# Physical Sciences Grade 11 Limpopo Prelim 2025 Paper & Memo

> Grade 11 Physical Sciences question paper and memo from the Limpopo Prelim Exam 2025, 25 pages. Read online or get the free PDF download.

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- Updated: 2026-09-28
- Subject: Physical Sciences
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2025
- Exam period: Limpopo Prelim Exam
- Curriculum: CAPS
- Pages: 25

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Physical Sciences · Grade 11 · Limpopo Prelim Exam · 2025. Question paper and memorandum, 25 pages. Read online or download the PDF.

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### Page 1

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LIMPOPO

PROVINCIAL GOVERNMENT
REPUBLIC OF SOUTH AFRICA

DEPARTMENT OF

EDUCATION

NATIONAL
SENIOR CERTIFICATE

MARKS: 100

TIME: 2 hours

This question paper consists of 10 pages, 4 data sheets and 1 graph paper.

Copyright reserved Please turn over

eee ee ene eee ee

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Physical Sciences/Gr 11 Limpopo DoE/September 2025

NSC
INSTRUCTIONS AND INFORMATION

1J Write your name and other applicable information in the appropriate
spaces on the ANSWER BOOK.

- The question paper consists of SEVEN questions. Answer ALL the
questions in the ANSWER BOOK, except question 7.2.2 which should
be answered on the provided graph paper on page 15.

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

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

5; Leave ONE line between two sub-questions, for example between QUESTION
3.1 and QUESTION 3.2.

6. You may use a non-progrs pocket calculator.

Ps You may use app atical instruments.

8. Show ALL formulae: tions in ALL calculations.

9. Round off your FINAL swers to a maximum of THREE

decimal places
10. Give brief motivations, discussions, etc. where required.
11. | You are advised to use the attached DATA SHEETS.

12. Write neatly and legibly.

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Physical Sciences/Gr 11 Limpopo DoE/September 2025
NSC

QUESTION 1: MULTIPLE-CHOICE QUESTIONS

Four 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
to the question number (1.1 — 1.5) in the ANSWER BOOK, e.g. 1.6 A.

1.1 The activation energy for the forward reaction of the reaction below is
230 kJ.mol™

A2 + B2 > 2C AH = + 150 kJ.mol™
What is the enthalpy change for this reaction?

A. 380

B. 230

Cc. 150

D. 80 (2)
1.2 ONE mole of wate ole of carbon dioxide (CO2z) will have

the same... sual

A. molar mass.

B. mass.

C. density.

D. number of molecules. (2)
1.3. Which of the following represents a neutralization reaction?

A. HCl+ Zn — ZnCte

B. H2NOs +. Cu — Cu(NOs)2=

c. 5 a2SO

D. ai (2)

14 the name of the salt formed in this reaction?

q) > CHsCOONa (aq) + H20 (/)

A. iur

B.  sodiur

C. sodium methanoate

D. sodium acetate (2)

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Physical Sciences/Gr 11 Limpopo DoE/September 2025
NSC

1.5 The pressure of a gas is 200 kPa when its volume is 3 000 cm‘. If the gas is
compressed to 2 000 cm* at constant temperature, what is its new pressure?

A. 250 kPa
B. 300 kPa
C. 350 kPa
D. 400kPa (2)

[10]

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Physical Sciences/Gr 11 5: Limpopo DoE/September 2025
NSC

QUESTION 2 (Start on a new page)

The following reaction between Ammonia and Oxygen takes place at constant
pressure and temperature:

2.1 Define the term acti Prgy (2)

2.2 Give a reason why t stion ic. (1)
2.3 Redraw the graph and a dotted line the effect of a catalyst on
the activation energy. (1)
2.4 Write down the values for the:
2.4.1 Energy of the products (1)
2.4.2 Energy of the reactants (1)
2.4.3 Activation energy for the reverse reaction (1)
2.4.4 Activation energy for the forward reaction. (1)
2.5 Calculate the heat of reaction,(AH), for the reverse reaction. (2)

2.6 What effect will the addition of the catalyst have on:
(Choose from INCREASES, DECREASES or REMAINS THE SAME)

2.6.1 The energy of the products? (1)
2.6.2 The heat of reaction? (1)
2.6.3 The activation energy for the reverse reaction? (1)

