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GAUTENG DEPARTMENT OF EDUCATION
GAUTENGSE DEPARTEMENT VAN ONDERWYS
PROVINCIAL EXAMINATION
PROVINSIALE EKSAMEN
JUNE / JUNIE 2017
GRADE / GRAAD 11
PHYSICAL SCIENCES
FISIESE WETENSKAPPE
PAPER / VRAESTEL 2
MEMORANDUM
14 pages / bladsye
1
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Physical Sciences Grade 11 Paper 2 Gauteng June 2017 Memo
Grade 11 Physical Sciences Paper 2 memorandum from the Gauteng June 2017, 14 pages. Read online or download the free PDF.
Physical Sciences · Grade 11 · Gauteng June · 2017. Memorandum, 14 pages. Read online or download the PDF.
- Subject
- Physical Sciences
- Grade
- Grade 11
- Document type
- Memorandum
- Year
- 2017
- Exam period
- Gauteng June
- Paper
- 2
- Curriculum
- CAPS
- Pages
- 14
- File size
- 1.3 MB
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Physical Sciences / Fisiese Wetenskappe
MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11
GAUTENG DEPARTMENT OF EDUCATION
GAUTENGSE DEPARTEMENT VAN ONDERWYS
PROVINCIAL EXAMINATION
PROVINSIALE EKSAMEN
PHYSICAL SCIENCES / FISIESE WETENSKAPPE
(Paper / Vraestel 2)
QUESTION 1: MULTIPLE-CHOICE QUESTIONS
VRAAG 1: MEERVOUDIGEKEUSE-VRAE
1.1 B ✓✓ (2)
1.2 B ✓✓ (2)
1.3 D ✓✓ (2)
1.4 C ✓✓ (2)
1.5 A ✓✓ (2)
1.6 D ✓✓ (2)
1.7 D ✓✓ (2)
1.8 C ✓✓ (2)
1.9 A ✓✓ (2)
1.10 B ✓✓ (2)
[20]
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Physical Sciences / Fisiese Wetenskappe
MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11
QUESTION 2 / VRAAG 2
2.1 Boiling point is the temperature when a liquid’s vapour pressure is equal to
its atmospheric pressure.✓✓
Kookpunt is die temperatuur wanneer die dampdruk van ʼn vloeistof gelyk is
aan die atmosferiese druk.✓✓ (2)
2.2.1 He ✓ (1)
2.2.2 NH3 ✓ (1)
2.2.3 NaCℓ ✓ (1)
2.2.4 CCℓ4 ✓ (1)
2.3 2.3.1 London / dispersion forces✓✓
London / dispersiekragte.✓✓ (2)
2.3.2 Dipole – dipole forces ✓✓
Dipool – dipoolkragte ✓✓ (2)
2.3.3 Ionic bonds ✓✓
Ioniese kragte ✓✓ (2)
[12]
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Physical Sciences / Fisiese Wetenskappe
MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11
QUESTION 3 / VRAAG 3
3.1 3.1.1 The sharing of one or more pairs of electrons between two non-
metals to form a molecule.✓✓
Die deel van een of meer elektronpare, tussen twee nie-metale (2)
om ʼn molekule te vorm.✓✓
3.1.2 A measure of an atom’s attractive force on bonding electrons to
form a molecule ✓✓
ʼn Aanduiding van die atoom se aantrekkingskrag op die
verbindingselektrone van ʼn molekule (2)
3.2
✓✓ (2)
3.3 The ∆EN between C and H is 2,5 -2,1 = 0,4 which gives it a polar bond, ✓✓
but the shape of the molecule is symmetrical non-polar molecule.✓✓
Die ∆EN tussen C en H is 2,5 -2,1 = 0,4 wat polêre binding is, ✓✓ maar die
vorm van die molekule is simmetries nie-polêre molekule.✓✓ (4)
3.4 3.4.1 Ammonia / NH3 ✓
Ammoniak / NH3 ✓ (1)
3.4.2 Methane / CH4 ✓
Metaangas / CH4 ✓ (1)
3.4.3 Sulphur dioxide / SO2 ✓
Swaeldioksied / SO2 ✓ (1)
3.4.4 Carbon dioxide / CO2 ✓
Koolstofdioksied / CO2 ✓ (1)
3.5 Ammonia / NH3 ✓ – polar substance will ionise in a polar liquid.✓
Ammoniak / NH3 ✓ – ʼn polêre molekule sal ioniseer in ʼn polêre vloeistof.✓ (2)
[16]
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Physical Sciences / Fisiese Wetenskappe
MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11
QUESTION 4 / VRAAG 4
4.1 H2O (1)
4.2
H2O EN = 3,5 – 2,1 = 1,4 ✓ polar covalent ✓ O attract shared
electron pair more than Hydrogen
