# Physical Sciences Grade 10 Paper 2 KZN November 2024 Paper & Memo

> Grade 10 Physical Sciences Paper 2 question paper and memo from the KZN November Exam 2024, 20 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-10-physical-sciences-2024-kzn-november-exam-question-paper-and-memorandum-paper-2/65b7d7f7-c8ea-40dc-b189-7f88e2629bc0
- Updated: 2026-09-28
- Subject: Physical Sciences
- Grade: Grade 10
- Document type: Question paper and memo
- Year: 2024
- Exam period: KZN November Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 20

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Physical Sciences · Grade 10 · KZN November Exam · 2024. Question paper and memorandum, 20 pages. Read online or download the PDF.

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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.

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INSTRUCTIONS AND INFORMATION

1.

10.

1s

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, discussions et cetera where required.

Write neatly and legibly.

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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;

1.1 Which of the following is an example of a chemical change?

A __ The melting of ice.
B The separation of Nitrogen from air using fractional distillation.
C__ Dissolving potassium iodide in water.

D Milk turning sour (2)

1.2 Which one of the following is the correct description of a substance in the
liquid phase?

A __ The particles of the substance are bonded in fixed positions.
B__ The substance has a fixed volume but not a fixed shape.

C__ The particles of the substance always occupy the total volume of the

container.
D__ The temperature of the substance is greater than its boiling point. (2)
1.3 Which one of the following has the largest atomic radius?
A Be
B Mg
Cc Ne
D Ar (2)

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1.4 Which one of the following pairs of ions and atoms have the same number of
electrons?

A S* and Ar
B  Mg** and Mg
GC) Mg** and Ar

D  N*and He (2)

1.5 il equation for the reaction below:
) — 2S0s(g)

the SOs formed when 22,4 dm® SOz reacts

B  2x6,02 x 10%

C 3x6,02 x 10%

D  6x6,02 x 1078 (2)
[10]

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QUESTION 2 (Start on a new page.)

2.1 A bag of cement has the following percentage composition by mass:

Ingredient

Percentage composition
(by mass)

an
ulphur

Jalcium,oxide 10) 62

Hum O}

2) 22
1203)

ulp! CaSO.)

ium. C

gO)

Alkalis

TOTAL

2.121

22

2.1.3

2.1.4

2.1.5

ls the cement a MIXTURE or a PURE SUBSTANCE?
Give a reason for the answer?

Identify the type of chemical bond that forms when calcium and
oxygen combine to form calcium oxide (CaO).

Draw the Lewis structure for calcium oxide (CaO).
Write down the name of the compound Fe2Os.

Determine the number of calcium oxide (CaO) particles in a 750 g
sample of the cement.

2:2 In a process called Calcination, Calcium carbonate (CaCOs) is heated to form
the calcium oxide found in cement.

This balanced chemical equation for this calcination is:

2.2.1

2.2.2

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CaCOz — CaO + CO2

Is the reaction described above a SYNTHESIS’ or

DECOMPOSITION reaction?

Is calcination of CaCO3 an example of a PHYSICAL or CHEMICAL
change?
Explain the answer.

November 2024

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2.3

24

Calcium carbonate reacts with Hydrochloric acid (hydrogen chloride) to form
Calcium chloride, Water and Carbon dioxide.

Ast Write a balanced chemical equation for the above reaction. (3)
2.3.2 Explain why tiny bubbles are noticed during this reaction. (1)
Equal volumes of oil, milk cylinder as
shown in the diagram be

2.4.1 Define the term de vords. (2)
2.4.2 Arrange the liquids from lowest to highest density. (1)

[22]

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QUESTION 3 (Start on a new page.)

3.1

3.1

3.2

3.3

3.4

3.5

The cooling curve of AMMONIA is represented in the graph below. The graph
is not drawn to scale.

November 2024

Cooling Curve of Ammonia

Time (min)

Temperature (°C)

Write down the boiling point of ammonia.

Explain in terms of temperature, energy and movement of particles, the
changes that takes place between Point C and Point D.

Explain the change taking place between Point D and Point E.

