# Physical Sciences Grade 12 P2 Free State Mock 2023 QP & Memo

> Grade 12 Physical Sciences Paper 2 question paper and memo from the Free State Mock Exam 2023, 32 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-12-physical-sciences-2023-free-state-mock-exam-question-paper-and-memorandum-paper-2/b952a904-7556-47f9-9fda-e2e283fde98c
- Updated: 2026-09-28
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Question paper and memo
- Year: 2023
- Exam period: Free State Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 32

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Physical Sciences · Grade 12 · Free State Mock Exam · 2023. Question paper and memorandum, 32 pages. Read online or download the PDF.

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### Page 1

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education

Department of
Education

FREE STATE PROVINCE

TIME: 3 HOURS

This question paper consists of 13 pages and 4 data sheets.

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Grade 12 Prep. Exam.

INSTRUCTIONS AND INFORMATION

1.

10.

11.

12.

Write your name and other applicable information in the appropriate spaces
on the ANSWER BOOK.

The question paper consists of TEN questions. Answer ALL the questions in the
ANSWER BOOK.

StartEACH question on a NEW page in the ANSWER BOOK.

Number the answers correctly according to the numbering system used in this
question paper.

Leave ONE line between two subquestions, for example between QUESTION
3.1 and QUESTION 3.2.

You may use a non-programmable pocket calculator.
You may use appropriate mathematical instruments.
Show ALL formulae and substitutions in ALL calculations

Round off your FINAL numerical answers to a minimum of TWO decimal places
where necessary.

Give brief motivations, discussions, et cetera where required.
You are advised to use the attached DATA SHEETS.

Write neatly and legibly.

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Grade 12 Prep. Exam.

QUESTION 1: MULTIPLE-CHOICE QUESTIONS

Various options are provided as possible answers to the following questions. Each
question has only ONE correct answer. Choose the answer and write only the letter
(A-D) next tothe question numbers (1.1 to 1.10) in your ANSWER BOOK, for example,

1.11 E.

1.1. Which-ONE of the following is a TERTIARY alcohol?

A

B

Cc

D

Butan-1-ol
Propan-2-ol
2-methylbutan-1-ol

Methylpropan-2-ol (2)

1.2 Consider the following organic compounds:

(i)

(ii)
(iii)
(iv)

CH3CH(CH3)CH(CH3)CHs3
CH3CH2CH2C(CH3)CH2
CH3CH2CH(CH3)CH2CHs

CH3sCCCH(CHs)2

Which of the above are chain isomers?

A

(i) and (iv).
(i) and (ii).
(ii) and (iv).

(i) and (iii). (2)

1.3. Which homologous series is the formyl group the functional group of?

A

B

Carboxylic acids
Ketones
Aldehydes

Haloalkanes (2)

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Grade 12 Prep. Exam.

1.4. Apiece of magnesium ribbon reacts with excess hydrochloric acid according to
the following equation:

Mg(s) + 2HC£ (aq) + MgCf2(aq) + H2(g)

Which/ONE of the following changes will NOT affect the reaction rate?

A Putting the reaction mixture in a hot water bath

B Using the same mass of powdered magnesium

Cc Increasing the volume of the hydrochloric acid

D Increasing the concentration of the hydrochloric acid (2)
1.5 Which ONE of the following statements is correct for a chemical reaction at

equilibrium?

A The concentration of reactants and products remains constant.

B The concentration of reactants is equal to the concentration of products.

Cc The reaction has stopped.

D The number of moles of reactants is equal to the number of moles

of products. (2)

1.6 Solid sulphur (Ss) reacts withyoxygem (Oz) to produce sulphur dioxide ...

"Sa(8) 802g) = 8S02(g)

What is the correct Ke expression for the equilibrium constant for the above

reaction?
— [$02]
A Ke soa
B «= Soul?
(O2)°
= [SgllOz]
Cc K [SO2]
— [S02]
D [= 2
“ [2] (2)

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Grade 12 Prep. Exam.

1.7 According to the Arrhenius theory, a base is ...

A a proton donor.

B acproton acceptor.

Cc a/Substance that dissociates in water to form H*.

D a substance that dissociates in water to form OH’. (2)

1.8 Consider the reaction below:
HsPOa(aq) + HCO; (aq) = H2PO; (aq) + H2COs(aq)
Which ONE of the following are two acids in the above reaction?
A HCO3 and H2PO;
B HsPO4 and HzPO3
Cc HsPO4 and H2COz
D HCO3 and H2CO3 (2)

1.9 A galvanic cell is constructed from cadmium and nickel. What is the
correct cell notation for this cell?

A Cd?*| Cd || Ni2*| Ni
B Nil Ni2*|| Cd2*| Cd
C Cdl Cd?*|| Ni2*| Ni
D Nil Nill Ca2*l Ca (2)

1.10 Which ONE of the following half-reactions takes place at the CATHODE
of an electrochemical cell used to electroplate an iron spoon with silver?

A Ag' +e —>Ag

B Ag Agi +e
Cc Fe2* + 2e — Fe
D

Fe — Fe** + 2e (2)
[20]

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Grade 12 Prep. Exam.

QUESTION 2 (Start on a new page.)

The letters A to E in the table below represent six organic compounds.