[13]

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Physical Sciences/Gr 11 Limpopo DoE/September 2025
NSC

QUESTION 3 (Start on a new page)

3.1 itric acid with ampholyte X to form
the reaction below. Study the flow
5 that follow.
+ | NOs
(2)
3.1.2 Write a chemical equation for the reaction above. (3)
pholyte X. (2)
3.2 base in the reaction below.
(2)
3:3 i e a ator. (2)
3.4 . is reaction and identify the acid and the base
required to prepare potassium chloride. (3)
[14]

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Physical Sciences/Gr 11 Limpopo DoE/September 2025
NSC

QUESTION 4 (Start on a new page)

A solution of calcium carbonate (CaCOs) was prepared by dissolving 20 g of
powdered CaCOs in 500 cm of water. a ial

This solution was then added to 500 cm? s
concentration of 0.25 mol.dm* as in the set u

— - / aas

ium
natte
ition
0,26
J cs

4.1 Define the term limiting agent. (2)
4.2 Write a balanced equation for the reaction in the set up above. (3)
4.3 Calculate the initial concentration of the CaCOs solution. (4)

4.4 Which substance is the limiting reagent? Show by means of a suitable

calculation. (4)
4.5 Calculate the percentage purity of the excess reagent. 3 (5)
[18]

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NSC

QUESTION 5 (Start on a new page)
5.1 Define the term molar mass in words. (2)

5.2 An organic acid was analysed and found that it has the following chemical
composition:

76.6%carbon(C) 12.05%Hydrogen (H) 11.35%oxygen(O)
The reaction of this acid with alcohols produces a compound that has a fruity
smell. During combustion, 0.01272g of this acid was used and it produced

0.01863g of carbon dioxide (COz) and 0.0075 g of water, according to the
following reaction:

+ $02(g)—»CO2 (g)+ H20(t)

5.2.1 Calculatelt ass of H2O (2)

5.2.2 Given the pere , determine the empirical formula for the

acid. (7)
5.3 Calculate the number of mo f carbon dioxide produced (3)
[14]

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Physical Sciences/Gr 11 9

Limpopo DoE/September 2025
NSC

QUESTION 6 (Start on a new page)

When sulphuric acid reacts with water, it ionises in two steps, as shown in the two balanced

equations below:

6.1
6.2

6.3
6.4

O(t) — Hs0*(aq) + HSO; (aq)
O(t) + Hs0*(aq) + SO%-(aq)

Define an acid < Bronsted theory.

Is the solution forme i ACIDIC or BASIC (ALKALINE)? Give a reason
for your answer.

Write down a conjugate base of H30*

A few drops of bromothymol blue indicator are added to a potassium hydroxide
solution in a beaker,Adilute sulphuric acid solution is now gradually added to this
solution until the ¢o dieator changes.

Write down the:
6.4.1 Type of rea
6.4.2 Balanced ct
6.4.3. Colour

for the reaction that takes place.

r. Choose from the following:
to blue OR Blue to purple

6.4.4 Write down a reason why bromothymol blue is the most suitable indicator
for this reaction.

(2)

(2)
(2)

(2)
(3)

(2)
(1)

[14]

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Physical Sciences/Gr 11 Limpopo DoE/September 2025
NSC

QUESTION 7 (Start on a new page)

(2)

occupies a volume of 5.0 L at a pressure of
decreased to 150 kPa at constant temperature,

the final volume? (4)

provided to draw a pressure vs volume graph
cale) and show by means of point X on the
rz the initial and the final value of the volume. (4)

TT A gas sample has a mass of 16 g and occupies a volume of 11.2 dcm*®
at standard temperature and pressure (STP).

