# Physical Sciences Grade 12 Paper 2 Free State Mock 2025 Memo

> Grade 12 Physical Sciences Paper 2 memorandum from the Free State Mock Exam 2025, 14 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physics-grade-12-nsc-p2-memo-september-2025-free-state/c322debe-a6c1-428e-86e5-a52989e91ba5
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2025
- Exam period: Free State Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 14

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Physical Sciences · Grade 12 · Free State Mock Exam · 2025. Memorandum, 14 pages. Read online or download the PDF.

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### Page 1

education

Department of
Education

FREE STATE PROVINCE

PREPARATORY EXAMINATION
VOORBEREIDENDE EKSAMEN

GRADE/GRAAD 12

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

SEPTEMBER 2025

MARKS/PUNTE: 150

MARKING GUIDELINES/NASIENRIGLYNE

These marking guidelines consist of 14 pages.
Hierdie nasienriglyne bestaan uit 14 bladsye.

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### Page 2

FS/VS/September 2025

QUESTION 1/VRAAG 7

11. Byy (2)
12 BY (2)
13°" BY (2)
14 Cv (2)
15 Dv (2)
16 Cvv (2)
17 Dvv (2)
1.8 Cv (2)
1.9 Avy (2)
110 Avv (2)

[20]

QUESTION 2/VRAAG 2

2.1.1 DY (1)
2.1.2 A&DY (1)
2.13 EV (1)
2.2 CnHane2 VV (2)
2.3.1 BY (1)
2.3.2 Chain (isomers)/Ketting (isomere) ¥ (1)
2.3.3. 2,2,4-trimethylpentane/trimetielpentaan Vv Vv (3)

Marking criteria/Nasienkriteria:

¢ Correct stem, i.e. pentane ¥
Korrekte stam, d.w.s. pentaan

e Correct substituents (methyl) identified. “
Korrekte substituente (metiel) geidentifiseer.

e IUPAC name entirely correct including numbering, sequence, hyphens and
commas. ¥
IUPAC-naam_ heeltemal korrek insluitend nommering, volgorde,
koppeltekens en kommas.

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### Page 3

FS/VS/September 2025

24 HOH HH Marking criteria/Nasienkriteria:
i th | tol ¢ Functional group/Funksionele groep V
H—C—C—C—C—C—H | e 5C-atoms in the longest chain Vv
| 1 tof 5 C-atome in die langste ketting
H H HH ¢ Whole structure correct ¥
Hele struktuur korrek
(3)
2.5.1 Esterification/condensation / Verestering/kondensasie Vv (1)
2.5.2 H2SO./sulphuric acid / swae/suur ¥ (1)
2.5.3 Propyl v ethanoate v / Propyl etanoaat (2)
[17]

QUESTION 3/VRAAG 3

3.1

3.2

Marking criteria/Nasienkriteria:

If any of the underlined key phrases in the correct context is omitted,
deduct 1 mark.

Indien enige van die onderstreepte sleutelfrases in die korrekte

konteks weggelaat is, trek 1 punt af.

The temperature at which the vapour pressure of a liquid is equal to the
atmospheric pressure. Vv ¥

Die temperatuur waarby die dampdruk van 'n vioeistof gelyk is aan die
atmosferiese druk. (2)

Compound C

e Is more branched/compact/spherical/smaller surface area V

e Weaker intermolecular forces Vv

e Less energy needed to break/overcome intermolecular forces ¥

Verbinding C

e — /s meer vertak/kompak/sferies/kleiner opperviakte

e  Swakker intermolekulére kragte

e Minder energie benodig om intermolekulére kragte te breek/oorkom

OR/OF

Compound B

e Js less branched/compact/spherical/larger surface area

e Stronger intermolecular forces

« More energy is needed to break/overcome intermolecular forces

Verbinding B

e — /s minder vertak/kompatl/sferies/groter opperviakte

e = Sterker intermolekulére kragte

e Meer energie benodig om intermolekulére kragte te breek/oorkom (3)

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### Page 4

3.3.1

3.3.2

3.4.1
3.4.2

FS/VS/September 2025

117,7 (°C) ¥ (1)