A | Butane B | CHsCH2CCCH3
fit
H—C—C—C—H
| | H H
HH | |
SC} wHuHaA | H D nn
birth I
eT yr yyy
H H H Ce H H
E Butanoic acid

2.1. Write down the letter(s) representing:

2.1.1. Acompound with the general formula CnHen-2 (1)
2.1.2 Aketone (1)
2.1.3. Acompound that can be used as fuel (1)
2.1.4 An unsaturated hydrocarbon (1)

2.2 For compound E, write down the:
2.2.1 NAME of the functional group (2)
2.2.2 Structural formula of its FUNCTIONAL ISOMER (2)
2.3. Write down the IUPAC name of:

2.3.1. Compound C (3)
2.3.2 Compound D (2)
2.4 Write down the IUPAC name of the functional isomer of Dj (2)

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Grade 12 Prep. Exam.

2.5 Write down the structural formula of:
2.5.1. ACHAIN ISOMER of compound A

2.5.2 ‘Compound E

QUESTION-3(Start on a new page.)

(2)

(2)
[19]

3.1. The vapour pressures of two organic compounds at a specific temperature are

compared.
MOLAR MASS
COMPOUND (g-mol")
A Butane 58
B Methylpropane 58

3.1.1 Define the term vapour pressure.

3.1.2 Which ONE of the two compounds has the highest vapour pressure?

Write down only A or B.

3.1.3 Fully explain the answer to QUESTION 3.1.2 (3)

3.1.4 Give a reason why the above comparison can be considered fair. (1)

3.2. The compounds propan-1-ol and propanone are compared. Which

compound has a higher boiling point? Fully explain the answer. (4)

3.3. The table below shows the results of an investigation regarding the melting

points of organic molecules.

Organic compound Boiling point (°C)
A 2,2-dimethylbutane 50
2,3-dimethylbutane 58
3-methylpentane 63
3.3.1 Write down the independent variable for this investigation. (1)
3.3.2 Explain the trend in boiling points of the above investigation. (3)

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[15]

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Grade 12 Prep. Exam.

QUESTION 4 (Start on a new page.)

The flow diagram below shows the preparation of different organic compounds from
CHsCH2CHs. x, y and z represent organic reactions.

CH3CH2CH3

Br2

NaOH
Compound A An alkene

y H2SOx4

v

Compound B
(Secondary
alcohol)

4.1. Write down the type of reaction represented by reaction:

414.1 x (1)
4.1.2 y (1)
413 2 (1)

4.2 Reaction x takes place in the presence of a strong base. Write down the:

4.2.1 IUPAC name of compound A (2)
4.2.2 Balanced equation for the reaction x, using STRUCTURAL
FORMULAE (4)
4.2.3, IUPAC name of the alkene produced (2)
4.3 State TWO differences in the reaction conditions for x and y= (2)

4.4. Compound B is a secondary alcohol. Define the term secondary alcohol. (2)
[15]

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Grade 12 Prep. Exam.

QUESTION 5 (Start on a new page.)

5.1. Glucose is metabolised in the human body through the following simplified
equation.

Ce6H1206 + 602 > 6CO2 + 6H20

The graph below shows the energy changes for the above reaction.

>

Potential energy
(kJ-mol")
o

(2)

Course of reaction

5.1.1 Is this reaction ENDOTHERMIC or EXOTHERMIC? Give a reason
for the answer. (2)

5.1.2 Express the following in terms of the energy values A, B and C
shown in the graph:

(a) The energy of the activated complex (1)
(b) The activation energy for the reverse reaction (2)
(c) AH for the forward reaction (2)

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Grade 12 Prep. Exam.

5.2 In an investigation of the rate of reaction, an EXCESS 200 cm solution of HC,
concentration 0,5 mol-dm* reacts with zinc powder at STP. The following
spontaneous reaction takes place:

Zn(s) + 2HCE (aq) > ZnCf2(aq) + Ha(g)

The volume of hydrogen gas produced is measured at regular time intervals
and the-following graph is drawn:

no}

8

3

2

ae A

a&

PS

Ke)

a

€

3s

3 >

Time(s)

5.2\4Define the term concentration. (2)
5.2.2) Write down the limiting reagent for the above reaction. (1)
5.2.8 Calculate the mass of the zinc used. (5)

5.2.4 The same mass of zinc granules is used instead of zinc powder.

(a) How will this affect the rate of reaction? Choose INCREASES,
DECREASES or STAYS THE SAME. (1)

(b) Use the collision theory to explain the answer to
QUESTION 5.2.4(a). (3)

5.2.5 Asuitable catalyst is added to the reaction mixture.
Redraw the graph above in your ANSWER BOOK and label the
given curve as A. On the same set of axes, sketch the curve that

will be obtained if a catalyst is added. Label this as curve B. (2)
[21]

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Grade 12 Prep. Exam.

QUESTION 6 (Start on a new page.)

Hydrogen lodide (HI) is allowed to decompose in a 5 dm® closed container at a
temperature of 425 °C, according to the following balanced equation:

2HI(g) = Ha(g) + l2(g)

When the mixture reaches equilibrium at 425 °C, it is found that 0,02 mol of Ha is
present in the container. The equilibrium constant for this reaction at 425 °C is 0,016.

