# 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. - Page: https://www.hlayiso.com/library/gr11-phy-p2-eng-afr-june-2017-possible-answers/305f5cfe-3c89-4d45-8559-35b65e4f133a - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 11 - Document type: Memorandum - Year: 2017 - Exam period: Gauteng June - Paper: 2 - Curriculum: CAPS - Pages: 14 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 11](https://www.hlayiso.com/library/grade-11) › [Physical Sciences](https://www.hlayiso.com/library/grade-11/physical-sciences) › [Physical Sciences Grade 11 Paper 2 Gauteng June 2017 Memo](https://www.hlayiso.com/library/gr11-phy-p2-eng-afr-june-2017-possible-answers/305f5cfe-3c89-4d45-8559-35b65e4f133a) ## About this document Physical Sciences · Grade 11 · Gauteng June · 2017. Memorandum, 14 pages. 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Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com 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 ### Page 2 Downloaded from hlayiso.com 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] 2 ### Page 3 Downloaded from hlayiso.com 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] 3 ### Page 4 Downloaded from hlayiso.com 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] 4 ### Page 5 Downloaded from hlayiso.com 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] 5 ### Page 6 Downloaded from hlayiso.com 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. 6 ### Page 7 Downloaded from hlayiso.com 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] 7 ### Page 8 Downloaded from hlayiso.com 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) 8 ### Page 9 Downloaded from hlayiso.com 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) [23] 9 ### Page 10 Downloaded from hlayiso.com 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) 10 ### Page 11 Downloaded from hlayiso.com 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] 11 ### Page 12 Downloaded from hlayiso.com 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] 12 ### Page 13 Downloaded from hlayiso.com 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 ### Page 14 Downloaded from hlayiso.com 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) 14 ## Related documents - [Gr11 Phy P2 June 2017 Question Paper](https://www.hlayiso.com/library/gr11-phy-p2-june-2017-question-paper/bdb6614f-dd32-4c01-90b9-46eded9ab82c) — Physical Sciences · QP · 2017 - [Gr11 Phy P2 (Afrikaans) June 2017 Question Paper](https://www.hlayiso.com/library/gr11-phy-p2-afrikaans-june-2017-question-paper/14dc2ba3-3d16-418a-a9c0-5897316247be) — Physical Sciences · QP · 2017 · Afr - [Gr11 Phy P1 (ENG AFR) June 2017 Possible Answers](https://www.hlayiso.com/library/gr11-phy-p1-eng-afr-june-2017-possible-answers/7dd664eb-b758-4f97-a529-32c91e5a9888) — Physical Sciences · Memo · 2017 · Eng - [JUNE PHY P2 QP MEMO 1](https://www.hlayiso.com/library/june-phy-p2-qp-memo-1/53a8d883-d5d6-4cce-9ed7-467601f0afcf) — Physical Sciences · Memo · 2017 - [Physical Sciences P2 Grade 11 Nov 2017 Memo Afr & Eng](https://www.hlayiso.com/library/physical-sciences-p2-grade-11-nov-2017-memo-afr-eng/47b92c2b-2ba4-45e4-8c1f-fee4087affae) — Physical Sciences · Memo · 2017 · Eng - [Gr11 Phy P2 June 2016 Possible Answer](https://www.hlayiso.com/library/gr11-phy-p2-june-2016-possible-answer/3bd7f864-d9b2-4099-b47e-495203010a0d) — Physical Sciences · Memo · 2016 ## More from Grade 11 Physical Sciences - [Grade 11 Physical Sciences P2 November 2026 Mock 2](https://www.hlayiso.com/library/grade-11-physical-sciences-2026-november-mock-exam-2-question-paper-paper-2/4bd46b50-41c8-4670-83a6-c99f40ba7aa1) — QP · 2026 · Eng - [PHYSICAL SC P1 QP GR11 NOV 2025 AFRIK watermark](https://www.hlayiso.com/library/physical-sc-p1-qp-gr11-nov-2025-afrik-watermark/3757d93e-3b4a-4e27-9a6b-69591d990a97) — QP · 2025 · Afr - [KZN Physical Sciences Grade 11 November 2025 P1 and Memo](https://www.hlayiso.com/library/kzn-physical-sciences-grade-11-november-2025-p1-and-memo/82b3d6a5-9e71-4efe-a8e0-f2a63d6aff58) — QP + Memo · 2025 - [KZN Physical Sciences Grade 11 November 2025 P2 and Memo](https://www.hlayiso.com/library/kzn-physical-sciences-grade-11-november-2025-p2-and-memo/f6f699c6-e985-44e2-9d36-d660697bd6fe) — QP + Memo · 2025 - [LP Physical Sciences Grade 11 June 2025 P2 and Memo](https://www.hlayiso.com/library/lp-physical-sciences-grade-11-june-2025-p2-and-memo/b76990f6-708a-47e3-a903-95cc3c58534d) — QP + Memo · 2025 - [MP Physical Sciences Grade 11 June 2025 P1 and Memo](https://www.hlayiso.com/library/mp-physical-sciences-grade-11-june-2025-p1-and-memo/4c5694d8-7cdc-4fd9-90b4-55bbb8969872) — QP + Memo · 2025 ## Collection - [Grade 11 past papers](https://www.hlayiso.com/grade-11/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. 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