secondary school improvement programme (ssip) 2019
TRANSCRIPT
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SECONDARY SCHOOL IMPROVEMENT
PROGRAMME (SSIP) 2019
GRADE 12
SUBJECT: PHYSICAL SCIENCES
TERM 01
EDUCATORS GUIDE
PAGE (1 OF 22)
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Organic Chemistry
Question 1
Multiple-choice Questions
1.1. A
(2)
1.2. B
(2)
1.3. C
(2)
1.4. B
(2)
1.5. A
(2)
1.6. B
(2)
1.7. A
(2)
1.8. D
(2)
1.9. C
(2)
1.10. A
(2)
1.11. C
(2)
1.12. B
(2)
1.13. B
(2)
1.14. D
(2)
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Question 2
2.1.
2.1.1. Alkynes
(1)
2.1.2. Hydroxyl group
(1)
2.1.3. C
(1)
2.1.4. 2-methylpentan-3-one
(2)
2.1.5.
C C
H
H H
H
(2)
2.1.6. 2πΆ4π»10 + 13π2 βΆ 8πΆπ2 + 10π»2π π΅ππ.
(3)
2.2. Same molecular formula, but different positions of the
functional group.
(2)
2.3
C C O
H
H
H H
H
H
+ C C C
O
O
H
H
H H
H
H
C C O C C C
OH
H
H
H
H H
H
H
H
H
+ O
H
H
(7)
[19]
Question 3
3.1.
3.1.1. B (1)
3.1.2. E (1)
3.1.3. F (1)
3.2.
3.2.1. 2-bromo-3-chloro-4-methylpentane (3)
3.2.2. Ethene (1)
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3.3.
3.3.1.
C C C C C C
C
CC
H
H
H H
H H
H
H
H
H
H
H
H
H
H
H
HH
H
H
(2)
3.3.2.
C C C C C
OH
H
H H
H
H
H
H
H
H
(2)
3.4.
3.4.1. Compounds with the same molecular formula
but different functional groups / different homologous series.
(2) 3.4.2. B & F (1)
[14]
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Question 4
4.1.
4.1.1. Carboxyl group (1)
4.1.2. Ketones (1)
4.1.3. Addition (1)
4.2.
4.2.1. Ethene (1)
4.2.2. 4-methylhexan-3-one (2)
4.2.3. 4-ethyl-2,2-dimethylhexane (2)
4.3. Carbon dioxide/CO2
Water / H2O (2)
4.4.
4.4.1.
C C C C
C
O
O
H
H
H H
H
H
H
H
H
H (2)
4.4.2.
C C C C
H
H
H H H H
H
H
OR/OF
C C C C
H
H H H
H
H
H
H
(2)
4.5.
4.5.1. E (1)
4.5.2. Substitution / halogenation / bromination (1)
4.5.3
C C C C
Br BrH
H
H H H H
H
H
(2)
[8]
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Question 5
5.1.
5.1.1. B (1)
5.1.2.
(1)
5.1.3. πΆππ»2πβ2 (1)
5.1.4. 4-ethyl-5-methylhept-2-yne (3)
5.1.5. Butan-2one (2)
5.2.
5.2.1. Alkanes / Alkane (1)
5.2.2. Methylpropane
C C C
C
H
H
H H H
H
H
H
H
H
(4)
5.3.
5.3.1. Haloalkanes / Alkyl halides (1)
5.3.2. Substitution / halogenation / bromination (1)
[6] Question 6
6.1Temperature at which the vapour pressure of the substance
Equals atmospheric pressure. (2)
6.2 .1 Boiling point increases as the chain length / molecular mass increases.
OR Boiling point increases from methane to butane. (1)
6.2.2
Chain length increases from methane to butane.
Strength of London forces / induced dipole forces increases
from methane to butane.
More energy needed to overcome intermolecular forces in
butane than in methane. (3)
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6.3 Between molecules of the alkanes are weak London forces or
induced dipole forces.
Between alcohol molecules are, in addition to weak London Forces or induced dipole forces, also strong hydrogen bonds. (2)
[8]
Question 7
7.1Alkanes have ONLY single bonds between C-atoms. (1)
7.2
7.2.1
OR (1)
7.2.2
C C C
C
O
H
H
H H
H
H
H
H
H
H (2)
7.3.1 What is the relationship between chain length / molecular size/
molecular structure / molecular mass / surface area and boiling point?
(2)
7.3.2
Structure:
The chain length / molecular size / molecular structure / molecular mass / surface area increases.
Intermolecular forces:
Increase in strength of intermolecular forces / induced dipole / London / dispersion / Van der Waals forces.
