# Physical Sciences Grade 11 P2 Gauteng June 2019 Memo (Afr)

> Grade 11 Physical Sciences Paper 2 memorandum from the Gauteng June 2019 (Afrikaans), 14 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/gr11-phy-p2-english-and-afrikaans-june-2019-possible-answers/0c1f524b-f055-4ce8-8969-8905ff39f1fa
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 11
- Language: Afrikaans
- Document type: Memorandum
- Year: 2019
- Exam period: Gauteng June
- Paper: 2
- Curriculum: CAPS
- Pages: 14

Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 11](https://www.hlayiso.com/library/grade-11) › [Physical Sciences](https://www.hlayiso.com/library/grade-11/physical-sciences) › [Physical Sciences Grade 11 P2 Gauteng June 2019 Memo (Afr)](https://www.hlayiso.com/library/gr11-phy-p2-english-and-afrikaans-june-2019-possible-answers/0c1f524b-f055-4ce8-8969-8905ff39f1fa)

## About this document

Physical Sciences · Grade 11 · Gauteng June · 2019 · Afrikaans. Memorandum, 14 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Matching paper and memo

- [Gr11 Phy P2 (Afrikaans) June 2019 Question Paper](https://www.hlayiso.com/library/gr11-phy-p2-afrikaans-june-2019-question-paper/67aec3a4-b25d-4e19-8631-45a2fa617cde) — QP · 2019 · Afr

## Document text

Text extracted from the PDF. Layout, diagrams and tables may not survive extraction.

### Page 1

Downloaded from hlayiso.com

              GAUTENG DEPARTMENT OF EDUCATION /
           GAUTENGSE DEPARTEMENT VAN ONDERWYS
      PROVINCIAL EXAMINATION / PROVINSIALE EKSAMEN
                            2019
                      GRADE / GRAAD 11

                      PHYSICAL SCIENCES
                     FISIESE WETENSKAPPE

                     PAPER 2 / VRAESTEL 2

                     MARKING GUIDELINES /
                       NASIENRIGLYNE

14 pages / bladsye

                              1

### Page 2

Downloaded from hlayiso.com
                                         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 (PAPER 2) /
                                          FISIESE WETENSKAPPE (VRAESTEL 2)

                    MARKING GUIDELINES / NASIENRIGLYNE

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

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

                                    2

### Page 3

Downloaded from hlayiso.com
                                                    Physical Sciences / Fisiese Wetenskappe
                            MEMORANDUM             (Paper / Vraestel 2)   GRADE / GRAAD 11

QUESTION 2 / VRAAG 2:
2.1   The bond energy of a compound is the energy needed to break one mole of its
      molecules into separate atoms. 
      OR
      The energy needed to break the bonds in a molecule.                                 (2)
      Die energie benodig om een mol van ŉ substans op te breek in atome. 
      OF
      Die energie benodig om ŉ binding te breek. 
2.2   -426 kJ.mol-1                                                                        (1)

2.3   Bond length is the average distance between the nuclei of two bonded atoms.
      Bindingslengte is die afstand tussen die kerne van die twee atome wat gebind
      is.                                                                                 (2)
2.4   74 x 10-12 m                                                                         (1)
2.5   At point 3 the force of attraction between the nucleus of the one atom and the
      electrons of the other atom balances out with the force of repulsion between the
      two nuclei, therefore making it more stable  (Potential energy is at the lowest).
      At point 4 the repulsion force between the nuclei is much stronger  than the
      attraction force between the nucleus of the one atom and the electrons of the other
      atom.  (Increased potential energy).                                                 (4)

      By punt 3 is die aantrekkingskrag tussen die kern van die een atoom en die
      elektrone van die ander atoom in balans met die afstotingskragte tussen die twee
      kerne.  Dus is die binding meer stabiel.  (Potensiële energie is op sy laagste).
      By punt 4 is die afstotingskragte tussen die twee kerne van die twee atome baie
      groter as die aantrekkingskragte tussen die kern van die een atoom en die
      elektrone van die ander atoom.  (Potensiële energie is verhoog).
                                                                                            [10]

                                               3

### Page 4

Downloaded from hlayiso.com
                                                    Physical Sciences / Fisiese Wetenskappe
                              MEMORANDUM           (Paper / Vraestel 2)   GRADE / GRAAD 11