6.1 Define the term closed system. (2)
6.2 — Calculate the initial number of moles of HI in the container. (8)

6.3. How will EACH of the following changes affect the amount of H2(g)
produced? Write down only INCREASES, DECREASES or REMAINS

THE SAME.
6.3.1 More l2(g) is added at 425 °C (1)
6.3.2 The pressure is increased (1)

6.4 At325 °C, the equilibrium constant (Kc) for the reaction above is 0,014.
Is the reaction EXOTHERMIC or ENDOTHERMIC? Fully explain the
answer. (4)
[16]

QUESTION 7 (Start on a new page.)

7.1. Two grams (2g) of magnesium hydroxide is reacted with 30 cm° sulphuric acid
solution of concentration 1,5 mol-dm, according to the following balanced
equation:

H2SO4(aq) + Mg(OH)2(aq) + 2H20(£) + MgSOa(aq)
7.1.1 Why is H2SO4 considered a STRONG ACID? (2)
7.1.2 Calculate the concentration of the final solution. (8)

7.2 Adilute HC2 solution has a concentration of 0,15 mol-dm’®. This dilute
solution reacts with a concentrated solution of NazCOs.

7.2.1. Define the term dilute acid. (2)
7.2.2 Calculate the pH of the HCf solution. (3)

7.3 Asolution is made by dissolving NazCOs (s) in water.

7.3.1 Is the solution of NazCO3 ACIDIC or BASIC? (1)
7.3.2 Write down a balanced chemical equation that explains the answer

to QUESTION 7.3.1. (3)

[19]

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Grade 12 Prep. Exam.

QUESTION 8 (Start on a new page.)

An electrochemical cell is set up under STANDARD CONDITIONS.

W)
WO)

Ccu—> Ag
f '

8.1 Write down ONE function of the part labelled X. (1)
8.2 Which electrode of the cell is the anode? Write down only Cu or Ag. (1)

8.3. Write down the name or formula of the electrolyte in:

8.3.1 The copper half-cell (1)
8.3.2 The silver half-cell (1)
8.4. Write down the balanced net (overall) equation for the above cell. (3)
8.5 Calculate the emf of the cell. (4)
8.6 State the standard conditions under which this cell operates. Hat

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Grade 12 Prep. Exam.

QUESTION 9 (Start on a new page.)

The simplified diagram below represents an electrochemical cell used for the
purification of copper.

I
Electrode B ——+
+— Electrode A

CuCl, —————_+—»
9.1 State the energy conversion that takes place in this cell. (2)
9.2 Define the term electrolyte. (2)
9.3. Which electrode contains impure copper? Write down Electrode AOR

Electrode B. (1)
9.4 Write down the half-reaction which takes place at electrode B. (2)
9.5 Is electrode A the ANODE or the CATHODE? (1)

9.6 The two electrodes are replaced with CARBON RODS.
9.6.1 Write down what will now be observed at electrode A. (2)
9.6.2 Write down a relevant half-reaction that explains the observation
made in QUESTION 9.6.1. (2)
[12]

TOTAL: 150

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Grade 12 Prep. Exam.

FS/September 2023

DATA FOR PHYSICAL SCIENCES GRADE 12
PAPER 2 (CHEMISTRY)

GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 12
VRAESTEL 2 (CHEMIE)

TABLE 1: PHYSICAL CONSTANTS / TABEL 1: FISIESE KONSTANTES

NAME / NAAM SYMBOL / S/MBOOL VALUE / WAARDE
Iolére gasvolume by STD Vm 22,4 dm?-mol
Avogadro's constant Na 6,02 x 1023 mot

TABLE 2: FORMULAE / TABEL 2: FORMULES

[H30*][OH~] = 1x107'* at/by 298 K

m N
n=a n= a
n
=F
OR/OF nae
m
“= Vv
ae =3 pH = —log [H30*]

@ _ 78 @
Evou _~ E Cathode ~ Eanode

OR

0 _ FO a
Eceu = Ereauction — Eoxiaation
OR

a0 _ Fé 6
Ecen = E, -£E

oxidising agent ‘redusing agent

6 _ po 6
Eset ~ Exatoac ~ Eanode
OF
6 _ 76 6
Eset = Ereauksie — Foxsidasie
OR
6 _ 76 a
Eset ~ Eksideermiddel a! Exeduseermiddel

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Grade 12 Prep. Exam.