Determine, b e ula of the gas. (5)

74 A bicycle ti
temperature rem
inside the tire? E

reducing its volume by half. If the
at happens to the pressure of the air

(2)
[17]
TOTAL: [100]

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Physical Sciences/Gr 11 11 LimpopoDoE/September 2025

NSC

TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES

NAME/NAAM SYMBOL/SIMBOOL VALUE/WAARDE
siete konstants Na 6,02 x 10 mot
pers x anion R 8,31 J:K*-molt
Sneoune p° 1,013 x 108 Pa
Molers gasvolume by STD Vi 22,4 dm?-mot"
Socereiener e 273 K

TABLE 2: FORMULAE/TABEL 2: FORMULES

P1V1 = P2V2

(oa OR/OF c=
Vv

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### Page 12

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NSC

TABLE 4A: STANDARD REDUCTION POTENTIALS

 - SiakeomailetiaT Falta | é D- TENSIALE
Half-reactions / Halfreaksies EP w

Increasing strength of oxidising agents / Toenemende sterkte van oksideermiddels

F2(g) + 2e-

Co** + e-

H2Oz + 2H* +2e-
MnO] + 8H* + 5e~
Cla(g) + 2e-
CrO2- + 14H* + 6e-
O2(g) + 4H* + 4e-
MnO2 + 4H* + 2e-
PI?* + 2e-

Bra(t) + 207

NO} + 4H’ + 3e-
Hg?* + 2e-

Ag* +o-

Cu?* +e
Sn** + 2e-

S + 2H* + 2e~
2H* + 20-
Fe* + 3e-
Pb?* + 2e-

Sn?* + 2e-
Ni?* + 2e-
Co** + 2e-
Cd?" + 2e~
cr? +e-
Fe** + 2e-
Cre* + 3e-
Zn?" + 2e-
2H20 + 2e-
Cr** + 2e-
Mn?* + 2e-
Al + 3e7
Mg?* + 2e—
Na* + e~
Ca?* + 2e-
Sr?* + 2e-
Ba?* + 2e-
Cs*+e
K* +e-

Lit + e-

PASE EE EEE CEE SE EEL ERE EP ECE ET ERED

2F-

Co**

2H20

Mn?" + 4420
2ce-

2Cr* + 7H20
2H20

Mn?" + 2H20
Pt

28r

NO(g) + 2H20
Hot)

Ag

NO2(g) + HzO
Fe**

H20O2

2I-

Cu

S+2H20

SO2(g) + 2H20
Cur

Sn?-

HeS(g)

H2(9)

Fe
Pb

Sn
Ni

H2(g) + 2OH-
Cr
Mn
At

+ 2,87
+ 1,81
+1,77
+ 1,51
+ 1,36
+ 1,33
+1,23
+1,23
+ 1,20
+ 1,07
+ 0,96
+ 0,85
+ 0,80
+ 0,80
+0,77
+ 0,68
+ 0,54
+ 0,52
+045
+0,40
+ 0,34
+0,17
+0,16
+0,15
+ 0,14
0,00

— 0.06
— 0,13

—0,14
—0,27
—0,28
—0,40
—0,41
—0,44
—0,74
—0,76
— 0,83
—0,91
=41,18
— 1,66
—2,36
—2,71
— 2,87
— 2.89
—2,90
-2,92
— 2,93
— 3,05

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Increasing strength of reducing agents / Toenemende sterkte van reduseermiddels

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Physical Sciences/Gr 11

TABLE 4B
T

physics.com

: STANDARD Ri

NSC

Half-reactions / Halfreaksies

Increasing strength of oxidising agents / Toenemende sterkfe van oksideermiddels

litte = Li —3,05
K'+e = K —2,93
Cs*+e- = Cs -—2,92
Ba?*+2e = Ba — 2,90
Sr*+2e- = Sr —2,89
Ca**+2e- = Ca — 2,87
Na’*+e- = Na —2.71
Mg**+2e- = Mg —2,36
AC+3e-— = AC — 1,66
Mn?* + 2e- = Mn —1,18
Cree+2e- = Cr —0,91
2H2O0+2e- = H2(g) + 20H- — 0,83
Zn?*+2e- = Zn —0,76
Cr*+3e5 .—= Cr —0,74
— Fe —0,44