QUESTION 3 / VRAAG 3:
3.1   Electronegativity is a measure of the tendency of an atom in a molecule to attract
      bonding electrons. 
      OR
      A measure of an atom’s attractive force on bonding electrons to form a
      molecule.                                                                          (2)
      Elektronegatiwiteit is ŉ aanduiding van die geneigdheid van ŉ atoom om die
      bindingselektrone van ŉ molekule aan te trek. 
      OF
      ŉ Aanduiding van die atoom se aantrekkingskrag op die verbindingselektrone
      van ŉ molekule. 
3.2   Aℓ EN = 1,5
                          
      S    EN = 2,5
       S has a higher EN and will thus form the negative ion 
       S het die hoogste EN en sal dus die negatiewe ioon vorm.                           (2)
3.3.1 O    EN = 3,5
                          
      Mg EN = 1,2
       EN = 2,3  ionic bond 
                       ioonbinding                                                        (2)
3.3.2 Cℓ    EN = 3,0
                          
      H     EN = 2.1
       EN = 0,9  polar covalent
                       polêr kovalent                                                     (2)
3.3.3 P     EN = 2,1
                          
      H     EN = 2.1
       EN = 0  non-polar covalent / pure covalent 
                   nie-polêr kovalent / suiwer kovalent                                   (2)

                                               4

### Page 5

Downloaded from hlayiso.com
                                                     Physical Sciences / Fisiese Wetenskappe
                                MEMORANDUM          (Paper / Vraestel 2)   GRADE / GRAAD 11

3.4     Valence electrons are the electrons in the highest energy level of an atom. 
        OR
        The electrons in the outer most orbital of the atom.                           (2)

        Valenselektrone is die elektrone wat in die hoogste energievlak van die atoom
        voorkom. 
        OF
        Die elektrone in die buitenste orbitaal van die atoom.
3.5.1
                                                                                         (2)
                

3.5.2
                                                                                         (2)
               

3.5.3
                                                                                         (2)
                

3.5.4

                              OR / OF                                                (2)

3.6     THREE / 3                                                                      (2)
        DRIE
3.7.1 Tetrahedral                                                                      (2)
        Tetraëdries
3.7.2 Trigonal planar                                                                  (2)
        Trigonaal planêr
3.7.3 Bent / Angular                                                                   (2)
        Hoekig / gebuig

                                                5

### Page 6

Downloaded from hlayiso.com
                                                    Physical Sciences / Fisiese Wetenskappe
                             MEMORANDUM            (Paper / Vraestel 2)   GRADE / GRAAD 11

3.8.1   Dative covalent bond / Coordinate bond                                         (2)
        Datief kovalente binding / koördinatiewe binding
3.8.2                                                                                 (3)

                                                                                         [33]

QUESTION 4 / VRAAG 4:
4.1.1   The temperature at which the vapour pressure of a substance equals
        atmospheric pressure.                                                          (2)
        Kookpunt is die temperatuur waneer die dampdruk van ŉ vloeistof gelyk is aan
        die atmosferiese druk.
4.1.2   There are strong hydrogen bonds between the water molecules. 
        There are weaker dipole – dipole forces between CH3Cℓ molecules. 
        Hydrogen bonds are stronger than dipole – dipole forces
        Therefore more energy is needed to overcome these forces and the boiling point
        is higher.                                                                      (3)
        Daar is sterk waterstofbindings tussen die molekules van die water  en swak
        dipool-dipool kragte tussen die CH3Cℓ molekules .
        Dus is meer energie nodig om die waterstofbindings tussen die watermolekules
        te breek. 

                                               6

### Page 7

Downloaded from hlayiso.com
                                                    Physical Sciences / Fisiese Wetenskappe
                              MEMORANDUM           (Paper / Vraestel 2)   GRADE / GRAAD 11

4.1.3   Benzene and chloroform have weak London forces between their molecules. 
        Benzene has a larger surface area therefore stronger intermolecular forces and
        thus a higher boiling point. 
        OR
        Chloroform has a smaller surface area therefore weaker London forces and a
        lower boiling pont.                                                              (2)

        Daar is swak Londonkragte ( geïnduseerde dipoolkragte) tussen die molekules
        van die chloroform en die bensien  maar die molekulêre massa van die
        bensien is baie hoër as die van die chloroform, dus is bensien se kookpunt
        hoër.
        OF
        Chloroform het ŉ kleiner oppervlakarea en dus het dit swakker Londonkragte en
        ŉ laer kookpunt.
4.2.1   MgCl2                                                                          (1)
4.2.2   Cl2  and / en CO2                                                              (2)
4.2.3   H2O  and / en HF                                                               (2)
4.2.4   HCl                                                                            (1)
4.2.5   MgCl2                                                                          (1)
4.3.1   Non-polar                                                                      (2)
        Nie-polêr
4.3.2   Polar                                                                          (2)
        Polêr
4.3.3   Non-polar                                                                      (2)
        Nie-polêr

