# Physical Sciences Grade 12 Mpumalanga Mock 2016 Memo

> Grade 12 Physical Sciences memorandum from the Mpumalanga Mock Exam 2016, 17 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-12-physical-sciences-2016-mpumalanga-mock-exam-memorandum/c71f1969-a9b9-4cbc-8b1d-988efa95dfc7
- Updated: 2026-09-28
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2016
- Exam period: Mpumalanga Mock Exam
- Curriculum: CAPS
- Pages: 17

Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Physical Sciences](https://www.hlayiso.com/library/grade-12/physical-sciences) › [Physical Sciences Grade 12 Mpumalanga Mock 2016 Memo](https://www.hlayiso.com/library/grade-12-physical-sciences-2016-mpumalanga-mock-exam-memorandum/c71f1969-a9b9-4cbc-8b1d-988efa95dfc7)

## About this document

Physical Sciences · Grade 12 · Mpumalanga Mock Exam · 2016. Memorandum, 17 pages. Read online or download the PDF.

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

## Document text

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

### Page 1

GRADE 12

                PHYSICAL SCIENCES: CHEMISTRY (P2)
                           SEPTEMBER 2016
                            MEMORANDUM

MARKS: 150

TIME: 3 hours

                 This memoranda consists of 15 pages

### Page 2

Physical sciences P2                       2                          MDE/September 2016
                                        NSC- Memorandum

GENERAL GUIDELINES

1         CALCULATIONS

1.1       Marks will be awarded for: correct formula, correct substitution, and correct
          answer with unit.

1.2       NO marks will be awarded if an Incorrect or inappropriate formula is used,
          even though there may be relevant symbols and applicable substitutions.

1.3       When an error is made during substitution into a correct formula, a mark
          will be awarded for the correct formula and for the correct substitutions, but no
          further marks will be given.

1.4       If no formula is given, but all substitutions are correct, a candidate will
          forfeit one mark.

          Example:

          No K c expression, correct substitution
                 (2)2              2
          K C = (2)(1)3 ✓ = 2 ✓   (3)

1.5       Marks are only awarded for a formula if a calculation has been attempted i.e.
          substitution have been made or a numerical answer given.

1.6       Marks can only be allocated for substitutions when values are substituted into
          formulae and not when listed before a calculation starts.

1.7       All calculations, when not specified in the question, must be done to two
          decimal places.

1.8       If a final answer to a calculation is correct, full marks will not automatically be
          awarded. Marks will always ensure that the correct/appropriate formula is used
          and that workings, including substitutions, are correct.

2         UNITS

2.1       Candidates will only be penalised once for the repeated use of an incorrect unit
          within a question.

2.2       Units are only required in the final answer to a calculation.

    Copyrights reserved                                                        please turn over

### Page 3

Physical sciences P2                         3                         MDE/September 2016
                                          NSC- Memorandum

2.3       Marks are only awarded for an answer and not for a unit per se. Candidates
          will therefore forfeit the mark allocated for the answer in each of the following
          situations:

          • Correct answer + wrong unit
          • Wrong answer + correct unit
          • Correct answer + no unit

2.4       Separate compound units with a multiplication dot, not a full stop, for example,
          mol.dm-3. Accept mol.dm-3 (mol/dm3) for marking purpose.

3         If one answer or calculation is required, but two are given by the candidate,
          only the first one will be marked, irrespective of which one is correct. If two
          answers are required, only the first two will be marked, etc.

3.1       When a chemical FORMULA is asked, and the NAME is given as answer the
          candidate forfeits the marks. The same rule applies when the NAME is asked
          and the FORMULA is given.

