# Physical Sciences Grade 10 Paper 2 KZN November 2023 Memo

> Grade 10 Physical Sciences Paper 2 memorandum from the KZN November Exam 2023, 19 pages. Read online or download the free PDF.

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- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 10
- Document type: Memorandum
- Year: 2023
- Exam period: KZN November Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 19

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## About this document

Physical Sciences · Grade 10 · KZN November Exam · 2023. Memorandum, 19 pages. Read online or download the PDF.

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               KWAZULU-NATAL PROVINCE
               EDUCATION
               REPUBLIC OF SOUTH AFRICA

                                   NATIONAL
                              SENIOR CERTIFICATE

        ,.
         /#
             �--------------                               -----------�.,
                 PHYSICAL SC                                                II
                                                                            II
                                                                            II
                                                                            II
                                                                            ,1
                                  NO EMBER 2 2                              II
                                                                            II
                                                                            II
                                                                            II
                                                                            II
        �                                                                   �
             �-------------------------------------��
                                                                        /1
        ��

MARKS: 100

DURATION: 2 hours

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

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INSTRUCTIONS AND INFORMATION

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

—
2. StartEACH question on a NEW page in the ANSWER BOOK.

3. uel e answers correctly according to the numbering system used in this
question paper.

4. Leave ONE line between two sub-questions, for example between
QUESTION 2.1 and QUESTION 2.2.

5. You may use a non-programmable calculator.

6. You may use appropriate mathematical instruments.

7. You are advised to use the attached DATA SHEETS.

8. Show ALL formulae and substitutions in ALL calculations.

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

10. Give brief motivations, discussions et cetera where required.

11. 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 1.6 D.

1.1 Whi the following is not a property of a metal?

A ny
B Ductile and malleable
C Thermal conductor

D Insulator of electricity (2)

1.2 An investigation is conducted to study the effect of temperature on the
conductivity of a substance. The results are represented in the graph below.

Which ONE of the following is the substance being investigated?

A Copper
B Aluminium
C__ Silicon

D~ Zinc (2)

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1.3 Which one of the following is an example of a chemical change?

A __ Separating oil and water mixture using a decanter.
QD

a

/Gendensation of water vapour.
Cc Magnesium burning in air.

B

D___Dissolving sugar in water. (2)

1.4 Consider the balanced chemical equation below:
AALS) + 302(g) — 2At20a(s)

Which one of the following is the correct number of oxygen atoms required to
form 4 moles of aluminium oxide (AtzO3z)?

A 6x6,02 x 107
B 12x 6,02 x 1073
C 3x6,02 x 10?

D  36x6,02 x 10? (2)

1.5 Consider the unbalanced chemical equation below:
NH3 + O2 > NO + H2O

Which ONE of the sets of coefficients will balance the equation?

A 2,2,2,3
B 2,3,2,3
C 5,4,5,6
D 4,5,4,6 (2)

[10]

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QUESTION 2 (Start on a new page.)

Four different substances, W, X, Y and Z are represented in the table below.

Substance W Substance X KEY

CO CO O - Hydrogen
CO - Lithium

Substance Y Substance Z &Y - Nitrogen

NW

2.1 Is Substance X an ELEMENT or a COMPOUND?

Give a reason for the answer. (2)
2.2 Write down the LETTER of the substance that represents a mixture. (1)
2.3 Write down the COMMON NAME of Substance Y. (1)
2.4 Draw the Lewis structure of Substance Y. (2)
2.5 Identify the type of chemical bond between atoms in Substance W. (1)
2.6 Hydrogen gas and Nitrogen gas reacts to produce Substance Y.

2.6.1 Write down a BALANCED chemical equation for the reaction that (3)

takes place.
2.6.2 Is this a synthesis or decomposition reaction?

Give a reason for the answer. (2)
2.6.3 Use the balanced chemical equation from Question 2.6.1 to prove

the law of conservation of mass in this reaction. (3)

[15]

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QUESTION 3 (Start on a new page.)

