# Physical Sciences Grade 12 Paper 2 North West Mock 2025 Memo

> Grade 12 Physical Sciences Paper 2 memorandum from the North West Mock Exam 2025, 15 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physics-grade-12-nsc-p2-memo-september-2025-north-west/dcdc3b5f-d332-4abd-ae1b-5d1b8a2e3099
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2025
- Exam period: North West Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 15

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## About this document

Physical Sciences · Grade 12 · North West Mock Exam · 2025. Memorandum, 15 pages. Read online or download the PDF.

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### Page 1

education

Department:

Education

North West Provincial Government
REPUBLIC OF SOUTH AFRICA

NATIONAL SENIOR CERTIFICATE
NASIONALE SENIOR SERTIFIKAAT

GRADE/GRAAD 12

SRSSSPS SSS SRC RRR RRR STARTER RESET RRR

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

MARKING GUIDELINES/NAS/ENRIGLYNE
SEPTEMBER 2025

MARKS/PUNTE: 150

These marking guidelines consist of 13 pages
Hierdie nasienriglyne bestaan uit 13 bladsy

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### Page 2

Physical Sciences/P2 NW/September 2025

QUESTION/VRAAG 1

1.D vv (2)
12 BO vv (2)
13 D vv (2)
14 A vv (2)
165 B vv (2)
16 A vv (2)
17° D vv (2)
18 C vv (2)
19 CC vw (2)
110 D vv (2)
[20]

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### Page 3

Physical Sciences/P2 NW/September 2025

QUESTION/VRAAG 2

24. A bond or an atom or a group of atoms that determine(s) the physical and
chemical properties of a group of organic compounds. v v (2 or 0)
‘n Chemiese binding of ‘n atom of ‘n groep atome wat die fisiese en chemiese
eienskappe van ’n groep organiese verbindings bepaal. (2 of 0) (2)

2.2 i
-¢€-0- G-— 4
| (1)

2.3
231 FY (1)
2.3.2 Butanal/Butanaal VV (2)

2:33) 2,3-dichloro-4,5-dimethylheptanev v ¥
2,3-dichloro-4,5-dimetielheptaan

Marking criteria/Nasienkriteria:

e Correct stem i.e. heptane. ¥

e All substituents (dichloro and dimethyl) correctly identified.”

e IUPAC name completely correct including numbering, sequence,
hyphens and commas. ¥
Korrekte stam d.i heptaan ¥

e Alle sunstituente (dichioro en dimetiel) korrek geindentifiseer
IUPAC naam heeltemal korrek insluitende nommering, volgorde,

koppeitekens en kommas. (3)
2.3.4 H/O\H H
pfu \r ot
H— r oF c— ro 4 Marking criteria/Nasienkriteria
H ¥v H Functional groupY
H=C =H Whole strucure corectY

Funksionele groep ¥
Hele struktuur korrek ¥ (2)

|
H

2.35 H nae H y
|

H=c —C. Cc— c-C-C-H
toot I |
H\WYFS/H H

Marking criteria/Nasienkriteria

Flourines on C3 ¥
Whole strucure correct ¥
Fluore op C3 ~

Hele struktuur korrek “

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### Page 4

Physical Sciences/P2 NW/September 2025

2.4.1 Organic molecules with the same molecular formula, “but different structural
formulae ~
Organiese molekules met dieselfde molekulére formule, maar verskillende
struktuurformules V (2)
2.4.2 Propan-1-ol/ 1-propanol Vv and Propan-2-ol/ 2-propanolY
Propan-1-ol/ 1-propanol en Propan-2-ol/ 2-propanol ¥ (2)
2.5 Aldehyde¥ and ketone “OR carboxylic acid and ester
Aldehied “en ketoon ~ OF karboksielsuur en ester (2)
[19]
QUESTION/VRAAG 3
3.1 The temperature at which the vapour pressure of a substance equals

3.2
3.2.1

3.2.2

3.3

atmospheric pressurev ¥
Die temperatuur waarby die dampdruk van ’n stof gelyk is aan die atmosferiese
druk.

