# Physical Sciences Grade 11 P2 2019 Memo (Eng) > Grade 11 Physical Sciences Paper 2 memorandum from the Eastern Cape November 2019 (English), 11 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/phys-sciences-p2-gr11-memo-nov2019-afr-english/227eeebb-5f49-4bbe-ac5c-1c0489760c7d - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 11 - Language: English - Document type: Memorandum - Year: 2019 - Exam period: Eastern Cape November - Paper: 2 - Curriculum: CAPS - Pages: 11 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 P2 2019 Memo (Eng)](https://www.hlayiso.com/library/phys-sciences-p2-gr11-memo-nov2019-afr-english/227eeebb-5f49-4bbe-ac5c-1c0489760c7d) ## About this document Physical Sciences · Grade 11 · Eastern Cape November · 2019 · English. Memorandum, 11 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 11 NOVEMBER 2019 PHYSICAL SCIENCES P2 FISIESE WETENSKAPPE V2 (CHEMISTRY/CHEMIE) MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 11 pages./ Hierdie nasienriglyn bestaan uit 11 bladsye. ### Page 2 Downloaded from hlayiso.com 2 FISIESE WETENSKAPPE V2 (EC/NOVEMBER 2019) QUESTION 1 / VRAAG 1 1.1 C ✓✓ (2) 1.2 B ✓✓ (2) 1.3 B ✓✓ (2) 1.4 A ✓✓ (2) 1.5 C ✓✓ (2) 1.6 C ✓✓ (2) 1.7 D ✓✓ (2) 1.8 B ✓✓ (2) 1.9 C ✓✓ (2) 1.10 B ✓✓ (2) [20] QUESTION 2 / VRAAG 2 2.1 A group of two or more atoms covalently bonded and it functions as a unit. ✓✓ ʼn Groep van twee of meer atome wat kovalent gebind en as ʼn eenheid funksioneer. (2) 2.2.1 Tetrahedral ✓ Tetraëdries (1) 2.2.2 Trigonal bipyramidal ✓ Trigonaal bipiramidaal (1) 2.3.1 O F ✓✓ F (2) 2.3.2 H C N ✓✓ (2) 2.4 The nitrogen (N) atom in NH3 contains a lone pair electrons. ✓ No lone pair in CCℓ4. Nitrogen (N) atom in NH3 can donate its lone pair into the vacant orbital of H+✓ Die stikstof (N) atoom in NH 3 bevat ʼn alleenpaar elektrone. Geen enkelpaar elektrone in CCℓ 4 nie Stikstof (N) atoom in NH 3 kan sy alleenpaar elektrone in die vakante wentelbaan van H + skenk (2) Kopiereg voorbehou Blaai om asseblief ### Page 3 Downloaded from hlayiso.com (EC/NOVEMBER 2019) PHYSICAL SCIENCES P2 3 2.5 Polar. ✓ • Sulphur atoms more electronegative than the hydrogen atom ✓ • Sulphur atom pulls the bonding electrons more towards itself. • (The change in electronegative is 2,5 – 2,1 = 0,4) • The sulphur atom has a partial negative charge and hydrogen atom has a partial positive charge. ✓ • The H2S molecule has an asymmetrical bent/angular shape. ✓ Polêr. • Swawelatome is meer elektronegatief as die waterstofatoom • Swawelatoom trek die bindingselektrone meer na hom toe. • (Die verskil in elektronegatief is 2,5 – 2,1 = 0,4) • Die swawelatoom het ʼn gedeeltelik negatiewe lading en waterstofatoom het ʼn gedeeltelik positiewe lading. • Die H2S-molekule het ʼn asimmetriese buiging / hoekige vorm. (4) [14] QUESTION 3/VRAAG 3 3.1 The temperature at which the vapour pressure of a liquid is equal to the external (atmospheric) pressure. ✓✓ Die temperatuur waarteen die dampdruk van ʼn vloeistof gelyk is aan die eksterne (atmosferiese) druk. (2) 3.2 Boiling point. Accept answers in the range (180 to 190) ✓ (K) Kookpunt. Aanvaar antwoorde tussen (180 tot 190) (K) (1) 3.3 • Group 4 hydrogen hydrides have London /dispersion/induced-dipole forces ✓ • Hydrogen halides have dipole-dipole forces ✓ • The dipole-dipole forces are stronger than the London/dispersion/induced-dipole forces ✓ • More energy will be required to overcome the dipole-dipole/ intermolecular forces in hydrogen halides ✓ • Groep 4 waterstofhidriede het London-/ verspreiding / geïnduseerde- dipool kragte • Waterstofhaliede het dipool-dipool kragte • Die dipool-dipoolkragte is sterker as die London-/verspreidingskragte/ geïnduseerde-dipool kragte. • Meer energie sal benodig word om die dipool-dipool / intermolekulêre kragte in waterstofhaliede te oorkom (4) 3.4 HF has hydrogen bonds ✓✓ HF het waterstofbindings (2) 3.5 GeH4 ✓.It has a lower boiling point. ✓ GeH4 .Dit het die laagste kookpunt (2) [11] Copyright reserved Please turn over ### Page 4 Downloaded from hlayiso.com 4 FISIESE WETENSKAPPE V2 (EC/NOVEMBER 2019) QUESTION 4/VRAAG 4 4.1.1 Boyle’s (law /wet) (1) 4.1.2 What effect will a (change in) pressure have on the volume of a fixed amount gas at constant temperature? ✓✓ Watter effek sal ʼn (verandering in) druk op die volume van ʼn vasgestelde gas by konstante temperatuur hê? (2) 4.1.3 Temperature. ✓ Accept mass / number of moles of gas Temperatuur. Aanvaar massa / aantal mol gas (1) 4.1.4 • According to the Kinetic Molecular Theory, the pressure exerted by a gas depends on the number of collisions per unit time per unit area. ✓ • The same number of particles in a smaller volume (area) leads to an increase in the number of collisions per unit volume (area) ✓ • The more collisions per unit volume (area) results in an increase in pressure. ✓ • Volgens die Kinetiese Molekulêre Teorie hang die druk wat ʼn gas uitoefen af van die aantal botsings per tydseenheid per eenheidsarea. • Dieselfde aantal deeltjies in ʼn kleiner volume (oppervlakte) lei tot ʼn toename in die aantal botsings per eenheid volume (oppervlakte) • Meer botsings per eenheid volume (oppervlakte) lei tot ʼn toename in druk. (3) 4.2.1 Experiment 2. ✓ • The product of pressure and volume (pV) is higher for the same amount of gas. ✓ • pV∝T ✓ Eksperiment 2. • Die produk van druk en volume (pV) is hoër vir dieselfde hoeveelheid gas. • pV∝T (3) 4.2.2 • The intermolecular forces thus increase and the gas liquifies. ✓ • The volume becomes constant at extreme pressure. ✓ • Die intermolekulêre kragte neem dus toe en die gas word 'n vloeistof. • Die volume word konstant by uiterste druk. (2) 4.3.1 Guy-Lussac (law/ wet) ✓ (1) Kopiereg voorbehou Blaai om asseblief ### Page 5 Downloaded from hlayiso.com (EC/NOVEMBER 2019) PHYSICAL SCIENCES P2 5 4.3.2 p1 V1 p V = 2 2✓ T1 T2 (101) p2 ✓= ✓ (V1 =V2) (25+273) (60+273) 𝑝2 = 112,86 kPa ✓ (4) 4.3.3 100 °C✓ or/of 373 K (1) [18] QUESTION 5/VRAAG 5 5.1 Ideal ✓ (gas) Ideale (gas) (1) 5.2 pV = nRT ✓ (100 × 103)(31,98) ✓ = n (8,31)(23 + 273) ✓ n = 1300,12 mol M = m/n✓ M = (2600)/(1300,12) ✓ M = 2 g·mol-1 ✓ H2 ✓ (7) [8] Copyright reserved Please turn over ### Page 6 Downloaded from hlayiso.com 6 FISIESE WETENSKAPPE V2 (EC/NOVEMBER 2019) QUESTION 6/VRAAG 6 6.1 Minimum energy required to start a chemical reaction ✓✓ Minimum energie benodig om ʼn chemiese reaksie te begin. (2) 6.2 Exothermic ✓ The total potential energy of the products is less than the