Physical Sciences Grade 12 P2 Limpopo November 2014 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 PHYSICAL SCIENCES: CHEMISTRY (P2) FISIESE WETENSKAPPE: CHEMIE (Vr2) SEPTEMBER 2014 MEMORANDUM MARKS: 150 This memorandum consists of 10 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Science/P2 2 LimpopoDoE/September 2014 NCS-Memorandum QUESTION/ VRAAG 1 11. Cvv (2) 12 Dvv (2) 1.3. Cvv (2) 14. Avy (2) 15 Avy (2) 16 Cvv (2) 1.7 Byy (2) 18 Byy (2) 19 Avy (2) 1.10 Dvv (2) [20] QUESTION/VRAAG 2 2.1 2.1.1 Uv (1) 2.1.2 Carboxyl groupY / karboksielgroep (1) 2.1.3 Aldehyde / alkanal v//aldehied (1) 2.1.4 2-bromo-1,4-dichloroY butane ¥ /2-bromo-1,4-dichworoburaan of 2-broom-1,4-dichloorobutaan (2) 2.1.5 polythenev /politeen ACCEPT: Poly(ethene) (1) 2.1.6. 2CgHiqg + 1902” > 12CO02+14H20Y = Balv ReactantsY ProductsY BalancingY (3) 2.1.7. 4-Methyl” hexan-1-ol” / 4-métielheksan-1-ol Y Accept: 4-Methyl-1-hexanol (2) 2.1.8. CnH2nOo” (1) 2.2 The reddish-brown (bromine water) solution decolourises/ The reddish-brown colour disappears” // die rooi-bruin kleur verdwyn/oplossing ontkleur (1) 2.3 H H HoH ‘cmc’ 4 Br-Br ——> H~C—CNH / \ | | H H Br sor Notes v alkene/alkene =v Bromine/ broom Y functional group correct (product)/funksionele group van produk Y whole structure correct (product)/ hele struktuur van produk Do not penalise for Bro (4) [17] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Science/P2 3 LimpopoDoE/September 2014 NCS-Memorandum QUESTION/VRAAG 3 3.1. A series of organic compounds that can be described by the same general formula

/’n reeks organiese verbindings wat dieselfde algemene formule het A series of organic compounds whose successive members differ by a CH2 unit only/ ’n reeks organiese verbindings wat met ‘n CH2 eenheid van mekaar verskil A series of organic compounds having the same functional group./ / ’n reeks organiese verbindings watdieselfde funksionele groep het (2) 3.2. Chloromethane
/ chloormetaan/chlorometaan Fluoromethane
/fluoormetaan/fluorometaan (2) 3.3. Boiling points increase
as the mass/size of the molecule increases. OR Boiling points decrease
as the mass/size of the molecule decreases
/ kookpunte neem toe soos die massa/grootte van die molekules toeneem OF kookpunte neem af soos die massa/grootte van die molekules afneem (2) 3.4. As the molecular mass increases
, the strength of Van Der Waals forces (IMFs) increases.
Hence more energy is needed to overcome the Van Der Waals forces (IMFs)
. Thus the boiling points increase./ Soos die molekulêre massa toeneem word die Van der Waalskragte sterker. Dus word meer energie benodig om die Van der Waalskragte (IMK) te oorkom, dus neem die kookpunte toe (3) 3.5. Criteria for investigative question/ kriteria Mark Refers to relationship between dependent and independent
variables/ verwys na verhouding tussen die afhanklike en onafhanklike veranderlikes It is a question – not stated as an aim/ is ‘n vraag
For example/ byvoorbeeld: What is the relationship between boiling point and type of organic compound/ functional group/ homologous series? // wat is die verhouding tussen kookpunt en die soort organiese verbinding/funksionele groep/ homoloë reeks?

