# Physical Sciences Grade 10 P2 KZN November 2025 Paper & Memo

> Grade 10 Physical Sciences Paper 2 question paper and memo from the KZN November Exam 2025, 18 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/kzn-physical-sciences-grade-10-november-2025-p2-and-memo/e9db82a3-3bf6-4d2b-ae6e-bec70c85d756
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 10
- Document type: Question paper and memo
- Year: 2025
- Exam period: KZN November Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 18

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## About this document

Physical Sciences · Grade 10 · KZN November Exam · 2025. Question paper and memorandum, 18 pages. Read online or download the PDF.

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## Document text

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### Page 1

Downloaded from Stanmorephysics.com

KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

MARKS: 100

DURATION: 2 hours

This question paper consists of 10 pages and 2 data sheets.

### Page 2

Physiewslance canon) Stanmorepbysics.com November 2025
2

INSTRUCTIONS AND INFORMATION

ds

10.

11.

This question paper consists of SEVEN questions. Answer ALL the questions
in the ANSWER BOOK.

Start EACH 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 sub-questions, for example between
QUESTION 2.1 and QUESTION 2.2.

You may use a non-programmable calculator.

You may use appropriate mathematical instruments.

You are advised to use the attached DATA SHEETS.
Show ALL formulae and substitutions in ALL calculations.

Round off your final numerical ansWers to a minimum of TWO decimal places.

Give brief motivations, ds u etera where required.

Write neatly and legibly.

h}
ee

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Physiowslonderaaom) Stanmorepbysics.com November 2025
3
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, for example 1.11
E.

14 Which one of the following is an example of a pure substance?

A Carbon dioxide

B_ Air

Cc Steel

D Aqueous sodium chloride (2)
1.2 Which one of the following has the highest first ionization energy?

A Li ]

B F Le —

C Na a

D cl srephi (2)
13 Which of the following is true for the process of condensation?

A __ The substance is heated and changes phase to gas.
B__ The substance is heated and changes phase to liquid.
C__ The substance is cooled and changes phase to gas.

D__ The substance is cooled and changes phase to liquid. (2)

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Physiewslandetrsaom) Stanmorepbgsics.com November 2025
1.4 Which one of the following contains 6,02 x 1023 ATOMS?

A. 10g of Ne

B 28g of N2 gas

C 4,48 dm of CHs gas at STP

D- 22,4 dm of NH3 gas at STP (2)

1.5 Consider the balanced chemical equation for the reaction below:
CHa(l) + H20(g) —> CO(g) + 3Ha(g)

The number of CH4 molecules required to form 22,4 dm? of H2 at STP is ...

A 6,02 x 1028

B  %x6,02 x 1078

C %x 6,02 x 10%

D %x6,02 x 10% (2)

[2x5=10]

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Physiewslance canon) Stanmorepbgsics.com November 2025
5

QUESTION 2 (Start on a new page.)

2:1 Two hypothetical gases, Xz and Y2, react in a closed container to form XY gas
according to the equation:

Xa(g) + Y2(g) > 2XY(g)

After the reaction, the container is found to contain both XY gas and some
unreacted Y2 gas.

24 Would you classify the contents of the container, after the reaction,
as a MIXTURE or a PURE SUBSTANCE?

2.1.2 To separate XY from Y2 the following steps take place:

STEP 1: The contents of the container is cooled to liquify BOTH
gases.

STEP 2: The liquid is thereafter gradually heated to a maximum
temperature of — 60 °C, to separate Y2 and XY

J
The boiling points of Y2 and XY is shown in the table below:

Substance | Boiling Point (°C)

Y2 a | - 190

XY norent fat es!)

(a) At which temperature will both gases be in their liquid
phase.
Choose from: 20 °C, -70 °C, -190 °C.

(b) Which of the two substances, Y2 or XY, will more easily
be converted to gas when the temperature is increased
in STEP 2.

(c) Does STEP 2 represent a PHYSICAL or CHEMICAL
change?

Explain the answer.

(d) The substance that is more easily converted to gas is
collected. It is observed that the collected gas is not
100% pure.

Give a reason for the observation.

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(1)

(1)

(1)

(2)

(2)

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6

2.2

2.3

2.4

QUESTION 3 (Start on a new pag!

3.1

Ammonium sulphate is one of the most common fertilizers used worldwide.

Ammonia reacts with sulphuric acid (hydrogen sulphate) to form ammonium
sulphate.

2.2/1 Write a balanced chemical equation for the above reaction.

23.2 Is the above reaction an example of a SYNTHESIS or
DECOMPOSITION reaction?

A farmer chooses to use ammonium nitrate (NH4NOs) as a fertilizer. He needs
to supply 30 kg of NITROGEN to a field.

2.311 Calculate the percentage nitrogen (N) in ammonium nitrate
(NH4NOs).
2.3:2 Determine the mass of ammonium nitrate (NH4NOs) required.