TABLE 4A: STANDARD REDUCTION POTENTIALS
TABEL 4A: STANDAARD-REDUKSIEPOTENSIALE

Increasing oxidising ability / Toenemende oksiderende vermoé

Copyright reserved

Half-reactions / Halfreaksies 59 (V)
Fag)+2e = 2F- + 2,87
Co*#+e = Co* + 1,81
H2O2 + 2H*+2e = 2H20 1,77
MnO, +8H*+5e- = Mn?*+4H20 +1,51
Chg) + 2e- = ACh + 1,36
Croz + 14H* + 6e 2Cr* + 7H20 + 1,33
Ox(g)+4H*+4e- = 2H20 +1,23
MnQ2+ 4H*+2e == Mn?* + 2H20 + 1,23
Pt! +2e = Pt + 1,20
Bro(t)+2e° = 2Br + 1,07
NO, +4H*+3e = NO(g)+2H20 + 0,96
Hg?*+2e- = Hg(t) + 0,85
Agtte = Ag + 0,80
NO, +2H*+e = NO2(g) + H20 + 0,80
Fe +e = Fe +0,77
Ox(g) + 2H* + 2e- = H2O2 + 0,68
k+2e = 2b +054
Cu*te- = Cu + 0,52
SO2+4H*+4e = S+2H20 +0,45
2H20+O2+4e = 40H + 0,40
Cu2*+2e° = Cu + 0,34
soz +4H*+2e = SO2a(g) + 2H20 +0,17
Cu*+e = Cut + 0,16
Sn +2e = Sn +0,15
S+2H*+2e = HeS(g) +0,14
2H*+2e- = Ha(g) 0,00
Fe +3e = Fe — 0,06
Pb**+2e- = Pb —0,13
Sn2*+2e- = Sn -0,14
Ni2*+2e-° = Ni - 0,27
Cu*+2e = Co - 0,28
Cd*+2e = Cd — 0,40
Crt+e- = Crt —0,41
Fe**+2e = Fe — 0,44
Crt+3e = Cr -0,74
Zn**+2e = Zn - 0,76
2H20 + 2e- == Ha(g) + 20H- - 0,83
Cr++2e = Cr - 0,91
Mn**+2e = Mn —1,18
Ali +3e = AL — 1,66
Mg?*+2e- = Mg — 2,36
Nat*+e = Na —2,71
Ca2*+2e- = Ca — 2,87
Sr*+2e- = Sr — 2,89
Ba*+2e = Ba — 2,90
Cs*+e@ = Cs - 2,92
Kite = K = 2,93
litte = Li = 3,05

FS/September 2023

Increasing reducing ability / Toenemende reduserende vermoé

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Grade 12 Prep. Exam.

TABLE 4B: STANDARD REDUCTION POTENTIALS
TABEL 4B: STANDAARD-REDUKSIEPOTENSIALE

Increasing oxidising ability / Toenemende oksiderende vermoé

Copyright reserved

Half-reactions / Halfreaksies EF Ww)
Lite = Li — 3,05
Kite = K — 2,93
Cst+e = Cs — 2,92
Ba’*+2e- = Ba — 2,90
Sr++ 2e = Sr — 2,89
Ca*+2e = Ca — 2,87
Nat+e = Na —2,71
Mg?* + 2e Mg — 2,36
ABbt+3e- =— Ae —1,66
Mn?*+2e- = Mn -1,18
Cr’*+2e- = Cr —0,91
2H20 + 2e- == Ha(g) + 20H” - 0,83
Zn*+2e = Zn -0,76
Cr++3e = Cr — 0,74
Fe**+2e = Fe — 0,44
Crt+e> = Crt -0,41
Cd**+2e- = Cd —0,40
Co?* + 2e- Co -0,28
Ni#*+2e° = Ni —0,27
Sn#*+2e = Sn — 0,14
Pb**+2e = Pb —0,13
Fe%+3e = Fe —0,06
2H*+2e- = Hag) 0,00
S+2H*+2e = H2S(g) +0,14
Sn**+2e- = Sn** +0,15
Cu*+e = Cut +0,16
SO +4H?+2e- = SO2(g)+2H20 | +0,17
Cu +2e = Cu + 0,34
2H20 +O2+4e- = 4OH- +0,40
SO2+4H*+4e- = S+2H20 +0,45
Cutte = Cu +0,52
kb+2e = ab +0,54
O2(g) + 2H*+2e = 202 + 0,68
Fe*+e = Fe +0,77
NO; +2H*+e- = NO2g)+H2O | +0,80
Agtte = Ag +0,80
Hg?*+2e- = Hg(é) + 0,85
NO, +4H'+3e = NO(g)+2H2O | +0,96
Br(f)+2e = 2Br + 1,07
Pet+2e = Pt + 1,20
MnOQ2+ 4H* + 2e- = Mn?* + 2H20 + 1,23
O2(g) + 4H*+4e- == 2H20 +1,23
CrO% +14H*+6e = 2Cr*+7HO | +1,33
Clog) +2e = 2Ce + 1,36
MnO, +8H*+5e- = Mn?*+4H20 + 1,51
H202 + 2H*+2e° = 2H20 +1,77
Co* +e = Co + 1,81
Fog)+2e = 2F- + 2,87

FS/September 2023

Increasing reducing ability / Toenemende reduserende vermoé

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education

Department of
Education

FREE STATE PROVINCE

PREPARATORY EXAMINATION
VOORBEREIDENDE EKSAMEN

GRADE/GRAAD 12

PHYSICAL SCIENCES: CH
FISIESE WETENSKAPPE: C}

marep

SEPTEMBER 2023
MARKS/PUNTE: 150

MARKING GUIDELINES/NASIENRIGLYNE

These marking guidelines consist of 15 pages.
Hierdie nasienriglyne bestaan uit 15 bladsye.

