# Physical Sciences Grade 12 Paper 2 Mpumalanga June 2025 Memo

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

- Page: https://www.hlayiso.com/library/physical-sciences-physics-grade-12-nsc-memo-p2-may-june-2025-mpumalanga/da672fd3-1de8-427d-ad4c-aecb6e56da1c
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2025
- Exam period: Mpumalanga June Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 15

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Physical Sciences · Grade 12 · Mpumalanga June Exam · 2025. Memorandum, 15 pages. Read online or download the PDF.

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### Page 1

education

MPUMALANGA PROVINCE
REPUBLIC OF SOUTH AFRICA

NATIONAL SENIOR CERTIFICATE
NASIONALE SENIORSERTIFIKAAT

GRADE/GRAAD 12

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

JUNE/JUNIE 2025

MARKING GUIDELINES/NAS/ENRIGLYNE

MARKS/PUNTE: 150

This marking guidelines consists of 15 pages.
Hierdie nasienryglyne bestaan uit 15 bladsye.

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NSC/NSS- Marking Guidelines

QUESTION 1 / VRAAG 1

110 BYY
12 Cvv
13° Dv
14. Avv
15 Cvv
16 Cvv
17 BYY
18 Dvv
19 Dvv
110 Dv¥v [20]

QUESTION 2 / VRAAG 2

2.1.1 Cv (1)
2.1.2 Av (1)
213 Dx (1)
2.2.1 5-ethyl-2-methylhept-3-yne /5-etiel-2-metielhept-3-yn
Marking criteria: Nasienkriteria:
e Correct stem i.e. heptyne. ¥ e Korrekte stam d.i. heptyn. ¥
e All substituents (ethyl and methyl) ¢ Alle substituente (etiel en metiel)
correctly identified. v korrek geidentifiseer. ~
e IUPAC name completely correct « /UPAC-naam _ heeltemal korrek
including numbering, sequence, insluitende volgorde, koppeltekens
hyphens and commas. ¥ en kommas. ¥ (3)

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2.2.2 HK
,  H-G-H
H H H Cl H | HH
I ft f tot | | v
“TTP rt tT
H Cl | H H HH H
H—C—H
H—C—H
H v

Marking criteria/Nasienkriteria

e Eight C atoms in longest chain/ Agt C-atome in langste kettingY

¢ One Cl atom on C2 and one on C4./Een Ci-atom op C2 en een op C4 Vv

e Ethyl! substituent on C3 and methy! substituent on C6./Etielsubstituent op C3
en metielsubstituent op C6V

IF/INDIEN
H-atom or bond omitted/H-atoom of binding uitgelaat. Max/Maks: 2/5 (3)
2.2.3. Aldehyde/Aldehied v (1)
2.2.4 O
| tl 4
on
(1)

2.3.1 | Marking criteria/Nasienkriteria
If any one of the underlined key phrases in the correct context is omitted, deduct

1 mark./indien enige van die onderstreepte frases in die korrekte konteks
uitgelaat is, trek 1 punt af.

The underlined phrases must be in the correct context. / Die onderstreepte
frases moet in die korrekte konteks wees.

Organic compound with the same molecular formula, but_different positions of

the side chain, substituents or functional groups on the parent chain. Vv
Organiese verbindings met dieselfde_molekulére formule, maar verskillende

posisies van die syketting. substituente of funksionele groepe op die stamketting (2)

2.3.2 H H H H Vv Marking criteria/Nasienkriteria
ee tog Ep ds e OH-group on C1 ¥
| | | |-—__— e OH-groep op C1
H H H H v e Whole structure correct. ¥
e Hele struktuur korrek (2)

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### Page 4

23.3

2.4.1

2.4.2

2.4.3

2.4.4

25

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NSC/NSS- Marking Guidelines

Secondary alcohol/Sekondére alkohol ¥

The —OH group is (covalently) bonded to a carbon that is bonded to two other

C-atoms “

Die —OH groep (kovalent) gebind is aan ‘n koolstof wat gebind is aan twee ander

C-atome

Esterification /Condensation/Esterifikasie/Kondensasie ¥

Marking criteria/Nasienkriteria

Functional group correct ¥
Funksionelegroep korrek
Whole structure correct. ¥
Hele struktuur korrek