(2) 3.6 Molecular mass
/molekulêre massa (1) 3.7 The carboxylic acids have higher boiling points than alcohols
which have higher boiling points than the aldehydes with the same molecular mass.
/ die karboksielsure het ‘n hoer kookpunt as die alkohole en hulle het ‘n hoer kookpunt as die aldehiedes (2) [14] QUESTION/VRAAG 4 4.1 Cracking
/Dehydrogenation/ elimination//kraking/eliminasie/dehidrogenasie (1) 4.2
H H H H H +
H C C C H O H H C C C H
H H H H H O H
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Downloaded from hlayiso.com Physical Science/P2 4 LimpopoDoE/September 2014 NCS-Memorandum 4.3 Propan-2-ol /2-propanol
(1) 4.4.1 Substitution reaction/ substitusie (1) 4.4.2 Warm (heat)
dilute NaOH or KOH
(under reflux)/ hitte, verdunde NaOH (2) 4.5.1 Polypropene
/ polipropeen (1) 4.5.2 Addition
/addisie (1) [11] QUESTION/VRAAG 5 5.1 CO2 gas generated escapes to the air
/ CO2 gas vrygestel ontsnap in die lug (1) ∆m 5.2. Average rate / gem tempo= ∆t 4,4 − 0 = 240 − 0
= 0,0183 g·s-1
(3)
5.3 4,4 g
(2) 5.4 All the calcium carbonate has reacted
/ al die CaCO3 het gereageer No more CO2 is being generated / geen CO2 word meer vrygestel nie The reaction has ceased (stopped)
/ reaksie het opgehou (1) m 5.5 n(CO2) produced/geproduseer =
M 4,4 =
44 = 0,1 mol From the balanced equation/ vanaf gebalanseerde vergelyking: n(CaCO3) = n(CO2)
= 0,1 mol
m(CaCO3) = nM = (0,1)(100)
= 10 g
(6) [13] QUESTION/VRAAG 6 6.1 The stage in a reversible reaction at which the rate at which A(g) and B(g) react to form C(g) equals the rate at which C(g) decomposes to reform A(g) and B(g)

// Dit is 'n dinamiese ewewig wanneer die tempo van die voorwaartse reaksie gelyk is aan die tempo van die terugwaartse reaksie. OR: The stage in a reversible reaction at which the concentrations of A(g), B(g) (reactants) and C(g) (products) remain constant.// die stadium in ‘n omkeerbare reaksie wanneer die konsentrasie van reagense en produkte constant bly (2) 6.2 H < 0 / H is negative
/ H is negatief (1) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Science/P2 5 LimpopoDoE/September 2014 NCS-Memorandum 6.3 Decrease the temperature
/ verlaag temperatuur The forward reaction is exothermic. According to Le Chatelier’s Principle, a decrease in temperature places a stress on the system.
To counteract the stress, the system reacts by favouring the exothermic reaction/ the reaction that releases heat
.The forward reaction is favoured / Equilibrium position shifts to the right/Product formation is favoured. Hence the concentration of C (g) increases.
Volgens Le Chatelier se beginsel word ‘n stress op die sisteem geplaas wanneer die temperatuur verlaag word. Die eksotermiese reaksie word dan bevoordeel, d.w.s die voorwaartse reaksie. Dus word meer produk gevorm en konsentrasie C neem toe (4) 6.4 CALCULATIONS USING MOLES: Mark allocation: N N • USING formula n =
/ gebruik formule n(A) = NA NA • Using ratio A(g) : B(g) : C(g) = 2 : 3 : 2
/ gebruik verhouding A : B : C = 2 : 3 : 2 • Equilibrium n(A) = initial – change Equilibrium n(B) = initial – change
Equilibrium n(C) = initial + change
Ewewig n(A) = aanvanklik - verandering Ewewig n(B) = aanvanklik - verandering
Ewewig n(C) = aanvanklik + verandering
• Divide n(A), n(B) & n(C) by 1 dm3
/deel deur 1 dm3 • Correct Kc expression (formulae in [ ])
/ korrekte Kc uitdrukking • Substitution of concentrations into Kc expression
vervang konsentrasies • Final answer/ finale antwoord : 0,064
OPTION / OPSIE 1 N 2,41× 1023 n(A) =
= = 0,4 mol NA 6,02 × 1023 R Ag(g) B(g) C(g) Initial (mole) 2 4 0 aanvanklik Change/ver- -0,4 -0,6 +0,4
ratio andering Equ. Mol 1,6 3,4 0,4