= cr —0,41
= Cd —0,40
= Co —0,28
- Ni -0,27
i= Sn —0,14
= Pb —0,13
- ‘Fe — 0,06

2H*+2e = H2(g) 0,00
S+2H*+2e- = H2S(g) +0,14
Sn**+2e7 = Sn? +0,15
Cu?*+e- = Cur +0,16
SO? +4H*+2e- = SO2x(g)+2H20] +0,17
Cu?*+2e- = Cu + 0.34
2H20 + O2+4e- = 40H + 0,40
SO2+4H*+4e- = S+2H20 + 0,45
curx+e- = Cu + 0,52
fa+2e- = 2k +0,54
O2x(g) + 2H*+2e- = H2O2 +0,68
Fe**+e- = Fe* +0,77
NO; +2H*+e- = NO2(g)+ HzO +0,80
Ag*+e- = Ag +0,80
Hg** + 2e- = Halt) +0,85
NO} +4H*+3e- = NO(g) + 2H20 + 0,96
Br(t)+2e- = 2Br- + 1,07
Pre+2e- =— Pt +1,20
MnO2z2+ 4H* + 2e- = Mn** + 2H2O + 1,23
O2x9)+4H*+4e- = 2H2O + 1,23
cro” +14H++60e- = 2Cr* + 7H20 + 1,33
C&(g)+2e- = 20% + 1,36
MnO, + 8H*+5e-— = Mn?* + 4H20 + 1,51
H202+2H*+2e- =— 2H20 +1,77
Co*+e- <= Co + 1,81
F2(9)+2e- = 2F- + 2,87

LimpopoDoE/September 2025

EDUCTION POTENTIALS
J

Increasing strength of reducing agents / Toenemende sterkte van reduseermiddels

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LimpopoDoE/September 2025

15
NSC

Physical Sciences/Gr 11

Downloaded from Stanmorephysics.com

SUBMIT THIS SHEET WITH THE ANSWER SHEET

QUESTION 7.2.2

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grea. LIMPOPO
oe PROVINCIAL GOVERNMENT

REPUBLIC OF SOUTH AFRICA

DEPARTMENT OF

EDUCATION

NATIONAL

SENIOR CERTIFICATE

| oRabeicbaao 11
: PHYSICAL SCIENCES: CONTROLLED TEST 2 ;
FISIESE WETENSKAPPE: KONTROLE TOETS 2 :
AMENDED MARKING GUIDELINE/AANGEPASTE
r NASIENRIGL YNE :
. SEPTEMBER 2025 J

MARKS/PUNTE: 100

This marking guideline consists of 11 pages/Hierdie nasienriglyne bestaan uit
11 bladsye

Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

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Physical Sciences/Gr 11 Limpopo DoE/September 2025
NSCINSS
Marking Guidelines/Nasienriglyne

QUESTION 1/VRAAG 1

1.14. Cvv (2)
12 Dvv (2)
13 Cvv (2)
14 Dvv (2)
15 Byv (2)
[10]
QUESTION 2/VRAAG 2
2.1. The minimum energy needed for a reaction to take place. ¥v/ Die minimum

energie wat be 'n'réaksie om plaas te vind

ased than absorbed. v ¥
rd vrygestel as geabsorbeer

(2)

The potential energy of the product is lower than that of the reactants. “/ Die
potensiéle energie van die produk is laer as dié van die reaktante. (1)

2.2

2.3

Ep (kJ.mol’)

=
&
i)
iw

eaction/Verloop van reaksie

Marking criteria/Nasienriglyne

e A dotted line under the activated complexv/
‘n Stippellyn onder die geaktiveerde
kompleks (1)

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Physical Sciences/Gr 11 Limpopo DoE/September 2025
NSCINSS
Marking Guidelines/Nasienriglyne

2.4

2.4.1, 183,3 kJ.mol" ¥ (1)
2.4.2 400 kJ.mol™ ¥ (1)
2.4.3. 2487.6 kJ.mol" v (1)
2.4.4 2670 kJ.mol" v (1)

2.5 AH = Energy of products - Energy of reactants/ AH = Energie van produkte —
energie van reaktante

= 400 - 183.3 V
= 216.7 kJ.mol” ¥ (2)
2.6
2.6.1
(1)
2.6.2 (1)
2.6.3 (1)
[13]