H2SO4/ Sulphuric acid/Swae/suur ~ (a)
Piping
a I} H e
H—C—C—C—C—O—C-+C—H
| | | “F-~ | vy
H H H H H i.
Butyl Yethanoate~/ Butieletancaat

Marking criteria / Nasienkriteria:

Substitute / Vervang 72 g:mol n = a v
Use mole ration/ gebruik molverhoudingv

Substitute number of moles and 22,4 dm?-mol™ in n = = v
m

a ‘, . actual
Substitute / Vervang 76% in %yield = acon

Final answer / Finale antwoord: 53,2 dm? v

v

_m
"=u
_ 45
~ 72v
= 0,625 mol
nCO, = 5nCsH12
= 5(0,625)
“en me NN
V=fVa ] ~~ ”
= 3,125 x 22,4V OR/OF ate Sonn
= 70 dm? Nactuatwerkiik = 2.375 mol
_x 'V=nVm
on tamenil = 2,375 x 22,4v
Vactualiwerkik = 53,2 dmv = 53,2 dmsv

Copyright Reserved/Kopiereg voorbehou

(2)
(1)
(1)

(2)
(2)

(5)
[28]

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QUESTION 3 / VRAAG 3

3.1

3.2

3.3

3.4.1

3.4.2
3.4.3
3.5
3.6

Copyright Reserved/Kopiereg voorbehou

Marking criteria/Nasienriglyne

lf any one of the underlined key phrases in the correct context is omitted,
deduct 1 mark./Indien enige van die onderstreepte frases in die korrekte
konteks uitgelaat is, trek 1 punt af.

The temperature at which the vapour pressure of a substance equals

atmospheric pressure. /Die temperatuur waarby die dampdruk van die stof gelyk
is aan atmosferiese druk. VV

IF Temperature is omitted 0/2.

Flamable/ViambaarY

Use straight chain v primary alcohols v
Gebruik requitketting primére alkohole

Chain length/number of C-atoms/molecular mass/surface area ¥
Kettinglengte/aantal C-atome/molekulére massa/opperviak

Hydroxyl (group)/Hidroksiel(groep) ¥
Ethanol/Etanol “

C/propan-1-ol/propan-1-ol ¥ ACCEPT/ANVAAR: Formula/formule

Marking criteria:
e Compare the structures of the compounds

e Compare the strength of intermolecular forces
e Compare the energy required to overcome intermolecular forces

e Vergelyk die struktuur van die verbindings
e Vergelyk die sterkte van intermolekulére kragte
e Vergelyk die energie benodig om intermolekulére kragte te oorkom

C/CH3CH2CH2OH/Propan-t1-ol
e Structure/Struktuur

LONGEST chain length/largest surface area (over which intermolecular
forces act ¥

e Intermolecular forces
STRONGEST/MOST intermolecular forces/ Van der Waals forces/London
forces/ dipole-dipole forces V

e Energy
Most/more_ energy is needed to overcome or break intermolecular
forces/Van der Waals forces/dipole-dipole forces VW

OR

(2)
(1)

(2)

(1)
(1)
(1)
(1)

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A/CH3;0H/methanol

« Structure
Shortest_chain length/smallest surface area (over which intermolecular
forces act ¥

« Intermolecular forces
Weakest/less intermolecular forces/ Van der Waals forces/London forces/
dipole-dipole forces ¥

e Energy
Least energy is needed to overcome or break intermolecular forces/Van der
Waals forces/dipole-dipole forces ¥

OR

From A to C

e Structure
Chain length/ surface area inreases. ¥

e Intermolecular forces
The intermolecular forces/ Van der Waals forces/London forces/ dipole-
dipole forces increase. ¥

e Energy
The energy require to overcome the intermolecular forces /Van der Waals
forces/London forces/ dipole-dipole forces increases ¥