[Eq] 1,6 3,4 0,4
(div) Kc = [C] 2
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Downloaded from hlayiso.com Physical Science/P2 6 LimpopoDoE/September 2014 NCS-Memorandum __ lo4F ~ fyeF 34h = 0,00159¥ (1,59 x 10°) (8) OR/of: CALCULATIONS USING CONCENTRATIONS /berekening met konsentrasies Mark allocation: ° Using the formula n(A) = v / gebruik formule n(A) ~ A A Divide n(A) & n(B) by 1 dm°v/ deel deur 1 dm?° USING ratio: A: B: C = 2:3: 2v/ gebruik verhouding A:B:C =2:3:2 ° Equilibrium n(A) = initial — change } Equilibrium n(B) = initial — change v Equilibrium n(C) = initial + changev Ewewig n(A) = aanvanklik - verandering Ewewig n(B) = aanvanklik - verandering v Ewewig n(C) = aanvanklik + verandering Correct K; expression (formulae in [ ])“/ korrekte Ke, uitdrukking Substitution of concentrations into K, expression” /vervang konsentrasies . Final answer /finale antwoord: 0,064v OPTION /opsie 2 [A]= n/v = 2/1 = 2 mol-dm® [B]= 4/1 = 4 mol: dm® N ,_ 241x 10% A) =— Vv = ———_,y;, = 0,4 n(A) Nh 60210" 0,4 mol A(g) Big) Cig) R 2 3 2 | (mol/dm*) 2 4 0 C (mol/dm*) -0,4 -0,6 +0,4Y _| ratio/verhouding E(mol/dm*) a6 | 34%/ 04v [] equ 1,6 3,4 0,4 v +1 c=", (mol-drv*) K, = [cf v [AF ier _ ~ (2@.5)3 = 0,064Y (8) 6.5 Remains the samev / bly dieselfde (1) [16] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Science/P2 7 LimpopoDoE/September 2014 NCS-Memorandum QUESTION/VRAAG 7 7A 7.2 7.2.1 7.2.2 7.3. 7A. A solution that contains a low (number of mole) amount of acid in proportion to the volume of water.v v//‘n oplossing wat min (aantal mol) suur in verhouding tot die volume water bevat (2) M(KOH) = 39 + 16 + 1 = 56 g:mol"’ m(KOH) = cMVY = (0,2)(56)v (0,3)Y = 3,36 gv OR/ OF n=c-V = (0,2)(0,3) = 0,06 mol m = nM = (0,06)(56) = 3,36 g (4) KOH “28k* 4 OH 0,2 mol KOH yield 0,2 mol OH” « [OH] = [KOH] = 0,2 mol-dm?v Kw =[HsO*][OH] = 1 x 10%4v « [HsO*]=1 x 107'4/0,2 =5x 10"4 mol-dm?’v pH = -log[H,O*]Y” =-log (5x 10*)v = -(-13,3) = 13,3v OR/ OF: pOH =-log [OH]Y¥ = - log (0,2)v =0,7¥ But pH + pOH = 14v «pH =14—pOH =14-0,7¥ = 13,3v (6) Bromothymol bluev/ broomtimolblou H2SOz is a strong acid and KOH is a strong basev. Therefore the equivalent point will be at approximately pH=7¥ //H2SO, is ‘n sterk suur en KOH is ‘n sterk basisv .Daarom is die ekwivalentpunt omtrent by pH = 7 (3) CaVa _ May CoVb Mb y _ C,(20,0) 1 : “ 2)050y =5 * (ratio) (0,2)(15,0) = SE! = 0,075 mol-dm?v 5 a=" (2)(2000) more °) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Science/P2 8 LimpopoDoE/September 2014 NCS-Memorandum OR: PaVaCa = PbVbCb
(2)(20,0)Ca
=(1)(15,0)(0,2)