3.2       When redox half- reactions are to be written, the correct arrow should be used.

          If the equation

          H 2 S → S + H+ +2e-   (2�2)

          Is the correct answer, the marks must be given as follows:

          H 2 S ⇌ S + H+ + 2e- (1�2)

          H 2 S ←S + 2H+ + 2e- (0�2)

          S + 2H+ + 2e- ← H 2 S (2�2)

          S + 2H+ + 2e-⇌ H 2 S (0�2)

3.4       One mark is forfeited when the charge of an ion is omitted per equation. (not
          for the charge on an electron)

3.6       The error carrying principle does not apply to chemical equations or half
          reactions. For example, if a learner writes the wrong oxidation/reduction half-
          reaction in the sub-question and carries the answer over to another sub-
          question (balancing of equations or calculation of Eº cell ) then the learner must
          not be credited for this substitution.

    Copyrights reserved                                                         please turn over

### Page 4

Physical sciences P2                          4                       MDE/September 2016
                                           NSC- Memorandum

3.7       In the structural formula of an organic molecule all hydrogen atoms must be
          shown. Marks must be deducted if hydrogen atoms are omitted.

          When a structural formula is asked, marks must be deducted if the learner
          writes the condensed formula.

3.8       When an IUPAC name is asked and candidate omits the hyphen (e.g. instead
          of pent-1-ene or 1-pentene the candidate writes pent 1 ene or 1 pentene),
          marks must be forfeited.

3.9       When a chemical reaction is asked, marks are awarded for correct reactants,
          correct products and correct balancing.

            If only a reactant(s) followed by an arrow, or only a product(s) preceded by an
          arrow, is/are written, marks may be awarded for the reactant(s) or products(s).
          if only a reactant(s) or only a product(s) are written, without an arrow, no marks
          are awarded for the reactants(s) or product(s).

           Examples: N 2 + 3H 2 ✓ → 2NH 3✓
                                       R              bal. ✓

           N 2 + H 2→ ✓                         1�
                                                  3

           → NH 3✓                              1�
                                                  3

           N2 + H2                              0�
                                                  3

           NH 3                                 0�
                                                  3
4         POSITIVE MARKING

4.1       Sub-question to sub-question:

          When a certain variable is calculated in one sub-question (e.g. 3.1) and needs
          to be substituted in another (3.2 or 3.3), e.g. if the answer for 3.1 is incorrect
          and is substituted correctly in 3.2 or 3.3 full marks are to be awarded for the
          subsequent sub-question.

4.2       A multi-step question: if the candidate has to calculate, for example, current
          in the first and gets it wrong due to a substitution error, the mark for the
          substitution and the final answer will be forfeited.

    Copyrights reserved                                                        please turn over

### Page 5

Physical sciences P2                     5                         MDE/September 2016
                                      NSC- Memorandum

5         NEGATIVE MARKING

           Normally an incorrect answer cannot be correctly motivated if on a conceptual
          mistake. If the candidate is therefore required to motivate in QUESTION 3.2
          the answer given to QUESTION 3.2. However, if the answer for e.g. 3.1 is
          based on a calculation, the motivation for the incorrect answer in 3.2 could be
          considered.

    Copyrights reserved                                                     please turn over

### Page 6

Physical sciences P2                  6                      MDE/September 2016
                                NSC- Memorandum

QUESTION 1 / VRAAG 1

1.1     C✓✓                                                                            (2)

1.2     B✓✓                                                                            (2)

1.3     C✓✓                                                                            (2)

1.4     B✓✓                                                                            (2)

1.5     D✓✓                                                                            (2)

1.6     C✓✓                                                                            (2)

1.7     A✓✓                                                                            (2)

1.8     D✓✓                                                                            (2)

1.9     B✓✓                                                                            (2)

1.10    A✓✓                                                                            (2)
                                                                                      [20]
QUESTION 2 / VRAAG 2

2.1.1 C ✓                                                                             (1)

2.1.2       O
            C   O       H✓
                                                                                       (1)
2.1.3 3-methyl✓propanal✓/3-metielpropanaal                                            (2)

2.1.4 E✓                                                                              (1)

2.1.5 D ✓ and F ✓                                                                     (2)

2.1.6 alkyne ✓                                                                        (1)

2.1.7 UNSATURATED /ONVERSADIG ✓
      There are multiple bonds✓ between the carbon atoms ✓/Daar is
      meervoudige bindings tussen die koolstofatome
      OR
      The carbon atom is not bonded to a maximum number of atoms / Die
       koolstofatoom is nie aan die maksimum aantal atome gebind nie                   (3)

