiseer inligting (som op; konstrueer; argumenteer; skep; formuleer)
           Evalueer inligting (beoordeel; assesseer; evalueer; kies; ondersteun; vergelyk; skat)

                                                                  14

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## Collection

- [Grade 11 past papers](https://www.hlayiso.com/grade-11/past-papers)

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