¢ Between molecules of compound A are 2 sites of hydrogen bonds (in
addition to London and dipole-dipole forces) “”

e Between molecules of compound B is 1 site of hydrogen bonds (in
addition to Lendon and dipole-dipole forces) “

e Between compound D are dipole-dipole forces (in addition to London
forces) “

e Intermolecular forces between molecules of B are weaker than those
between molecules of A “ but stronger than those between molecules of
Dv

e Less energy needed to overcome intermolecular forces of B compared to
that of A v/ More energy needed to overcome intermolecular forces of B
compared to that of D

e Tussen molekules van verbinding A is 2 plekke van waterstofbindings
(bykomend tot Londen en dipool-dipoolkragte)

e Tussen molekules van verbinding B is 1 plek van waterstofbindings
(bykomend tot Londen en dipool-dipoolkragte)

e Tussen verbinding D is dipool-dipoolkragte (bykomend tot Londen-kragte)

e Intermolekulére kragte tussen molekules van B is swakker as dié tussen
molekules van A, maar sterker as dié tussen molekules van D

e Minder energie benodig om intermolekulére kragte van B te oorkom in
vergelyking met dié van A/ Meer energie benodig om intermolekulére kragte
van B te oorkom in vergelyking met dié van D (6)

Butane/Butaan vv (2)

From compound E to G:

e Chain length increases Vv

e Strength of intermolecular forces increases V

e More energy needed to break the intermolecular forces ¥

Vanaf verbinding E na G:

e Kettinglengte neem toe

e  Sterkte van intermolekulére kragte neem toe

e Meer energie benodig om die intermolekulére kragte te breek

OR/OF

From compound G to E

e Chain length decreases

e Strength of intermolecular forces decreases

e Less energy needed to break the intermolecular forces

Van verbinding G na E

e Kettinglengte neem af

e Sterkte van intermolekulére kragte neem af

e Minder energie benodig om die intermolekulére kragte te breek (3)
[17]

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FS/VS/September 2025

QUESTION 4/VRAAG 4

41 Marking criteria/Nasienkriteria:

lf any of the underlined key phrases in the correct context is omitted, deduct
1 mark,

Indien enige van die onderstreepte sleutelfrases in die korrekte konteks
weggelaat is, trek 1 punt af.

A chemical process in which longer chain hydrocarbon molecules are broken
down to shorter, more useful molecules. vv

'n Chemiese proses waarin langerketting koolwaterstofmolekules in korter,
meer bruikbare molekules afgebreek word. (2)

4.2.1 Exothermic/Eksotermies ¥ (1)

4.2.2 CsHs + 502Y” — 3CQ2¥;+ AH20 ¥

Marking criteria/Nasienkriteria:
e Reactants ¥ Products “ Balancing v
Reaktanse Produkte Balansering
« Ignore = and phases
ignoreer = en fases
« Marking rule 6.3.10/Nasienreé! 6.3.10

(3)
4.2.3 Carbon dioxide/Koolstofdioksied V (1)
4.3.1 Addition/hydrohalogenation/hydrobromination v
Addisie/hidrohalogenering/hidrobromering (1)
4.3.2 No water/use unreactive solvent
Geen water/gebruik onreaktiewe oplosmiddel (1)
4.3.3. 2-bromo “propane Y /2-broom propaan (2)
4.4.1
v Marking criteria:
e Functional group correct. Vv
H H e Whole structural formula correct. Vv
| | Nasienkriteria:
H—C C—HY  Funksionele groep korrek.
| | e Hele struktuurformule korrek.
H H H
(2)
4.4.2 Substitution/hydrolysis / Substitusie/hidrolise ¥ (1)
4.5.1 Dehydration / Dehidrasie ¥ (1)
4.5.2 Propene/prop-1-ene/1-propene / Propeen/prop-1-een/1-propeen Vv ¥ (2)
[17]

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FS/VS/September 2025

QUESTION 5/VRAAG 5

5.1. Gas syringe/measuring cylinder / Gasspuit/maatsilinder V (1)

5.2 ¢ Change in concentration/mass/volume/amount of reactants/
products per unit time v “
« Mass/volume/amount/concentration of reactants used/products
formed per unit time

« Rate of change in concentration/mass/volume/amount of reactants
or products (2 or/of 0)

e Verandering in konsentrasie/massa/volume/hoeveelheid reaktanse/
produkte per tydseenheid

e Massa/volume/hoeveetheid/konsentrasie van reaktanse gebruik/
produkte gevorm per tydseenheid

e Tempo van verandering in konsentrasie/massa/volume/hoeveelheid (2)
van reaktanse of produkte

5.3 Marking criteria/Nasienriglyne:

(a) Substitute 0,05 in rate formula /Vervang 0,05 in tempoformule Vv
(b) Substitute 40 in rate formula / Vervang 40 in tempoformule v
(c) Final answer/Finale antwoord: 1,25 x 10° dm*-s!