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NSCINSS
Marking Guidelines/Nasienriglyne

QUESTION 3/VRAAG 3

3.4
3.1.1. A base is a substance that produces hydroxide ions (OH") when it dissolves
in water. v/'n Basis is 'n stof wat hidroksiedione (OH) produseer wanneer

dit in water oplos. (2)
3.1.2--HNOs (aq) ¥+H2O0(t) ¥—> H30*(aq) + NOs (aq)¥ (3)
Notes/notas:
e Reactants ¥ V/reaktante
« Products “/produkte
e Ignore phases/ignoreer fases
e Ignore double arrows/ignoreer dubbel pyle

(2)

3.2 Water/H20(2) identi a base in the reaction because it accepts a
Hydrogen ion/prote idrochloric acid HC£(aq) to form H3
O*(aq).¥“// Water/H t eras 'n basis in die reaksie omdat dit 'n
waterstofioon/proi soutsuur HCf(aq) aanvaar om H30 “aq) te vorm. (2)

or the OH" ion concentration in a solution
indikator is 'n swak suur, of 'n swak basis, waarvan
“ioonkonsentrasie of die OH -ioonkonsentrasie in

(2)
p(s) + H20(f)v’ (3)

Reactants v V/reaktante

Products Y/produkte

Ignore phases/igoneer fases

Ignore double arrows/ignoreer dubbel pyle

[14]

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Physical Sciences/Gr 11

NSCINSS

Marking Guidelines/Nasienriglyne

QUESTION 4/VRAAG 4

Limpopo DoE/September 2025

4.1 Areagent that gets used up completely in a chemical reactiony “/ 'n Reagens

wat volledig in 'n chemiese reaksie opgebruik word

4.2 CaCOs(s) + H2SOa (aq) Y— CaSOa(aq) + H20(2) + CO2(g)” Balv

Notes/Notas:

e Reactantsv /reaktante Products ¥/produkte Balancing “/balansering
e Ignore phases/ignoreer fases
e Ignore double arrows/ignoreer dubbel pyle

OPTION 2/OPSIE 2

(100) (0,5)v

= 0.4 mol.dm=v

n(CaCOs) = “~ v
M

4.4 POSITIVE MARKING FROM 4.3/POSITIEWE NASIEN VANAF 4.3

OPTION 1/OPSIE 1
0.2mol n(CaCOs) x 0.1mol HzO

0.2mol (CaCO3)

= 0.1 mol H2O produced by CaCOs ¥ = 20.
(100)
n(H2SOs) = CV =0.2 mol.dm=v
= (0.25) (0.5) ¥ =
= n(H2SO4) = CV
0.125 mol = (0.25) (0.5) ¥
= 0.125 mol

OPTION 2/OPSIE 2
n(CaCOs) = “— v
M

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(2)
(3)

(4)

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Limpopo DoE/September 2025

NSCINSS
Marking Guidelines/Nasienriglyne

0.425mol (H2SO4) x_0.1mol H20
0.125mol (H2SOs)

= 0.1 mol H2O produced by H2SO ¥

Ratio/ Verhouding
CaCOs: H2SO4
44

0.2mol : 0.125 mol

Since you will need 1mol of CaCO3 to
react with 1mol H2SO4, to produce
0.1mol H2O. H2S04 is less and it will

Ratio/verhouding
CaCOs3: H2SO4
31

0.2 mol : 0.125 mol

Since you will need 1mol of CaCOs to
react with 1mol H2SO4, to produce
0.1mol HzO. H2SOs is less and it will
run out first. ./ Aangesien jy 1 mol
CaCO3 benodig om met 1 mol H2SO4

run out first./ Aangesien j' yy 1 mol oI CaCOs te reageer om 0.1 mol H20 te (4)
benodig om te) produseer, is H2SO4 minder en dit sal
reageer om 0.1 ni eerste opraak.
is H2SO4 mind
opraak. H2SOsis a limiting agent v / H2SOz
HeSOais a limitinc is 'n beperkende middel
'n beperkende middel~
4.5 POSITIVE MARKING FROM 4.3 and 4.4/POSITIEWE NASIEN VANAF 4.3
EN 4.4
Mol ratio 1:1 only 0.125 mol CaCOscan react./ Molverhouding 1:1, slegs
0.125 mol CaCO3 kan reageer.
CaCOs : H2SO4
Suiwerheid = Sulwer stof
onsuiwer stof
_ 125 100 v
20
= 62.5 %v (5)
[18]