H2O EN = 3,5 – 2,1 = 1.4 ✓ polêr kovalent ✓ O trek die gedeelde
elektronpaar meer as die Waterstof. (2)
4.3 Big spaces between molecules ✓
Positive side of one molecule aligned with
negative side of next molecule✓
Groot spasies tussen molekules ✓
Positiewe kant van een molekule
georiënteer met negatiewe kant van
volgende molekule.✓ (2)
4.4 Ice has a very regular pattern with the molecules rigidly apart from one
another, connected by the hydrogen bonds that form a crystalline lattice.✓
These crystals have a number of open regions and pockets making ice less
dense than liquid water.✓
Watermolekules in ys het ʼn presiese, rigiede patroon in ʼn kristallyne vorm.✓
Die kristallyne vorm het ʼn groot aantal oop ruimtes wat ys minder dig as
water maak.✓ (2)
4.5 As water evaporates from leaves, it tugs on the water molecules below
-Cohesion and adhesion pull water up and replace missing water molecules
✓
Capillarity: Water molecules will “tow” each other along when in a thin tube. ✓
Soos wat die water deur die blare verdamp, trek dit die watermolekules op in
die plant in deur middel van kohesie en adhesiekragte om die watermolekules
te vervang. ✓
Kapillêre kragte: watermolekules trek mekaar aan beweeg op in ʼn dun
buisie.✓ (2)
[9]
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Physical Sciences / Fisiese Wetenskappe
MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11
QUESTION 5 / VRAAG 5
5.1
P1V1 = P2V2 ✓
T1 T2
102 x 31,8 ✓ = 75 x 34.5 ✓
302 ✓ T2
T2 = 240,91 K also accept / aanvaar ook - 32,09 ° C ✓ (5)
5.2 As the temperature decreases the average kinetic energy of the molecule
decreases.✓ pressure of the molecule on each other and on the sides of
the container decreases as there will be fewer collisions.✓
Soos die temperatuur verlaag sal die gemiddelde kinetiese energie van die
molekules afneem ✓ die druk wat die molekules op mekaar en op die
wande van die houer uitoefen sal verminder aangesien daar minder botsing
sal wees.✓
(2)
5.3 5.3.1 It is a hypothetical gas that will obey all the gas laws under all
conditions of pressure and temperature.✓✓ (Two marks or none)
ʼn Hipotetiese gas wat al die gaswette nakom onder alle
(2)
omstandighede van temperatuur en druk.✓✓ (Twee of geen
punte)
5.3.2 • Particles are in continual motion in all directions.
• Particles do not contribute to the volume of the gas. Any 3
Enige 3
• There are no forces between the particles or the particles
and the wall of the container, except during collisions.
• Collisions are perfectly elastic with no loss of total energy
of the molecules.
• All molecules are identical.
• The temperature of the gas is a measure of the average
kinetic energy of the particles.
• Collisions of particles on the surface cause pressure.
• There is no motion and therefore no pressure at 0 K.
• Deeltjies is konstant in beweging in alle rigtings.
• Deeltjies dra nie by tot die volume van die gas nie.
• Daar bestaan geen kragte tussen die deeltjies onderling en
die wande van die houer nie, behalwe gedurende botsings.
• Botsings is volkome elasties met geen verlies aan die
totale energie van die molekules nie.
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Physical Sciences / Fisiese Wetenskappe
MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11
• Alle molekules is identies.
• Die temperatuur van die gas is ʼn maatstaf vir die
gemiddelde kinetiese energie van die deeltjies.
• Botsings van deeltjies op die oppervlak veroorsaak druk.
• Daar is geen beweging en dus ook geen druk by 0 K.