At which region of the graph, A to B or C to D, are the particles closer
together. Explain your answer.

The melting and boiling points of three substances is indicated in the table
below.

Substance Water Ethanol Ethanoic acid
Melting Point (°C) 0 -114,1 16,6

Boiling Point (°C) 100 78,37 117,9

3.5.1 Define the term melting point in words.

3.5.2 Which substance from the table has the weakest forces between

its particles? Explain your answer.

3153 What is the phase of Ethanoic acid at room temperature?

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QUESTION 4 (Start on a new page.)

November 2024

4.1 The three stable isotopes of Magnesium and their percentage abundance is
indicated below:
Isotope Percentage Abundance
Mg - 24 78,99 %
Mg - 25 10%
Mg - 26 REMAINDER
4.1.1 Why does the atomic mass of Mg — 24 differ from that of Mg — 26? (1)
4.1.2 Determine the relative atomic mass of magnesium (3)
4.2 The following is an extract from the periodic table (refer to page 12).
TABLE 3: THE PERIODIC TABLE OF ELEMENTS
4 2 3 4 5 6 7 8 3 10 11 12 13 14 16 16 7 18
() (il) - ay (iy) (Yt) (ts (vty
1 KEY/SLEUTEL A onasete 2
Ix H He
; a8 5 6 7 8 9 +
o Symbol
Li i>} e CU Simboot ZBIECENB OF] Ne
7 = 11 12 14 16 19 20
13 14 16 16 17 18
Approximate relative atomic mass 2aee silk P |S s Sce) Ar
Benaderde relatiewe atoommassa 27 28 31 32 35,5 40
e term electronegativity in words. (2)
’ e the trend in electronegativity from left to right across a
period. (1)
4.2.3 Explain the trend described in QUESTION 4.2.2. (2)
4.2.4 No electronegativity value is indicated for Neon (Ne).
Provide a reason for this observation. (2)
4.3 Use the balanced chemical equation below to prove the law of conservation of
mass.
2Cas(POs)2 + 6SIO2 + 6CaSiO3+ PaO10 (3)
[14]

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QUESTION 5 (Start on a new page)

Aluminium powder (Al) reacts with Sulphuric acid (H2SO4) according to the balanced
equation below.

2Al(s)+ 3H2SO4(aq) — Al2(SOa«)3(aq) + 3H2(g)

In one such reaction exactly 5 g Aluminium (Al) reacts with Sulphuric acid (H2SOs) of
concentration 0,1 mol.dm* at STP.

5.1 Define the term concentration in words. (2)

5.2 Calculate the :

5.2.1 Number of moles of Al that reacted (3)

52:2 Volume of H2SOa (in dm®) that reacted (4)

§.2.3 Volume of He gas (in dm’) that formed at STP (4)
[13]

QUESTION 6 (Start on a new page)

6.1 A compound is made up of Sodium (Na), Phosphurus (P) and Oxygen (O)
only. A 10 g sample of this compound consists of 4,21 g Sodium (Na) and
1,89 g Phosphurus (P).
Determine the empirical formula of this compound. (6)

6.2 15,75 g of hydrated oxalic acid, H2C2O04.xHz0 is dissolved in water to make a
250 cm’ solution of concentration 0,5 mol.dm*.

6.2.1 Calculate the molar mass of H2C204.xH20 (3)

62.2 Determine “x”, the number of moles of water of crystallization. (3)

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6.3 The reaction between lead (I) nitrate (Pb(NOs)2) and potassium iodide(KI) is
given by the balanced chemical equation below:

Pb(NOs)a(s)+ 2KI(aq) — Pbl2(s) + 2KNOs(aq)
The percentage yield of lead (II) iodide (Pbl2) is 75 %.

Determine the mass of potassium iodide (KI) required to form 80 g of Pble. (6)
[18]

QUESTION 7 (Start on a new page)

Hydrogen gas (Hz) is prepared from the reaction between methane (CHa) and
steam (H20).

The reaction is represented by the following balanced equation:
CHa(g) + H20(g) > CO(g) + 3Hea(g)

The graph below (not drawn to scale), shows the volume of hydrogen (H2) formed
when different volumes of methane (CHs) eactswith excess steam (H20) at STP.

