# Physical Sciences Grade 12 Paper 1 Limpopo Mock 2022 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Limpopo Mock Exam 2022, 34 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/limpopo-physical-science-grade-12-pre-trial-2022-p1-p2-and-memo/3cf69f2a-97bf-46f2-b72e-2c2596150018
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2022
- Exam period: Limpopo Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 34

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### Page 1

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                                    NATIONAL

                              SENIOR CERTIFICATE

                                  GRADE 12

      PHYSICAL SCIENCES: PHYSICS (P1) AND CHEMISTRY (P2)

                               PRE-TRIAL EXAM

                               22 AUGUST 2022

MARKS: 150
TIME: 3 hours

            This question paper consists of 16 pages and 5 data sheets.

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                                   NSC

INSTRUCTIONS AND INFORMATION

1.    Write your NAME in the appropriate space on the ANSWER BOOK.

2.    This question paper consists of 7 questions. Answer ALL the questions in the
      ANSWER BOOK.

3.    Start EACH question on a NEW page in the ANSWER BOOK.

4.
                                                                  s.co
      Number the answers correctly according to the numbering system used in this
      question paper.
                                                                      m
5.                                                           si
      Leave ONE line between two subquestions, for example between
                                                                c
      QUESTION 2.1 and QUESTION 2.2.
                                       st
6.                                        an
      You may use a non-programmable calculator.
                                            mor
7.    You may use appropriate mathematical instruments.
                                                ep
8.                                                hy
      You are advised to use the attached DATA SHEETS.

9.                                om
      Show ALL formulae and substitutions in ALL calculations.

10.                           fr
      Round off your FINAL numerical answers to a minimum of TWO decimal
      places.
                     oa
11.                    de
      Give brief motivations, discussions, et cetera where required.
                         d
12.     do
      Write neatly and legibly.
          wn
            l

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QUESTION 1 (Start on a new page)

1.1            ∆n
      If X =        applies to chemical reactions, where ∆n = change in number of
               ∆t
      moles and ∆t = time interval, then X probably represents …

      A        equilibrium constant.

      B        yield of products.

      C        heat of reaction.                                     s.co
      D        rate of reaction.                                         m                   (2)
                                                                 si c
1.2   Consider the following hypothetical reaction which is at equilibrium:
                                            st
                                               an
                                       2XY(s) ⇌ 2X(l) + Y2(g)

      The pressure of the system is INCREASED.
                                                 mor
                                                     ep
      How will the CONCENTRATION of Y2(g) and the value of KC be affected (at
                                                       hy
      constant temperature) when the new equilibrium is established?

                     CONCENTRATION of Y2(g)                     KC VALUE
                                       om
          A                   Unchange
                                    fr                          Unchange

          B            oa     Decrease                          Decrease

          C
                         de   Increase                          Increase
                           d
          do
          D                   Decrease                          Unchange
            wn
              l                                                                              (2)

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1.3   Consider the following equilibrium reaction equation:

                                  C(s) + O2(g) ⇌ CO2(g)

      Which ONE of the following graphs BEST represents the relationship
      between [CO2] and [O2] at CONSTANT TEMPERATURE?

      A                                          C

      [CO2]                                     [CO2]

                           [O2]                                       [O2]

      B                                          D

      [CO2]                                     [CO2]

                                                                                         (2)
                           [O2]                                       [O2]

1.4   The engine of a car does work, W, to increase the velocity of the car from v to
      2v. The work done by the engine of the same car to increase the velocity from
      2v to 4v, is …

      A       W

      B       2W

      C       3W

      D       4W                                                                         (2)

                                                                                         [8]

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QUESTION 2 (Start on a new page)

2.1    Two experiments are performed, with everything the same EXCEPT that one
       experiment is done at 20°C, and the other is done at 60°C. It is found that the
       experiment at 20°C takes 320 s and the experiment at 60°C takes 80 s. For
       each question, write down the LETTER from the LIST below which represents
       the BEST possible answer.

        V      The rate of reaction increases when temperature is increased.

        W      The rate of reaction is higher at 60°C than at 20°C.