Energy:
More energy needed to overcome / break intermolecular forces. OR
Structure:
From propane to methane the chain length / molecular size / molecular structure / molecular mass / surface area decreases.
Intermolecular forces:
Decrease in strength of intermolecular forces / induced dipole forces /
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London forces / dispersion forces.
Energy:
Less energy needed to overcome / break intermolecular forces. (3) 7.4
Between propane molecules are London forces/dispersion forces/
induced dipole forces.
Between propan-1-ol molecules are London forces/dispersion
forces/induced dipole forces and hydrogen bonds.
Hydrogen bonds / Forces between alcohol molecules are stronger or
need more energy than London forces / dispersion forces / induced dipole forces.
OR Between propane molecules are weak London forces / dispersion forces / induced dipole forces and between propan-1-ol molecules are strong hydrogen bonds. (3) [12]
Question 8
8.1 Alkenes / Alkene (1)
8.2
8.2.1 Addition / Hydrohalogenation / Hydrochlorination (1)
C C C
O
H
H
H
H
H
H
H
H
Propan-2-ol (3)
8.2.2 Elimination / Dehydration (1)
8.2.3 Catalyst (1)
8.3
8.3.1 Sodium hydroxide / Potassium hydroxide (1)
8.3.2 Dissolve base in ethanol / Concentrated (strong) base
Heat strongly (2)
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8.3.3
C C C
Cl
H
H
H
H
H
H
H
+ ONa
H
C C C
H
H H H
H
H
+ ClNa + O
H
H
(5)
[15]
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Question 9
9.1
9.1.1 Substitution / chlorination / halogenation (1)
9.1.2 Substitution / hydrolysis (1)
9.2
9.2.1 Hydrogenation (1)
9.2.2
C C C
H
H
H H H
H
+ H H C C C
H
H
H H
H
H
H
H
(3)
9.3
C C C
Cl
H
H
H
H
H
H
H
(2)
9.4
9.4.1 Esterification / Condensation (1)
9.4.2 Concentrated H2SO4 / Concentrated sulphuric acid (1)
9.4.3
C C C O C C
OH
H
H H
H H
H H
H
H
(2)
9.4.4 Propyl ethanoate (2)
9.5 Sulphuric acid / H2SO4 / Phosphoric acid / H2PO4 (1)
[15]
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NEWTON MEMO
Question 1 Multiple choice questions
1.1. D 1.2. A 1.3. A 1.4. A 1.5. C 1.6. B 1.7. D 1.8. D 1.9. B 1.10. D
[ππ Γ π = ππ]
Question 2
2.1 When a resultant / net force acts on an object, the object will accelerate in the direction of the force. This acceleration is directly proportional to the force and inversely proportional to the mass of the object. (2)
2.2 REMAINS THE SAME / BLY DIESELFDE (1)
2.3 . Accepted Labels FG Weight, gravitational force fk Friction N Normal force FT Tension
(4)
2.4 2.4.1. πΉπππ‘ = ππ = πΉπ + ππ + πΉπΊβ
ππ = πΉπ + πππππππ 30Β° + πππ ππ30Β° {ππ = πππ}
FG FGβ
FGβ«
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{ π = β πΉπΊβ« =πππππ 30Β°} (6)(4) = πΉπ β (0,2)(6)(9,8)(πππ 30Β°)β (6)(9,8)(π ππ30Β°) β΄ πΉπ = 63,58 π (5)
2.4.2. πΉ + ππ 3ππ + πΉπΊβ = ππ
πΉ β (0,2)(6)(9,8)πππ 30Β°β (0,1)(3)(9,8)πππ 30Β°β (3 + 6)(9,8)π ππ30Β° =0 πΉ = 56,83 π (6)
2.5 DECREASES (1)
[19]
Question 3
3.1. When a resultant / net force acts on an object, the object will accelerate in the direction of the force. This acceleration is directly proportional to the forceand inversely proportional to the mass of the object. (2)
3.2. . (3)
3.3. πΉπππ‘ = ππ
5kg π2 + πΉπΊ + π1 = ππ 250 β (5)(9,8) β π1 = 5π 201 β π1 = 5π π1 = 201 β 5π β¦β¦..(1) 20kg π1 + πΉπΊ = ππ π1 β [(20)(9,8)] = 20π π1 = 196 + 20π β¦β¦..(2)
(1) = (2)
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201 β 5π = 196 + 20π π = 0,2 π β π β2upwards β΄ π1 = 201 β (5)(0,2) β΄ π1 = 200 π (6)
3.4. Q
(1) [12]
Question 4
4.1. When a body exerts s force on a second body, the second body exerts a force of
equal magnitude in the opposite direction on the first body. (2)
4.2. . Accepted Labels FG Weight, gravitational force FA Applied force N Normal force FT Tension f Friction
(5)
4.3.