                                               7

### Page 8

Downloaded from hlayiso.com
                                                     Physical Sciences / Fisiese Wetenskappe
                             MEMORANDUM             (Paper / Vraestel 2)   GRADE / GRAAD 11

4.4.1   Vapour pressure is the pressure exerted by a vapour at equilibrium with its liquid   (2)
        in a closed system. 
        Dampdruk is die druk van gasmolekules van ŉ stof wat in kontak is met sy
        vloeistof, in ewewig, in ŉ geslote sisteem. 
4.4.2   PH3 has a higher vapour pressure. 
        PH3 has London forces between its molecules. 
        NH3 has hydrogen bonds between its molecules.       
        Hydrogen bonds are stronger than London forces, thus less energy is required
        to weaken the London forces in PH3                                                  (4)
        PH3 het ŉ hoër dampdruk. 
        PH3 het Londonkragte tussen die molekules. 
        NH3 het waterstofbindings tussen die molekules. 
        Waterstofbindings is sterker as Londonkragte, dus word minder energie benodig
        om die swak Londonkragte in PH3 te breek.
                                                                                             [26]

                                                8

### Page 9

Downloaded from hlayiso.com
                                                     Physical Sciences / Fisiese Wetenskappe
                             MEMORANDUM             (Paper / Vraestel 2)   GRADE / GRAAD 11

QUESTION 5 / VRAAG 5:
5.1     Oil is a non-polar molecule and water is a polar molecule. 
        Oil is less dense than water.                                                   (2)
        Olie is ŉ nie-polêre molekule waar water polêr is. 
        Olie is minder dig as water. 
5.2     Density is the number of molecules per unit volume.                            (2)
        Digtheid: die aantal molekules per volume eenheid. 
5.3.1   London forces / dispersion forces.                                             (2)
        Londonkragte / dispersiekragte 
5.3.2   Electrostatic (Ionic bond)                                                     (2)
        Ioonbindings (Elektrostatiese aantrekkingskragte)
5.3.3   Water                                                                          (2)
5.3.4   Water is polar and has hydrogen bonds (dipole – dipole forces) between
        molecules. 
        KMnO2 is an ionic solid. 
        Water and KMnO4 therefore have intermolecular forces of similar strength , so
        KMnO4 can dissociate in the water, turning it purple.                           (4)

        Water is polêr en het waterstofbindings (sterk dipool-dipoolkragte) tussen die
        molekules. 
        KMnO2 is ŉ ioniese vastestof. 
        Water en KMnO4 het dus dieselfde sterkte intermolekulêre kragte , dus kan
        KMnO4 in water dissosieer om dit pers te kleur. 
                                                                                         [14]

                                                9

### Page 10

Downloaded from hlayiso.com
                                                       Physical Sciences / Fisiese Wetenskappe
                             MEMORANDUM               (Paper / Vraestel 2)   GRADE / GRAAD 11

QUESTION 6 / VRAAG 6:
6.1.1   ANY THREE (3) OF THE FOLLOWING: 
            Particles are all identical.
            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.
            Collision is perfectly elastic, with no loss of total energy of the molecules.
            Obey gas laws in all temperatures and pressures.                                 (3)

        ENIGE DRIE (3) VAN DIE VOLGENDE: 
           Alle molekules / deeltjies is identies.
           Deeltjies is konstant in beweging in alle rigtings.
           Deeltjies dra nie by tot die volume van die gas nie.
           Daar bestaan geen kragte tussen die deeltjies onderling, of tussen die
            deeltjies en die wande van die houer nie behalwe gedurende botsings.
           Botsings tussen deeltjies is volkome elasties met geen verlies aan die
            totale energie van die molekules nie.
           Gehoorsaam gaswette in alle temperature en druk.
6.1.2   High Temperatures 
        Low pressures                                                                        (2)
        Hoë Temperature
        Lae druk

                                                10

### Page 11

Downloaded from hlayiso.com
                                                      Physical Sciences / Fisiese Wetenskappe
                             MEMORANDUM              (Paper / Vraestel 2)   GRADE / GRAAD 11

