# Physical Sciences Grade 11 P2 2016 Memo

> Grade 11 Physical Sciences Paper 2 memorandum from the National Exemplar November Exam 2016, 10 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physical-sciences-p2-grade-11-nov-2016-memo-eng-afr/c76a3a49-f2df-48f4-997a-23de3158d918
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 11
- Document type: Memorandum
- Year: 2016
- Exam period: National Exemplar November Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 10

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## About this document

Physical Sciences · Grade 11 · National Exemplar November Exam · 2016. Memorandum, 10 pages. Read online or download the PDF.

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### Page 1

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                                            -

                                       NATIONAL
                                  SENIOR CERTIFICATE/
                                      NASIONALE
                                  SENIOR SERTIFIKAAT

                                        GRADE/GRAAD 11

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

                                        NOVEMBER 2016

                                        MEMORANDUM

MARKS/PUNTE: 150

                        This memorandum consists of 10 pages.
                      Hierdie memorandum bestaan uit 10 bladsye.

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Physical Sciences P2/Fisiese Wetenskappe/V2       2                       DBE/November 2016
                             CAPS/KABV – Grade/Graad 11 – Memorandum

QUESTION 1/VRAAG 1

1.1       B                                                                                      (2)

1.2       D                                                                                      (2)

1.3       D                                                                                      (2)

1.4       C                                                                                      (2)

1.5       A                                                                                      (2)

1.6       A                                                                                      (2)

1.7       B                                                                                      (2)

1.8       A                                                                                      (2)

1.9       D                                                                                      (2)

1.10      C                                                                                      (2)
                                                                                                   [20]

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Physical Sciences P2/Fisiese Wetenskappe/V2       3                          DBE/November 2016
                             CAPS/KABV – Grade/Graad 11 – Memorandum

QUESTION 2/VRAAG 2

2.1       Electronegativity is 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                                     Marking Criteria/Nasienkriteria
                                      • O atom shown with 8 electrons around it.
                                          O-atoom met 8 elektrone rondom getoon
                                        • Two electron pairs on O atoms shared
                                          with two F atoms as shown.
                                          Twee elektronpare op O-atome gedeel
                                          met twee F-atome, soos getoon
                                                                                                      (2)

2.3       ΔEN (O – F) = 4 – 3,5 = 0,5
          0 < ΔEN < 1, the bond is weakly polar/die binding is swak polêr                            (2)

2.4       • The bonds in both molecules are polar due to the difference in
            electronegativity between O and F and C and O.
          • The shape of the OF 2 molecule is angular and because the charge
            distribution is asymmetrical around the central atom the molecule is polar.
          • The shape of the CO 2 molecule is linear and because the charge
            distribution is symmetrical around the central atom the molecule is non-
            polar.
          • Die bindings in albei molekules is polêr as gevolg van die verskil in
            elektronegatiwiteit tussen O en F en tussen C en O.
          • Die vorm van die OF 2 -molekule is hoekig en omdat die ladings
            asimmetries versprei is rondom die sentrale atoom is die molekuul polêr
          • Die vorm van die CO 2 -molekuul is lineêr en omdat die ladings simmetries
            versprei is rondom die sentrale atoom is die molekuul nie-polêr                           (4)

2.5.1     X = bond energy 
          Y = bond length 
          X = bindingsenergie
          Y = bindingslengte                                                                          (2)

2.5.2     The energy needed to break one mole of its molecules into separate
          atoms.
          Die energie benodig om een mol van sy molekules in aparte atome op
          te breek                                                                                    (2)

2.5.3     The higher the bond order, the shorter the bond length, the stronger the
          bond and the bond energy increases.
          Hoe hoër die bindingsorde, hoe korter is die bindingslengte, hoe sterker die
          binding en die bindingsenergie neem toe                                                     (3)
                                                                                                      [17]

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Physical Sciences P2/Fisiese Wetenskappe/V2       4                       DBE/November 2016
                             CAPS/KABV – Grade/Graad 11 – Memorandum