Use the balanced chemical equation below to prove the law of conservation of
mass.
4NH3.+, 502. — 4NO + 6H20

A teacher performs a demonstration using ammonia (NH3s) and hydrochloric
acid (HCI) GASES in a long glass tube.

Ammonia and hydrochloric acid are simultaneously released from opposite
ends of a long glass tube. A white ring of ammonium chloride (NH4Cl) form

where the gases react.

The position of the white ring formed is shown on the diagram below.

werk
\
NHs (g) ——> ' 1 +— HCl (9)
/
s Z

| a ring
Middle of | ‘rmed
glass tube

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November 2025

(3)

(1)

(3)

(3)

[20]

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Physiowslonderaaom) Stanmorepbysics.com November 2025
7

3.1.1 Define the term diffusion in words. (2)

3.1.2 Provide a reason why the white ring formed closer to HCI than (1)
NHs.

3.1.3 The temperature of the gases in the glass tube is now increased.

How would this effect the TIME taken for the white ring to appear?
Choose from INCREASE, DECREASE or REMAIN THE SAME.
Explain the answer using the kinetic molecular theory. (3)

3.2 The heating curve of Hydrochloric acid (HCI) is represented by the graph
below. The graph is not drawn to scale.

The Heating Curve of Hydrochloric Acid
- 85,1
oO
Q
2
>
©
oO
a
£
oO
Fe
-114,2
ti te ts t4
Time (min)
3.2.1 Define the term boiling point in words. (2)
3.2.2 Write down the boiling point of hydrochloric acid. (1)
3.2.3 At which point, A or B, are the forces between HCI particles (1)
stronger?
3.2.4 Refer to the temperature, energy and movement of particles to
explain the changes that take place between tz and ts. (3)

[13]

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Physiewslance canon) Stanmorepbesics.com
8

QUESTION 4 (Start on a new page.)

4.1 Silicon occurs naturally as three stable isotopes:
e Si-28
e Si-29
e Si-30

November 2025

The percentage abundance of Si - 28 is 92,2 % and the relative atomic mass
of silicon is 28,1 AMU.

4.1.1 Define the term isotope in words. (2)
4.1.2 Determine the percentage abundance of Si — 29. (5)
4.1.3 Fill in the table below for the most abundant isotope of silicon.
Write down only the letter (a — c) and the correct answer.
Number of protons (a)
Number of neutrons (b)
Number of valence =. (c)
a @)
4.2 Magnesium combines rine Biorm magnesium chloride.
4.2.1 Identify the type id that forms when magnesium and chlorine
combine. (1)
4.2.2 Draw the Lewis dot diagram for magnesium chloride. (2)
4.2.3 Draw the Aufbau diagram for the magnesium ion. (2)
4.24 Write down the name of the noble gas that has the same electronic
configuration as the chloride ion. (1)
4.3 How does the atomic radius of Chlorine (Cl) compare to Magnesium (Mg)?
Choose from GREATER THAN, LESS THAN or EQUAL TO.
Explain your answer by referring to the atomic structure. (4)
[20]

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Physiowslonceraaom) Stanmorepbysics.com November 2025
9

QUESTION 5 (Start on a new page)

Magnesium nitride (MgsNz) reacts with hydrochloric acid (HCI) according to the
balanced equation below:

MgsNa2(s)+ 6HCl(aq) — 3MgCle(aq) + 2NHs(g)

In one'such reaction a learner completely reacts 10 g magnesium nitride (MgsNz2) with
1,2 dm? hydrochloric acid (HCI) at standard temperature and pressure (STP).

54 Define the term concentration in words. (2)
5.2 Calculate the number of moles of MgsNz that reacted. (3)
5.3 Determine the:
5.361 Concentration of HCI (4)
5.3.2 Volume of NHs gas (in dm°) that formed at STP (4)
5.3.3 Number of chloridesions (GI’) present in the MgClz produced (5)
ky \ — S|
ie [18]

| :
QUESTION 6 (Start ona new page)

6.1 Analysis of a compound that is made up of Magnesium (Mg), Phosphorus (P)
and Oxygen (O) only, provided the following percentage composition by
mass.

e Magnesium: 27,7 %
e Phosphorus: 23,6 %

e Oxygen: 48,7 %

(6)
Determine the empirical formula of this compound.
6.2 Determine the mass of hydrated potassium sulphate(K2SO4.7H2O) needed to
prepare 100 ml of a 0,2 mol-dm’? potassium sulphate solution. (4)
[10]

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### Page 10

Physiewslendetranoin) Stanmorepbysics.com November 2025
10
QUESTION 7 (Start on a new page)

A metal ¥ reacts with copper (ii) oxide (CuO) according to balanced chemical below:
2Y + 3CuO — Y203 + 3Cu

A 5g sample of Y reacts with excess copper (ii) oxide (CuO) to produce 15 g of copper
(Cu).