2.2.1 Distinct/pleasant aroma smell ✓ /Aangename geur                                 (1)

2.2.2 Ethanol/Etanol✓✓                                                                (2)
                                                                                      [14]

 Copyrights reserved                                               please turn over

### Page 7

Physical sciences P2                     7                       MDE/September 2016
                                   NSC- Memorandum

QUESTION 3 / VRAAG 3

3.1     The temperature at which the vapour ✓ pressure of the substance is equal to
        the atmospheric pressure. ✓
        Die temperatuur waar die dampdruk gelyk is aan die omliggende atmosferiese
        druk.                                                                             (2)

3.2     The more branched an alkane , the lower the boiling point. ✓ ✓
        Hoe meer vertak ‘n alkaan is, hoe laer word die kookpunt.
        Criteria
        Independent and dependent variable has to be mentioned in the hypothesis.         (2)

3.3      Pentane has no branches/is a straight chain hydrocarbon. ✓
         It has the largest surface area. ✓
         The London forces/ Dispersion forces/induced dipole forces are the
         strongest✓
         More energy is needed to overcome the intermolecular forces✓, therefor it
         has the highest boiling point.
         Pentaan het geen vertakkings nie.
         Dit het die grootste kontakoppervlakte.
         Die Londonkragte/dispersiekragte/geïnduseerde dipoolkragte is die sterktste.
         Meer energie word benodig om die intermolekulêre kragte te oorkom.
         Die kookpunt is dus die hoogste.                                                 (4)

3.4     2,2-dimethylpropane✓ It has the lowest boiling point ✓/
        2,2-dimetielpropaan. Dit het die laagste kookpunt.                                (2)

3.5     2 C 5 H 12 + 8 O 2 ✓ → 5 CO 2 + 6 H 2 O ✓ (bal ✓)
        Accept multiples/Aanvaar veelvoude                                                 (3)
                                                                                          [13]

 Copyrights reserved                                                   please turn over

### Page 8

Physical sciences P2                          8                MDE/September 2016
                                        NSC- Memorandum

QUESTION 4 / VRAAG 4

4.1.1    Heat ✓/Hitte                                                                   (1)

4.1.2   Elimination✓ / Dehydrohalogenation/debromonation
        Eliminasie/Dehidrohalogenering                                                  (1)

4.1.3        H   H
         H   C C ✓H      H

             H  C C      C   H ✓
                H H      H
        Functional group ✓/Funksionele groep                                            (2)
        Whole structure correct ✓

4.1.4   Water ✓or dilute sodium hydroxide /Water of verdunde natriumhidroksied          (1)

4.1.5   Substitution / Hydrolysis /Substitusie/hidrolise ✓                              (1)

4.1.6   Elimination/ Dehydration /Dehidrasie ✓                                          (1)

4.2.1   Esterification/ Esterifikasie    ✓                                              (1)

4.2.2   Sulphuric acid / swawelsuur H 2 SO 4 ✓                                          (1)

 Copyrights reserved                                                 please turn over

### Page 9

Physical sciences P2                              9                  MDE/September 2016
                                            NSC- Memorandum

4.2.3    OPTION1/OPSIE 1
                    m
          n =
                    M        ✓
                    68
                =
                    74

                0,92 mol ✓                        1 : 1 ✓          = 0,92 mol ✓

                              m = n x M = 0,92 x 46✓ = 42,32 g ✓
                        42,32       100
         %:                     ×         = 70,53%
                         60           1

         OPTION 2/OPSIE 2

                    m
          n =
                    M✓
              68
            =
              74
            = 0,92 mol ✓

         n(C3 H6 O) 1
                     =
         n(C3 H6 O2 ) 1
                              1
         n(C3 H6 O)= 1✓(0,92)✓
         =0,92 mol

         m = n x M = 0,92 x 46✓ = 42,32 g

                        42,32         100
         %:                       ×         = 70,53% ✓
                         60            1
                                                                                             (6)