Markingcriteria/Nasienkriteria

If any one of the underlined key phrases in the correct context is
omitted, deduct 1 mark./ndien enige van die onderstreepte frases in die
korrekte konteks uitgetaat is, trek 1 punt af. (2)

Carboxylic acid/Karboksielsuurv (1)

Hydrogen bond or dipole-dipole bond.“
Waterstofbinding of dipoo!l-dipool binding (1)

From A to D boiling point increasesY

e Structure:
The chain length/number of carbon atoms/molar mass/ surface area/
contact area increases V from compound A to D.

e Intermolecular forces:
The strength of the London forces/induced dipole/Van der Waals/
intermolecular forces increase. Vv

e Energy:
More energy needed to overcome/break the intermolecular forces from A
to D.v
OR

Boiling point decrese from D to AY

e Structure:
The_chain length/number of carbon atoms/molar mass/ surface area/
contact area decreases from compound D to A.”

e Intermolecular forces:
The strength of the intermolecular forces/London forces decrease from D
toAv

« Energy:
Lesser energy needed to weaken/overcome/break the intermolecular
forces from D to A.”

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Physical Sciences/P2 NW/September 2025

© =©Struktuur:
Die kettinglengte/aantal koolstowwe/molére
massa/reaksieopperviak/kontakarea neem toe van verbinding A na D.
« Intermolekulére kragte:
Die sterkte van die Londonkragte/geinduseerde dipoolkragte/Van der
Waalskragte neem toe.

e Energie:
Meer energie benodig om intermolekulére kragte te oorkom/breek van A

na D.
Van D na A neem die kookpunt af

e = Struktuur:
Die kettinglengte/ aantal koolstowwe/molére
massa/reaksieopperviak/kontakarea neem af van verbinding D na A..
e  Intermolekulére kragte:
Die sterkte van die Londonkragte/geinduseerde dipoolkragte/Van der
Waalskragte neem af.

e =©Energie:
Minder energie benodig om intermolekulére kragte te oorkom/breek van D (4)
naA.
3.4 C/butanoic acid vit has the highesi/higher boiling pointY
C/butanoésuur ¥ dit het die hoogste/hoér kookpunt ¥ (2)
3.5 Pentanoic acid/Pentanoésuur v ¥ (2)

3.6 e Compound D/Pentanoic acid/carboxylic acid has two sites for hydrogen
bond v
Esters has dipole-dipole forcesY
Intermolecular forces in compound D/pentanoic acid are stronger than
Intermolecular forces in esters V
OR
Intermolecular forces in ester are weaker than in compound D
e Less energy needed to overcome/break the intermolecular forces in ester
than in compound D/pentanoic acid.”
OR
More energy needed to overcome/break the intermolecular forces in D
than in ester.

e Verbinding D/Pentanoésuur/carboksielsuur het twee plekke vir
waterstofbinding ~
Ester het dipool-dipoolkragte ~
Intermolekulérekragte in verbinding D/pentanoésuur is sterker as die

intermolekulérekragte in esters ~

OF

Intermolekulérekragte in ester is swakker as in vebinding D

Minder energie benodig om die intermolekulére kragte te oorkom/breek in

esters as in verbinding D/pentanoésuur.¥

OF

Meer energie benodig om die intermolekulére kragte te oorkom/breek in

verbinding D/pentanoésuur as in esters.v (4)
[16]

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Physical Sciences/P2 NW/September 2025

QUESTION/VRAAG 4

4.1.1

4.1.2

4.2

4.3.4
4.3.2
44

4.5.1

4.5.2

Copyright reserved/Kopiereg voorbehou

Substitution/Hydrolysis reaction. ¥
Substitusie/Hidrolise reaksie

(1)

NOTE/NOTA
IPenalise 1 mark if a hyphen is missing/Penaliseer as daar ‘n koppelteken
luitgelaat is.

