Download PHYSICAL SCIENCES GRADE 11 JUNE P2 MARKING GUIDLINE 2

B 33
B 33
A 33
D33
D 33
C33
Positive marking from question 2.1.3
+5+x=─3 3
OR
x = ─ 8 nC
2.5
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= 5 x 1010 3 electrons
ܳ
݊ൌ
݁
ͺ‫ିͲͳݔ‬ଽ
ൌ
3
ͳǡ͸‫ିͲͳݔ‬ଵଽ
Q3
3
2.4
= ( + 5 + (─ 11)
2
Q = ─ 3 nC3
ܳଵ ൅ ܳଶ
݂ܳܽ‫ ݎ݁ݐ‬ൌ
ʹ
─11 ─ x = ─ 33
x = ─ 8 nC
[10]
(3)
(1)
(2)
(2)
Please turn over
The net/total charge of an isolated system remains constant during any
physical process33
2.3
They attract each other3because they have opposite charges. 3
2. 2
(2)
[12]
June 2022 Common Test
2.1
QUESTION 2
1.1
1.2
1.3
1.4
1.5
1.6
QUESTION 1
2
NSC- Grade 11
଼ି଴
౳
3
NSC- Grade 11
݇ܳଵ ܳଶ
‫ݎ‬ଶ
3
9
163 =
9 x10 x Q
Q = 4,22 x10-5 C3
‫ܨ‬ൌ
2
3
3
Marking Rubric : Triangle of forces
Criteria
Mark allocation
Forces correctly drawn and
3x1=3
labelled
angle correctly shown
1
NB : Ignore the sizes of the force vectors.
F / Fattraction / Felectrostatic
FTension / T
60
0
The net force acting on sphere B is zero.
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4.1.2
4.1.1
3
଴ǡହି଴
Positive marking from question 3.3
QUESTION 4
3.4
=
= 16 3 (Accept range 15-17)
Please turn over
(4)
FTension /T
300
F / Fattraction / Felectrostatic
(1)
(4)
[9]
(2)
(1)
(2)
June 2022 Common Test
QUESTION 3
3.1
The magnitude of the electrostatic force exerted by one point charge
(Q1) on another point charge (Q2 ) is directly proportional to the product
of the magnitudes of the charges 3and inversely proportional to the
square of the distance (r) between them 3.
3.2
They are directly proportional 3
௱ி
3.3
Gradient = భ
௱ మ
Physical Sciences
Fgravity / Fg / W
Physical Sciences
Fgravity / Fg / W
4
NSC- Grade 11
ி௚௥௔௩௜௧௬
݇ܳ
‫ݎ‬ଶ
3
3
ENet = 45000 ─ 33 7503
= 11 250 NC-1 3
Accept : 33 750 ─ 45 000 3
= ─ 11 250 NC-13(Ignore sign of answer)
= 9x109 x 2 x 10-9
(0,020)2
= 45000 NC-1
‫ܧ‬ሺܳJܴሻ ൌ
= 9 x109 x 6 x 10-9
(0,040)2
= 33750 NC-1
݇ܳ
‫ܧ‬ሺܲJܴሻ ൌ ଶ
‫ݎ‬
Please turn over
(5)
(3)
Marking Rubric : Sketch of Electric field
Criteria
Mark
allocation
Correct direction of field lines
1
Shape of electric field
1
No field lines crossing each other/
1
or in the sphere
NB: if shape incorrect
0/3
(2)
(5)
June 2022 Common Test
The electric field at a point is the electrostatic force experienced per unit
positive charge placed at that point 33 ( 2 or 0)
Ͳǡʹ͹͸ͷ
–ƒ ͸Ͳ଴ 3 ൌ
3
‫݃ܨ‬
‫ ݃ܨ‬ൌ Ͳǡͳͷͻܰ3
–ƒ ߠ ൌ
ி
݇ܳଵ ܳଶ
‫ݎ‬ଶ
ሺͻ ൈ ͳͲଽ ሻሺ͵ǡʹ ൈ ͳͲି଻ ሻሺ͸ ൈ ͳͲି଼ ሻ3
ൌ
ሺͲǡͲʹͷሻଶ 3
ൌ 0,2765 N
‫ܨ‬ൌ
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4.2.3
4.2.2
4.2
4.2.1
4.1.3