Rate/ Te 7
ate/Tempo = =
_0.05 ¥

40
= 1,25 x 10° (dm?-s*) v

(3)

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FS/VS/September 2025

5.4 Marking criteria/Nasienriglyne:

(a) Formula/Formule: n = ~! n =o v

(b) Substitute/Vervang 22,4 in n= = v

(c) Using ratio/Gebruik verhouding n(Mg): n(H2) = 1:1 ~

(d) Substitute/Vervang 24 inn =o v

(e) Divide/Deel m(Mg)reactedigereageer Dy M(MQ)initiat/begin/divide/deel n(Mg)reacted/gereageer DY
N(MQ)initiavbegin

(f) Multiply by/Vermenigvuldig met 100 v

(g) Final answer/Finale antwooord: 2,5 % ¥

RANGE/GEBEID: 2,5 % to 2,79 %

OPTION/OPSIE 1 OPTION/OPSIE 2
Hy) = H)= v
n( 2) vm m 2)
005 _ 005
~ 22,4 224 %
= 2,232 x 103 mol = 2,232 x 10% mol
n(Mg) = n(H2) N(MQ)keactewgereageer = N(H2)
= 2,232 x 10° mol “ = 2,232 x 10° mol Vv

m (Mg)reactea =nxM - 8
= (2,232 x 103)(24) v | M(Ma)intiainegin =
= 0,054 g

% purity/suiwerheid =

v

Rl wsts

4

x 100 ¥
au = 0,083 mol
=2,5%V

2,232%10°°
Ee = Ov

= 2,69 % Vv

% purity/suiwerheid =

5.5.1 Temperature/Temperatuur ¥ (1)
5.5.2 AV (1)

5.5.3 Curve A:
e Fewer particles with enough/sufficient kinetic energy v
OR

Fewer particles with Ex > Ea
e Less effective collisions per unit time v

Kromme A:
e Minder deeltjies met genoeg/voldoende kinetiese energie
OF
Minder deeltjies met Ex = Ea
e Minder effektiewe botsings per tydseenheid (2)
[17]

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FS/VS/September 2025

QUESTION 6/ VRAAG 6

6.1 Yes/Ja v (1)
6.2 The system doesn't exchange matter with the surroundings/sealed container.
vv
Die stelsel ruil nie materie uit met die omgewing nie/verseélde houer. (2)

6.3

CALCULATIONS USING NUMBER OF MOLES

Marking criteria:

(a) — Multiplying [CO] and dividing [COz] by 2. ¥

(b) Change moles CO = equilibrium moles CO ¥

(c) USING ratio COz: CO=1:2v

(d) Equilibrium moles CO2 = initial moles COz - change moles COz2 Vv
(e) Correct K, expression (formulae in square brackets) v

(f) Substituting K, value into K; expression Vv

(g) Substituting concentrations into the correct K. expression ¥

(h) Final answer = 0,056

BEREKENINGE MET GEBRUIK VAN AANTAL MOL

Nasienkriteria:

(a) Vermenigvuldig {CO} en deel [CO2] deur 2. ”

(b) Verander mol CO = ekwilibrium mol CO ~~

(c) GEBRUIK verhouding CO2: CO=1:2”

(d) Ekwilibrium mol CO2 = aanvanklike mol CO2 - verander mol CO2 “
(e) Korrekte K--uitdrukking (formules tussen vierkantige hakies) ~

(f) | Vervanging van K.-waarde in K.-uitdrukking ~

(g) Vervang konsentrasies in die korrekte K.-uitdrukking ~

(h)__ Finale antwoord = 0,056

OPTION/OPSIE 1 “CO, | CO

Initial quantity/Aanvanklike hoeveelheid (mol)/ 0,056” 0 ratio)
Change/Verandering (mol) 0,04 0,08 | roudin av
Quantity at equilibrium/Hoeveelheid by ewewig 0,016“ '° 0.08