6.2.1     Boyle’s Law apparatus                                                           (1)
          Apparaat vir Boyle se Wet
6.2.2     The pressure of certain mass of an enclosed gas is inversely proportional to
          the volume it occupies at a constant temperature.                              (2)
          Die druk van ŉ sekere massa van ŉ ingeslote gas is omgekeerd eweredig aan
          die volume, by ŉ konstante temperatuur.
6.2.3.1   Pressure                                                                        (1)
          Druk
6.2.3.2   Volume                                                                          (1)
          Volume
6.2.3.3   Mass 
          Temperature                                                                     (2)
          Massa
          Temperatuur
6.2.4     120 kPa                                                                        (2)
6.2.5     The shape would be the same  but it would be higher up  on the graph
          (Higher volume at higher pressure).                                              (2)
          Die vorm van die grafiek sal dieselfde bly, maar dit sal hoër op beweeg / hoër
          op die grafiek lê (Hoër volume teen ŉ hoër druk). 
6.3.1
                        
              

          V2 = 4,84 dm3 / ℓ 
                                                                                           (5)
6.3.2     t = T – 273
           = 282 – 273 
           = 9o C                                                                         (2)
6.3.3     His lungs will expand / might burst. 
          As the pressure decreases and the temperature increases the volume will
          increase, as V ∝ 1/P and V ∝ T.                                                 (3)

          Sy longe sal uitsit en mag dalk selfs bars  aangesien die volume van ŉ
          ingeslote gas verhoog soos die druk verminder. Die volume sal dus drasties
          toeneem aangesien V ∝ 1/P en V ∝ T. 
                                                                                           [26]

                                                11

### Page 12

Downloaded from hlayiso.com
                                                       Physical Sciences / Fisiese Wetenskappe
                               MEMORANDUM             (Paper / Vraestel 2)   GRADE / GRAAD 11

QUESTION 7 / VRAAG 7:
7.1.1   6H2O  + Al2S3  3H2S  + 2Al(OH)3                                              (3)
7.1.2   M(Al(OH)3) = 27 + 3(16 +1)
                    = 78 g.mol-1

             
        = 0,1282 … mol

                                     
                    = 0,0641 … mol
        m = nM
         = (0,0641 …)[1(27)+3(32)] 
         = 7,88 g                                                                       (4)

7.1.3
                                                                                         (3)
7.2.1   n(HCl) = cV 
              = (0,4)(0,2) 
              = 0,08 mol

                        

               
        = 0,2 mol
                                         HCl     : CaCO3
                                           2     :     1
                                     0,08 : 0,04
                                               OR / OF
                                          0,4 : 0,2 

        Therefore CaCO3 is in excess   CaCO3 is in oormaat                             (6)

                                                 12

### Page 13

Downloaded from hlayiso.com
                                                       Physical Sciences / Fisiese Wetenskappe
                               MEMORANDUM             (Paper / Vraestel 2)   GRADE / GRAAD 11

7.2.2   n(CaCl2) = 0,08 x ½ 
                 = 0,04 mol

                     
        m = nM
          = (0,04)(40 + 2(35,5)
          = 4,44 g                                                                      (3)
7.2.3                 actual                           reël                              (2)
        % yield                x 100 / % opbrengs            x 100
                    theoretical                      teoreties

                         
        = 90,09% 
                                                                                         [21]

                                                 13

### Page 14

Downloaded from hlayiso.com
                                                                              Physical Sciences / Fisiese Wetenskappe
                                           MEMORANDUM                        (Paper / Vraestel 2)   GRADE / GRAAD 11

Taxonomy Grid
                                   Recall               Comprehension                  Analysis                   Evaluation
                               Q no:     Mark           Q no:   Mark               Q no:     Mark              Q no:      Mark
                                1.1.      2             1.2. &    4                 1.4.       2
                                                         1.3.
                                                        1.5. –    6                  1.8           2
                                                          1.7
                                                        1.9 &     4
                                                         1.10
                                 2.1.          2         2.2.     1                 2.5.           4
                                 2.3.          2         2.4.     1
                                 3.1.          2         3.3.     6                 3.2.           2
                                 3.4.          2          3.5.           8          3.6.           2
                                3.8.1.         2          3.7.           6
                                                         3.8.2.          3
                                4.1.1.         2                                   4.1.2.          3           4.1.3.            2
                                4.4.1.         2          4.3.           6          4.2.           7
                                                                                   4.4.2           4
                                 5.2.          2          5.1.           2         5.3.4.          4
                                                          5.3.           6
                                6.1.1.         3         6.1.2.          2         6.2.4.          2           6.3.3.            3
                                6.2.1.         1         6.2.3.          4         6.2.5.          2
                                6.2.2.         2         6.3.2.          2         6.3.1           5
                                                          7.1.           2          7.2.           5            7.3.             3
                                                                                   7.4.2.          3           7.4.1.            6
                                                                                                               7.4.3.            2
   Total mark                            24                     63                        47                               16
   Total %/100%                  P1&2: 15%              P1:35%/P2:40%              P1:40%/P2:35%                   P1&2: 10%