4-chloropropane/1-chioropropaan v ¥

Primary alcohol “

Alcohol in which the carbon atom bonded to the functional group(-OH) is
attached to one carbon atom. ¥ “

Prim6re alkohol

Alkohol waar die koolstofatoom gebind aan die funksionele groep(-OH) gebind
is aan een ander koolstofatoom.

Dehydration reaction/Dehidrasie reaksie V
CH3CH2CH20H ¥ > CH3CHCH2Y + H20 ¥

Propane/Propaan V¥

(2)

(3)
(1)
(3)
(2)

INOTE/NOTA
INo mark for dilute H2SO4/Geen punt vir verdunde H2SO4

Sulphuric acid/Hydrogen sulphate/H2SO4/Phosphoric acid/HsPO4 ¥
Swawelsuur/Waterstofsulfaat/H2S O4/Fosforsuur/ H3PO4

Marking criteria/Nasienkriteria:

Correct structural formula for methanoic acid ¥
Correct structural formula for alcohol ¥

The whole structural formula for ester correct ¥
H20 ¥

Korrekte struktuurformule vir metanoésuur ¥
Korrekte struktuurformule vir alkohol ¥

Die hele struktuurformule vir die ester is korrek ¥
H20 ¥

IF/AINDIEN
Condensed structural formula used/Gekondenseerde struktuur formule
gebruik Max/Maks %

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(1)

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Physical Sciences/P2 NW/September 2025

H H H v fe)
I | | ll v
H-C- C-— CC —O-H + H-C —0O-H—>
| { [
H H H
HOT Ir + HoOv
H-C -C —-C -O-C-H¥Y
| | I
H H H (4)
[17]

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### Page 8

Physical Sciences/P2 NW/September 2025

QUESTION/VRAAG 5
a4 ANY ONE:

Ei
e

Change in concentration Y of products/reactants per unit time.v

Change in amount/number of mole/volume/mass “of products/reactants per
unit time.v

Amount/number of mole/volume/mass of products formed/reactants usedv

er unit time.”
Rate of change in concentration/amount/number of moles/volume/mass.v ¥

(2 or 0)

NIGE EEN:
Verandering in konsentrasie “van produkte/reaktanse per eenheidtyd. ~
Verandering in hoeveelheid/getal mol/volume/massa “van
produkte/reaktanse per eenheidtyd. ~
Hoeveelheid/getal mol/volume/massa produkte gevorm/reaktanse gebruik~

per eenheidtyd. ~
Tempo van verandering in konsentrasie/hoeveelheid/getal (2)

mol/volume/massa. “(2 or 0)

NOTE/NOTA
Give the mark for per unit time only if in correct context of reaction rate
Gee die punt vir per eenheidtyd slegs indien in korrekte konteks met reaksietempo.

5.2 Surface area/State of divisionv

Kontakopperviak/Toestand van verdeeldheid (1)
5.3. Experiment/Eksperiment Iv (1)
5.4 e The surface area increase/Experiment 2 has a larger surface area/_more

particles exposed (to oxygen gas)”

More particles will have enough kinetic energy”

More effective collision per unit time/second/Higher frequency of effective
collisions ¥

Die kontakoppervlak vergroot/Eksperiment 2 het ‘n groter
kontakopperviak/meer deeltjies word blootgestel (aan suurstofgas)

Meer deeltiies het genoeg/voldoende kinetiese energie

Meer effektiewe botsings per eenheidtyd/sekonde/hoér frekwensie van
effektiewe botsings

(3)

5.5.1 Marking criteria/Nasienkriteria:

® Am (7,36-0) v & At (5,2-0) ¥

e _Answer/Antwoord 1,42 (g:s"') Vv

Rate/Tempo =Am/At

= (7,36-0) v/ (5,2-0) v
= 1,42 (g-s1) v (3)

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### Page 9

Physical Sciences/P2

5.5.2

5.6

NW/September 2025

Marking criteria:
e Formula nav
M

e Substituting 94 g-mol' vin

n=
M
USE mol ratio n{K) = 2n(K20) ¥

Substitution of 39 g-mol"’ and
‘ . m
reacting mole in n “" v

6,11

« Substitution of —— (100)”
6,92

e Answer 88,25%~”
Range: 88,25% - 89,05%

Nasienkriteria:

© Formule n= ~
M

e Vervanging van 94 g-mol? vin

n=—
M
e GEBRUIK molverhouding
n(K) = 2n(K2O) ~
e = Vervanging van 39 g:mor

4 m
en reagerende mol in n= oe

e Vervanging van oa (100) “

e Antwoord 88,25% ~“
Range: 88,25% - 89,05%

m
n=—v
M
7,36
eo
n= 94
= 0,0783 mol

\

n(K) = 2 x 0,783 ¥

= 0,1566 mol
n(K) =m/M
m(K) = 0,1566 x 39”
=6,11g

% purity/suiwerheid = c= (100) v

= 88,25% ¥

Negative Y Exothermic reaction/energy of the product < energy of the reactant/

energy of the reactant > energy of the product “

Negatief ¥ Eksotermiese reaksie/energie van die produkte < energie van die

reaktanse/ energie van die reaktanse > energie van die produkte ¥

ACCEPT/AANVAAR

More energy is released than absorbed/Meer energie word vrygestel as wat

geabsorbeer word.

(6)

Combustion reaction. It releases energy/Verbrandingsreaksie. Stel energie vry (2)

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[18]

### Page 10

Physical Sciences/P2 NW/September 2025

QUESTION/VRAAG 6

6.1

6.2

6.3

6.4

The dynamic equilibrium when) the rate of the forward reaction equals the rate of
the reverse reaction.v v (2 or 0)

OR

The stage in a chemical reaction when the concentration of reactants and
products remain constant

(Die dinamiese ewewig wanneer) die tempo van die voorwaartse reaksie gelyk is
aan die tempo van die terugwaartse reaksie. (2 of 0)

OF

Die stadium in ’n chemiese reaksie waar die konsentrasie van die reaktanse en

produkte konstant bly. (2)
Endothermic/Eksotermies VV

e Decrease in temperature favours the the reaction that decreases
temperature (exothermic reaction) ¥
e The reverse reaction is favouredY
¢ Concentration of reactant/NO2 increases ¥
OR
Concentration of products/NO and Oz decreases
e Afname in temperatuur bevoordeel die eksotermiese reaksie “
e Die terugwaartse reaksie word bevoordeel~
e Konsentrasie van reaktanse/NO2 neem toe ~
OF (3)
Konsentrasie van produkte/NO en O2 neem af

CALCULATIONS USING NUMBER OF MOLES/BEREKENING DEUR
MOL TE GEBRUIK

Marking criteria:
a) Change in n(NOz) = equilibrium nNOz) - initial n(NOz) = x - 0,6 ¥
b) Using ratio: NO: N20: O2=2:2:1v4
c) Equitibrium n(NO) = intial n{NO) + change in n(NO)
Equilibrium n(Oz) = intial n(Oz) + change in n(O2) v
d) Dividingby2 v
e) Correct Ke expression in square bracketsY
f) Substitution of equilibrium constant’
g) Substitution of equilibrium concentrations into Kc expressionsY
h) Final answer. 1,48 mol“
Nasienkriteria:
a) Verandering in n(NO2) = ewewig nNO2) - aanvanklike n(NOz) = x -0,6”
b) gebruik verhouding: NO : N20 ; O2 = 2: 2:1 Vv
c) Ewewig n(NO) = aanvanklike n(NO) + verandering in n(NO)
Ewewig n(O2) = aanvanklike n(O2) + verandering in n(O2) v7
d) Korrekte Ke uitdrukking in vierkantige hakkiesY
e) Deeldeur2 v
f) Vervanging van ewewigskonstante ¥
g) Vervanging van ewewigskonsentrasies in Ke uitdrukkingY
i) _ Finale antwoord. 1,48 mol~”

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### Page 11

Physical Sciences/P2

Mol 2NOz2 2NO O2
Initial mole x 0 0
Aanvanklike mol =
Change in mole x-0,6V C_| x-0,6 | x-0,6/2> v
Verandering in mol ———
Equilibrium mole 0,6 Cc x-0,6 (x-0,6/2) |v
| by ewewig ———o
ncentration 0,3. C¢ | (x-0,6)/2 x-0,6/2 + 2 DV