        X      Is the rate of a reaction higher at 60°C than at 20°C?

        Y      How does increasing temperature affect the rate of a reaction?

        Z      The rate of reaction is 4 times faster at 60°C than at 20°C.

       2.1.1      Investigative question.                                                    (1)

       2.1.2      Hypothesis.                                                                (1)

       2.1.3      Conclusion.                                                                (1)

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2.2    A learner conducts a practical investigation in order to test whether the
       dissolution of solid ammonium chloride is exothermic or endothermic. The
       apparatus used include a beaker, a salt and a certain measuring instrument.
       The graph below shows the energy changes that occur when ammonium
       chloride dissolves in water.

                               200

       Potential energy (KJ)
                               170

                               50

                                                   Course of Reaction

       2.2.1                         Name the measuring instrument that must be used during this
                                     investigation.                                                       (1)

       2.2.2                         Will the reading on the instrument used in QUESTION 2.2.1
                                     INCREASE, DECREASE or REMAIN THE SAME during the course
                                     of the reaction? Explain the answer with reference to the graph.
                                                                                                          (2)
       For this reaction, calculate:

       2.2.3                          The heat of reaction                                                (2)

       2.2.4                          Activation energy                                                   (1)

       In another experiment, a learner adds a catalyst. On addition of a catalyst,
       state whether the following will INCREASE, DECREASE or REMAIN THE
       SAME:

       2.2.5                         Potential energy of the products                                     (1)

       2.2.6                         Activation energy                                                    (1)
                                                                                                          [11]

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QUESTION 3 (Start on a new page)

A group of chemistry learners use the reaction between magnesium and sulphuric
acid to investigate one of the factors that affects reaction rates.

3.1            Define the term reaction rate in words.                                            (2)

They add 2,4 g of magnesium ribbon to EXCESS dilute sulphuric acid and measure
the mass of magnesium used per unit time.

                             Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)

3.2            Suggest a reason as to why the acid must be in excess.                             (1)

The learners repeat the experiment using EXCESS concentrated sulphuric acid.

3.3            Provide a reason why the learners use the SAME amount (in grams) of
               magnesium ribbon in BOTH experiments.                                              (1)

The results obtained for the reaction using DILUTE sulphuric acid are represented in
the graph below.

               p

  n(Mg)(mol)

           0,08

               0
                                       160                              Time (s)

3.4            Determine the value of p (shown in the graph).                                     (3)

3.5            Using the graph, calculate the mass of magnesium used from t = 0 s to
               t = 160 s.                                                                         (3)

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3.6    Hence, find the average reaction rate (in gꞏs-1) during the FIRST 160 s.               (2)

3.7    Copy the graph above in the ANSWER BOOK (Label it R). On the same set
       of axes, use a dotted line to show the curve that will be obtained when
       CONCERNTRATED sulphuric acid is used. (Label it S). No numerical values
                                                                                              (2)
       needed.

3.8    Use the collision theory to explain the difference in the two graphs.                  (3)
                                                                                              [17]

QUESTION 4 (Start on a new page)

4.1    Consider the following equilibrium reaction equation:

                                 A2(g) + 3B2(g) ⇌ 2AB3(g), ∆H<0

       4.1.1         Write down ONE condition necessary for a chemical system to be
                     in equilibrium.                                                          (1)

       A graph showing the concentration of AB3(g) produced versus time is plotted
       below.

      [AB3(g)]

                 0           P               Q          R      S   Time         T

       How does the rate of the FORWARD REACTION compare to the rate of the
       REVERSE REACTION during the following intervals? (Write down only:
       GREATER THAN, EQUAL TO or LESS THAN).