(3)
5.3.3 At high temperatures and low pressures✓✓
Teen hoë temperature en lae druk. ✓✓ (2)
5.4 22,4 dm3 at STP ✓✓
22,4 dm3 by STP ✓✓ (2)
[16]
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Physical Sciences / Fisiese Wetenskappe
MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11
QUESTION 6 / VRAAG 6
6.1 Gay Lussac’s Law: ✓ The pressure of an enclosed gas is directly
proportional to its temperature if the amount of a gas is at a constant
volume.✓✓
Gay Lussac se wet: ✓ Die verwantskap tussen druk en temperatuur van ʼn
spesifieke hoeveelheid gas is direk eweredig aan mekaar mits die volume
konstant gehou word.✓✓ (3)
6.2 6.2.1 Pressure ✓✓
Druk ✓✓ (2)
6.2.2 Temperature ✓✓
Temperatuur ✓✓ (2)
6.2.3 Volume ✓✓
Volume ✓✓ (2)
6.3 What is the relationship between the temperature and the pressure of an
enclosed gas of a spesific volume?
Wat is die verwantskap tussen die temperatuur en die volume van ʼn
ingeslote gas by ʼn spesifieke volume?
Criteria for investigative Question / Kriteria vir ondersoekende Mark/
vraag: Punt
The dependent and independent variables are stated.
Die afhanklike en onafhanklike veranderlikes is genoem.
Ask a question about the relationship between dependent and
independent variables.
Vra ʼn vraag oor die verwantskap tussen die afhanklike en onafhanklike
veranderlikes. (2)
6.4 According to the graph the temperature is directly porportional to the
pressure of the enclosed gas.✓✓ (As temperature increases, pressure
increases.)
Volgens die grafiek is die temperatuur en die druk van ʼn ongeslote gas direk
verwant.✓✓ (Soos die temperatuur styg sal die druk ook toeneem.) (2)
6.5 Extrapolate the line: At zero pressure temp = 273 K. ✓✓
Ekstrapoleer die lyn van die grafiek: By ʼn druk van Nul kPa sal die
temperatuur = 273 K wees. ✓✓ (2)
6.6 Absolute Zero ✓✓
Absolute zero / nul ✓✓ (2)
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Physical Sciences / Fisiese Wetenskappe
MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11
6.7 Pi Pf
Ti = Tf ✓ Ti = 25 °C + 273
= 298 K
500 Pf Tf = 40°C +273
298 ✓ = 313 ✓ = 313 K
Pi =500 kPa
Pf = ?
Pf = 527,17 kPa ✓
Yes, ✓he will reach his destination safely as the final pressure is less than
the maximum pressure. ✓
Ja, ✓ hy sal sy bestemming veilig bereik aangesien die finale druk kleiner is
as die maksimum toegelate druk. ✓ (6)
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Physical Sciences / Fisiese Wetenskappe
MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11
QUESTION 7 / VRAAG 7
7.1 Sodium carbonate ✓✓
(2)
Natriumkarbonaat ✓✓
7.2 𝑚
c= ✓ M = Na2CO3
𝑀𝑉 = 2(23) + 12 + (16 x 3)
𝑚 = 106 g.mol-1
0,25 ✓ = ✓
106 ✓ x 5𝑥10−1
(4)
m = 13,25 g ✓
7.3 𝑚
n= ✓ Positive marking from 7.2
𝑀
Positiewe nasien vanaf 7.2
13,25
= ✓
106
= 0,125 mole of Na2CO3✓
But in every 1 mole of Na2CO3 = 2 mole of Na ✓ /
Maar in elke 1 mol Na2CO3 = 2 mol Na ✓
0,125 x 2 = 0,25 mole Na ions / mol Na ione ✓
𝑛
c=
𝑉
= 0,25
0,5
= 0,5 mol . dm-3 (6)
7.4.1 Na2SO4 + CaCO3 + 2✓C Na2CO3 + 2✓CO2 + CaS (2)
7.4.2 The substance that will be used up first ✓✓
Die reagens wat eerste opgebruik sal word ✓✓ (2)