Copyright reserved/Kopiereg voorbehou

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### Page 19

Paventeade dafaaenedarmovephysics.com
Grade/Graad 12 Prep.Exam./Voorb.Eksam.
Memorandum/Nasienriglyne

QUESTION 1/VRAAG 1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

1.10

QUESTION 2/VRAAG 2

Dvv
Dvv
Cv¥
Cvv
AvY
Byy
Dvv
Cvv
Cvv

Avy

2.1.1 BY

2.1.2 DY

2.1.3 A/JBY

2.14 BY

2.2.1 Carboxyl/Karboksiel Vv

2.2.2

Copyright reserved/Kopiereg voorbehou

x
~——s

x=I—o—r

I—o—_ i

FS/VS/September 2023

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### Page 20

Paunisade dafecmuGdatramovephysics.com FS/VS/September 2023
Grade/Graad 12 Prep.Exam./Voorb.Eksam.
Memorandum/Nasienriglyne

Marking criteria/Nasienkriteria: - “«

e Whole structure correct/Hele struktuur korrek: 7)

e Only functional group correct/single bonds between carbon atoms: Ye
Slegs funksionele groep korrekte/enkelbindings tussen koolstofatome: ‘%

2.3.1 3,5-dichloro/dichloor” -4-methyl/metiel Yoctane/oktaan “

Marking criteria/Nasienkriteria:
e 3,5-dichloro/dichloor ¥

e 4-methyl/Metiel “

¢ Octane/Oktaan ¥

(3)
2.3.2 Propanone/propan-2-one/Propanoan/propaan-2-een Vv ¥ (2)
2.4 Propanal/Propanaal vv (2)

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### Page 21

Paventeade dafaaenedarmovephysics.com
Grade/Graad 12 Prep.Exam./Voorb.Eksam.
Memorandum/Nasienriglyne

2.5.1 Hy
H—C—

|
fa
H

—H v

=mI—oO—-=z x=

FS/VS/September 2023

Marking criteria/Nasienkriteria:
e Whole structure correct/Hele struktuur korrek:
e One methyl substituent/Een metielsubstituent:

4/2

Notes/Aantekeninge:
Condensed formulae or semi-structural formula:

Gekondenseerde formules of semi-struktuurformule:

Max./Maks.

Molecular formula/Molekulére formule: Vy
(2)
PEG oY
2.5.2 H—C—C—C
| | | v
H H H
Marking criteria/Nasienkriteria:
e Whole structure correct/Hele struktuur korrek: 2/o
e Only functional group correct/Slegs funksionele groep korrek: %
(2)
[19]

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### Page 22

Paunisade dafecmuEdatamovephysics.com FS/VS/September 2023

Grade/Graad 12 Prep.Exam./Voorb.Eksam.

Memorandum/Nasienriglyne

QUESTION 3/VRAAG 3

3.1.1

3.1.2

3.1.3

3.2

Marking criteria/Nasienkriteria:

If any-oflthe underlined key phrases in the correct context is omitted,
deduct! mark.

Indiéhenige van die onderstreepte sleutelfrases in die korrekte
konteks weggelaat word, trek 7 punt af.

The pressure exerted by vapour at equilibrium with its liquid in a closed

system. vv
Die druk wat uitgeoefen word deur damp by ewewig met sy vioeistof in_'n

geslote sisteem. (2)
BY (1)

B/methylpropane/metielpropaan
e Smaller surface area/Kleiner opperviakte W
e Weaker intermolecular forces/Swakker intermolekulére kragte V

e Less energy needed to break the intermolecular forces/Minder energie
benodig om die intermolekulére kragte te breek ¥

Albutane/butaan

e Larger surface area/Groter opperviakte v

e Stronger/more intermolecular forces/Sterker/meer intermolekulére kragte V

e More energy needed to break the intermolecular forces/Meer energie
benodig om die intermolekulére kragte te breek V (3)

One independent variable/same homologous series and are (chain)
isomers/same molecular mass and are (chain) isomers. ~

Een onafhanklike veranderlike/dieselfde homoloé reeks en is (ketting)
isomere/dieselfde molekulére massa en is (ketting) isomere. (1)

propan-1-ol “

¢ propan -1-ol has hydrogen bonding, (dipole-dipole and London forces)
between molecules propanone has dipole-dipole forces (and London
forces). v

e Intermolecular forces in propan-1-ol are stronger than intermolecular

forces in propanone. ¥

¢ propaan -1-ol

¢ propaan -1-ol het waterstofbinding, (dipool-dipool en Londen-kragte)
tussen molekules propanoon het dipool-dipool kragte|(en;Londen-
kragte).

e Intermolekulére kragte in propaan-1-ol is sterker as intermolekulére
kragte in propanoon.

OR/OF

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### Page 23

Paunisade dafecmnEdaramovephysics.com FS/VS/September 2023

Grade/Graad 12 Prep.Exam./Voorb.Eksam.

Memorandum/Nasienriglyne

Intermolecular forces in propanone are weaker than intermolecular forces
in propan-1-ol.

More energy needed to overcome or break intermolecular forces/van der
Waals forces in propan-1-ol v than propanone.

OR

Less energy needed to overcome or break intermolecular forces/Van der.
Waals forces in compound D than compound C.

Intermolekulére kragte in propanoon is swakker as intermolekulére
kragte in propaan-1-ol.

Meer energie benodig om intermolekulére kragte/Van der Waals kragte
in propaan-1-ol te oorkom of te breek as propanoon.