QUESTION 3/VRAAG 3

3.1       The temperature at which the vapour pressure of a substance equals
          atmospheric pressure.
          Die temperatuur waarteen die dampdruk van 'n stof aan die atmosferiese
          druk gelyk is                                                                            (2)

3.2       What is the relationship between intermolecular forces and boiling point?
          Wat is die verband tussen intermolekulêre kragte en kookpunt?                            (2)

3.3       Glycerine,It has the highest boiling point.
          Glisirien, dit het die hoogste kookpunt                                                  (2)

3.4       No, boiling point is only affected by the atmospheric pressure.
          Nee, kookpunt word slegs deur die atmosferiese druk beïnvloed.                           (2)

3.5       Avoid direct heating with open flame
          Work in a well-ventilated room/use a fume cupboard 
          Vermy dit om naby 'n oop vlam te werk
          Werk in 'n goed geventileerde vertrek/gebruik 'n dampkas                                 (2)

3.6       Nail polish remover, lowest boiling point, weakest intermolecular forces,
          less energy is required to overcome intermolecular forces and can easily
          change to vapour.
          Naellakverwyderaar, laagste kookpunt, swakste intermolekulêre kragte,
          minder energie is nodig om die kragte te oorkom en kan maklik in 'n damp
          verander.                                                                                (3)

3.7       Sunflower oil has a large molecular mass.
          Sonneblomolie het 'n grote molekulêre massa                                              (1)
                                                                                                   [14]

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Physical Sciences P2/Fisiese Wetenskappe/V2       5                                             DBE/November 2016
                             CAPS/KABV – Grade/Graad 11 – Memorandum

QUESTION 4/VRAAG 4

4.1       When the temperature of a gas increases, the average speed of the particles
          also increases. The molecules collide with the walls of the container more
          often with greater impact. These collisions will push back the walls, so that
          the gas occupies a greater volume than it did at the start.
          Wanneer die temperatuur van 'n gas toeneem, sal die gemiddelde spoed van
          die deeltjies ook toeneem. Die molekules bots meer dikwels en heftiger met
          die kante van houer. Hierdie botsings sal die kante van die houer wegdruk
          sodat die gas 'n groter volume beset as voorheen.
          OR/OF
          When the temperature of a gas decreases, the average speed of the
          particles also decreases. Number of collisions decreases and the volume
          decreases
          Wanneer die temperatuur van 'n gas afneem, sal die gemiddelde spoed van
          die deeltjies ook afneem. Die aantal botsings neem af en die volume neem af.                                   (3)

4.1.2
                                      Graph of volume against temperature
                                     Grafiek van volume teenoor temperatuur
                           160
                           140
                           120

            Volume (cm3)
                           100
                           80
                           60
                           40
                           20
                            0
                                 0        20          40         60           80        100        120
                                                           Temperature (˚C)

                                                   Marking Criteria/Nasienkriteria
                                               •   Labelling x-axis with correct unit
                                                   Benoem x-as met korrekte eenheid
                                               •   Labelling y-axis with correct unit
                                                   Benoem y-as met korrekte eenheid
                                               •   Plotting points 
                                                   Stip van punte
                                               •   Shape of the graph
                                                   Vorm van die grafiek
                                                                                                                         (4)

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Physical Sciences P2/Fisiese Wetenskappe/V2       6                          DBE/November 2016
                             CAPS/KABV – Grade/Graad 11 – Memorandum

4.1.3                              Graph of volume against temperature
and/en                            Grafiek van volume teenoor temperatuur
                                                     180
4.1.6                                                          P
                                                     160
                                                     140
                                                     120

                Volume (cm3)
                                                     100
                                                      80
                                                      60
                                                      40
                                                      20
                                                       0
                    -300       -200        -100            0           100           200
                                            Temperature (˚C)
          x-intercept = -273 ºC 
          x-afsnit = -273 ºC                                                                         (1)