4.3.1. ππ = πππ
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ππ = πππππππ π ππ = (0,29)(1)(9,8)πππ 30Β° ππ = 2,46 π (3)
πΉπππ‘ = ππ 1kg πΉπ΄ + π + π + πΉπΊβ = ππ
40 β π β 2,46 β (1)(9,8)π ππ30Β° = 1π 40 β π β 2,46 β 4,9 = π 32,64 β π = π β¦β¦..(1) 4kg π + πΉπΊβ + π = ππ
π β (4)(9,8)π ππ30Β° β 10 = 4π π β 19,6 β 10 = 4π π β 29,6 = 4π β¦β¦..(2) (1) +(2) 32,64 β π = π π β 29,6 = 4π 3,04 = 5π π = 0,608 π β π β2 β΄ π β 29,6 = (4)(0,61) β΄ π = 32,04 π (6)
[16]
Question 5
5.1. 5.1.1. When a body exerts s force on a second body, the second body exerts a force of
equal magnitude in the opposite direction on the first body. (2)
5.1.2. 2,5kg πΉπππ‘ = ππ = πΉπ + πΉπΊ πΉπ β (2,5)(9,8) = (2.5)(0) πΉπ = 24,5 π (3)
5.1.3. ππ = ππ π 24,5 = 0,2π
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π = 122,5 π π = βπΉπΊ π = ππ 122,5 = π(9,8) π = 12,5 ππ (5)
5.1.4. 5kg ππ = πππ ππ = (0,15)(5)(9,8) ππ = 7,35 π πΉπππ‘ = ππ = πΉπ + ππ 5π = π β 7,35 β¦β¦..(1) 2,5kg πΉπππ‘ = ππ = πΉπΊ + πΉπ 2,5π = (2,5)(9,8) β πΉπ 2,5π = 24,5 β πΉπ β¦β¦..(2) (1) + (2) 5π = π β 7,35 2,5π = 24,5 β πΉπ 7,5π = 17,15 π = 2,29 π β π β2 (5)
πΉπΊ = πΊπ1π2
π2
πΉπΊ = (6,67 Γ 10β11)(6,5 Γ 1020)(90)
(550 Γ 103)2
πΉπΊ = 12,899 π (4)
+
+
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MOMENTUM AND IMPULSE Question 1
Multiple choice questions
1.1. D
1.2. C
1.3. B
1.4. D [4 2 = 8]
Question 2
2.1. The total linear momentum in a closed system remains constant / is
conserved.
(2)
2.2. Ξ£ππ = Ξ£ππ
(π1 + π2)π£π = π1π£1π + π2π£2π (2π + 4π)(0) = (2π)(2) + (4π)( π£2π)
β4π = 4ππ£2π
β΄ π£π = β1 π β π β1
β΄ π£π = 1 π β π β1; in the opposite direction to that of the boys
(5)
2.3. GREATER THAN .
(1)
[8]
Question 3
3.1. π = ππ£
π = (50)(5) π = 250 ππ β π β π β1 , (downwards) (3)
3.2. The product of the net force and thee time interval (during which the force acts)
.
(2)
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3.3. βπ = πΉπππ‘βπ‘
0 β 250 = πΉπππ‘(0,2) πΉπππ‘ = β1 250 π πΉπππ‘ = 1 250 π (3)
3.4. GREATER THAN
(1)
3.5. For the same momentum change , the stopping time (contact time) will be
smaller (less) . β΄ the (upward) force exerted (on her) is greater . (3)
[12]
Question 4
4.1. Ξ£ππ = Ξ£ππ
(π1 + π2)π£π = π1π£1π + π2π£2π (3 + 0,02)(0) = (3)(β1,4) + (0,02)π£2π
π£2π = 210 π β π β1
(4) 4.2. π£π
2 = π£π2 + 2πβπ₯
(0) = 2102 + (2)(π)(0,4) π = β55 125 π β π β2 πΉπππ‘ = ππ πΉπππ‘ = (0,02)(β55 125) πΉπππ‘ = β1 102,5 π β΄ πΉπππ‘ = 1 102,5 π (5)
4.3. THE SAME .
(1)
[10]
Question 5 5.1. The total linear momentum in a closed system remains constant/is
conserved.
(2)
5.2.