OF
e Minder energie benodig om intermolekulére kragte/Van der Waals kragte
in verbinding D te oorkom as in verbinding C. (4)
3.3.1 Branching/Vertakking Vv (1)

3.3.2

Copyright reserved/Kopiereg voorbehou

e From A to CG: less branching/greater surface area ¥

e Stronger/more intermolecular forces (London forces) ~“

e More energy needed to break the intermolecular forces
(London forces) ¥

e Vanaf A tot C: minder vertakking/groter opperviakte

e = Sterker/meer intermolekulére kragte (Londen-kragte)

e Meer energie benodig om die intermolekulére kragte te
breek (Londen-kragte)

(3)
[15]

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### Page 24

Paunisade dafesmuEdatamovephysics.com FS/VS/September 2023

Grade/Graad 12 Prep.Exam./Voorb.Eksam.
Memorandum/Nasienriglyne

QUESTION 4/VRAAG 4

4.1.1. Elimination/dehydrohalogenation/dehydrobromination
Eliminasie/dehidrohalogenering/dehidrobromering

4.1.2 Substitution/hydrolysis/Substitusie/hidrolise Vv
4.1.3 Elimination/dehydration/Eliminasie/dehidrasie V

4.2.1 2-bromo Ypropane/propaan ¥

4.2.2
| i’ N iv
H—C C—H + NaOH —> c==c—c—H + NaBr ¥ +H20 ¥
| f 4d
H HH H H

Notes/Aantekeninge:
e Ignore//gnoreer =
e Any additional reactants and/or products./Enige addisionele reaktanse

en/of ander produkte. Max./Maks. x
« Accept coefficients that are multiples./Aanvaar koéffisiénte wat

veelvoude is.
e Condensed or semi-structural formulae/Gekondenseerde of semi-

struktuurformules: Max./Maks. 2/4
e Molecular formulae/Molekulére formules: Max./Maks. 2/4

4.2.3, prope-1-ene/propene/prope-1-een/propeen ¥ ¥

4.3. X-—concentrated strong base, heatigekonsentreerde sterk basis, hitte V
Y - dilute strong base, mild heat/verdunde sterk basis, matige hitte “

4.4 Alcohol where the C atom bonded to hydroxyl/functional group (-OH) is
bonded to two other carbon atoms. Vv

(4)
(2)

(2)

Alkohol waar die C-atoom gebind aan hidroksiel/funksionele groep (-QH) aan

twee ander koolstofatome gebind is.

OR/OF
|
—C—O—H
The functional group ( i )is bonded to two other carbon atoms.
Die funksionele groep is aan twee ander koolstofatome gebind. (2)

[15]

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### Page 25

Paunisade dafecmuEdaramovephysics.com FS/VS/September 2023

Grade/Graad 12 Prep.Exam./Voorb.Eksam.

Memorandum/Nasienriglyne

QUESTION 5/VRAAG 5

5.1.1

5.1.2

5.2.1

5.2.2
5.2.3

5.2.4

5.2.5

Exothermic/Eksotermies Vv

Energy-of products is less than that of reactants/energy is given off/

AH < Da

Energié-van produkte is minder as dié van reaktanse/energie wat afgegee

wordi4H)< 0. (2)
(a) AY (1)
(b) A-Cvv (2)
(c) C-BYY (2)
The amount of a substance per volume of water/solution Vv”
Die hoeveelheid van 'n stof per volume water/oplossing (2)
Zinc/Zn ¥ (1)

_VvV
n(H2) = Tam v

= 04144

22,4
= 0,0185 mol

n(Zn) = n(H2) = 0,0185 mol v
m(Zn) = n(Zn) x M

= 0,0185 x 65 ¥

=1,2025g v
(RANGE/GEBIED: 1,19 g to 1,3) (5)
(a) DECREASES/AFNEEM “ (1)

(b) e Decrease in surface area/Afname in opperviakte V
e Fewer particles with correct orientation/Minder deeltjies met korrekte
oriéntasie V
e| Fewer effective collisions per unit time/Minder effektiewe botsings
per eenheidstyd v (3)

Note: If both graphs are not labelled
Let Wel: Indien beide grafiek nie benoem is nie %, (2)
[21]

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### Page 26

Pauntaads dzfrseuSdarimovephysics.com

Grade/Graad 12 Prep.Exam./Voorb.Eksam.

Memorandum/Nasienriglyne

QUESTION 6/ VRAAG 6

6.1

6.2

FS/VS/September 2023

Asystem that is isolated from its surroundings./A system where substances

cannotleave/escape the container. vv

'n Sisté6m wat van sy omgewing geisoleer is./'n Sisteem waar stowwe nie die

houerkan verlaat/ontsnap nie.
OPTION/OPSIE 1

(2)

CALCULATIONS USING NUMBER OF MOLES/
BEREKENINGE MET GEBRUIK VAN AANTAL MOL
Mark allocation/Puntetoekenning:
(a) Change n(H2) = equilibrium n(Hz) = 0,02
Verandering n(H2) = ewewig n(H2) = 0,02
(b) USING ratio HI:H2:l2 = 2:1:1
GEBRUIK verhouding HI!:H2:l2 = 2:1:1
(c) Equilibrium mole of l2 = Change mole lz ”
Ewewig mol van I2 = Verander mol 2
(d) Divide 0,02 by 5 AND multiplying 0,0316 by 5 v
Deel 0,02 deur 5 EN vermenigvuldig 0,0316 met 5
(e) Correct Ke expression (formulae in square brackets) v
Korrekte Ke uitddrukking (formules tussen vierkantige hakies)
(f) Substitution of Kc 0,016
Vervanging van K- 0,016
(g) Substitution of concentrations into Ke expression
Vervanging van konsentrasies in K- uitdrukking
(h) Initial mole of HI = Equilibrium + Change = 0,198 mol v
Range/Gebied: 0,19 — 0,2 mol

HI Ha

Initial quantity (mol)
Aanvanklike hoeveelheid (mol) VA 0,198” (h)

Change/Verander (mol) _
0,04 0,02 Y (a)

0,02

Quantity at equilibrium (mol) 5
ne ov 0,02
Hoeveelheid by ewewig (mol) 0,158 ,
n=cv

0,02 v

Equilibrium concentration (mol-dm)
Ewewig konsentrasie (mol-dnv3) 0,0316 0,004

0,004

v(f) 0,016

Copyright reserved/Kopiereg voorbehou

= Holla) y
Ke THI (e)

(0,004)? v(g)
[HI]?
[Hl] = 0,0316 mol.dm*?

