# Physical Sciences Grade 11 2023 QP & Memo

> Grade 11 Physical Sciences question paper and memo from the Eastern Cape September Test 2023, 18 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/ec-amathole-west-physical-science-grade-11-september-2023-qp-and-memo/5f8a52c4-ec7a-45e2-a007-695acafbba8b
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2023
- Exam period: Eastern Cape September Test
- Curriculum: CAPS
- Pages: 18

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## About this document

Physical Sciences · Grade 11 · Eastern Cape September Test · 2023. Question paper and memorandum, 18 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

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### Page 1

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ead Province of the =e
EASTERN CAPE
EDUCATION

MARKS: 100

TIME: 2 HOURS

This question paper consists of 11 pages including Data Sheet

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INSTRUCTIONS AND INFORMATION

1. Start EACHiquestion on a NEW page.

2. Number yout;answers correctly according to the numbering system used in this
question;paper.

3. Leave ONE line between two sub-questions, e.g., between QUESTION 2.1 and

QUESTION 2.2.

A non-programmable calculator may be used.

Appropriate mathematical instruments may be used.

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, etc. where required.

You are advised to use the attached data sheets.

SDP NOMS

0.Write neatly and legibly.

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QUESTION 1: MULTIPLE-CHOICE QUESTIONS

Various options are provided as possible answers to the following questions. Each question has only

ONE correct a . Write only the letter (A-D) next to the question number (i.e 1.1 D) in the
ANSWER BO
Ll The minimum amount of energy needed for a reaction to start is called......
A Activation energy
B Catalyst
Cc Heat of reaction
D Bond energy
(2)
1.2... Which ONE of the following statements about a chemical reaction is CORRECT?
The actual yield of a chemical reaction is usually.......
A ge yield
B : ntage yield
Cc Less theoretical yield
D Greaterithan the:thedretical yield
(2)

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Which ONE of the following statements is CORRECT for an endothermic

reaction?

A
B
Cc
D

GaN WU >

GaN Ww >

The temperature of the surroundings increases
The enthalpy change for the reaction is negative
Heat flows-from the surroundings into the system

The enthalpy of products is less than the enthalpy of reactants

Q)
Consider an incomplete chemical equation below:
X + 2HNO3 — Zn(NO3)2 + H2O0 + CO
Which ONE of the following is represented by X in the above equation?
ZnCO3
ZnlICO;
ZnCOz
Zn(OH)2
(2)

The number of ions present in 3 moles of MgCl is..........

3, 612 X 10%
5, 418 x 10%
1, 08 x 104
7, 22 x 10°4
(2)

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According to the kinetic-molecular theory, molecules of different gases at the same

temperature always have the same....

Pressure

Volume

Kinetic’ y

Average kinetic energy
(2)

Two different gases of the same volume at STP will have the same......

Mass

Density

Molar mass

Number of molecules
(2)
[14]

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QUESTION 2 (Start on a new page)

40

_— GRAPH OF VOLUME VERSUS PRESSURE

fry:

50 100 150 200 250

Pressure (kPa)

A fixed mass of oxygen is used to verify one of the gas laws. The results obtained are shown in the

graph below.

2.1 Write down a mathematical expression, in symbols, for the relationship between

the variables shown in the graph. (1)
2.2. Give the name and state the gas law investigated. (3)
2.3 Explain the relationship in QUESTION 2.1 in terms of the kinetic theory of gases. (2)
24 Write down TWO variables that must be kept constant during this investigation

and briefly describe how this is done. (4)
2.5 From the graph, write down the volume of oxygen, in cm?, whi pressure is

120 kPa. (2)
2.6 Using the graph and the information provided, calculate the p e, in kPa,

exerted on the gas when it is compressed to 5 cm*. (4)

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2.7 Write down TWO conditions under which oxygen gas will deviate from ideal gas
behaviour. (2)

[18]

QUESTION 3-Start on a new page.)

Consider the reaction represented by the equation below.