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Physical Sciences / Fisiese Wetenskappe
MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11
7.4.3 Na2SO4 CaCO3 2C Na2CO3 2CO2 CaS
1 1 2 1 2 1✓
52,54 g 45 g
𝑚 𝑚
n= n= ✓
𝑀 𝑀
52,54 𝟒𝟓
= ✓ = ✓
142 𝟏𝟎𝟎
= 0,37 = 0,45
mole / mol mole / mol✓
But ratio :
1
1
Limiting excess /
reactant / oortollig
Beperkende
reagens ✓ (6)
7.4.4 Na2CO3 M = (23 x 2) + (12) + (16 x 3) = 106 g.mol-1
Mole ratio / Molverhouding: 1 Na2SO4 : 1 Na2CO3 ✓
𝑚
n= ✓
𝑀
0,37 = m / 106 ✓
m Na2CO3 = 39,22 g ✓ (4)
[26]
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Physical Sciences / Fisiese Wetenskappe
MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11
QUESTION 8 / VRAAG 8
8.1. 2 LiOH (s) CO2 ( g ) Li2CO3 (s) H 2O(l )
✓ ✓ ✓ ✓ (4)
8.2 𝑚
n= ✓
𝑀
= 800
24 ✓
(3)
n = 33,33 mole / mol ✓
8.3 From balanced equation: LiOH : CO2 ✓ Positive marking from 8.2
Vanaf gebalanseerde vergelyking: Positiewe nasien vanaf 8.2
Mole ratio / molverhouding 2 : 1 ✓
33,33 mole / mol : 16,67 mole / mol CO2 ✓
(3)
8.4 𝑚
n= ✓ Positive marking from 8.3
𝑀
Positiewe nasien vanaf 8.3
16,67 ✓ = m
44 ✓
m = 733,48 g ✓ (4)
[14]
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Physical Sciences / Fisiese Wetenskappe
MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11
QUESTION 9 / VRAAG 9
9.1 𝑚
n=
𝑀
= 12
342 ✓
n = 0,035 mole / mol ✓
In one mole of Al2(SO4)3 : 12 mole of O atoms✓ /
In een mol van Al2(SO4)3 : 12 mol van O atome✓
0,035 mole / mol x 12
= 0,42 mole / mol O ✓
Natoms = n x NA ✓
Natome = 0,42 x 6,02 x 1023 (6)
= 2,53 x 1023 O atoms / atome ✓
9.2 Al S
36% 64 % ✓
36g 64g
27 32 ✓
1,33 2
1,33 1,33 ✓
1 : 1,5 ✓
X2 2 3 ✓
Emperical formula: Al2S3
Empiriese formule (5)
9.3 6 H2O + Al2S3 → 3 H2S + 2 Al(OH)3
✓ ✓ ✓ (3)
[14]
TOTAL / TOTAAL: 150
13
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Physical Sciences / Fisiese Wetenskappe
MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11
Taxonomy Grid
Recall Comprehension Analysis Evaluation
Q no: Mark Q no: Mark Q no: Mark Q no: Mark
1.1 2 1.3 2 1.7 2 8.1 4
1.2 2 1.4 2 1.10 2 9.2 7
2.1 2 1.5 2 3.3 4 9.3 3
3.1 4 1.6 2 3.5 2
4.1 1 1.8 2 4.4 2
5.3.1 2 1.9 2 4.5 2
5.3.2 2 2.2 4 5.1 5
5.3.3 2 2.3 6 5.2 2
5.4 2 3.2 2 6.4 2
6.1 3 3.4 4 6.5 2
7.4.2 2 4.2 2 6.7 5
4.3 2 7.2 3
6.2 6 7.4.3 6
6.3 2 8.3 3
6.6 2 8.4 4
7.1 2 9.1 6
7.3 6
7.4.1 2
7.4.4 4
8.2 3
Total mark 16% 24 39,33% 59 34,67% 52 9,3 % 14
Total % / 100% P1&2: 15% P1:35% / P2:40% P1:40% / P2:35% P1&2: 10%
Correct application of Bloom’s / Barrett’s Taxonomy:
Level 1: Recall of information (what? which? when? list ; label; name; define; give; describe)
Level 2: Understanding and using information (summarize; classify; apply rules; discuss)
Applying information (distinguish; specify; compare; design; explain; investigate; interpret; calculate; give your
input)
Level 3: Analysis of information (classify; explain; identify; interpret; compare; give reasons; prove; give causes and
effects)
Level 4: Synthesize information (summarize; construct; argue; create; relate; design; formulate)
Evaluate information (judge; assess; evaluate; choose; support; compare; estimate)
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