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### Page 22

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Physical Sciences/Gr 11 Limpopo DoE/September 2025
NSCINSS
Marking Guidelines/Nasienriglyne

QUESTION 5/VRAAG 5

5.1 The mass of one mole of a substance measured in g:mol™.v “/ Die massa
van een mol van 'n stof gemeet in g:‘mol™ (2)

2
5.2.1 M (H2O) = 1(2) + 16%
= 18 g.mol'"v (2)

5.2.2 Percentage composition in mass/ Persentasie samestelling in massa

Element | Percentage Divide all mole

ition i | values with 0.709/
Deel alle
molwaardes met
0.709

6.383/0.709
= Ov

H 12.05% 12.05/1 12.05/0/709
= 12.05V =17vV

fe} 11.35% 11,35/16 = 0.709/0.709
= 0.709 ¥ =1v

The empirical formula is/Die empiriese formule is: Co3H170Y

(7)

5.3 Calculate the number of moles of carbon dioxide produced/ Bereken die
aantal mol koolstofdioksied wat geproduseer word

m
n=-—v
M

_ 0.01863
44

v

= 0.0004 mol¥ (3)
[14]

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### Page 23

Downloaded from Stanmorephysics.com
Physical Sciences/Gr 11 8

Limpopo DoE/September 2025
NSC/NSS
Marking Guidelines/Nasienriglyne

. Vv I 'n Suur is 'n proton skenker. (2 orlof 0) (2)

ed in water.”/ Hidroniumione word in water gevorm. (2)

(2)

(2)
(3)
(2)
6.4.4 _ It’s areaction between strong acid and strong base /neutralisation occurs at
almost pH = 7 ¥ / Dit is 'n reaksie tussen 'n sterk suur en 'n sterk basis /
neutralisasie vind plaas by amper pH = 7. (1)
[14]

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### Page 24

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Physical Sciences/Gr 11 Limpopo DoE/September 2025

NSCINSS
Marking Guidelines/Nasienriglyne

QUESTION 7/VRAAG 7

7.1. The pressure of an enclosed gas is inversely proportional to the volume it
occupies at constant temperature. v “/ Die druk van 'n ingeslote gas is
omgekeerd eweredig aan die volume wat dit by konstante temperatuur beslaan

7.2
7.2.1 PiV1= P2V2¥

(200) (5.0) “= (150) V2 ¥
V2=6,67L vv

E22 eal gas at constant temperature./ Druk

ale gas by konstante temperatuur

Notes/Notas

X and Y axis labelled “/X en Y as byskrifte
Curved linew’/geboé lyn
Initial volume on graph/ Aanvanklike volume op grafiek

e Point marked x on graph (200:5)“/ Punt gemerk met x op grafiek (200:5)
Final volume on graph/Finale volume op grafiek

e Point marked x on graph (140 -160: 6-7) (Any reasonable range point)
Punt gemerk x op grafiek (140 -160: 6-7) (Enige punt met redelike bereik)

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(2)

(4)

(4)

### Page 25

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Physical Sciences/Gr 11 Limpopo DoE/September 2025
NSCINSS
Marking Guidelines/Nasienriglyne
13 Vy
Vin
= 12
22,4
= 0,5 mol
m
n=—
M
0,5=16v/M
=32,0 g. mol’ ¥
Molar mass of apps ly $2,0)g. mol correspond to that of O2/Oxygen ¥
/ Molére massa V g. mol™ stem ooreen met dié van
O2/suurstof (5)
7.4 The pressure doub @ same number of molecules are now

confined to a smalle ding to more frequent collisions with the

tire.v/ Die druk verdubbel omdat dieselfde aantal molekules nou tot 'n kleiner

volume beperk is, wat lei tot meer gereelde botsings met die band. (2)
[17]
TOTAL/ITOTAAL [100]

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