       4.1.2         OP                                                                       (1)

       4.1.3         QR                                                                       (1)

       4.1.4         ST                                                                       (1)

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       Changes are made to the graph from Q to S due to changes in the
       temperature.
       4.1.5      At which point (Q, R or S) will the temperature be GREATEST?
                  Justify the answer.                                                      (2)

       4.1.6          At which point (Q,R or S) will the value of KC be the SMALLEST?      (1)

       4.1.7          If the changes in the graph from Q to S are due to pressure
                      changes, at which point (Q,R or S) will the pressure be the
                      HIGHEST? Explain the answer.                                         (2)

4.2    Initially, 0,086 mol of Br2 is placed in a 1,26 dm3 flask and heated to 1 680 K,
        a temperature at which the halogen dissociates to atoms according to the
        following equation:

                                    Br2(g) ⇌ 2Br(g), ∆H>0

       Br2(g) is 4,8% dissociated at 1 680 K.

       Calculate the equilibrium constant for this reaction at 1 680 K.                    (6)
                                                                                           [15]

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QUESTION 5 (Start on a new page)

5.1   Hydrochloric acid (HCl) is added to distilled water.

      5.1.1      Write down the balanced equation for the ionisation of hydrochloric
                acid in water.                                                                    (2)

      5.1.2     Classify hydrochloric acid as a STRONG acid or WEAK acid.                         (1)

                                                             -
5.2   Calculate the concentration of the hydroxide (OH ) ions in a sulphuric acid
      solution with a pH = 3,5 at 25°C.                                                           (3)

5.3   A standard solution of oxalic acid ((COOH)2) is prepared by dissolving 3,8 g of
      the acid in 250 cm3 of distilled water. The acid is titrated against an impure
      solution of sodium hydroxide (NaOH). The impure solution of sodium hydroxide
      is prepared by dissolving 5,0 g of the impure mixture in 250 cm 3 of distilled water.
      (Assume that the impurities do not react). In a titration,         25,0 cm3 of the
      impure sodium hydroxide is neutralized by 30 cm 3 of the acid.

                  (COOH)2 (aq) + 2NaOH(aq) → (COONa)2(aq) + 2H2O(l)

      5.3.1     Is an aqueous solution of (COONa)2 acidic, basic or neutral? Explain
                the answer with the aid of a chemical reaction equation.                          (3)

      5.3.2     Calculate the concentration of the oxalic acid solution.                          (3)

      5.3.3    Calculate the mass of the impurities contained in the 5,0 g of the impure
                sodium hydroxide.                                                                 (6)
                                                                                                  [18]

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     QUESTION 6 (Start on a new page.)

     The diagram below represents a simplified electrolytic cell used to
     electroplate a spanner with chromium. The spanner is continuously
     rotated during the process of electroplating.

                                                DC power supply

                                                                            Electrode X

                            Spanner

             Cr(NO3)3(aq)

     A constant current passes through the solution and the concentration
     of Cr(NO3)3(aq) remains constant during the process. In the process,
     a total of 0,03 moles of electrons is transferred in the electrolytic cell.

     6.1        Define the term electrolysis.                                        (2)

     6.2         State the energy conversion that takes place in the above cell.     (1)

     6.3        Write down the:

           6.3.1 Half-reaction that occurs at the spanner                            (2)

           6.3.2 NAME or FORMULA of the metal of which electrode X is made           (1)

            6.3.3 NAME or FORMULA of the oxidising agent                             (1)

     6.4        Give a reason why the concentration of the Cr(NO3)3(aq) remains
                constant during the process.                                    (2)

     6.5.      State on reason why the spanner is electroplated.                     (1)

                                                                                    [10]

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     QUESTION 7( Start on a new page)

     A standard electrochemical cell is set up, using silver and an
     unknown electrode, D, as shown in the sketch below,. The initial
     reading on the voltmeter is 2,46 V. After the cell was connected for
     some time, the mass of the silver electrode increased.