Physical Sciences
(Positive marking from question 4.2.3)
F = E·Q3
= (11 250)(1,6x10-19) 3
= 1,8 x 10-15 N3
4,76
4,76
0,68
7
14
No of moles
Divide by smallest
Ratio
9
18
6,16
0,68
6,16
6,16
1
H
6,16g
1
2
0,68
0,68
0,68
9,52
14
M = 14(12) + 18(1) + 2(14) + 5(16) = 294 g.mol-1
Empirical Formula: C14 H18 N2 O5 9
57,14
12
C
57,14g
n = m/M
Mass
Simplest ratio of elements in a compound. 99
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6.3
6.2
6.1
N
9,52g
9
[8]
(1)
(5)
9
X2
9
9
9
(2)
(2)
[8]
(4)
Please turn over
2,5
5
1,70
0,68
1,70
27,18
16
O
27,18g
Water. 3
Higher boiling point. 3
5.3
QUESTION 6
Ammonia has ONE site for hydrogen bonding 3while water has TWO
sites for hydrogen bonding.3
Therefore, force of attraction between water molecules are stronger than
that of ammonia molecules. 3
Thus more energy is needed to overcome intermolecular forces in
water, 3 therefore higher boiling point.
5.2
(2)
[23]
(3)
June 2022 Common Test
The temperature at which the vapour pressure of a substance equals
atmospheric pressure. 33
5
NSC- Grade 11
5.1
QUESTION 5
4.2.4
Physical Sciences
6
NSC- Grade 11
: NO(g) :
1
0,01
=
=
x
75% 3
0,48
0,64
100
1
actual yield
theoretical yield
Mass of O2 = nM
= (0,02)(32) 3
= 0,64g 3
Positive marking from 7.3
Percentage yield =
O2 (g)
1
0,02
= 0,75
30
= 0,03mol3
n = m/M
NO
3
x 100
1
= nNO + nO2 + nNO2
= 0,01 3 + 0,02 + 0,02 3
= 0,05 mol 3
O3(g)
1
0
Total moles of gas
at end
Ratio
Moles at end
n O3
:
nO2
1
:
1 3
Therefore nO2 = n O3
= 0,02mol3
Therefore O3 is limiting3
= 0,8
48
= 0,02mol 3
n = m/M 3
O3
:
NO2(g)
1
0,02
Please turn over
[15]
(4)
(3)
(2)
(4)
(2)
June 2022 Common Test
The reactant that is completely used up in a chemical reaction.33
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7.5
7.4
7.3
7.2
7.1
QUESTION 7
Physical Sciences
7
NSC- Grade 11
= C x V
= 1,0 x V 3
= 0,016 dm-3 3
% purity = Pure sample x 100
Impure sample
1
= 1,10 x 100
3
1,5
1
= 73,33% 3
m = nM
= (0,011)(100) 3
= 1,10g
n CaCO3 : nCO2
1
:
1 3
Therefore: n CaCO3 = 0,011mol
n CO2 = m/M
= 0,5 3
44
= 0,011 mol
n NaHCO3 : nCO2
1
:
1 3
n = m/M
0,016 = m 3
44
m = 0,70g 3
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8.2
8.1.3
n
0,016
V
8.1.2 n NaHCO3 = m/M
= 1,35 3
84
= 0,016 mol
n NaHCO3 : nHCI
1
:
1 3
Therefore, nHCI = 0,016 mol 3
(3)
(5)
(2)
Please turn over
TOTAL MARKS: 100
M = 40 + 12 + 3(16)
= 100 g.mol-1
M = 12 + 2(16)
= 44 g.mol-1
M = 23 + 1 + 12 + 3(16)
= 84 g.mol-1
(5)
[15]
June 2022 Common Test
8.1.1 The number of moles per unit volume of a substance. 33
QUESTION 8
Physical Sciences