total potential energy of the reactants. ✓ OR More energy is released than the energy taken in. OR The heat of the reaction is less than zero/negative. Eksotermies Die totale potensiële energie van die produkte is minder as die totale potensiële energie van die reaktante OF Meer energie word vrygestel as die energie wat ingeneem word OF Die reaksiewarmte is minder as nul / negatief. (2) 6.3 679,1 kJ·mol-1 ✓ The energy needed to break all the bonds✓✓ / Activation energy Die energie wat benodig word om al die bindings te breek / Aktiveringsenergie (3) 6.4 Bond formation/Bindingsvorming = 184,7 + 679,1✓ Bond formation/Bindingsvorming = 863,8 kJ·mol-1 863,8 kJ·mol-1 is the energy released for two HCℓ molecules/is die energie wat vrygestel word vir twee HCℓ- molekules Bond energy for each/ Bindingsenergie vir elke HCℓ = 863,8 / 2 ✓ Bond energy for each/ Bindingsenergie vir elke HCℓ = 431,9 kJ·mol-1✓ (3) 6.5 No effect. ✓ Catalyst only has an effect on the activation energy and no effect on the heat of the reaction ✓ Geen effek. Katalisator het slegs ʼn invloed op die aktiveringsenergie en het geen invloed op die hitte van die reaksie nie. (2) [12] Kopiereg voorbehou Blaai om asseblief ### Page 7 Downloaded from hlayiso.com (EC/NOVEMBER 2019) PHYSICAL SCIENCES P2 7 QUESTION 7/VRAAG 7 7.1 OPTION 1/ OPSIE 1 𝑛= ✓ 𝑚 𝑚 𝑚 𝑀 𝑛= 𝑛= 11,79 𝑀 𝑀 𝑛= ✓ 𝑛= 69,57 ✓ 𝑛= 18,64 ✓ 12 35,5 19 𝑛 = 0,9825 𝑚𝑜𝑙 𝑛 = 1,9597 𝑚𝑜𝑙 𝑛 = 0,9811 𝑚𝑜𝑙 0,9825 1,9597 0,9811 = = ✓ 0,9811 0,9811 0,9811 Ratio/Verhouding = 1:2:1 Empirical formula/ Empiriese formule: CCℓ2F Relative formula mass/ Relatiewe formulemassa = 12 +2(35,5) + 19 = 102 Ratio/Verhouding = 204/102 = 2✓ Molecular formula/ Molekulêre formule: C2Cℓ4F2✓(Order of elements not important/ Volgorde van elemente nie belangrik nie) OPTION 2/OPSIE 2 11,79 m(C) = 204 × 100 ✓= 24,05 g 69,57 m(Cℓ)= 204 × 100 ✓= 141,92 g 18,64 m(F) = 204 × 100 ✓= 38,03 g 24,05 n(C) = = 2 mol✓ 12 141,92 n(Cℓ) = 35,5 = 4 mol✓ 38,03 n(F) = 19 = 2 mol✓ Molecular formula/ Molekulêre formule: C2Cℓ4F2✓ (Order of elements not important/Volgorde van elemente nie belangrik nie) (7) 7.2.1 Limiting reagent is the substance that is completely used up during a chemical reaction ✓✓ Die beperkende reagens is die stof wat tydens ʼn chemiese reaksie volledig opgebruik word. (2) Copyright reserved Please turn over ### Page 8 Downloaded from hlayiso.com 8 FISIESE WETENSKAPPE V2 (EC/NOVEMBER 2019) 7.2.2 𝑛(𝐿𝑖) = 𝑚 ✓ 𝑛(𝑁2 ) = 𝑚 𝑀 𝑀 12,3 33,6 𝑛(𝐿𝑖) = ✓ 𝑛(𝑁2 ) = ✓ 7 28 𝑛(𝐿𝑖) = 1,76 𝑚𝑜𝑙 𝑛(𝑁2 ) = 1,20 𝑚𝑜𝑙 6 𝑚𝑜𝑙 𝐿𝑖 n(N2) required if ALL 1,76 mol of Li react. Stoichiometri ratio = ✓ 1 𝑚𝑜𝑙 𝑁2 n (N2) benodig as AL 1,76 mol Li reageer 1 n(N2) =1,76 × 6 = 0,29 𝑚𝑜𝑙 ✓ 1,76 1,47 Available ratio = = ✓ required/benodig 1,2 1 Therefore Li is limiting 1,2 mol is available ✓ reagent✓ 1,2 mol is beskikbaar Daarom is Li die beperkende Therefore Li is the limiting reagent ✓ reagens Daarom is Li die beperkende reagens n(Li) required if ALL 1,20 mol of N2 react. 