                                                 2 ,46 V
                                            V

                     Ag                                           D

                                  A                        B

     7.1      What process takes place at the silver electrode? Choose from
              OXIDATION or REDUCTION.                                               (1)

     7.2      Define an oxidising agent in terms of electron transfer.              (2)

     7.3      Write down the NAME of a solution which can be used in                (1)
              beaker A.

     7.4     Determine the standard electrode potential for electrode D and
             identify D from the standard potential table.                          (5)

     7.5      Write down the cell notation for this cell.                           (3)

                                                                                   [12]

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QUESTION 8 (Start on a new page)

During an investigation to verify the speed of sound, Mr Chiv use a siren with a single
frequency of 400 Hz, mounted on a remote-controlled toy car. With the siren
activated, he move the car along a 5 m track at a constant velocity of 7,50 mꞏs-1. One
of the learners stands halfway down the track and records the sound emitted by the
siren for the duration of the motion on the cellphone. They are expecting to record
two different pitches of the sound emitted by the siren.

8.1    For this investigation, write down the:

       8.1.1     Independent Variable                                                      (1)

       8.1.2     Dependent Variable                                                        (1)

       8.1.3    ONE control Variable                                                       (1)

8.2    Write down the NAME of and state in words, the phenomenon on which the
       learners depends in order to calculate the speed of sound.                          (3)

When playing back the sound recorded on the cellphone in the presence of a
frequency meter, two different notes were registered: One with a frequency of
409 Hz and the other with a frequency lower than 400 Hz.

8.3    Using the frequency of one of the recorded notes, calculate the speed of
       sound in air.                                                                       (3)

One of the uses of this phenomenon is in ultrasound.

8.4    State ONE application of ultrasound in the medical field.                           (1)
                                                                                           [10]

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QUESTION 9 (Start on a new page)

9.1 A crate of mass 10 kg slides FROM REST down path ABC, as shown in the diagram below.

                    
                    A

                                 35°
                                           
                                       h

                                                                
                                                                C
    Along the portion A to B, the coefficient of kinetic friction between the crate and the surface is
    0,16. The distance from A to B is 5 m.

    9.1.1       Define the term conservative force.                                                      (2)

    9.1.2       Draw a labelled free-body diagram for the crate as it slides along portion AB.
                                                                                                         (3)
    9.1.3       Write down the work-(kinetic) energy theorem in words.                                   (2)

    9.1.4       Use the work-(kinetic) energy theorem to calculate the speed of the crate as it
                reaches point B.
                                                                                                         (5)
    The crate reaches point C with a speed of 8,5 mꞏs-1.

    9.1.5       Assume portion BC is smooth (frictionless) and calculate the vertical height, h, of B
                above C, using energy principles.                                                        (4)

9.2 Water is pumped from a borehole 100 m deep. A pump with a power output of 2,3 kW (2 300
    W) is used.

    9.2.1       Define the term power as applied in physics.                                             (2)

    9.2.2       Calculate the time (in seconds) taken for the pump to bring the 500 kg of water to
                the surface.                                                                             (3)

                                                                                                         [21]

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QUESTION 10 (Start on a new page)

Two charged spheres, S and P, on insulated stands, with charges of 5 µC and -6 µC
respectively, are placed 0,02 m apart, as shown in the diagram below.

                                  0 , 02 m
                        S                         P     -6 µC
               5 µC

10.1    State Coulombs’ Law in words.                                                          (2)

10.2      Calculate the magnitude and direction of the electrostatic force that sphere S exerts on
         sphere P.                                                                             (4)

Spheres S and P are brought into contact with each other and then returned to their original
positions. A third, negatively charged, sphere R with an UNKNOWN charge is now placed
0,01 m to the right of sphere P as shown below.

                 S           0 , 02 m         P       0 , 01 m   R

10.3     Calculate the magnitude of the NEW charge on S after being in contact with P.         (1)

10.4     Draw a free-body diagram showing ALL the electrostatic forces experienced by P. (2)

10.5     The magnitude of the net electrostatic force experienced by P due to S and R is
          84,375 N. Calculate the magnitude of the charge on R.                          (5)

10.6     Calculate the magnitude of the net electric field at a point 0,01 m to the LEFT of R
         due to the charges on S and R.                                                        (3)
                                                                                              [17]

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QUESTION 11 (Start on a new page.)

11.1 A teacher demonstrates how current can be obtained using a bar magnet, a
coil and a galvanometer. The teacher moves the bar magnet up and down, as shown
by the arrow in the diagram below.