4.3.1   H               Cl
                ✓
            C       C
        H               H ✓                                                                  (2)
4.3.2   Addition ✓/Addisie                                                                    (1)
                                                                                             [18]

 Copyrights reserved                                                      please turn over

### Page 10

Physical sciences P2                         10                  MDE/September 2016
                                       NSC- Memorandum

QUESTION 5 / VRAAG 5

5.1      The change in the concentration/mass/mol/volume of the reactants / products
         ✓ per unit time.✓
         Die verandering in die konsentrasie van die reaktante/produkte per                (2)
         tydseenheid.

5.2     The concentration of the acid.✓/Die konsentrasie van die suur                      (1)
5.3.1    Change in moles = 0,08 - 0,1 = -0,02 mol ✓
                     Δm = Δn x M
                         = -0,02 x 65 ✓
                         = -1,3 g     ✓
                               mass of Zn used
         Rate of reaction =
                                     time
                         =     − 1,3       ✓
                                60
                        =      0,02 g.s – 1 ✓
        OR

        Change in moles = 0,1 – 0,08 = 0,02 mol ✓
                              moles of Zn used    0,02
         Rate of reaction =                    =−      ✓
                                     time          60
                          = 3,33 x 10-4mol.s-1✓
                          = (3,33 x 10-4 x 65✓) g.s-1
                          = 0,02 g.s-1 ✓
                                                                                           (5)

5.3.2   Less than/Kleiner ✓                                                                (1)

 Copyrights reserved                                                    please turn over

### Page 11

Physical sciences P2                          11                     MDE/September 2016
                                        NSC- Memorandum

5.4

                                    ✓
                                        ✓

              Number of particles
                                                   Higher peak at the same point/Hoër
                                    A              piek op dieselfde punt ✓✓

                                            B

                                                Kinetic energy
                                                                                              (2)

5.5      If we increase the concentration of the acid,
         the number of moles increases. ✓
         More effective collisions per second. ✓
         Higher reaction rate. ✓
         As die konsentrasie van die suur verhoog word,
         neem die aantal mol deeltjies toe.
         Meer effektiewe botsings per sekonde.
         Hoër reaksietempo.                                                                   (3)

                                                                                              [14]

 Copyrights reserved                                                       please turn over

### Page 12

Physical sciences P2                      12                          MDE/September 2016
                                    NSC- Memorandum

QUESTION 6 / VRAAG 6

6.1     CALCULATIONS USING NUMBER OF MOLES
        Mark allocation/Puntetoekenning
        • USING ratio / Gebruik die verhouding: 1:1:1:1 ✓
        • Equilibrium: n(CO 2 ) = 1.276 mol ;n(H 2 ) = 0,276 mol (initial - change) ✓
        • Equilibrium: n(CO) = 0,724 mol; n(H 2 O) = 0,724 mol (initial + change) ✓
        • Divide by volume = 2,0 dm3 ✓
        • Correct K c expression (formulae in square brackets) ✓
        • Correct substitution in K c expression ✓
        • Final answer/Finale antwoord: 1,49 ✓

                                 CO 2         H2         CO                 H2O

         Initial moles                  2          1           0,2             0

         Moles used/reacted        -0,724      -0,724     +0,724     ✓      +0,724

         Moles at equilibrium     1,276✓       0,276        0,924 ✓          0,724

         [..] at equilibrium       0,638       0, 138         0,462✓         0,362
                           n
         (mol∙dm-3 )     c=v

              [CO][H2 O]
        KC =              ✓
              [CO2 ][H2 ]
              (0,462)(0,362)
        KC =                 ✓
              (0,638)(0,138)
        K C = 1,49✓

 Copyrights reserved                                                       please turn over

### Page 13

Physical sciences P2                         13                        MDE/September 2016
                                       NSC- Memorandum