4.1.4     It is the absolute zero 0 K (-273ºC), which is the coldest possible
          temperature at which the particles will have no kinetic energy.
          Dit is die absolute nulpunt 0 K (-273 ºC) wat die koudste moontlike
          temperatuur is waarby die deeltjie geen kinetiese energie sal hê nie                       (2)

4.1.5     V1 V2
              =     
          T1 T2
          155     V
                = 2 
          373 393
          V 2 = 163,31 cm3                                                                          (3)

4.1.6     straight line above graph (steeper gradient) drawn in QUESTION 4.1.2
          reguitlyn bo grafiek (groter helling) in VRAAG 4.1.2 geteken                               (2)

4.2.1     pV = nRT
          (96 x 103)(0,32 x 10-3)= n(8,31)( 300)
          n = 0,012 mol
              m
          n=
              M
                   0,77
          0,012 =       
                    M
          M = 64,17 g∙mol-1                                                                         (5)

4.2.2     SO 2                                                                                      (1)
                                                                                                     [21]

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Physical Sciences P2/Fisiese Wetenskappe/V2       7                          DBE/November 2016
                             CAPS/KABV – Grade/Graad 11 – Memorandum

QUESTION 5/VRAAG 5

5.1       Amount of solute per litre of solution.
          Hoeveelheid oplosmiddel per liter oplossing                                                 (2)

5.2            m
          c=        
              MV
                  8
          c=              
              (156 )(0,5)
          c = 0,10 mol.dm-3                                                                          (3)

5.3.1     Empirical formula is the simplest whole number ratio between the elements in
          a compound.
          Empiriese formule is die eenvoudigste heelgetalverhouding tussen die
          elemente in 'n verbinding                                                                   (2)

5.3.2       Element          g                    m             Simplest ratio/
                                             n=                 Eenvoudigste
                           100g                   M
                                                                 verhouding
               K          26,58         26,58/39 = 0,68          1 x 2 = 2
               Cr         35,35         35,35/52 = 0,68          1 x 2 = 2
               O          38,07         38,07/16 = 2,38         3,5 x 2 = 7

          Empirical formula/Empiriese formule= K 2 Cr 2 O 7                                          (7)
                                                                                                      [14]

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Physical Sciences P2/Fisiese Wetenskappe/V2       8                       DBE/November 2016
                             CAPS/KABV – Grade/Graad 11 – Memorandum

QUESTION 6/VRAAG 6

6.1       Limiting reagent in a reaction is the reactant that is consumed completely.
          It determines the maximum amount of product that is made.
          Beperkte reagens in 'n reaksie is die reagens wat volledig opgebruik word.
          Dit bepaal die maksimum hoeveelheid produk wat gemaak word.                              (2)

6.2       n (CH 3 COOH) = cV
                         = (0,2)(0,1)
                         = 0,02 mol
                         m
          n(NaHCO 3 ) =
                         M
              10
          n=      
              84
          = 0, 12 mol
          Mole ratio/Molverhouding 1 : 1
          0,02 mol CH 3 COOH reacts with/reageer met 0,02 mol NaHCO 3 
          CH 3 COOH is the limiting reagent./is die beperkte reagens                              (6)

6.3       Mass of NaHCO 3 in excess:/Massa NaHCO 3 in oormaat:
          m = nM 
          m = (0,12 – 0,2)(84) 
          m = 8,4 g                                                                               (3)

6.4       0,02 mol CH 3 COOH → 0,02 mol CO 2 
          V = nV m 
            = (0,02)(22,4)
            = 0,448 dm3                                                                           (4)
                                                                                                   [15]

QUESTION 7/VRAAG 7

7.1       Exothermic, energy is given out.
          Eksotermies, energie word afgegee.                                                       (2)

7.2       Energy is needed to break the bond between atoms/ions in molecules.
          Energie is nodig om die bindings tussen atome/ione te breek                              (2)