                                                    galvanometer

    coil

11.1.1. Briefly describe how the magnet must be moved in order to obtain a LARGE
deflection on the galvanometer.                                                                (2)

The two devices, A and B, below operate on the principle described in QUESTION 11.1.1
        above.

           Output
                                                 Output

           11.1.2     Write down the name of the principle.                                     (1)

           11.1.3     Write down the name of part X in device A.                                (1)

11.2       A 240 V, AC voltage is supplied from a wall socket to an electric kettle of resistance
           40 Ω. Wall sockets provide rms voltages and currents.

           Calculate the:

           9.2.1 Electrical energy consumed by the kettle per second                            (4)

           9.2.2      Maximum (peak) current through the kettle                                 (3)
                                                                                               [11]
                                         Total Marks: 150

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                                   NSC

DATA FOR PHYSICAL SCIENCES GRADE 12
                               PAPER 1 (PHYSICS)

                         TABLE 1: PHYSICAL CONSTANTS:
                NAME                  SYMBOL                 VALUE

Acceleration due to gravity              g                  9,8 m·s-2

Gravitational constant                   G          6,67 x 10-11 Nm2kg-2

Radius of Earth                          RE               6,38 x 106 m

Mass of the Earth                        ME              5,98 X 1024 kg

Coulomb’s constant                       k              9,0 x 109 Nm2·C-2

Plank’s contant                          h               6,63 x 10-34 Jꞏs

Speed of light in a vacuum
                                         c               3,0 x 108 m·s-1

Charge on electron                       e                -1,6 x 10-19 C

Electron mass                            me              9,11 x 10-31 kg

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TABLE 2: FORMULAE
MOTION

 v f  v i  a t                          Δx  v i Δt  21 at 2 or/of Δy  v i Δt  21 at 2

                                                 v vf                 v  vf 
 v f  vi  2 ax or/of v f  v i  2ay   Δx   i      Δt or/of Δy   i       Δt
   2      2                 2       2

                                                 2                     2 

FORCE

 Fnet  ma                                 p = mv

 fsmax = µsN                               Fk = µkN

 Fnet t  m v
                                           w  mg
 p  mv f  mv i

      m 1 m2     m 1 m2                         GM
 F = G 2 or F = G 2                        g
       d           r                             r2

WORK ENERGY AND POWER

 W  Fx cos                              U  mgh or Ep  mgh

                                           Wnet  K or Wnet  Ek
    1            1
 K  mv 2 or Ek  mv 2
    2            2                         K  K f  K i or E k  E kf  E ki

                                                W
 Wnc  K  U or Wnc  E k  E p        P
                                                t

 Pave  Fv ave / pgemid = Fvgemid

WAVES, SOUND AND LIGHT

                                                1
 v  f                                    T=
                                                f

        v  vL             V ± VL                             hf
 fL           f s or FL = V ± V fb        E  hf or/of E 
        v  vs                  b                             

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ELECTROSTATICS
          KQ1 KQ2                              KQ
 F=                                       E=
                  r2                           r2

          W                                    F
 V=                                       E=
          q                                    q

       Q                    Q
 n=           or n =
          e                 qe

ELECTRIC CIRCUITS
      V                                   Emf (ε) = I(R + r)
 R=I

 Rs = R1 + R2 + …                         q = I∆t
 1            1        1
      =            +        +…
 Rp       R1           R2

 W = Vq                                        W
                                          P=
                                               ∆t
 W = VI∆t
                                          P = Vi
 W = I2R∆t

      V2 ∆t                               P = I 2R
 W=
           R
                                               V2
                                          P=
                                                R

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                                   NSC       Mogodumo Circuit /22 August 2022

ATERNATING CURRENT
          1max                               Pave = Vrms Irms
 Irms =
          √2
                                            Pave = I2rmsR
          Vmax
 Vrms =                                               v2
           √2                               Pave = rms
                                                   R