CO2() + 2H20 0) > CHaie) + 2029)
During the reaction the temperature of the reaction mixture DECREASES
Bond energies (D) in (kJ.mo!"):
C=0 D=602 H-O D=459 C-HD=411 O=O D = 494

(2)

3.1 Define the term enthalpy change.

3.2 Does the enthalpy change (AH) for this reaction have a positive or negative value?

Explain the answer by referring to the energy involved.

3.3. Sketch a labelled potential energy versus course of reaction graph for this reaction. On

the graph, show the position of the reactants, products, AH and activation energy.

(3)

(6)
[11]

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QUESTION 4 (Start on a new page)

The chemical reaetion for the production of the drug, aspirin, from two compounds, X and Y, is

represented by thé/balanced equation below.

267H6O3 + CaH6O3 + 2CoHgO4 + H2O
x Y aspirin
A chemist reacts 14 g of compound X with 10 g of compound Y.
4.1 Define the term limiting reagent in a chemical reaction. (2)
4.2 Perform the necessary calculations to determine which one of compound X or compound

Y is the limiting reagent. (5)
The actual mass of aspirin obtained is 11, 5 g.

4.3 Calculate the percentage yield of aspirin (5)
[12]

QUESTION 5 (Start on a new page)

In order to determine the empirical formula and molecular formula of a compound, CxHy, a certain

mass of the compound is burnt completely in excess oxygen to produce 47, 1 g CO> and 19, 35 g

H20 as the only products.
5.1.1 Define the term empirical formula. (2)
5.1.2. Use relevant calculations to determine the empirical formula of the compound. (8)

5.1.3. The molar mass of the compound is 28 g.mol"!. Determine by using calculations the

value of x and y. (3)

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A sample of IMPURE calcium carbonate (limestone) of unknown mass required a continuous supply

of strong heat to decompose according to the following equation:

CaCO) > CaO) + CO2@)

After the completion of reaction, 11, 76 g CaO was produced.

The percentage purity of calcium carbonate is found to be 80%.

5.2 Calculate the mass of the impure calcium carbonate. (6)
5.3. In an experiment, a learner added 1, 5 g of sodium carbonate (Na2CO3) to hydrochloric acid

(HCl). A volume of 306 cm? of carbon dioxide gas was formed and collected under standard

pressure at room temperature. Take the molar gas volume at room temperature (Vm) as 24,45 dm.

The unbalanced equation for the reaction is:

NaxCOxs) + 2HCl(aq) + 2NaCl(ag) + H20q) + COxz)

5.3.1 Calculate the mass of sodium carbonate that reacted (6)
5.3.2 Calculate the percentage of sodium carbonate in excess (4)
[29]

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QUESTION 6 (Start on a new page)

A solution of potassium hydroxide (KOH) is prepared by dissolving 3, 36 g crystals of KOH in 250

cm’ of solution.

6.1 Define the-term concentration. (2)

6.2 Calculate the concentration of potassium hydroxide solution. (4)

25 cm’ of a potassium hydroxide solution of concentration 0, 25 mol.dm* completely neutralises a

dilute solution of sulphuric acid (H2SOx,) in a flask.

The incomplete equation below represents the reaction that takes place:

2KOH + H2SO4 — K2SO,4 + H20

6.3.1 Define the term standard solution (2)

6.3.2 Calculate the number of moles potassium hydroxide (KOH) (3)

6.3.3 Calculate the mass of sulphuric acid in the flask (5)
[16]

TOTAL: 100

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TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES

NAME/NAAM SYMBOL/SIMBOOL VALUE/WAARDE
Avogadro’ stant 23 a
Avogadro/konstante Na 8,02 x 10° mol
oe oe R 8,31 J-K*-mot"

jolére stante
Standard_pregsure a 5
Standaarddruk P 1,013 x 10° Pa
Molar gas volume at STP saat
Molére gasvolume by STD Vim 22,4 dm*mor"
Standard temperature e
Standaardtemperatuur r 273 K