7.3       Produces a greenhouse gas and which can contribute to global warming.
          Verwaardig 'n kweekhuisgas wat tot aardverwarming kan bydra.                             (2)
                                                                                                   [6]

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Physical Sciences P2/Fisiese Wetenskappe/V2       9                          DBE/November 2016
                             CAPS/KABV – Grade/Graad 11 – Memorandum

QUESTION 8/VRAAG 8

8.1       Acid is a proton donor         
          Suur is 'n protonskenker                                                                    (2)

8.2.1     conjugate pairs/gekonjugeerde pare

                  H 2 PO 4 -    +    NH 3     →      HPO 4 2-        +    NH 4 +
                    a1               b1             b1                  a 2
                                                                                                      (4)

8.2.2     A substance which can act as an acid or a base.
          'n Stof wat as 'n suur of 'n basis kan optree                                               (1)

8.2.3     H 2 PO 4 -/HPO 4 2-                                                                       (2)

8.3.1     Na 2 CO 3 + 2HCℓ → NaCℓ + H 2 O + CO 2     balancing/balansering                           (1)

8.3.2     n = cV
            = (0,2)(0,1) 
            = 0,02 mol                                                                               (3)

8.3.3     n Mg(OH)2 = (0,02)(0,1) 
                       = 0,002 mol

          n HCℓ(excess) = 2(0,002) = 0,004 mol

          n HCℓ react with Na 2 CO 3 = 0,02 – 0,004 = 0,016 mol

          n Na2CO3 = 0,008 mol

          m Na2CO3 = nM = (0,008)(106) 
                      = 0,848 g

                      0,848
          %purity =         × 100  = 8,48%
                        10                                                                            (8)
                                                                                                      [21]

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Physical Sciences P2/Fisiese Wetenskappe/V2     10                     DBE/November 2016
                             CAPS/KABV – Grade/Graad 11 – Memorandum

QUESTION 9/VRAAG 9

9.1       Oxidation is the increase in oxidation numbers 
          Oksidasie is die toename in oksidasiegetalle                                     (2)

9.2.1     Cu(s)                                                                          (2)

9.2.2     Silver ion/Silwerioon                                                          (2)

9.2.3     Cu → Cu2+ + 2e-                                                                (2)

9.3.4     Cu → Cu2+ + 2e-
          Ag+ + e-→ Ag 
          Cu + 2Ag+ → Cu2+ + 2Ag balancing/balansering                                  (4)

9.3       Cu + 2AgNO 3 → Cu(NO 3 ) 2 + 2Ag
                +1                      0
            +
          Ag oxidation number changes from +1 to 0, gaining electrons, reduction. 
          Ag+ se oksidasiegetal verander van +1 na 0, wen elektrone, reduksie              (2)
                                                                                           [14]

QUESTION 10/VRAAG 10

10.1      Open cast mine/Oopgroefmyn   
          Underground mine/Ondergrondse myn                                               (2)

10.2      Create job opportunities/Local communities benefit from infrastructure /
          Provides opportunities for other industries
          Skep werksgeleenthede/Plaaslike gemeenskappe verkry voordeel ten opsigte
          van infrastrukture
          Verskaf geleenthede aan ander nywerhede                                          (2)

10.3      Changing landscape
          Air pollution
          Resource consumption                 any two
          Water poisoning
          Veranderende landskap
          Lugbesoedeling
          Hulpbronverbruik                       enige twee
          Watervergiftiging                                                                (2)

10.4      It is too soft 
          Dit is te sag                                                                    (1)

10.5      Used in computer circuits /Used as a heat protector in space suits/any other
          relevant answer
          Gebruik in rekenaarstroombane/Gebruik as 'n hittebeskermer in ruimtepakke/
          enige ander relevante antwoord                                                   (1)
                                                                                           [8]

                                                                       TOTAL/TOTAAL:       150

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