OR

From C to A

e Structure
Chain length/ surface area decreases. ¥

e Intermolecular forces
The intermolecular forces/ Van der Waals forces/London forces/ dipole-
dipole forces decrease. ¥

e Energy
The energy require to overcome the intermolecular forces /Van der Waals
forces/London forces/ dipole-dipole forces decreases V

C/CH3CH2CH20H/Propan-1-ol
e = Struktuur

Langste_kettinglente/grootste opperviak (waaroor intermolekulére kragte
werk) ¥

e Intermolekulére kragte
Sterkste/meeste _intermolekulére__kragte/ Van der Waalskragte/
Londonkragte/ dipool-dipoolkragte ¥

e Energie
Meeste energie benodig om _ die __intermolekulére__kragte/Van der
Waalskragte/ dipool-dipoolkragte te oorkom/breek. ¥

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OF

C/CH30H/Metanol

«© = Struktuur
Kortste_kettinglente/kleinste opperviak (waaroor intermolekulére kragte
werk) ¥

e Intermolekulére kragte
Swakste/minste__intermolekulére__kragte/ Van der  Waalskragte/
Londonkragte/ dipool-dipoolkragte ¥

« Energie
Minste energie benodig om die intermolekulére kragte/Van der Waalskragte/
dipool-dipooikragte te oorkom/breek. ¥

OF

Van A tot C

e Struktuur
Kettinglengte/ opperviakte(waaroor intermolekulére kragte werk) verhoog.”

e Intermolekulére kragte
Die intermolekulére kragte/ Van der Waalskragte/ Londonkragte/ dipool-
dipoolkragte verhoog.“

e Energie

e Die energie benodig om die intermolekulére kragte/Van der Waalskragte/
dipool-dipoolkragte te oorkom/breek verhoog. ¥ (3)

OF
Van C totA
e Struktuur
Kettinglengte/ opperviakte(waaroor intermolekulére kragte werk) verlaag.~”
e Intermolekulére kragte
Die intermolekulére kragte/ Van der Waalskragte/ Londonkragte/ dipool-
dipoolkragte verlaag.¥
e Energie
Die energie benodig om die intermolekulére kragte/Van der Waalskragte/
dipool-dipoolkragte te oorkom/breek verlaag. V

3.7.1. The two compounds have the same molecular mass/Die twee verbindings het
dieselfde molekulére massa (1)

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Boke

Marking criteria:

State that carboxylic acids have two sites for hydrogen bonding
State that alcohols have one site for hydrogen bonding
Compare the strength of intermolecular forces

Compare the energy required to overcome intermolecular forces
Nasienkriteria:

Benoem die intermolekulére kragte

Ste! dat karboksielsure twee plekke het vir waterstofbindings
Stel dat aikohole een plek het vir waterstofbinding

Vergelyk die sterkte van die IMK's / energie benodig om IMK's te
oorkom

Vergelyk die energie benodig om intermolekulére kragte te oorkom

Copyright Reserved/Kopiereg voorbehou

Compound C/ CH3CH2CH:OH/ propan-t-ol/ alcohol has one site for
hydrogen bonding

Compound X/CHsCOOH/ ethanoic acid/carboxylic acid has two/more sites
for hydrogen bonding. “

Intermolecular forces in Compound X/CH3COOH/ ethanoic acid / carboxylic
acid are stronger than the intermolecular forces in compound _ C/
CH3CH2CH2OH/ propan-1-ol/ alcohol. ~

More energy is needed to overcome/break intermolecular forces in
compound Compound X/CHsCOOH/ ethanoic acid / carboxylic acid than
Compound C/ CH3CH2CH20H/ propan-1-ol/ alcohol “

Verbinding C/ CH3sCH2CH2O0H/propan-t-ol/alkohol het een punt_vir
waterstofbindings en verbinding X/CH3COOH/etanoésuur / karboksielsuur
het twee/meer punte vir waterstofbindings. “