        CALCULATIONS USING CONCENTRATION

        Mark allocation/Puntetoekenning

        •     Divide by volume = 2,0 dm3 ✓
        •     USING ratio / Gebruik die verhouding: 1:1:1:1 ✓
        •     Equilibrium: [CO 2 ] = 1.276 mol/dm3 ;[H 2 ] = 0,276 mol/dm3 (initial - change✓
        •     Equilibrium: [CO] = 0,724 mol/dm3; [H 2 O] = 0,724 mol/dm3(initial+change)✓
        •     Correct K c expression (formulae in square brackets) ✓
        •     Correct substitution in K c expression ✓
        •     Final answer/Finale antwoord: 1,49 ✓

                                   CO 2          H2         CO                H2O

            Initial [ ]                   1           0,5         0,1           0✓

            [ ] used/reacted          −0,362     −0,362✓      +0,362 ✓        +0.362

            Equilibrium [ ]           0,638        0,138       0,462 ✓         0,362

              [CO][H2 O]
        KC =              ✓
              [CO2 ][H2 ]
              (0,462)(0,362)
        KC =                 ✓
              (0,638)(0,138)
        K C = 1,49✓                                                                              (7)

6.2    EXOTHERMIC ✓
       If the Kc value decreases, it means that the concentration of the products
       decreases. ✓ The reverse reaction is favoured. According to Le Chatelier, the
       endothermic reaction will be favoured when the temperature is increased. ✓
       The reverse reaction is endothermic. ✓The forward reaction is exothermic.

       EKSOTERMIES
       As die Kc waarde afneem, beteken dit dat die konsentrasie van die produkte
       afneem. Die terugwaartse reaksie word dus bevoordeel. Volgens Le Chatelier
       sal die endotermiese reaksie bevoordeel word as die temperatuur toeneem. Die
       terugwaartse reaksie is endotermies en dus is die voorwaartse reaksie
                                                                                                 (4)
       eksotermies.

 Copyrights reserved                                                          please turn over

### Page 14

Physical sciences P2                              14                MDE/September 2016
                                            NSC- Memorandum

6.3.1 DECREASES/AFNEEM                      ✓                                                (1)

6.3.2 NEGATIVE MARKING
      According to Le Chatelier, if the number of moles of water vapour is increased,
      the system will oppose the change by lowering ✓ the number of moles of water
      vapour. The reverse reaction is favoured. ✓ The concentration of the products          (3)
      decrease. ✓
      As die aantal mol waterdamp verhoog word, sal volgens Le Chatelier, die
      sisteem die verandering teenweerk deur die aantal mol waterdamp te verlaag.
      Die terugwaartse reaksie word bevoordeel. Die konsentrasie van die produkte
      neem af.
                                                                                             [15]

QUESTION 7 / VRAAG 7

7.1     A strong acid is an acid that ionizes completely in water ✓ to form a high
        concentration of H 3 O+ ions.                                                        (2)
        ’n Sterk suur ioniseer volledig in water.

7.2.1   Ampholyte ✓/Amfoliet                                                                 (1)
7.2.2   HSO 4 - + H 2 O ⇌ H 3 O+                +   SO 4 2- ✓✓                               (2)
7.3     OPTION 1 / OPSIE 1

        [OH-] = 0,4 mol∙dm-3 ✓ (ratio)
        [H 3 O+] [OH-]     = 1 x 10-14 ✓
           [H 3 O+] (0,4) = 1 x 10-14

                               1×1014
                   [H 2 O] =            P

                                0,4
                          = 2,5 x 10-14 mol.dm-3
                       pH = – log [H 3 O+] ✓
                          = – log (2,5 x 10-14)
                          = 13,6 ✓

        OPTION 2

        [OH-] = 0,4 mol∙dm-3 ✓ (ratio)

        pOH = -log[OH-]✓= -log(0,4) = 0,4

        pH + pOH = 14 ✓

        ∴pH = 13,6 ✓                                                                         (4)

 Copyrights reserved                                                      please turn over

### Page 15

Physical sciences P2                             15                        MDE/September 2016
                                            NSC- Memorandum