3.1

3.1

3.2

3.3

3.4

3.5

3.6

3.7

The cooling curve of ETHANOIC ACID is represented in the graph below. The

grap! ot drawn to scale.

ira C The Cooling Curve of Ethanoic Acid
B

Temperature (°C)

10 20 _. . 80 90 95
Time (min)
Define the term melting point in words. (2)
What is the melting point of ethanoic acid? (1)

Are the forces between particles stronger in ETHANOIC ACID or in WATER? (2)
Give a reason for the answer.

Write down the phase of ethanoic acid at 5 minutes. (1)

Write down the name given to the phase change occurring between 10 and 20
minutes. (1)

Explain why the change referred to in QUESTION 3.5 is a Physical Change. (2)

At which points, A, B or C are the following statements applicable?

3.7.1 The forces between particles are the strongest. (1)

3.7.2 The spaces between the particles are the largest (1)

3.7.3 The kinetic energies of the particles remain constant. (1)
[12]

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QUESTION 4 (Start on a new page.)

November 2023

Some information for the substances IRON and the SULPHIDE ION are given in the

table below.
Number of | Number of | Number of
Name 10 protons electrons neutrons
IRON | Fe 4.1.1 26 4.1.2
SULPHIDE
ION 4.1.3 16 4.1.4 16
4.1 Complete the table by providing the answers to questions 4.1.1 to 4.1.4.
Do not redraw the table. Write the numbers (4.1.1— 4.1.4) down and the correct
answer next to it. (4)
4.2 Draw the aufbau diagram and write down the electronic configuration
(sp-notation) for the sulphide ion. (3)
4.3 How many valence electrons does the sulphide ion have? (1)
4.4 Iron (Fe) has three common isotopes as indicated below
Isotope Percentage Abundance
Fe - 54 5,8 %
Fe - 56 91,7%
Fe - 57 REMAINDER
4.4.1 Define the term isotope in words. (2)
4.4.2 Determine the relative atomic mass of Iron (Fe). (3)
[13]

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QUESTION 5 (Start on a new page.)

5.1 The table below shows the first and second ionisation energies of elements in
PERIOD 2 of the periodic table.

<4cFIRST IONISATION ENERGY | SECOND IONISATION ENERGY
‘2lti.3.mol) (kJ.mo!*)
Li }~520 7 297
Be 899 1757
B 801 2 427
Cc 1086 2 352
N 1 402 2 854
fe) 1214 3391
F 1681 3381
Ne 2 080 3 964
5.1.1 Explain why there is a general increase in the first ionisation energy
on going from left to right across the period. (2)
5.1.2 It is observed that the second ionisation energy of Li (Lithium) is
considerably higher than Be (Beryllium). Explain this observation. (2)

5.1.3 Write down the NAME of the group whose elements have the highest
first ionisation energy in the period. (1)

5.2 Write down the CHEMICAL FORMULA for:

5.2.1 Sodium oxide (1)
5.2.2 Calcium phosphate (1)
5.2.3 Copper (Il) carbonate (1)

5.3 Write down the CHEMICAL NAME for:

5.3.1 MgBrz2 (1)
5.3.2 (NHa4)2SO4 (1)
[10]

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QUESTION 6 (Start on a new page)

Sulphur dioxide (SOz) reacts with oxygen (Oz) to form sulphur trioxide (SO3), as shown
in the balance: uation below.

2SO02(g) + O2(g) > 2SO0s(g)
In one such reaction 2450 cm? sulphur trioxide (SOs), is formed at STP.
6.1 State Avogadro’s law in words. (2)

6.2 Calculate the:

6.2.1 Number of moles of SO3 that formed. (4)

6.2.2 Mass of SOz that reacted. (4)

6.2.3 Number of oxygen (Oz) molecules that reacted (4)
[14]

QUESTION 7 (Start on a new page)

rok The analysis of a compound made up of the elements sodium (Na), sulphur
(S) and oxygen (O) provides the following percentage composition, by mass.