                                PAPER 2 (CHEMISTRY)
      m                                           N
 n=                                         n=N
      M                                           A
      n               m                           V
 c=         OR   c=                         n=
      V               MV                          Vm
 Cb Vb      n                               pH = -log[H3O+]
          = b
 Ca Va      na
 Kw = [H3O+][OH-] = 1 x 10-14 at 298 K

          NAME                           SYMBOL                     VALUE
 Avogadro’s constant                       NA                   6,02X1023 mol-1

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                       NATIONAL SENIOR CERTIFICATE

                                      GRADE 12

        PHYSICAL SCIENCES: PHYSICS (P1) AND CHEMISTRY (P2)

                                   22 AUGUST 2022

                                   MEMORANDUM

MARKS: 100

                     These marking guidelines consist of 10 pages.

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 QUESTION 1

 1.1     D                                                                                (2)

 1.2     A                                                                                (2)

 1.3     B                                                                                (2)

 1.4     D                                                                                (2)

                                                                                            [8]

 QUESTION 2

 2.1.1 Y (Ask general question)                                                            (1)

 2.1.2 V (Predict)                                                                         (1)

 2.1.3 V  (State)                                                                          (1)

 2.2.1 Thermometer                                                                         (1)

 2.2.2 Decrease
             One mark the first bullet
                      OR
             One mark for the second bullet
                                                                                            (2)
          The potential energy for the products is higher than that of the reactants/.
           The potential energy for the reactants is lower than that of products. 
          This shows that the reaction is endothermic/heat absorbing. 

 2.2.3       ∆H = EP(P) – EP(R)
                = 170 – 50
                = 120 KJ                                                                   (2)

 2.2.4       Ea = EP(Ca) – EP(R)
                = 200 – 50
                = 150 KJ                                                                   (1)

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 2.2.5 Remain the same.                                                                 (1)

 2.2.6 Decrease.                                                                        (1)
                                                                                         [11]

 QUESTION 3

 3.1   The change in concentration of products or reactants per unit time. 
       OR:
       The change in number of moles/ mass/ volume of reactants/products per
       unit time.                                                                      (2)

 3.2   To ensure that all the magnesium gets used up.                                   (1)

 3.3   To make it a fair comparison/test.                                               (1)

 3.4
                     m
         ni(Mg) = M
                     2,4
                = 24 
               = 0,10 mol
         ∴P = 0,10                                                                      (3)

 3.5   Positive marking from 3.4
        n(Mg) used = ni – nf
                    = 0,10 –0,08             Option 2: 𝒎(𝑴𝒈) = 𝒏. 𝑴
                    = 0,02 mol                            = (0,1)(24)

         m(Mg)used= nM                        = 2,4 𝑔 (𝑖𝑡 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑟𝑒𝑎𝑚𝑏𝑙𝑒)
                  = (0,02)(24)               𝒎(𝑴𝒈) = (𝟎, 𝟎𝟖). (𝟐𝟒) 
                  = 0,48 g                                                              (3)
                                                           = 𝟏, 𝟗𝟐 𝒈
                                                  𝒎(𝑴𝒈)𝒖𝒔𝒆𝒅 = 𝟐, 𝟒  − 𝟏, 𝟗𝟐
                                                                = 𝟎, 𝟒𝟖 𝒈  

 3.6   Positive marking from 3.4 & 3.5

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                               m(Mg)
                              used
         Average rate = Time taken
                            0,48
                                                                                     (2)
                          = 160 
                          = 3 x 10-3 gꞏs-1

 3.7

           n(Mg)(mol)
                                        R
                               S

                   0
                                                         Time (s)
        Steeper gradient
        X – intercept for S smaller than for graph R                               (2)

 3.8    Higher concentration for curve S means:
         More reactant particles will be present in a given volume /
          More particles will have the correct orientation. 
         More effective collisions per unit time/Higher frequency of effective
          collisions. 
         Increased reaction rate. 

        OR: Lower concentration for graph R means:
         Less reactant particles will be present in a given volume /
          Less particles will have the correct orientation. 
         Less effective collisions per unit time/ Lower frequency of effective
          collisions. 