Intermolekulére__kragte in _verbinding — X/CH3COOH/__ etanoésuur/
karboksielsuur_is_ sterker_ as die intermofekulére kragte in verbinding C/
CH3CH2CH20H/propan-1-ol/alkohol. “

Meer energie word benodig om _intermolekulére_kragte in _verbinding
X/CHsCOOH/etanoésuur/__karboksielsuur__as__in__verbinding _C/
CH3sCH2CH20H/propan-1-ol/alkohol “

(4)
[17]

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NSC/NSS- Marking Guidelines

QUESTION 4/ VRAAG 4

4.1.1

4.1.2

4.1.3

4.2.1

4.2.2

4.2.3

4.3.1

4.3.2

4.4

Copyright Reserved/Kopiereg voorbehou

Addition/hydrogenationY
Addisie/hidrogenasie/hidrogenering

Substitution”
Substitusie

Elimination/dehydrationY
Eliminasie/dehidrasie

Butan-2-ol/2-Butanolv

Marking criteria:

¢ Correct stem i.e.butanol ¥

e |UPAC name completely correct including numbering and hyphens.”
Nasienkriteria

e Korrekte stam d.i. butanol

e /UPAC-naam heeltemal korrek insiuitende nommering en koppeltekens.

Water/H20 ¥

Hydration/Hidrasie. ¥

Saturated/Versadig ¥

Compounds in which there are no multiple bonds between C atoms in their
hydrocarbon chains. “

Verbindings waarin daar geen meervoudige bindings tussen C-atome in hul
koolwaterstofkettings is nie

Heat/UV light/hf/ Hitte/UV lig/hf ~

H /Cl\ H_ H H oH V H

lit} 1 | % vo strongly | jets 1 ¥
H—C—-C+-C—C—H + NaOH(c) ——* H—-C-*C==C-+-C—H + NaCl + H20

1 Yvyod heated J. oe

H H HH H H H

Marking Criteria/Nasienriglyne
e Chlorine on the second C for 2-chlorobutaneY
Chloor atoom op die tweede C-atoom vir 2-chiorobutaan
e Whole structural formula correct ¥
Hele struktuurformule vir 2-bromobutaan korrek
e NaOH as reactant ¥
NaOH as ‘n reaktant
e Water(H20) as part of the productsY¥
Water (H2O) as deel van die produkte
e The functional group on the second carbon for but-2-eneY
Die funksionele groep op die tweede koolstof atoom vir but-2-een
e Whole structural formula correct “
Hele struktuur korrek

(1)

(1)

(1)

(2)
(1)
(1)

(2)
(1)

(6)

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NOTE/LET WEL

molecular formulae
e Penaliseer slegs

« Penalise only once for the use of condensed structural formulae or

een keer vir die gebruik van gekondenseerde

Struktuurformules of molekulére formule

4.5 No water (H2O)” and a non-reactive solvent/dissolve in ethanolY
Geen water (H20) en onreaktiewe oplosmiddel/ los op in etanol

QUESTION 5/ VRAAG 5

5.14 _m,
n(Q2) =

_ 36,8
~ 32v
= 1,15 mol”

Marking criteria/Nasienriglyne

® Formula /Formule
¢ Substitute 32 in formula / Vervang 32 in formule
« Answer / Antwoord

5.1.2 POSITIVE MARKING FROM QUESTION 5.1.1

n(PbS) = = n(02)

n(PbS) =F (1,15)

= 0,7666 mol
m
n=
m
0,7666 = 2307
m = 183,23 g Vv

Marking criteria/Nasienriglyne

e Mol ratio/ mol verhouding
e Substitute 239 in formula / Vervang 239 in formule
e Answer / Antwoord

5.1.3 POSITIVE MARKING FROM QUESTION 5.1.2

83,23

%Purity = a x 100%

00
=22,9%V

5.2.1 H20 v

5.2.2 2(min) ¥

5.2.3 The reaction stopped/The reaction reached completion/The limiting reactant

has been used up.”