Volume of CHa (cm°)

7.1 Calculate the gradient of the graph (2)
72 What does the gradient of the graph represent? (2)
73 Determine the volume of Hz (in cm®) formed when 72 cm* of methane (CHa)

reacted with excess steam. (2)

74 The TOTAL VOLUME of gas formed when x cm? of CHs reacts with excess
H20 to form y cm’ of H2 is 420 cm’.
Determine the values of x and y on the graph. (4)
[10]

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DATA FOR PHYSICAL SCIENCES GRADE 10

CHEMISTRY
TABLE 1: PHYSICAL CONSTANTS
NAME SYMBOL VALUE

Standard pressure p° 1,013 x 10° Pa
Molar gas volume at STP Vin 22,4 dm*-mol
Standard temperature T? 273K
Avogadro's constant Na 6,02 x 1023 molt
TABLE 2: FORMULAE

m N
n=— fie —

M Na
(a an or c= im n= a

Vv MV Vi

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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) ; (iy (IV) (VM) (VI) (Vs)
Atomic number
8 KEY/SLEUTEL Atoomgetal hs
e e
N
1 7 29 4
3 4 Electronegativity Symbol 5 6 7 8 9 10
‘i —» 2 Cu
2 Li |2% Be Elektronegatiwiteit “| - paige Simbool = e¢CiENis SF Ne
7 9 , 11 12 14 16 19 20
1 12 t 13 14 15 16 17 18
s Na|= Mg Approximate relative atomic mass 2A@|2 SiIS P [IS 2Ce! 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
s K |2 Ca[? Sc |2 Ti |S 2 Cr |2Mn|2 Fe |2 Co |Z Ni |2 Cu /2 Zn |2 Ga|2 Ge |S 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
s Rb/2 Sr | Y |2 Zr| Nb\2Mo(2 Tc\s Ru|& Rh|s Pd |Z Ag |= Cd |= In |2 Sn |2 Sb|& Te |g | 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|3 Ba} La|2 Hf) Ta| W/ Re| Os| Ir| Pt| Au}-Hg/2 Te |2 Pb|2 Bi |S Po|G At| Rn
133| 137| 139| 179| 181; 184| 186; 190} 192| 195] 197} 201| 204| 207| 209
87 88 89
ns o
o Fr js pel Ac 58 | 59 | 60 | 61 | 62 | 63 | 64 | 65 | 66 | 67 | 68 | 69 | 70 | 7A
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

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1.1 Dvv
72 Byv
1.3 Byv
1.4 Avy
1.5 Cvv
QUESTION 2

2.1.1 Mixture. ¥
It consists of different substances that are not chemically combined. “

2.1.2 — lonic bond¥

* Multiply 750 by 0.62 ¥

e Substitute 465 and 56 into formula ¥

e n=-v

¢ Substitute 6,02 x 10% v
¢ Final answer: 5 x 10% particles V

m(CaO) = 750 x 0.62 Vv
= 465 g
n(Cad) = =

ane ys
56
= 8,3 mol

N
n(CaO) = A v
‘A

_ N
~ 6,02 x 1023 ¥

N(CaO) = 5 x 10” particlesY

8,3

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2.2.1. Decomposition. ¥

2.2.2 Chemical change. ¥
New substances are formed / A chemical reaction took place. “

2.3.1 ~ CaCO3 + 2HCI— CaClz+H20+COz LHS v¥
RHS v
Balancing ¥

2.3.2 COz gas is formed ¥
2.4.1. The mass per unit volume of a substance Vv

2.4.2 Oil, milk, honey ¥

QUESTION 3
3.1 - 33,34 °C v¥
3.2 The temperature of the substance decrease. Y The average kinetic energy of

the particles decrease. V Particles move slower. ¥

3.3 The phase is changing from liquid to solid. ¥

liquid phase from C to D but in a gas phase from A
» together in liquids than gas ¥

a solid, given sufficient heat, becomes a liquid. Vv“

st melting and boiling pointsY

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QUESTION 4
4.1.1 Mg — 26 has 2 more neutrons that Mg — 24. OR