Ke = [NO}x [Oa _”
[NO2}2

= (x= 0,62)? x (x-0,6/4) “ x= 1,48 molv
(0,3

0,48 %

QUESTION/VRAAG 7

iaiA

dete

Copyright reserved/Kopiereg voorbehou

Solution with known concentration vv
Oplossing met bekende konsentrasie

[Marking criteria/Nasienkriteria:
Formula c = oY or n=n/M or c=n/V

Dividing by/dee/ deur 0,5dm3 ¥

Substituting/vervanging van 40 g-mol' and 4gv
Answer/AntwoordY

OPTION 1/OPSIE 1 OPTION 2/OPSIE 2

Ic = m/MV ¥ n=n/M

c= 4/(40x0,5) v vw = 4/40 ¥

c= 0,2 mol+dm? ¥ =0.1 mol _ |Any one/Enige een ¥

c=n/V
= 0.1/0.5 ¥
= 0.2 mol:dm? ¥

NW/September 2025

(8)

[15]

(2)

(4)

Marking criteria/Nasienkriteria:
e Formula 22Va _ Ma v
C.V, n,
e Substitution on CaVaand Substitution on CeVp~/ Substitution on n= cV and
n=cV
e Ratio 1:1”
e Final answer 0,17 mol: dm3’v”
e Formule £2Va _Ma vw
ChVp any
Vervanging in CaVa & vervanging in CpVp” Vervanging van n=cV enn =cV
Verhouding 1:1”

Finale antwoord 0,17 mol :dnv3

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### Page 12

Physical Sciences/P2 NW/September 2025
OPTION 4/OPSIE 1: OPTION 2/OPSIE 2:
CV, _ Ny n (NaOH) = cV ¥
8. 2G.
CaX35 ly = 0,2 xX 0,03
02x30 2
ca = 0,17 mol«dm3 v = 0,006 mol
Ratio: 1:1
n (HCl) = 0,006 mol ¥
c=2 v
4
= 2,006
~ 0,035
= 0,17 mol-dm3 ¥ (4)

7.14 NOv

It is a salt of strong acid and a strong base. v

Dit is ‘n sout van ‘n sterk suur en 'n sterk basis

7.2.1 Neutral ¥

[H3O*] = [OH] = 1x107?’moldm* or pH=7  ¥

ACCEPTIAANVAAR
[H3sO*] = [OH]

7.2.2 |OPTION 1/OPSIE 1: OPTION 2/OPSIE 2:
pOH = -log [OH] ¥ [HsO*][OH}] = 1x10-4
pOH = -log (3x 10-7) v| |[HsO*][3x10°7] = 1x10-4
pOH=6,52 v¥ [H30*] = 3.33x10°% (moledm*) | “~~ Any one
pH + pOH = 14
pH +6,52=14 v pH = -log [H30*]
pH = 14-6,52 = -log [3.33x10*] v
pH=7,48 ¥ =7.48 V

QUESTION/VRAAG 8
8.1 e Concentration must be 1 mol-dm?v
« Temperature must be 25°C/298KY
e Konsentrasie moet 1 mol-dnr? wees~
e Temperatuur moet 25°C/298K wees”

8.2 Sodium chloride/NaCf/Potassium chloride/KC? or any suitable electrolyte”
Natriumchloried/ NaCt/Kaliumchloried/ KC#/ of enige geskikte elektroliet

ACCEPT/AANVAAR

Any metal nitrates/Enige metal nitrate

QR2 Flactrada wihara radiictinn tavanlarca/ /

(2)

(2)

(5)
[19]

(2)

(1)

### Page 13

Physical Sciences/P2

8.4

8.5

8.6

8.7

Copyright reserved/Kopiereg voorbehou

NW/September 2025

Marking guidelines/Nasienkriteria:
© Pb*%+2e= Pb (1/,)