6 n(N2) =1,20 × 1 = 7,2 𝑚𝑜𝑙 ✓ required/benodig Only 1,76 mol is available ✓ Therefore Li is the limiting reagent ✓ Slegs 1,76 mol is beskikbaar Daarom is Li die beperkende reagens (6) Positive marking from 7.2.2/ Positiewe nasien vanaf 7.2.2 7.2.3 Theoretical yield/Teoretiese opbrengs n(Li) : n (Li3N) 6 : 2✓ 2 n(Li3N) = 1,76 × 6 ✓ n(Li3N) = 0,59 mol 𝑚 𝑛= 𝑀 𝑚 0,59 = 35 ✓ m = 20,65 g 5,89 %yield/opbrengs = 20,65 × 100 % ✓ %yield/ opbrengs = 28,52 % ✓ (5) [20] Kopiereg voorbehou Blaai om asseblief ### Page 9 Downloaded from hlayiso.com (EC/NOVEMBER 2019) PHYSICAL SCIENCES P2 9 QUESTION 8/VRAAG 8 8.1.1 Acid is a substance that donates protons (H+) ‘n Suur is ‘n stof wat protone (H+) skenk (2) 8.1.2 Base. ✓ It accepts protons (H+) in both reactions ✓ Basis. Dit aanvaar protone (H + ) in albei reaksies (2) 8.1.3 HSO4-✓✓ (2) 8.1.4 H2SO4 + 2 NaHCO3 ✓→ Na2SO4 + 2 H2O + 2 CO2 ✓ (✓ Balanced/ Gebalanseerd) Accept/Aanvaar H2SO4 + NaHCO3 → NaHSO4 + H2O + CO2 (3) 8.2.1 A standard solution is a solution of which the concentration is exactly known. ✓✓ ʼn Standaardoplossing is ʼn oplossing waarvan die konsentrasie presies bekend is . (2) 8.2.2 Reaction 2/ Reaksie 2 n(NaOH) = cv ✓ n(NaOH) = (0,968)(0,025) ✓ n(NaOH) = 0,0242 mol Mole Ratio/Verhouding CH3COOH : NaOH 1 : 1 n(CH3COOH) = 0,0242 mol ✓ Original/Oorspronlik (CH3COOH) n(CH3COOH) = cv n(CH3COOH) = (0,5)(0,25) ✓ n(CH3COOH) =0,125 mol n(reacted) = 0,125-0,0242 ✓ n(reacted) = 0,1008 mol Reaction 1/Reaksie Mole Ratio CH3COOH : CaCO3 2 : 1✓ n(CaCO3) = 0,1008/2✓ n(CaCO3) = 0,0504 mol m(CaCO3) = nM m(CaCO3) = (0,0504)(100) ✓ m(CaCO3) = 5,04 g 5,04 % purity / suiwerhede= 56 × 100% ✓ %purity/ suiwerhede = 9%✓ (10) [21] Copyright reserved Please turn over ### Page 10 Downloaded from hlayiso.com 10 FISIESE WETENSKAPPE V2 (EC/NOVEMBER 2019) QUESTION 9/VRAAG 9 9.1 Oxidation is the loss in electrons ✓✓ Oksidasie is die verlies in elektrone (2) 9.2 W Cℓ- ✓✓ (2) 9.3 Mg ✓ Mg oxidation number increases from 0 ✓ to +2 ✓ Mg oksidasiegetal neem toe vanaf 0 na +2 (3) 9.4 2 H+(aq) + 2 e- → H2 (g) ✓✓ (2) 9.5 6✓ (Fe2+ → Fe3+ + e-) ✓ 14H+✓ + Cr2O72- + 6e- ✓→ 2 Cr3+✓ + 7 H2O ✓ 6 Fe2+ + 14 H+ + Cr2O72- → 6 Fe3+ + 2 Cr3+ + 7 H2O ✓ Marking guideline/Nasienriglyne • Correct oxidation half reaction/ Korrekte oksidasie-halfreaksie • 7 H2O in the reduction half reaction/reduksie-halfreaksie • 14 H+ in the reduction half reaction/reduksie-halfreaksie • 2 Cr3+ balancing the Cr3+ ions/Balansering van die Cr3+ ione • 6e- in reduction half reaction/ reduksie-halfreaksie • ×6 the oxidation half reaction/ oksidasie-halfreaksie • Correct final balanced equation/Korrekte finale gebalanseerde vergelyking (7) [16] Kopiereg voorbehou Blaai om asseblief ### Page 11 Downloaded from hlayiso.com (EC/NOVEMBER 2019) PHYSICAL SCIENCES P2 11 QUESTION 10/VRAAG 10 10.1 Witwatersrand ✓ (1) 10.2 Redox reaction ✓ Oxidation number of gold changes from 0 to +1 ✓ OR Oxidation number of oxygen decreases from 0 to -2. Redoksreaksie Oksidasiegetal van goud verander vanaf 0 na +1 OF Oksidasiegetal van suurstof verminder vanaf 0 to -2. (2) 10.3 Zinc ✓✓ Sink (2) 10.4 Activated carbon ✓✓ Geaktiveerde koolstof (2) 10.5 Process Z is the smelting process of gold. ✓ Gold has a very high boiling point. ✓ Large amount of energy is needed for gold to change state. ✓ Proses Z is die smeltproses van goud. Goud het ʼn baie hoë kookpunt. ʼn Groot hoeveelheid energie is nodig om die fase van goud te verander. 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