=

Element ercente osition

Sodium (Na) lang

Sulphur (S) 4

Oxygen (O) 30.
7.1.1 Define the term empirical formula in words. (2)
7.1.2 Determine the empirical formula of this compound (5)

7.2 Determine the percentage composition of water (H2O) in hydrated copper

sulphate (CuSO4*5H20). (4)

[11]

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QUESTION 8 (Start on a new page)

Calcium carbonate (CaCOs) reacts completely with excess phosphoric acid (HsPOs) of
concentration mol-dm*.

The reaction is! presented by the following balanced equation:
3CaCOx(s) + 2HsPO«(aq) — Cas(POa)2(aq) + 3CO2(g) + 3H20(2)

The graph below (not drawn to scale), shows the mass of the calcium carbonate
(CaCOs) as the reaction progresses

0123 4 Time (min)
8.1 Determine the number of moles of Cas(PO4)z2 FORMED AFTER 1 MINUTE. (5)

The reaction is thereafter allowed to reach completion.

8.2 Determine the number of moles of calcium carbonate (CaCOs) used. (2)
8.3 Calculate the volume of phosphoric acid that reacts with the calcium
carbonate after 4 MINUTES. (4)
8.4 Calculate the percentage yield of carbon dioxide (COz), if 4 dm? of COz2 was
formed at STP. (4)
[15]
TOTAL: 100

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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 107 mol
TABLE 2: FORMULAE
n=™ n=
M Na
n m Vv
c=— or c=— n=——
Vv MV Vin

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Physicebeuenaadedefiopm Stanmorephysics.com NSC June 2023
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) . ay (iv) (VY) (VI) s(t) Ss ()
Atomic number
4 1 KEY/SLEUTEL Atoomgetal 2
; H He
N
1 29 4
3 4 Electronegativity || o, Cu ¢_Symbol 5 6 7 8 9 10
3 Li |/% Be Elektronegatiwiteit “| 4 63,5 | Simbool = Bt Nie ¢ Ne
7 9 , 11 12 14 16 19 20
11 12 f 13/ 14/ 15| 16/ 17| 18
o Na|S Mg Approximate relative atomic mass 2 Al\2 Si ls P IS &Ce! 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
e K |3 Cal Se |S Ti |S & Cr |3Mn|& Fe |3 Co/S Ni |G Cu/S Zn |S Ga|S 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
6 Rb|S Sr |S Y |S Zr| Nb|SMo|3 Tc|s Ru|s Rh|a Pd /3 Ag |S Cd/5 In (8 Sn/3Sb\5 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
5 Cs|2 Ba} La|S$ Hf) Ta| W]| Re| Os| Ir| Pt| Au| Hg/% Te |2Pb/% Bi /3Po/s At| Rn
133| 137| 139| 179| 181| 184| 186/] 190| 192| 195; 197| 201| 204| 207| 209
87 88 89
cS Fr |G Ra| Ac
S © 396 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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KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

MARKS: 100

DURATION: 2 hours

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det Dvv
12 Cvv
1.3 Cvv
1.4 Byv
1.5 Dvv
QUESTION
2.1 Element. ¥

It consists of one type of atom only. ~
2.2 (Substance) Z ¥
2.3 Ammonia ¥
2.4

vv

2.5 lonic bond ¥
2.6

2.6.1 N2 + 3H2 > 2NH3 LHS v

2.6.2

2.6.3

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RHS v¥
Balancing ¥

Synthesis. ¥
Smaller elements are combining to form a larger compound ¥

POSITIVE MARKING FROM 2.6.1
Mass of reactants = 14x2+ 3 (1x2) v
= 34 g.mol?
Mass of products = 2(14 + 1x3) v
= 34 g.mol?
Mass of reactants is equal to the mass of products. Therefore mass
is conserved. ¥

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

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

(1)

(3)

(2)

(3)

[15]

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QUESTION 3
3.1 The temperature at which a solid, given sufficient heat, becomes a liquid. VV
3.2 16°C”
3.3 Etha

It ha: gher boiling point/melting point than water. ¥
3.4 Gas
3.5 Condensation ¥
3.6 No new substance is formed / No chemical bonds are broken. Vv
3.7 3.7.1 Cv