Die reaksie het gestop/Die reaksie het voltooing bereik/ Die beperkende

reaktant is opgebruik

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(2)
[17]

(3)

(3)

(2)
(1)
(1)

(1)

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5.2.4 The change in concentration v of reactants or products per unit time”
/Die verandering in konsentrasie V van reaktante of produkte per eenheid tydv . (2)

5.2.5 Am
Rate = — Wt

25 — 50v

=—€ 2(60)” )
= 0,2083 (g:s*)v

(3)

5.2.6

Marking criteria/Nasienriglyne

« Both curves start at the same
point and have the same
endpoint/ Beide kurwes begin by
dieselfde punt en het dieselfde
eindpunt ¥

e Steeper gradient / steiler hellingY

Mass /Massa SO3

Time /Tyd (min) (2)

5.2.7 e Increase in temperature increases the average kinetic energy/molecules
move faster. Vv
Toename in temperatuur verhoog die gemiddelde kinetiese energie/molekule
beweeg vinniger.
e More molecules have enough/sufficient kinetic energy/More molecules have
E.2E.v
Meer molekule het genoeg/voldoende kinetiese energie/Meer molekule het
E,2E..
e More effective collisions per unit time/second. /Frequency of effective
collisions increases. v
Meer _effektiewe _botsings per __eenheidtyd/sekonde./Frekwensie van
effektiewe botsings neem toe. (3)
[21]

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QUESTION 6 / VRAAG 6

6.1.1 Close system is isolated from its surroundings./ ‘n Geslote sisteem is geisoleer
van die omgewingY (1)

6.1.2 e From green to yellow./ Van groen na geelv
e When concentration is increased the reaction that decreases the
concentration is favoured / Wanneer konsentrasie verhoog sal die reaksie wat
die konsentrasie verlaag bevoordeel word. ¥
e [NO] increases/the reverse reaction is favoured/ [NO] verhoog/die
terugwaartse reaksie word bevoordeel. ¥
e [NOC{] will increase / [NOCE] sal toeneemv (4)

6.1.3 | Marking criteria/Nasienriglyne

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

When the equilibrium in a closed system is disturbed, the system will re-instate
a new equilibrium by favouring the reaction that will oppose the disturbance. /
Wanneer die ewewig in 'n geslote sisteem versteur word, stel die sisteem 'n

nuwe ewewig in deur die reaksie wat die versteuring teenwerk, te bevoordeel.
vv (2)

DECREASED / VERLAAG. Vv

The concentration of Cfz inceases. / Die konsentrasie van Cf2 verhoog. ¥

The forward reaction is favoured.Die voorwaartse reaksie is bevoordeel”

The reaction that forms the most/greatest number of moles was favoured,

thus the pressure was decreased. /Die reaksie wat die meeste/grootste aantal

mol vorm was bevoordeel.v (4)

6.2.1 Exothermic / Eksotermies V (1)

6.2.2 e From the graph, we can see that an increase in temperature decreases the

yield of ammonia/ Vanaf die grafiek kan afgelei word dat ‘n toename in
temperatuur die hoeveelheid ammoniak laat toeneem. V

e Anincrease in temperature favours the endothermixc reaction/ ‘n toename in
temperatuur sal die endotermiese reaksie bevoordeel.

e Increase in temperature favours the reverse reaction/Verhoging in
temperatuur bevoordeel die terugwaarste reaksie. ¥

e Thus, the forward reaction is exothermic / die voorwaartse reaksie is
eksotermies. ¥ (4)

6.2.3 50% (1)

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6.3. Marking criteria / Nasienkriteria:

a.

b

oa

Substitute 4 in nCuC4* initial and 2,2 in NCu(H2O)e6"* final in table

Vervang 4 in nCuC#é aanvankik @N 2,2 in NCu(H2O)6*finaal in tabel

Use mol ratio / Gebruik mol verhouding : 1:4:1 ¥

n Equilibrium/ n Ewewig = nReactants / nReaktanteinitiaybegin + NReactants /
nReaktante reacted! reageer a