Mg — 24 has 2 less neutrons that Mg — 26. v
4.1.2 — % Mg -26=100-78.99-10=11,01% Vv

78,99 10 11,01
i i = v
Relative atomic mass = 24x 700 + 25x 700 * 26” 700
= 24,32 AMU ¥

4.2.1. A measure of the tendency of an atom in a molecule to attract bonding

electrons VV

4.2.2 The electronegativity increases v (from left to right across a period)

4.2.3 e As you move from left to right across a period the effective nuclear
charge increase OR As you move from left to right across a period the
atomic radius decreasesY

e The force of attraction between the nucleus and the bonding electrons
increase. ¥

4.2.4 e Neon has a complete valence shell / The outermost energy level in
neon is full. ~

e It will not bond with other elements. ~
4.3 Mass of reactants = 2[40x3 + 2(31 + 4x16)] + 6[28 + 2x16] v = 980 g.mol'

Mass of products = 6[40 + 28 + 16x3] + 31x4 +16x10 v = 980 g.mol"'
Mass of reactants is equal to the mass of products. Therefore mass is
conserved. ¥

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QUESTION 5
5.1 ‘The number of moles of solute per cubic decimetre of solution. Vv (2)
5:2.1 m

an

n(Al) uM

aw.

~ 27

= 0,19 mol ¥ (3)

ITIVE MAR

(4)

POSITIVE MARKING FROM 5.2.1 AND 5.2.2

Al: He OR H2SO, ‘ H2
2:3 3 23 (or 1:1) ¥
n(H2) = 0,285 mol

Vv
n(He) = G~ ¥

m

Vv
See of”

0,285 A

V(H2) = 6,38 dm3 ¥ (4)

[13]

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QUESTION 6

6.1 Marking criteria:

e Calculate mass of oxygen ¥

e Calculate the number of moles of each element (3 x 1 = 3) Vvvv
e Determine the simplest ratio of all elements ¥

e Final answer: Na3PO, ¥

m(O) = 10- 4,21 - 1,89=3,9gv

mass _m ‘ :
Element (g) n= Simplest Ratio
4,21 0,18
pce v —_=
Na 4,21 = 0,18 0,06 V
8 0,06 = (Obtaining
0,06 all simplest
0,24 ratios)
0,06 —
6.2.1 OPTION 2
v
15,75 n
ee =—_
05 = 0.28) 05 = O25

M(H2C204.xH20) = 126 g.mol* ¥ — n(H2C204.xH20) = 0,125 mol >
m=nxM
15,75 = 0,125 x M
M(H2C204.xH20) = 126 g.mol! ¥

6.2.2 POSITIVE MARKING FROM 6.2.1
M(H2C204.xH20) oF M(H2C204) € xM(H20)
126 Vv =2 + 2x12 + 4x16 + x(18) Vv

x=2Vv

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Marking criteria:

e n= wd
Substitute 461 ¥

Substitute n(Kl) and 166”

2:1 ratio between KI and Pblo¥

e Substitute into formula to determine theoretical mass of Pbl2”

Final answerin?6

- Ie
n(Pbl2) = 0,23 mol
KI: Pblz

2:1¥

n(Kl) = 0,46 mol

n(Kl) = a

m
0,46 = see

m(KI) = 76,36 g ¥

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OR

m(Pbl)actual = m(Pble)theoretical x 0,75

80 = m(Pbl2)theoretical X 0,75 Y

(6)

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wt

7.2

7.3

7.4

_,_ 120-0
Gradient = 70.0

=3Vv
The volume/mole ratio V between hydrogen and methane ¥

V(H2) = 72x 3 ¥
= 216 cm° Vv

Marking Criteria:
¢ Separating the 420 cm’ in the ratio 1:3 to determine the V(H2) ¥
© y=315v
e Dividing y by 3¥
e x=105V¥

CO: H2
1: 3
3

V(Hz) = 5 * 4207

= 315 cm®
y=315v
x=315+3Vv
= 105 v

TOTAL:

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100

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