5.2.1. ππ = ππ
π1π£π1 + π2π£π2 = π1π£π1 + π2π£π2
+
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(π1 + π2)π£π = π1π£π1 + π2π£π2
0 = (0,4)π£π1 + (0,6)(4)
π£π1 = β6 π β π β1
= 6 π β π β1 π‘π π‘βπ ππππ‘ (4)
5.2.2. βπ = πΉπππ‘βπ‘
(0,6)(4 β 0) = πΉπππ‘(0,3) πΉπππ‘ = 8 π (4)
5.3. NO
(1)
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SESSION 06
Doppler effect MEMO
Question 1 1.1. A 1.2. C 1.3. A 1.4. A 1.5. C
Question 2
2.1. Smaller than (1)
2.2. Doppler effect (1)
2.3. π£ = ππ 345 = π0.55 π = 627,27 π»π§
ππΏ = π£ Β± π£πΏ
π£ Β± π£π ππ
ππΏ = 345Β± 0
345β33,33 (627,27)
ππΏ = 694,35 π»π§ (7)
2.4. Decrease (1) [10]
Question 3
3.1.1. An (apparent) change in observed/detected frequency (pitch), (wavelength) as a result of the relative motion between a source and an observer (listener). (2)
3.1.2. Towards
Observed/detected frequency is greater than the actual frequency. (2)
3.1.3. ππΏ = π£ Β± π£πΏ
π£ Β± π£π ππ
1200 = 343 Β± 0
343β π£π 1130
π£π = 20,01 π β π β1 (5)
3.2. The star is approaching the earth.
OR
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The earth and the star are approaching (moving towards) each other. The spectral lines in diagram 2 are shifted towards the blue end / blue shifted. (2)
[11]
Question 4
4.1.1. π£ = ππ 340 = 520 π π = 0,65π (2)
4.1.2. ππΏ = π£ Β± π£πΏ
π£ Β± π£π ππ
ππΏ = 340Β± 0
340β 15 (520)
ππΏ = 544 π»π§
π£ = ππ 340 = 544π π = 0,63 π (6)
4.2. The wavelength in QUESTION 4.1.2 is shorter because the waves are compressed as they approach the observer. (2)
4.3. The red shift occurs when the spectrum of a distant star moving away from the earth is shifted toward the red end of the spectrum. (2)
[12]
Question 5 5.1.1. Frequency (of sound detected by the listener (observer)) (1)
5.1.2. The apparent change in frequency or pitch of sound (detected (by a listener) because the sound source and the listener have different velocities relative to the medium of sound propagation. (2)
5.1.3. Away Detected frequency of source decreases (2)
5.1.4. OPTION 1 EXPERIMENT 2
ππΏ = π£ Β± π£πΏ
π£ Β± π£π ππ
874 = π£ Β± 0
π£+10(900)
π£ = 336.15 π β π β1 (323.33 β 336.15) (5)
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EXPERIMENT 3
ππΏ = π£ Β± π£πΏ
π£ Β± π£π ππ
850 = π£ Β± 0
π£+20(900)
π£ = 340 π β π β1 (313.33 β 340) (5) EXPERIMENT 4
ππΏ = π£ Β± π£πΏ
π£ Β± π£π ππ
827 = π£ Β± 0
π£+30(900)
π£ = 339,86 π β π β1 (339,86 β 345) (5) OPTION 2 EXPERIMENT 2 AND 3
ππΏ = π£ Β± π£πΏ
π£ Β± π£π ππ
874(π£+10)
π£ =
850(π£+20)
π£ both frequencies
874π + 8740 = 850π + 1700 π£ = 344,17 π β π β1
EXPERIMENT 2 AND 4
ππΏ = π£ Β± π£πΏ
π£ Β± π£π ππ
874(π£+10)
π£ =
827(π£+30)
π£ both frequencies
874π + 8740 = 827π + 24810 π£ = 341,91 π β π β1 EXPERIMENT 3 AND 4
ππΏ = π£ Β± π£πΏ
π£ Β± π£π ππ
850(π£+20)
π£ =
827(π£+30)
π£ both frequencies
850π + 1700 = 827π + 24810 π£ = 339,57π β π β1
5.2. Away from the Earth (1)
[11] Question 6
6.1. π£ = ππ
π£ = (222 Γ 103)(1,5 Γ 10β3) π£ = 333π β π β1
(3)
6.2. 6.2.1. Towards the bat (1)
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6.2.2. POSITIVE MARKING FROM QUESTION 6.1
ππΏ = π£ Β± π£πΏ
π£ Β± π£π ππ
230,3 = 333Β± 0
333β π£π (222)
π£π = 12 π β π β1 (6)
[10]