° Pb < Pb*+2e (2/5)
¢ Pb = Pb%+2e (%/,)
© Pb Pb%+2e- — (0/5)

e Ignore charges omitted on electron//Ignoreer indien lading
weggelaat is op elektron

e If charge (+) omitted on Pb** / Indien (+) lading weggelaat is op
Pb?* Max. (1/5)

Example/Voorbeeld: Pb? + 2e° > Phv¥

Pb** + 2e° > Ph vv

E®cei! = E® reduction — E®oxidationY
E®eu = -0,13¥ — (-0,91) ~
= 0,78V¥

Notes/Aantekeninge:

e Accept any other correct formula from the data
sheet./Aanvaar enige ander korrekte formule
vanaf gegewensblad

e Any other formula using unconventional
abbreviations, ¢.g E®csn = E° oa — E® ra
followed by correct substitutions/Enige ander
formule wat onkonvensionele afkortings
gebruik bv.

E®sei = E° om — E® rm gevolg deur korrekte
vervangings: %

arking criteria/Nasienkriteria:
° oxidation’ double lineY —_ reduction V
e oksidasie ¥ dubbellyn“ —_ reduksie ¥

Cr(s)/ Cr2* (aq)(1 mol-dm3)VIl ¥ Pb2*(aq)(1 mol-dm3)/ Pb(s)”

Increases/VerhoogY
Cr is oxidised to Cr?*v
Cr is geoksideer na Cr?*

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(2)

(4)

(3)

(2)
[16]

### Page 14

Physical Sciences/P2

QUESTION/VRAAG 9
9.1 Cell in which electric energy is converted to chemical energy.v Y (2 or 0)
Sel waarin elektriese energie omgeskakel is na chemiese energie .v V (2 or 0)
9.2 Copper/Koper (Cu)v
9.3 Xv
Oxidation takes place at X/ X loses electrons V
Oksiedasie vind by X plaas/X verloor elektrone
9.4

Copyright reserved/Kopiereg voorbehou

NW/September 2025

Marking guidelines/Nasienkriteria:
© Cu +2e = Cu (1/5)

* Cus Cu*+2e (2/5)
¢ Cu = Cu%*+2e (9/,)
* CuCu*+2e = (%,)

e Ignore charges omitted on electron/fgnoreer indien lading weggelaat is
op elektron

e If charge (+) omitted on Cu?*
Indien lading (+) lading weggelaat is op Cu2* Max. (1 />)

Example/Voorbeeld: Cu? + 2e > Cuv

Cu?* + 2e > Cuvv

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(2)
(1)

(2)

(2)

### Page 15

Physical Sciences/P2

9.5 Marking criteria/Nasienkriteria:
« ReactantsY ProductsY
Ignore double arrows
Ignore phases

Marking rule 6.3.10

{gnoreer dubbel pyle
{gnoreer fases
Nasienreél 6.3.10

BalancingY

Reaktanse~  Produkte~  Balansering~

Cu**(aq) + 2C£(aq)” > Cu(s)“+ Cha balv

ACCEPT/AANVAAR
CuClz (aq) ¥ > Cu(s)¥+ Clo balv

Copyright reserved/Kopiereg voorbehou

NW/September 2025

(3)
[10]

TOTAL/ITOTAAL: 150

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- [Physics Grade 12 NSC P2 QP September 2025 Gauteng](https://www.hlayiso.com/library/physics-grade-12-nsc-p2-qp-september-2025-gauteng/4d3cbea6-10ec-47ff-98ee-50646165ae9f) — Physical Sciences · QP · 2025
- [Physical Sciences P2 May-June 2025 Eng](https://www.hlayiso.com/library/physical-sciences-p2-may-june-2025-eng/efdef1a0-c94f-4d02-954a-5262f99ca4af) — Physical Sciences · QP · 2025 · Eng
- [NW Phys .Fis Memo](https://www.hlayiso.com/library/nw-phys-fis-memo/c47091be-07ff-46dd-b160-ce6147c6ea06) — Physical Sciences · Memo · 2016

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

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