The percentage yield of copper (Cu) in this reaction is 85 %.

TA Determine the theoretical mass of copper (Cu) formed. (3)
7.2 Use relevant calculations to determine the NAME of metal Y. (6)
[9]

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### Page 11

Physiowslonderaaom) Stanmorepbysics.com November 2025
1
DATA FOR PHYSICAL SCIENCES GRADE 10

CHEMISTRY
TABLE 1: PHYSICAL CONSTANTS
NAME SYMBOL VALUE
Standard pressure p° 1,013 x 105 Pa
Molar gas volume at STP Voi 22,4 dm3-mol
Standard temperature T° 273K
Avogadro's constant Na 6,02 x 1023 mol

TABLE 2: FORMULAE

or c=

JM,
MV

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PhysicdDanenagdedefiopm Stanmorephysics.com NSC November 2025
12
TABLE 3: THE PERIODIC TABLE OF ELEMENTS
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
(I) (Il) 7 | qm) (IV) (Vt sy
‘ Atomic number
7 4 KEY/SLEUTEL Atoomgetal i
i e
N
1 29 4
7 3 . 4 Electronegativity __ | 2 Cy «Symbol 7 5 bs 6 co 7 o 8 cs 9 10
= Li |? Be Elektronegatiwiteit “| - 63.5 Simbool S aCcis s OS Ne
7 9 Z 11 12 14 16 19 20
11 12 tT 13 14 15 16 17 18
ce Na|S Mg Approximate relative atomic mass 2 Ag |= SIS aSi¢Cce| Ar
23 24 Benaderde relatiewe atoommassa 27 28 31 32 35,5 40
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
& K [2 Cal? Sc |e Ti |? 2 Cr |2Mn|2 Fe |2 Co|2 Ni |2 Cu|2 Zn |2 Ga/2 Ge |& As |S Se |S Br| Kr
39 40 45 48 51 52 55 56 59 59 63,5 65 70 73 75 79 80 84
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
¢ Rb|2 Sr |S Y |= Zr| Nb|2Mo/2 Tc|s Ru |e Rh | Pd |2 Ag |= Cd |= In [2 Sn |Z Sb/& Te|S 1 | Xe
86 88 89 91 92 96 101 103 106 108 112 115 119 122 128 127 131
55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86
S Cs |S Ba La|2 Hf} Ta W)} Re| Os Ir Pt) Au! Hg/2 Te |2 Pb|2 Bi |G Po|& At} Rn
133 137 139 179 181 184 186 190 192 195 197 201 204 207 209
87 88 89
n oO
s Fr js yea Ac 58 59 60 61 62 63 64 65 66 67 68 69 70 71
Ce | Pr | Nd | Pm| Sm| Eu | Gd Tb | Dy | Ho} Er | Tm | Yb | Lu
140 141 144 150 152 157 159 163 165 167 169 173 175
90 91 92 93 94 95 96 97 98 99 100 101 102 103
Th | Pa | U | Np | Pu | Am | Cm! Bk | Cf | Es | Fm/| Md | No | Lr
232 238

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### Page 13

KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

MARKS: 100

DURATION: 2 hours

Copyright reserved

### Page 14

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14 Avv

Pas
1.4
15

QUESTION 2
Zac] Mixture ¥
2.1.2a) - 190°C v

2.1.2b) Yo ¥

2.1.2c) Physical (change). ~

No new substance is formed. OR No chemical reaction took place. W

2.1.2d) Some of the XY would evaporate below its boiling point.v v

LHS v
RHS v¥

2.2.1 2NHs + H2SO4 — (I

2.2.2 Synthesis Vv

14«x2 x
14x2+1x4+16x3v
=35%V

100

23.1 ‘N=

2.3.2 POSITIVE MARKING FROM 2.3.1
m (NH4NOs) = 30 ¥ + 0,35 ¥
= 85,71 kg ¥

2.4 Mass of reactants = 4[14 + 1x3] + 5[16x2] Vv = 228g
Mass of products = 4[14 + 16] + 6[1x2 + 16] Vv = 228g
Mass of reactants is equal to the mass of products. Therefore mass is

conserved. ¥

Copyright reserved

Balancing ¥

(2)
(2)
(2)
(2)
(2)
[10]

(1)
(1)
(1)

(2)
(2)

(3)
(1)

(3)

(3)

(3)
[20]

### Page 15

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QUESTION 3

3.1.1

The movement of atoms or molecules Y from an area of higher concentration
to an area of lower concentration. v