. n Equilibrium/ n Ewewig = n Reactants initiabegin - N Reactants reacted/ reageer

dq

‘e. Divide by the volume (2)/ Deel deur die volume (2) V
f. Correct Ke expression / Korrekte Ke uitdrukking v
g
he

. Correct substitute into Kc expression / Korrekte invervanging in

Ke — uitdrukking ¥

. Final answer / Finale antwoord (0,00218) v

Cu(H20)2"(aq) 40 (aq) CuCl? (aq)

Initial mol 0 0 4
Aanvanklike mol

Change in mol +2,2 - + -2,2 :
Verandering in mol ae Ba Ratio v (b)

Equilibrium mol C227 8.8 1.8

Ewewigs mol .
Equilibrium concentration 11 ve 44 0.9 vig)! .
Ewewigskonstante ( ( ) Y Divide by 2 (e)
_ [cue] ee
= [Cu(H,0)2"}/cr}* (f) NB: If the chemical equation has
(0.9) been reversed in the table, award
= ———; 7/8 marks, no mark for the answer.
(1,1)(4,4)
=0,00218Y (h) (8)
[25]

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NSC/NSS- Marking Guidelines

QUESTION 7 / VRAAG 7

7.1. An Acid is a proton donor (H* -ion donor) /'n Suur is 'n protonskenker (H* -ioon-

skenker). vv (2)
7.2.1 Asubstance that can act as either acid or base. vv

‘n Stof wat as of 'n suur Of 'n basis kan reageer (2)
7.2.2 HSO3 ¥ (1)

7.2.3 H2SOs, HSO3 Yand/en | ACCEPT: if arrows on the equation were used.
H20, H30* ¥ | AANVAAR: Indien pyle op die vergelyking gebruik is (2)

7.3. Bromothymol blue / Broomtimoibiou “
e ls most suitable for a solution with a pH = 7/ Mees geskik vir oplossing wat
’n pH =7 het. ¥
e Reaction of a strong acid and a strong base/The equivalence pont is equal
to pH 7/ Reaksie van 'n sterk suur en 'n sterk basis/Die ekwivalente punt is

gelyk aan ’n pH van 7” (3)
7.4.1. n(NaOH) = cV
= (1,2)(0,2)”
= 0,24 mol ¥ (2)

POSITIVE MARKING FROM QUESTION 7.4.1
POSITIEWE NASIEN VAN VRAAG 7.4.1

74.2 Marking criteria / Nasienkriteria:

a) Formula c=o v
b) Substitute 0,85 and 0,02367 in correct formula / Vervang 0,85 en 0,02367
in regte formule”
c) Mol ration (NazCOsz) :n (HNO) 1:2 ¥
d) V(HNOs) in 0,05 dm? to 0,2 dm3v
e) 1 initial (HNOs) ~ N excess (HNOs) vv
f) | Substitude 56 in correct formula
4,426464

g) purity = —="— x 100% %
h) Final answer (34,0 — 34,5) v

n (NazCOs) = cVY (a) CaVa_ Ne
= (0,85) (0,02367) v(b) | GV, ny ;
= 0,0201195 mol c,x50 2 fa
1

n(HNOs) = 2n(Na2COs) aes ¥(c)

=0,0201195 x 2 (c)

=| aa
= 0,040239 mol in 50cm? —| =2:80478 mot-dmn

n(HNOzs)=cV
=0,80478x0,05
=0,040239 mol

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### Page 15

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NSC/NSS- Marking Guidelines

Therefore (4 x 0,040239 mol) = 0,160956 mol in 200 cm? v (d)

used (HNO3) = Ninita| (HNO3) — Nexcess(HNO3)
Nuseg(HNO3) = 0,24 — 0,160956 vv (e)
Nusedf HNO) = 0,079 mol

m
n(KOH) = uM
0,079044 = sav 8
m = 4,426464 g
4,426464
%purity/suiwerheid = —3,— 100¥ (g)

= 34,05% v (h) (9)
[21]

GRAND TOTAL/GROOTTOTAAL [150]

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