Correct application of Bloom’s / Barrett’s Taxonomy:
Level 1: Recall of information (what? which? when? list ; label; name; define; give; describe)
Level 2: Understanding and using information (summarize; classify ; apply rules; discuss)
               Applying information (distinguish; specify; compare; design ;explain; investigate ;interpret; calculate; give your input)
Level 3:       Analysis of information (classify; explain; identify; interpret; compare; give reasons; prove; give causes and effects)
Level 4:       Synthesize information (summarize; construct; argue; create; relate; design; formulate)
               Evaluate information (judge; assess; evaluate; choose; support; compare; estimate)

                                                                        14

## Related documents

- [PHYS SCIENCES P2 GR11 QP NOV2019 Afrikaans](https://www.hlayiso.com/library/phys-sciences-p2-gr11-qp-nov2019-afrikaans/92935de9-5cb8-467d-83dd-c7a72d4cd8c3) — Physical Sciences · QP · 2019 · Afr
- [Gr11 Phy P2 (Afrikaans) June 2018 Question Paper](https://www.hlayiso.com/library/gr11-phy-p2-afrikaans-june-2018-question-paper/fd9f92c8-f4d1-4104-926b-48e832bea015) — Physical Sciences · QP · 2018 · Afr
- [Gr11 Phy P2 (Afrikaans) June 2017 Question Paper](https://www.hlayiso.com/library/gr11-phy-p2-afrikaans-june-2017-question-paper/14dc2ba3-3d16-418a-a9c0-5897316247be) — Physical Sciences · QP · 2017 · Afr
- [Gr11 Phy P2 (Afrikaans) June 2016 Question Paper](https://www.hlayiso.com/library/gr11-phy-p2-afrikaans-june-2016-question-paper/91bc2d21-0fb6-44a8-8053-f5fcb970d501) — Physical Sciences · QP · 2016 · Afr
- [Gr11 Phy P2 (AFR & ENG) November 2019 Possible Answers](https://www.hlayiso.com/library/gr11-phy-p2-afr-eng-november-2019-possible-answers/510ad17e-b55d-4bbe-86a4-8e89e1bb424d) — Physical Sciences · Memo · 2019 · Eng
- [PHY G11-P2-Jun-2019-and-Memo](https://www.hlayiso.com/library/phy-g11-p2-jun-2019-and-memo/f102fd1e-4a1e-4d83-b1d8-7760f4784c21) — Physical Sciences · Memo · 2019

## More from Grade 11 Physical Sciences

- [PHYSICAL SC P1 QP GR11 NOV 2025 AFRIK watermark](https://www.hlayiso.com/library/physical-sc-p1-qp-gr11-nov-2025-afrik-watermark/3757d93e-3b4a-4e27-9a6b-69591d990a97) — QP · 2025 · Afr
- [KZN Physical Sciences Grade 11 November 2025 P1 and Memo](https://www.hlayiso.com/library/kzn-physical-sciences-grade-11-november-2025-p1-and-memo/82b3d6a5-9e71-4efe-a8e0-f2a63d6aff58) — QP + Memo · 2025
- [KZN Physical Sciences Grade 11 November 2025 P2 and Memo](https://www.hlayiso.com/library/kzn-physical-sciences-grade-11-november-2025-p2-and-memo/f6f699c6-e985-44e2-9d36-d660697bd6fe) — QP + Memo · 2025
- [LP Physical Sciences Grade 11 June 2025 P2 and Memo](https://www.hlayiso.com/library/lp-physical-sciences-grade-11-june-2025-p2-and-memo/b76990f6-708a-47e3-a903-95cc3c58534d) — QP + Memo · 2025
- [MP Physical Sciences Grade 11 June 2025 P1 and Memo](https://www.hlayiso.com/library/mp-physical-sciences-grade-11-june-2025-p1-and-memo/4c5694d8-7cdc-4fd9-90b4-55bbb8969872) — QP + Memo · 2025
- [PHYS SCIENCES P1 QP GR11 NOV 2024 AFR D watermark](https://www.hlayiso.com/library/phys-sciences-p1-qp-gr11-nov-2024-afr-d-watermark/918f95ae-174f-4a62-9066-40676642795b) — QP · 2024 · Afr

## Collection

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

---

Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.