7.4.1    CH 3 COOH + NaOH → CH 3 COONa + H 2 O

                          c = 0,309 mol⋅dm-3
                          V = 0,01957 dm-3

                     n NaOH = c x V✓
                            = 0,309 x 0,01957
                            = 6,05 x 10-3 mol✓
                         m = n x M CH3COOH ✓
                            = (6,05 x 10-3) x 60 ✓
                            = 0,363 g ✓
                                                                                                    (5)
7.4.2 POSITIVE MARKING FROM 7.4.1
      % = 0,363 / 10 x 100 = 3,63% ✓ Yes ✓                                                          (2)

7.4.3    Phenolphtaleien✓ /Fenolftaleïen
         Weak acid and strong base titration ✓ /Swak suur en sterk basis titrasie.
         OR
                             6 + 13
         pH of solution ≈           ≈ 9,5
                               2                                                                    (2)
7.4.4    Hydrolysis ✓/ Hidrolise                                                                    (1)

                                                                                                    [19]

QUESTION 8 / VRAAG 8

8.1     Concentration of electrolytes: 1 mol.dm-3 ✓
        Pressure of gases: 1 atm / 101,3 kPa ✓                                                      (2)

8.2     Pt is an inert electrode/very weak reducing agent ✓
        Pt is ‘n onaktiewe metal/Baie swak reduseermiddel

        It is a (good) conductor of electricity ✓ /Dit gelei elektrisiteit                          (2)

8.3     O 2 + 4H+ + 4e- →               2H 2 O ✓✓                                                   (2)

8.4     Fe(s) | Fe2+ (aq) or (1 mol.dm-3) ✓ ||✓ O 2 (g) (1 atm) | H 2 O () | Pt(s)✓                (3)

8.5     Eθcell = Eθcathode -Eθanode ✓
               = 1,23✓-(-0,44)✓
               = 1,67 V✓                                                                            (4)

8.6      The cell reaction reaches equilibrium. ✓ /Die selreaksie bereik ewewig.                    (1)

                                                                                                    [14]

  Copyrights reserved                                                            please turn over

### Page 16

Physical sciences P2                        16                      MDE/September 2016
                                      NSC- Memorandum

QUESTION 9 / VRAAG 9

9.1    Cathode ✓/Katode                                                                      (1)

9.2      A solution / liquid that conducts electricity through the movement of ions. ✓✓
        ‘n Oplossing wat elektrisiteit gelei deur die beweging van vry ione.
       OR
       a substance whose aqueous solution contains ions OR a substance that                  (2)
       dissolves in water to give a solution that conducts electricity

9.3    Silver nitrate / Ag NO 3 ✓                                                            (1)
       Accept any soluble Ag+ salt /Aanvaar enige oplosbare Ag+ sout

9.4    Ag+ +    e-      →   Ag                                                             (2)

9.5    Remains the same./Bly dieselfde                                                      (1)

9.6    The rate at which the Ag+ ions is reduced, is equal to the rate at which the Ag
       is oxidised. 
       Die tempo waarteen die Ag+ ione gereduseer word, is gelyk aan die tempo
       waarteen die Ag geoksideer word                                                       (2)
                                                                                              [9]

 Copyrights reserved                                                      please turn over

### Page 17

Physical sciences P2                       17                        MDE/September 2016
                                     NSC- Memorandum

QUESTION 10 / VRAAG 10

10.1.1     Nitrogen /    N2    ✓                                                               (1)

10.1.2     N 2 + 3H 2 ✓ ⇌       2NH 3 ✓ (bal ✓)                                                (3)

10.1.3     Ammonia/Ammoniak ✓                                                                  (1)

10.1.4     Ostwald process       ✓                                                             (1)

10.1.5     Ammonium nitrate / NH 4 NO 3       ✓                                                (1)

10.1.6     Eutrophication ✓/Eutrifikasie                                                       (1)

10.2.1     A ✓, Lower nitrogen prevents too much leaf growth at the expense of
           fruit growth. ✓/Minder stikstof sal die oormatige groei van die blare ten
           koste van die vrugte, voorkom.
           OR
           It has a higher phosphorus content to promote seed and fruit
           development. /Dit het ‘n hoër fosforinhoud wat die ontwikkeling van sade
           en vrugte bevoordeel.
           OR
           It has a higher potassium content for seed and fruit quality./Dit het ‘n hoër
           kaliuminhoud wat die gehalte van die sade en vrugte verhoog.                        (2)