Ammonia diffuses faster than hydrochloric acid ¥

3.1.2
3.1.3. Decrease. ¥
At a higher temperature the particles move faster OR have more kinetic
energy v.
Therefore, the gases diffuse faster. VW
S24 The temperature of a liquid at which its vapour pressure equals the external
(atmospheric) pressure. Vv”
3.2.2  -85,1°C VY
3.2.3 (point) AY
3.2.4 The temperature of HC! increase. Y The average kinetic energy of the
particles increases. V Particles move faster. “
|
QUESTION 4 i eal |
—\Kiee |
4.1.1 Atoms of the same element having the same number of protons, but different
number of neutrons. VY
4.1.2 Let the % of Si- 29 be X.
The % of Si — 30 will be 7,8 — X
92,2 Xx 7,8-X
Vi= Vv v v
aR 28 eg ag? on
X=5,6% ¥
4.13  (a)=14v
(b) = 14 ¥
()=4v
4.2.1 lonic bonds ¥
4.2.2
OPTION 1 OPTION 2
co 4 eo a ‘ -
xcs 2+ x e 2+
a e Mg 3 Cl e Vv Mg 2 scle Vv

(2)
(1)

(3)
(2)

(1)
(1)

(3)
[13]

(2)

(5)

(3)
(1)
(2)

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4.2.3 (2)

ap |tyItyity
ae tL

1s ‘1 fy

4.24  Argonv (1)

4.3 Less than. ¥
¢ Both chlorine and magnesium are in the same period OR have the
same number of energy levels. ¥
e The effective nuclear charge of chlorine is higher than magnesium. “
e The force of attraction between the nucleus and the outer orbital is

stronger in chlorine. ¥ (4)
[20]
QUESTION 5 ee |
51 The number of moles of solutelper cubic decimetre of solution. Vv (2)
5.2 m rd
n(Mg,No) = 9G ¥ 4

_ 10

~ 24x3+14x2

= 0,1 mol ¥ (3)

5.3.1 POSITIVE MARKING FROM 5.2
Mg3Nz2 : HCI
1:6Vv
n(HCl) = 0,6 mol

c(HCl) = “ v

oO

08 |
1,

2
0,5 mol-dm3 v (4)

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### Page 17

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5.3.2 POSITIVE MARKING FROM 5.2 AND 5.3.1
MgsNz : NH3 OR HCI : NH3
1 v2 6 :2 (or3:1)v¥

n(NHs3) = 0,2 mol
Vv
n(NH3) = —— ¥
(NHs) Ve

Vv
=— V
tar 22,4

V(H2) = 4,48 dm? v

5.3.3 POSITIVE MARKING FROM 5.2, 5.3.1 AND 5.3.2
Mg3Nz2 : MgCle HCI : MgCl
1 Rn A oe 6 :3 (or21)v ee 2 13%
n(MgClz) = 0,3 mol

n(MgCl,) = = v
_ N

~ 6,02 x 1079
N(MgClz) = 1,806 x 1023 formula units
N(CI’) = 2.x ¥1,806 x 107°

QUESTION 6 ee 4]

NHsz : MgCl

6.1 Marking criteria:
Calculate the number of moles of each element (3 x 1 = 3) ¥Vv

Determine the simplest ratio of all elements ¥
Multiplying by 2 W
Final answer: MgsP2Os ¥

Element ) n= a Simplest Ratio

Mg 27,7 eal =4115 on =9151 3
P 23,6 ie = 076 wa sae 1 Y |x2v /|2
fe) 48,7 * =304 Vv a =4 8

Empirical Formula: MgsP20s ~

Copyright reserved

(4)

(5)

[18]

(6)

### Page 18

6.2Dowm|acsda.dHfopsses-eamoraphysies,com
= 300 g.mo!l
OPTION 1 OPTION 2

v

<I5

=)—— Vv =
Cc MV Cc

m

— er 4
0.2 = 300(0,1)

0,

N
I

n
0,1
m=6g v n(K2S04.7H20) = 0,02 mol >

m=nxM
= 0,02 x 300
=6gv (4)
[10]

QUESTION 7

71 actual yield
oY Vi = ——
eieh theoretical yield
15v
=D *100
theoretical yield 15¥ = m(Cu)theoretical X 0,85 W
m(Cu)theoretical = 17,659 v (3)

100

OR m(Cu)actual = m(Cu)theoretical x 0,85

85v

7.2 POSITIVE MARKIN

Marking criteria:
e Divide by 63,5 v
e Ratio (2:3) ¥
e Formula: n(Y) = Frerephy
e Substitute n(Y) ¥
e Substitute 5g Vv
Final answer: Aluminium ¥
_m
n(Cu) = uM
_ 17,65
~ 63,5 ¥
= 0,278 mol

Y:Cu
2a v
n(Y) = 0,185 mol

ny) = oY

5v
=
0,185) M

M(Y) = 27,03 g

Y is Aluminium ¥ (6)
[9]

TOTAL: 100

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