(mol) : mg

Equilibrium concentration/Ewewigs- 8x103 | 004 +&x2v
konsentrasie (mol-dm) a

[cor
K.= ——— Vv F 5
[CO] Wrong K- expression Max/Maks: °/,
» 0,042 Verkeerde Ke-uitdrukking
O2v=
[C02] No Ke expression Max/Maks: UP
[CO,] = 8 x 10% Geen Ke-uitdrukking 8)

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FS/VS/September 2025

OPTION/OPSIE 2 CO2 CO
Initial quantity/Aanvanklike hoeveelheid (mol) | [x 0 ;
Change/Verandering (mol) | 0,04 0,08 V_| Ratio/

ngv
Quantity at equilibrium/Hoeveeiheid by vamouding

- v
ewewig (mol) x — 0,04 0,08

Equilibrium concentration/Ewewigs- x- 0,04 004 |zax2v
konsentrasie (mol-dm) 2 ,
[cor
Ke= AW v
[COz] Wrong K. expression Max/Maks: 5/,
0,042 Verkeerde Ke-uitdrukking

0,2 v =xo0r ¥
2
x = 0,056 (mol) v

No Ke expression Max/Maks: ty.
Geen Ke-uitdrukking

6.4 Low/Laag v (1)
6.5 Low kK value/Lae Kc-waarde v v (2)
6.6.1 Endothermic/Endotermies V (1)

6.6.2 Decrease in temperature favours exothermic reaction v
Reverse reaction is favoured/concentration of products decreases/
concentration of reactants increases V

Verlaging in temperatuur bevoordeel eksotermiese reaksie
Omgekeerde reaksie word bevoordeel/konsentrasie van produkte neem af/
konsentrasie van reaktanse neem toe (2)

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FS/VS/September 2025

6.7
(a)
(b)

(c)
(d)

Marking criteria/Nasienkriteria:

Correct shape of the graph/Korrekte vorm van die grafiek Vv

Ea (reactants = CO) correctly indicated/Ea (reaktanse = CO) korek
aangedui ¥

AH correctly indicated/Korrek aangedui

Energy of reactants with label CO and energy of products with label
CO2 + C indicated/Energie met reaktanse met byskrif CO en energie
van produkte met byskrif CO2 + C aangedui Vv

Potential energy/Potensiéle energie

Reaction coordinate/Reaksiekoérdinaat

OR/OF

Potential energyPotensiéle

AH

CO2+C

Reaction coordinate/Reaksiekoérdinaat

(4)
[21]

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### Page 11

QUESTION 7/VRAAG 7

7.1.1 Asolution of known concentration vv

FS/VS/September 2025

‘n Oplossing met 'n bekende konsentrasie (2)
7.1.2 Strong/Sterk v (1)
7.1.3. Dissociates/ionises completely in water to form a high concentration of
hydroxide/OHions v ¥
Dissosieer/ioniseer heeltemal in water om 'n hoé konsentrasie van hidroksied/
OH-ione te vorm (2)
7.1.4 Marking criteria/Nasienriglyne:
(a) Formula/Formule pH = -log[H30*]/Kw = [H3O*][OH"] ~
(b) Correctly substituting 13,4 in pH formular “
Korrekte vervanging 13,4 in pH formule
(c) Substituting 10° in formula Kw = [H30*][OH]”
Vervang 10° in formule Kw = [H30*][OH]
(d) Substitute 3,98 x10°* in formula Kw = [H30*][OH] ”
Vervang 3,98 x10"4 in formule Kw = [H3O*][OH]
(e) _ Final answer/Finale antwoord: 0,25 mol-dm> Vv
PH = -log(HsO*] “
13,4 Vv = -log[H30*]
[H30*] = 3,981 x 10°4 mol-dm?
Kw = [H30*][OH]
10% v =3,981 x 10° v [OH]
[OH"] = 0,25 mol-dm3 v (5)
1.15

Marking criteria/Nasienriglyne:

CaVa _
(a) Formula/Formule =i

min =v

(b) Multliply/ Vermenigvuldig 0,4 by/met 3,125 in
0,4 by 3,125 x 10° inn = cVY

(c) Using ratio/Gebruik verhouding: n(H2SO4) : n(NaOH) = 1:2 ¥

(d) Multiply/ Vermenigvuldig V by 0,25 v
(e) Final answer/Finale antwoord:

CaVa _ na
CbVb ny OR/OF

0,01 dm?/10 cm? Vv

OPTION/OPSIE 1
CaVa _ na
CbVb/y np

(0,4\(3,125)” 4 7

(0,25)(Vb) 2

Vb = 10cm$ ¥

OPTION/OPSIE 2
n(H2SOa4) = cV ¥

= (0,4093,05 x 10°) v
= 1,25 x 10° mol

n(NaOH) = 2n(H2SOa)
=2x 1,25 x 103 v
=2,5x 103

n(NaOH) = cV
2,5 x 10° = 0,25 Vv

### Page 12

FS/VS/September 2025

22 Marking criteria:
m
(a) Substituting 5 & 84 in Formula n = M v
(b) — Multiplying 1,6 by 0,05 v
(c) Using ratio: n(HC2) : n(NaHCOs) = 1:1 v
(d) Subtracting n(HC?) reacted from n(HC2)initial v
(e) Dividing n(HC2)left by 0,05 v
(f) Substituting c(HC2) in pH formular ¥
(g) Final answer = 0,4 v

Nasienkriteria:
m
(a) Vervang 5 & 84 in Formule n = rv} al

(b) — Vermenigvuldiging 1,6 met 0,05 ~

(c) Gebruik verhouding: n(HCé} : n(NaHCOs) = 1:1 ~~
(d) Trek n(HCé) gereageer van n(HC£) begin ~~

(e) Deel n(HCé) links deur 0,05 ~”

(f) Vervanging van c(HC?) in pH-formule ~

(g) Finale antwoord = 0,4 ~

m
n(NaHCOs)initiarbegin = =

M
|
=v
84
= 0,06 mol
N(HCE)reacted/gercageer = M(NaHCOs) v
= 0,06 mol
N(HC8)initia/begin =cV
= (1,6)(0,05) v
= 0,08 mol
N(HC£)unreacted/ongereageerde = 0,08 - 0,06 v
= 0,02 mol
HC?) ==
(HC) =
0,05
=0,4 mol-dm®

[H3O*] = c(HC2) = 0,4 mol-dm?

PH = -log [H30*]
=-log 0,4 v
=04Vv

RANGE/GEBIED: 0,39 - 0,4 (7)
[22]

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FS/VS/September 2025

QUESTION 8/VRAAG 8
8.1. Asolution that conducts electricity through the movement of ions V ¥
‘n Oplossing wat elektrisiteit gelei deur die beweging van ione (2)
v
8.2 Pb(s) | Pb**(aq) v || Ag*(aq) | Ag(s) ¥ (3)
8.3 Potassium chloride/KCt vv
Any other soluble salt, excluding lead and silver salts
Kaliumchloried/KCf
Enige ander oplosbare sout behalwe lood- en silwersoute (2)
8.4 | OPTION/OPSIE 1 Notes/Aantekeninge:
Eceuse = Eri eel poe “i al sotrect formate
_A , Aanvaar enige ander korrekte
=0,93Vv
formule uit die datablad.
Any other formula using
unconventional abbreviations, e.g.
Fe o7®S, KER. followed by correct
substitutions:
Max: 3/ ‘4
Enige ander formule wat
onkonvensionele afkortings gebruik,
by. F.7ES KER.
gevolg deur korrekte vervangings:
Maks: 3/ 4
OPTION/OPSIE 2
Agt +e — Ag lw E® = +0,80 (V) v
Pb — Pb’ + 2e" E° = +0,13 (V) Vv
2Ag* + Pb — 2Ag + Pb?* E° = 40,93 Vv
(4)
[11]

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FS/VS/September 2025

QUESTION 9/VRAAG 9

9.1. (Gas) mask/work in fume cupboard ¥

(Gas)-masker/werk in dampkas (1)
9.2 Chlorine gas is poisonous/Chloorgas is giftig v ¥ (2)
9.3 Qto/tofP  v (1)
9.4 Copper/Koper ¥ (1)

9.5 2Cf(aq) + Cu**(aq) Y — Ch(g)+Cu(s)~  Balancing/Balansering Vv

Marking criteria/Nasienkriteria:
e Reactants ¥ Products ¥ Balancing ¥
Reaktanse Produkte Balansering
e Ignore = and phases
Ignoreer = en fases
e Marking rule 6.3.10/Nasienreé! 6.3.10

(3)
[8]

TOTAL/TOTAAL: 150

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