Please turn over/Blaai om asseblief

ratio ¥(b)
verhouding

(c)

v(d)

### Page 27

Paunisade dafesmuEdatamove physics.com FS/VS/September 2023

Grade/Graad 12 Prep.Exam./Voorb.Eksam.
Memorandum/Nasienriglyne

Wrong Ke expression/Verkeerde Ke uitdrukking Max/Maks: 5/e

No Ke expression followed by correct substitutions/Geen K.-uitdrukking nie gevolg
deur korrektevérvangings Max/Maks: UP
OPTION/OPSIE 2

CALCULATIONS USING CONCENTRATIONS/
BEREKENINGE MET GEBRUIK VAN KONSENTRASIES
Mark allocation/Puntetoekenning:
(a) Change [Hz] = equilibrium [Hz] = 0,04
Verandering [H2] = ewewig [H2] = 0,04
(b) USING ratio HI:Ha:l2 = 2:1:1
GEBRUIK verhouding HI:H2:l2 = 2:1:1
(c) Equilibrium concentration of lz = Change concentration |2 ¥
Ewewig konsentrasie van I2 = Verander konsentrasie |2
(d) Correct Ke expression (formulae in square brackets)
Korrekte K- uitddrukking (formules tussen vierkantige hakies)
(e) Substitution of concentrations into Ke expression ¥
Vervanging van konsentrasies in Ke uitdrukking
(f) Substitution of Ke 0,016
Vervanging van K- 0,016
(g) Initial concetration of HI = Equilibrium + Change “
Aanvanklike konsentrasie van HI = Ekwilibrium + Verandering
(h) Devide 0,2 by SAND multiplying ,0396 by 5
Deel 0,2 deur 5 EN vermenigvuldig ,0396 met 5
Range/Gebied: 0,19 — 0,2 mol

Equilibrium/Ewewig [Hz] = <= 0,004 mol-dm

HI Ha
Initial quantity (concentration)
Aanvanklike hoeveelheid 0,0396(g)
(konsentrasie) fo
Change (concentration)
Verander (konsentrasie) 0,008 0,004 0,004 ratio v’(b)
¥ (a) verhouding
Equilibrium concentration (mol-dm’?) |
Ewewig konsentrasie (mol-dm*) 0,0316 0,004 0,004 |~
Ko = a d) v n(Hl)initial/aanvanklike;=-0,0396 x 5
v(f) 0,016 = (0,004)? » (e) =-0,198 mol
[HI]?
[Ht] = 0,0396 mol.dm?

(8)

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### Page 28

Paunisade dafecmuEdaramove phiysics.com FS/VS/September 2023
Grade/Graad 12 Prep.Exam./Voorb.Eksam.
Memorandum/Nasienriglyne

Wrong K- expression/Verkeerde uitdrukking: Max/Maks: 5/e

No K, expression followed by correct substitutions/Geen Kc-uitdrukking nie gevolg

deur korrektevérvangings: Max/Maks: UP
6.3.1 Decreases/Neem af v (1)
6.3.2 Remains the same/Bly dieselfde Vv (1)

6.4. Endothermic v
e Kc decreases with a decrease in temperature v”
e Reverse reaction is favoured/concentration of reactants increases/
concentration of products decreases/yield decreases V
e Decrease in temperature favours an exothermic reaction V

Endotermies

e K-neem af met 'n afname in temperatuur

e Omgekeerde reaksie word bevoordeel/konsentrasie van reaktanse neem
toe/konsentrasie van produkte neem affopbrengs neem af

e Afname in temperatuur bevoordeel 'n eksotermiese reaksie (4)
[16]

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### Page 29

Paunisade dafecmuEdatamove physics.com FS/VS/September 2023
Grade/Graad 12 Prep.Exam./Voorb.Eksam.
Memorandum/Nasienriglyne

QUESTION 7/VRAAG 7

7.1.1 It dissociates/ionises completely” in water.
Dit dissosieer/ioniseer heeltemalin water. (2)
7.1.2

Marking criteria/Nasienriglyne:

n
a) Formula/Formule n ="/ c= 7 v
M Vv

b) Substitute/Vervang 58 inn == v

¢) USING ratio/GEBRUIK verhouding n(H2SOa) : n(Mg(OH)2) = 1:1”
d) Substitute/Vervang 1,5 and/en 0,03448 inn =cV v
e) n(H2SOa)final = Ninitial - NreactedY” V

v

<|5

f) Substitute/ Vervang n(H2SOs4)rinal and/en 0,03 in c =

g) Final answer/Finale antwoord: 0,5 mol-dm? ”
Range/Gebied: 0,5 to 0,67 mol:'dm?

n(MgOH):
n==v(a)
M

= = V(b)