(ratio) (0,2)(15,0)(0,2) Ca = (2)(20) = 0,075 mol·dm-3
OR/ OF n(KOH) = cV = (0,2)(15,0x 10-3)
= 3 x 10-3 mol But from the balanced equation:/ vanaf gebalanseerde vergelyking n(H2SO4)=½ x n(KOH) = ½(3x10-3)
= 1,5 x 10-3 mol [H2SO4] = n/V
1,5 × 10−3 =
20,0 × 10−3 = 0,075 mol·dm-3
(5) [20] QUESTION/VRAAG 8 8.1 Salt bridge
/ soutbrug It maintains electrical neutrality in the cell

/It completes the circuit
(ANY ONE) Dit handhaaf elektriese neutraliteit/ dit voltooi die stroombaan (enige een) (2) 8.2 The reducing ability of zinc is greater than that of chromium, which is in turn greater than that of silver.

// die reduseervermoë van sink is beter as die van chroom en dit is weer beter as silwer s’n (2) 8.3 Voltmeter’s terminals have been connected incorrectly (+ to anode and – to cathode)

// die voltmeter se terminale is verkeerd om gekoppel (+ aan die anode en - aan die katode) (2) 8.4 Zinc is a better reducing agent than chromium
and chromium is a better reducing agent than silver.
/ Sink is ‘n beter/sterker reduseermiddel as chroom en chroom is ‘n beter reduseermiddel as silwer (2) 8.5.1 Ecell = Ecathode - Eanode / Esel = Ekatode - Eanode = -0,74
– (-0,76)
Ecell = + 0,02 V
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Downloaded from hlayiso.com Physical Science/P2 9 LimpopoDoE/September 2014 NCS-Memorandum OR: Cr** +3e° > Cr f E'= -0,74v Zn >Zn** 426 Jv E* = 40,76” E®..= 0,02Ww (4) 8.5.2. Yesv /Ja Ece > 0 / Ecen is positive / Ege: is positief (2) 8.5.3 Decreases vv /afneem (2) 8.6 Measurements not done under standard conditions of temperature of 25°CVY and concentration of electrolytes of 1 mol:dm°v (Accept: internal resistance from the salt bridge) // Lesing nie onder standaard toestande van temperatuur 25°C en konsentrasie van elektroliete 1 mol-dm™® geneem (Aanvaar interne weerstand van soutbrug) (2) [19] QUESTION/VRAAG 9 9.1 BY (1) 9.2 Mass deposited is directly proportional to currentv/ massa neergeslaan is direk eweredigg aan die stroom (1) 9.3 9.3.1 The same asv /dieselfde as (1) m =—Vv 9.3.2 n(Cu) M 32 = = v 63,5 0,504 mol Cu (2) 9.3.3 32,0 g Cu = 0,504 moles of Cu/ mol Cu 32 g CU — 0,504 mol electrons/elektrone -. 63,5 g Cu — 2 mol electrons/ elektrone Q =ne = (1,008 x 6,02 x 107%)(1,6 x 10°)” =9707C But/maar Q. =I-Atv 9707 = 1-(3600)” “1 =2,70 AY (5) [10] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Science/P2 10 LimpopoDoE/September 2014 NCS-Memorandum QUESTION/VRAAG 10 10. 1 By the fractional distillation of liquid air
/fraksionale distillasie van lug (1) 10.2 10.2.1 Ammonia/ NH3
/ ammoniak (1) 10.2.2 Sulphur/ S
/swael (1) 10.2.3 Ammonium sulphate/ (NH4)2SO4
/ ammoniumsulfaat (1) 10.3 Vanadium pentoxide/ Divanadium pentoxide/ V2O5
/ vanadiumpentoksied (1) 10.4 Nitrogen
/ stikstof (1) 10.5 2SO2(g) + O2(g) 2SO3(g) (Reactants
Products
Balancing
) (3) 10.6 (ANY ONE) • Soil cannot replenish nutrients at a fast enough rate to sustain plant growth.

/ grond kan nie vining genoeg voedingstowwe vrystel om groei van plante te onderhou • Globally a bigger demand for food. .

/groter wêreldwye behoefde aan kos (1) [10] GRAND TOTAL/ Groot Totaal: 150 Copyright reserved Please turn over