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           Decreased reaction rate. 
                                                                                           (3)
                                                                                           [17]

 QUESTION 4

 4.1.1     ANY ONE:
          Closed system                                                                (1)
          Reversible reaction
          Accept isolated system

 4.1.2 Greater than                                                                     (1)

 4.1.3 Less than                                                                        (1)

 4.1.4 Equal to                                                                         (1)

 4.1.5 R
        The reverse reaction is endothermic and so the temperature is greatest
         where [AB3 (g)] is lowest.                                                     (2)

 4.1.6 R                                                                                (1)

 4.1.7    Q
          An increase in pressure will the favour the forward reaction, resulting in
           the [AB2] increasing.
         Therefore, high pressure corresponds to a high [AB3].                          (2)

 4.2           Equation                         Br2(g)           ⇌ 2Br(g)
               Ratio                              1                  2
               Initial amount (mol)             0,086                0
               Change in amount (mol)        -4,128x10-3       +8,256x103
               Equilibrium amount (mol)       0,081872          8,256x10-3                    
                     n
               C = v (molꞏdm-3)                0,06498         6,5524x10-3

                                            𝑘𝑐 =
                           4,8
         n(Br2)decomposed = 100(0,086)
                         = 4,128x10-3 mol

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                [𝐵𝑟]2
         𝑘𝑐 = [𝐵𝑟 ] 
                    2

             (6,5524×10−3 )2
         =                                                                                    (6)
                (0,06498)
                       −4
         =6,61 × 10 

                                                                                               [15]

 QUESTION 5

 5.1.1 HCl(aq) + H2O(l) ⇌ H3O+(aq) + Cl-(aq)  Accept single arrow                              (2)

 5.1.2 Strong acid                                                                              (1)

 5.2      OPTION 1:                             OPTION 2:
          pH = -log[H3O+]                       pH + pOH = 14
          3,5 = -log[H3O+]                     pOH= 14 - pH = 14 - 3,5 = 10,5
          -3,5 = log[H3O+]                      pOH = -log[OH ] -

          [H3O+]= 10-3,5                        10,5 = -log[OH-]
          [H3O+]= 3,1622 X 10-4 molꞏdm-3        [OH-] = 10-10,5
                                                [OH-] = 3,16228 x 10-11 molꞏdm-3
          Kw = [H3O+][OH-]
          1 X 10-14 = (3,16228 X 10-4) [OH-]
          [OH-]= 3,16228 X 10-11 molꞏdm-3                                                       (3)

 5.3.1 Basic
        (COO-)2(aq) + 2H2O(l) ⇌ (COOH)2(aq) + 2OH-(aq) 
        Excess OH- ions are formed
        Therefore the solution is basic
        Hence the pH > 7                                                                        (3)

                m                          m
 5.3.2    c = MV                  OR: n = M
                                                                              n
                    3,8                  3,8                          ∴c = V 
             =(90)(0,25)               = 90
                                                                     0,0422222
             =0,169 molꞏdm-3           = 0,0422222 mol          =    0,25
                                                                = 0,169 molꞏdm-3                (3)

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 5.3.3   OPTION 1:                     OPTION 2:

         PacaVa = PbcbVb               1 mol (COOH)2 reacts with 2 mol NaOH
              cb Vb   nb               na= cV = (0,169)(0,03) 
          OR c V = n
               a a     a
                                              = 5,07x10-3 mol
         (2)(0,169)(30) =(1)cb(25) 
                                       Pbnb = Pana
         Cb = 0,4056 molꞏdm-3
                                       (1)nb = (2)( 5,07x10-3)
                                       ∴ nb = 0,01014 mol in 25 cm3
         n(NaOH)=cV
                                       ∴nb =(10)(0,01014) 
                =(0,4056)(0,25) 
                                            = 0,1014 mol in 250 cm3
                =0.1014 mol
                                       ∴m(NaOH) = Nm 
                                                    = (0,1014)(40) = 4,056 g
         m(NaOH) = nM 
                                       ∴m(impurities) = 5,0 –4,056 = 0,944 g
                 =(0,1014)(40) 
                 = 4,056 g

         m(impurities) = 5,0 
                             – 4,056
                                                                                       (6)
                       = 0,944 g
                                                                                       [18]

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 QUESTION 8

 8.1.1    The direction of the car with the siren relative to the observer/listener.        (1)

 8.1.2    The observed/detected frequency of the siren.                                     (1)

 8.1.3     ANY ONE:
          The actual frequency of the siren.                                               (1)
          The velocity of the car.