= 0,03448 mol
Nreacted(H2SO4) = n(MgOh)
n(H2SOa) = ee
Ninitia(H2SO4) = c x V
= 1,5 x 0,03 v(d)

= ct. mol
Nfinal(H2SO4) = 0,05 — 0,03 v ¥(e)
= 0,02 mol

n
[H2SO4] c Vv

= 0,67 mol-dm° v’(g) (8)

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### Page 30

Paunisade dafecmnEdaramove physics.com FS/VS/September 2023
Grade/Graad 12 Prep.Exam./Voorb.Eksam.
Memorandum/Nasienriglyne

7.2.1. Contain a small amount (number of moles) of acid Vin proportion to the
volume of water. Vv

Bevat 'n klein hoeveelheid (aantal mol) suur in verhouding tot die
volume=water. (2)

7.2.2 pH =/log[H3O*]

pH =slog(0,15) v

pH 20,82 v (3)
7.3.1 Basic/Basis Vv (1)

7.3.2

Marking criteria/Nasienkriteria:

a) Reactants/Reaktanse v Products/Produkte v
Balancing/Balansering v

b) Ignore single arrows and phases//gnoreer enkel pyle en fases

c) Marking rule/Nasienreé! 3.10

CO$ (aq)+ H20(2) ¥ = HCO3(aq) + OH (aq) YY balancing/balansering
OR
NazCOs(aq) + H2O(t) “ = H2CO3(aq) + NaOH(aq) “ “ balancing/balansering

(3)
[19]

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### Page 31

Paunisade dafecmnEdaramove physics.com FS/VS/September 2023

Grade/Graad 12 Prep.Exam./Voorb.Eksam.

Memorandum/Nasienriglyne

QUESTION 8/VRAAG 8

8.1. Completes the circuit/maintains electrical neutrality/provides path for
movement of ions. W
Voltooidie stroombaan/handhaaf elektriese neutraliteit/verskaf pad vir
beweging van ione. (1)
82 Cuv (1)
8.3.1 CuSOa/copper(Il) sulphate/koper(Ii)sulfaat ¥ (1)
Accept: Salt that contains Cu?* ions/Aanvaar: Sout wat Cu?' ione bevat
8.3.2 AgNOs/silver nitrate/silwernitraat ¥ (1)
Accept: Salt that contains Ag* ions/Aanvaar: Sout wat Ag* ione bevat
8.4 Cut 2AgtY — Cu + 2Ag ¥ ¥ balancing/balansering (3)
Marking criteria/Nasienkriteria:
e Reactant/Reaktanse ¥ Products/Produkte ~
Balancing/Balansering ¥
e Ignore double arrows/Ignoreer dubbe! pyle
e Marking rule/Nasienreé/ 6.3.10
8.5 OPTION/OPSIE 1 Notes/Aantekeninge:
Eo, =6o, . eo Accept any other correct formula from the
cell reduction” oxidation data sh
=08v -034Vv lata sheet.
_ 0 46 V Y Aanvaar enige ander korrekte formule vanaf
7" gegewensblad.
Any other formula using unconventional
abbreviations, e.g. Bw ES AER followed
by correct substitutions: 3/4
Enige ander formule wat onkonvensionele
atkortings gebruik, bv.
Fer «x FRegevolg deur korrekte
vervangings: 3/ 4
OPTION/OPSIE 2
(pu — Cu + 2e E° = -0,34(V) v
“ lAgt +e Ag E°= 08 (Vv)
Cu + 2Ag* — Cu?* + 2Ag E°= 046Vv (4)
8.6 Temperature/Temperatuur. 25 °C/ 298 K v
Concentration/Konsentrasie: 1 mol-dm? ¥ (2)
[13]

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### Page 32

Paunisade dafecmuEdatamovephysics.com FS/VS/September 2023

Grade/Graad 12 Prep.Exam./Voorb.Eksam.
Memorandum/Nasienriglyne

QUESTION 9/VRAAG 9
9.4 Electrical to chemical/Elektries tot chemies vv

9.2 Asolutiorthat conducts electricity through the movement of ions vv”
'n Oplossing wat elektrisiteit gelei deur die beweging van ione

9.3 (Electrode/Elektrode) A v

9.4 Cut +2e—>Cuvv

Marking criteria/Nasienkriteria:

© Cu +2e = Cu % Cu = Cu?! + 2e %
Cu — Cut + 2e % Cu — Cu?* + 2e° %

e Ignore if charge omitted on electron/Ignoreer as lading op elektron
weggelaat is

e If charge (+) omitted on Cu?*/Indien lading (+) weggelaat is op Cu?*
Max./Maks: Ms

e Example/Voorbeeld: Cu? + 2e° + Cu

9.5 Anode v

9.6.1 Ayelow green gas/chlorine / Cf, gas will form vv
'n Geelgroen gas/chloor/ Cf, gas sal vorm

9.6.2 2Cf- +Cé, +2e vv

Marking criteria/Nasienkriteria:

© 2Cf =Ch+2e 1 Cl, + 2e= 2C#- 9%
Cl2+ 2e — 2C2- % Cl, +2e — 208 %

e Ignore if charge omitted on electron/ignoreer as lading electron op
weggelaatis —

© If charge (+) omitted on C £/Indien lading (+) weggelaat is op Ct:

e _Example/Voorbeeld: 2C ¢ > Ce, + 2e

(2)
[12]

TOTAL/TOTAAL: 150

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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.