10.2.2        1
                ✓×20% = 2,222 %✓
              9
            2,222
                  ×50✓= 1,11 kg✓
             100

                         OR

                    1
                      ✓×50 = 5,5556 Kg✓
                    9
             20
                ×5,5556✓= 1,11 kg✓
            100                                                                                 (4)

                                                                                               [14]

                                                                      TOTAL/TOTAAL: 150

 Copyrights reserved                                                        please turn over

## Related documents

- [Physical Sciences P2 Memo A E](https://www.hlayiso.com/library/physical-sciences-p2-memo-a-e/b5728283-415c-47c5-8b2e-77517aba9088) — Physical Sciences · Memo · 2016
- [FS Memo Prep Sept 2016](https://www.hlayiso.com/library/fs-memo-prep-sept-2016/b00f1996-569c-4623-960f-26833a4b3c7e) — Physical Sciences · Memo · 2016
- [Lim Chemistry Chemie Memo](https://www.hlayiso.com/library/lim-chemistry-chemie-memo/150a3b71-e772-47db-a2b8-0fcbccedbe4c) — Physical Sciences · Memo · 2016
- [NW Phys .Fis Memo](https://www.hlayiso.com/library/nw-phys-fis-memo/c47091be-07ff-46dd-b160-ce6147c6ea06) — Physical Sciences · Memo · 2016
- [PHYSICAL SC P2 SEPT 2016 QP ENG](https://www.hlayiso.com/library/physical-sc-p2-sept-2016-qp-eng/7f759a6e-8648-4f8e-9bd0-01e989bc56c8) — Physical Sciences · QP · 2016 · Eng
- [Phsc P1 Backup Qp Eng June 2016](https://www.hlayiso.com/library/phsc-p1-backup-qp-eng-june-2016/4e8a907f-6a4c-4951-a82f-057b082aad28) — Physical Sciences · QP · 2016 · Eng

## More from Grade 12 Physical Sciences

- [PHYSICAL SCIENCE P1 QP JCT 2025 AFR D watermark](https://www.hlayiso.com/library/physical-science-p1-qp-jct-2025-afr-d-watermark/b9429ac5-acbb-4fae-989d-af9e1478a96b) — QP · 2025 · Afr
- [PHYSICAL SC P2 QP GR12 SEPTEMBER 2025 AFR1 watermark](https://www.hlayiso.com/library/physical-sc-p2-qp-gr12-september-2025-afr1-watermark/37cff2db-7c1a-4ec8-8ac5-07ec5bcf87a9) — QP · 2025
- [Physics Grade 12 NSC P2 QP September 2025 Gauteng](https://www.hlayiso.com/library/physics-grade-12-nsc-p2-qp-september-2025-gauteng/4d3cbea6-10ec-47ff-98ee-50646165ae9f) — QP · 2025
- [Physics Grade 12 NSC P1 QP September 2025 KZN](https://www.hlayiso.com/library/physics-grade-12-nsc-p1-qp-september-2025-kzn/6d225dc8-24c6-4c28-a181-df906d5a2c0e) — QP · 2025
- [Physical Sciences Physics Grade 12 PS01 QP March 2025 Limpopo](https://www.hlayiso.com/library/physical-sciences-physics-grade-12-ps01-qp-march-2025-limpopo/482ee9af-e88a-4818-9120-11ab9e045a1c) — QP · 2025
- [Physics Grade 12 NSC P1 QP September 2025 Limpopo](https://www.hlayiso.com/library/physics-grade-12-nsc-p1-qp-september-2025-limpopo/b65e9832-a6be-4312-8f5a-57425d647121) — QP · 2025

## Collection

- [Grade 12 past papers](https://www.hlayiso.com/grade-12/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.