 8.2      The Doppler Effect.
          The apparent change in frequency (or pitch/wavelength) of the sound
           detected by the listener because the sound source and the listener have
           different velocities relative to the medium of sound propagation.               (3)

 8.3             v±v
           fL = ( v ± vL )fs
                       s
                       v+0
             409 = (v - v )(400) 
                           s
           (409)(v-7,5) = (v)(400)
           V = 340,83 mꞏs-1                                                                 (3)

 8.4       ANY ONE:
          To monitor blood flow.                                                           (1)
          To detect the heartbeat of a foetus.
                                                                                             [10]

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 QUESTION 9

 9.1.1 A force for which the work done (in moving an object between two points) is
       independent of the path taken. 
       OR: A force for which the work done on a particle moving through any closed path
       is zero.                                                                             (2)

 9.1.2                               N/FN

         fk / Ff/f
                                                       f:    Frictional force/friction
                                                      N:    Normal force
                                                       Fg/w: Gravitational force/weight

                                   w/Fg
                                                                                              (3)

 9.1.3 The net/ total work done on an object is equal to the change in the object’s kinetic
       energy.
       OR: The work done on an object by the net force is equal to the change in the
       object’s kinetic energy.                                                             (2)

 9.1.4   Wnet = ∆Ek
         WN + Wfk + W w = ∆Ek
                                            1
         0 + fkꞏ∆xꞏcosꞵ+ mgꞏ∆xꞏcosα = 2m(v2f -v2i )                       Any one
                                            1
         µk ꞏNꞏ∆xꞏcosꞵ+mgꞏ∆xꞏ cosα = 2m(v2f -v2i )
                                                1
         µk ꞏmgcosθꞏ∆xꞏcosꞵ+mgꞏ∆xꞏ cosα = 2m(v2f -v2i )
                                                                            1
         (0,16)(10)(9,8)(cos 35°)(5)(cos180°)+(10)(9,8)(5)(cos 55°/305°)= (10) (v2f -02 )
                                                                            2
         -64,22152027 + 281,0524538 = 5v2f
         216,8309335 = 5v2f
         v2f = 43,36618671
         vf = 6,5853 mꞏs-1
         ∴The speed is 6,5853 mꞏs-1                                                          (5)

                    𝑊𝑤 𝑐𝑎𝑛 𝑎𝑙𝑠𝑜 𝑏𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑏𝑦 𝑢𝑠𝑖𝑛𝑔 1) 𝑊𝑤 = 𝐹𝑔‖ × ∆𝑥 × 𝑐𝑜𝑠𝜃

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                                                     2) 𝑊𝑤 = −∆𝐸𝑝 = −𝑚𝑔(ℎ𝑓 − ℎ𝑖 )

 9.1.5    (EP + EK)B = (EP + EK)C
                      1             1
                                            Any one
          (mgh + 2mv2)B = (mgh + 2mv2)C
                               1                        1
          (10)(9,8)h + 2(10)(6,5853)2 = (10)(9,8)(0) + 2(10)(8,5)2
          (98)h + 216,8308805 = 0 + 361,25
          (98)h = 144,4191195
          ∴h = 1,4737 m
          ∴The height is 1,4737 m
                                                                                           (4)

 9.2.1 The rate at which work is done.
       OR: The rate at which energy is expended/transferred.                             (2)

 9.2.2 W = EP
         = mgh
         = (4500)(9,8)(100) 
         = 4410000 J

                W
         P = ∆t
                     4410000
         2300 =                
                     ∆t
                4410000
         ∆t =       2300
           = 1917,39 s                                                                    (3)
                                                                                           [21]

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                         GRAND TOTAL: 150 MARKS

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