TABLE 2: FORMULAE/TABEL 2: FORMULES

PVs _ PoVo
= pV =nRT
Tj,
N
ne anal.
M N,
Vv n m
=— c=— OR/OF c=
Vn V Ww
TABLE 3: THE PERIODIC TABLE OF ELEMENTS
1 2 3 4 15 6 TtB
o m * 4 § 8 FF FeO Pe wy mM Mo mM vy
5 FA KEVISLEUTEL oa ie
1 4
a [2 sL elvis) ©
2 LIS Be Electronegativity —& = Cu le Simpel a BRCRNIS O'S F| Ne
7 9 | Miskoronegaiemen 635 1) 12| 14| 16] 19) 20
WW) 4 3) 4) 15] 16) 7] 18
3 Na > Mg Approximate relative stomic mass PACE SIG PIE SF Ce Ar
23 | 24 Benaderde reiatiews 3 27 | ae! 31 | 32] 355) 40
19] 2 | a] 2] | 2| 25| 2| 2| 2] 2] %0| 31| $2] 33| 34] 35] 36
s K = Cal® Se |S Ti |= V [2 Cr |S Mnj= Fe |= Co|= Ni |= Cu/= Zn |= Ga = Ge|5 As|= Se = Br) Kr
so} aol 45/ 48| 51/52! 55| 56| s0| 50| 635| 65| 70! 73| 75| 79| 90| 84
37] 38 | 39) 40) 41] 42 | 43| 44] 45] 46] 47] 48] 40] 80] 51] 52] 83] 54
= Rb S Sr it Y |= Zr) Nb/= Molt Te |S Ru Rhis Pd |S Ag|=Cd/> In |S Sn\FSbie Tels | Xe
|_s6, s8| eo| o1/| 92! 96 401| 103| 106| 108] 112| 415| 149| 122| 128, 127, 131
35 | 56 | 8?) 72 | 73] 74 | | 76] 77] 78] 79] 80 | 81 | 82] 83 | ea | 85] 66
5 Csi2 Ba| Laj/2 Hf) Ta| W/| Re| Os| Ir| Pt} Au) Hg/S Te 2 Pb/2 Bi/R Pos At Rn
133) 137, 139 179 181 184| 186| 190 192 195 197 201 204 | 207 209 i l
_ ai ey 88
is Fr is a) Ae % ] |) 8) | | | | 6) | oe) Oo) |
1 Ce | Pr | Nd | Pm | Sm/| Eu | Gd | Tb | Dy | Ho Tm) Yb Lu
tao | 141 | 148 150 | 152 | 157 | 159 | 163 185 169 | 173°) 175
a for | 702) 403
Th | Pa | U | Np) Pu| Am/|Cm| Bk | Cf Md | No) Lr
232 238 a

11

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wi Province of the =e
Yen EASTERN cre

iz
AMATHOLE WEST DISTRICT

MARKS: 100

This MARKING GUIDELINE consists of 7 pages lie cover page

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QUESTION 1

11 A VV
1.2 ¢ VV
1.3¢ VV
14A VV
1.5 A VV
1.6 D VV
1.7 D VV

[14]

QUESTION 2

21

1 1
— ORIOF V a— v¥
pay a

(1)

22

Boyle’s law V, The pressure of an enclosed gas is inversely proportional to the
volume it occupies at constant temperature VV

(3)

23

When the volume of the container is decreased, the particles will be closer
together V, hence the particles will collide more with each other and the wall of
the container leading to increase in the pressure. Vv

(2)

2.4

Mass of the gasV — ensure there’s no leakage.V Temperature v- gradually
increase the pressure without having much effect on the temperature.V

(2)

25

30 cm3 VV

(2)

2.6

Pav = P2v2 Vv
120(30) V = p2(5) V
P2 =720 kPa V

Note
May use any set of values from the graph that
gives the correct answer

(4)

2.7

High pressures Vv
Low temperatures V

(2)

[18]

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Potential energy
Potensiéle energie)

>
Progress of the reaction/Reaksieverioop

Marking criteria/Nasienriglyne:

Labelled axes/Benoemde asse ¥

Shape/Vorm ¥

Position of reactants/Posisie van reaktanse ¥
v

QUESTION 3
——
RB! The a of heat/energy released or absorbed in a chemical reaction. (2)
3.2 Positi
The e of the products is greater than energy of the reactants vv (3)
3.3

(6)

(11)

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QUESTION 4
4.1 Limiting reagent is the substance that is totally used up in a chemical reaction. V¥ (2)
42 From the equation;
n(CoHeO: n) = 2/1=2V¥
From supplied;
n{CrHeOs) = m/M
= 14/138
=0,1014mol ¥
n(CcHeQa) = m/M
= 10/102
= 0,098mal Vv
n{CrHeOa)/n{C4HeOs) = 0, 1014/0, 098
=1,034¥
Therefore, CeHeOa(Y) is the limiting reagent..V (5)
43 CaHeOs : CoHeOy
42 2
0,098: x
ni(CuHs0,) = 0, 196 mol
m=nM
m= 0, 196 x 180 Vv
m = 35, 28g
percentage yield = —vielt + 190% V

theoretical yield

_ 4s
= sey x 100%

= 32,60%V
(5)

{12)

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QUESTIONS
$11 formuta: Smallest whole number ratio of elements that make up | (2)
vane WW
faYaYay
s1i2]0 1 OPTION 2 (8)
=m/M ¥ 96H in H20 = 2/18 x 100 ¥
= 19,35/18 ¥ = 11,11%
= 1,075 mal m{H) in #0
n(H) = 2(H20)
=2x1075 =11,11% of 19,35eV
=2,15 mol ¥ =2,15¢
Poy = 21541
=2,15mol ¥
(COs) = m/M %C in CO) = 12/44 x 100 ¥
=47,1/44 ¥ =27,27%
= 1,07 mol m(C) in CO: = 27,27% of 47,1¢ V
n(C) =n (COs) = 12,84 g (to 12,85 g)
=1,07 mol ¥ Macy = 12,84 + 12
=1,07mol¥
mol C: mol H mol C: mol H
1,07:2,15 ¥ 1,07:2,15 ¥
1:2 Vv 1:2
Empirical formula: CHa ¥
5.1.3 | POSTIVE MARKING FROM QUESTION 5.1.2 (3)
M{CH2) = 1(12) + 2(1) = 14g.mot*
(14)n= 28 ¥
n=2
molecular formula: (CH2)2
GH. ¥ x=2y=4v
52 (6)
n(Ca0)}=m/M Vv
11,76/56=0,21 mol
(CacOs) : (Cad)
+s 2
Micacoa = 0, 21 mol ¥
m (CaCO3) = nM
= a (100) Vv
=
x pure sample
mapurity = impure sample 100%
== x 100% V = 26.25% V

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$.3.1

= n(NazCOs) = 0,0125 mal ¥ (mole ratio 1:1)
0) =nM ¥

= 0,0125 vx (23(2) + 12 +3(16)) ¥
=133gV

(6)

$.3.2

Mass (Na:CO,) excess = 1,5 ~ 1,330
=O,17gV¥
% excess = (0,17/1,5) x 100 ¥
= 11, 33%V

(4)

[30]

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QUESTION 6
6.1 | Concentration is the amount of solute per litre of solution. | (2)
OR
Con tion is the number of moles of a substance per dm? of solution
6.2 opTiON-E OPTION 2 (4)
c=m/Mv Vv n=m/M =3,36/56V
= 3,36 V/ (56) (0,25) ¥ = 0,06 mo!
= 0,24 mol.dm? ¥ c=nvv¥
=0,06/0,25 ¥
= 0,24 mol.dm? ¥
6.3.1 Solution of known concentration VV (2)
63.3 |n=cv¥ (3)
n(KOH) = (0,25) (0,025) ¥
= 6,25x 107 mol ¥
6.3.4 1i(H2SOs) = 0,5n(KOH) ¥ (mole ratio 1:2) (4)
= 0,5 (6,25x10")
= 3,125 x 10°mol
m(H2SO.) = nM Vv
= 3,125 x10 (98) V
=0,31gV
[15]

Total =

[100]

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