3.7.2 Av

3.7.3 BY
QUESTION 4
4.1

4.2.1 26 ¥

4.2.2 30 ¥

4.2.3 stv

4.2.4 18 v
4.2

as |ty

sp-notation: 1s?2s?2p°3s73p° v

4.3 8v

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(2)
(1)

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

(1)
(1)

[12]

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

(3)
(1)

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4.4
441
4.4.2
QUESTION 5
5.1
5.1.1
5.1.2
5.1.3
5.2
5.2.1
5.2.2
5.2.3
5.3
5.3.1
5.2.2

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Atoms of the same element having the same number of protons but
different numbers of neutrons. VW

% Fe-57 = 100-5,8-91,7=2,5% Vv

. : 7 5,8 91,7 2a
Relative atomic mass = 54x 700 + 56x 700 + 57x 700 V
= 55,91 AMU ¥

As you move from left to right across a period the effective nuclear
charge increase ¥

The force of attraction between the nucleus and the electrons in the
outer orbital increases. ¥

Losing a second electron will result in an unstable electron state in
Li Y, whilst losing a second electron results in a stable electron state
in Bev

The second electron in Li fi
close to the nucleus, compar
attraction between the nucleus an
Li. Hence more energy is needed-
Li compared to Be V

the energy level very
herefore the force of

iter electron is stronger in
ve the second electron in

Noble gas ¥

NazO ¥
Cas(POa)2 ¥

CuCOs ¥

Magnesium bromide ¥

Ammonium sulphate ¥

(2)

(3)
[13]

(2)

(2)
(1)

(1)
(1)
(1)

(1)
(1)

[10]

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QUESTION 6

6.1 One mole of a gas occupies the same volume at the same temperature and
pressure. V¥

6.2
6.2.1
Vv v
n(SOs) = 7
m
2,45
n(S0s)= 554”

n (SOs) = 0,11 mol ¥

6.2.2 POSITIVE MARKING FROM 6.2.1

SO2 : SOs
2 :2v
n(SOz) = 0,11 mol
m v
n(SO2) = 7
=. v
0,11 = 64

m (SO2) = 7,04 g ¥

6.2.3 POSITIVE MARKING FROM 6.2.1
O2: SOs
+2"

- v
ao 6,02 x 1023

N(Oz) = 3,31 x 10? O2 molecules ¥

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(2)

(3)

(4)

(4)

[13]

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QUESTION 7

Tal TAL The simplest whole number ratio of elements in a compound. v¥ ¥

4.2 Consider 100g

mass m . .
Ele! n=— Simplest Ratio
ment) (g) M
Na 2o11 | 29-11 _ 127. v
23 v
40,51 (Obtaining
ie v
7 40,51 32. 1,27 all simplest
ratios)
fe) 30,38 a -19

Empirical Formula: NazS2O3 ¥

7.2 M(CuSO«.5H20). = 63,5 + 32 + 4x16 + 5(2+16) ¥
= 249,5 g.mol?

. s x18 wv
% H2O 49,5 x100

= 36,07 %V¥

QUESTION 8

8.1 m(CaCOs) = 25-10 ¥
=15g

n(CaCO3) = 7 v

_ 15 VS.
~ 100
= 0,15 mol
CaCOs : Cas(POa)2
3 oe -T v

n [Caa(POx)2] = 0.05 mol ¥

8.2 _m
n(CaCQ3) = M
_ 25 4
~ 400
= 0,25 mol ¥

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(2)

(5)

(4)

[11]

(5)

(2)

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8.3 POSITIVE MARKING FROM 8.2

CaCOs : H3PO4
| 1 2v
n(HsPOa) = 0,17 mol

(4)

8.4 POSITIVE MARKING FROM 8.2
CaCOs : CO2
3.:3v
n(COz) = 0,25 mol

Vv
n(COz) = va

m

v

0,25 = 224

V (COz) = 5,6 dm?

actual yield

theoretical yield nad

% yield=
4v
% yield= = x100 ¥
% yield = 71,43% Vv (5)
[16]

TOTAL: 100

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