Suggested Stage 2 Chemistry 2008 SACE Board of SA Exam ...

204 2009 SASTA Chemistry Study Guide 

Suggested Stage 2 Chemistry 2008 SACE Board of SA Exam Solutions 

Question 1 Comments 

(a) (i) [Any one:] 

• N2 and O2 present in air 

• N2 reacts with O2 

[Any one:] 

• high temperature in engine 

• Ea supplied in engine. 

(ii) [Any one:] 

• Limited O2 present in air/engine. 

• Incomplete combustion of fuel. 

(iii) Disadvantage: [Any one:] 

• Less energy available per volume of fuel. 

• Lots of soot produced. 

Explanation: [Any one:] 

• travel less distance per unit volume of fuel 

• must buy more fuel for same trip. 

• visual pollutant/respiratory problems 

(b) (i) 2 NO + 2 CO → 2 CO2 + N2 

Note: 

Must have two distinct 

clear points for full 

marks. 

Note: 

Needs one points. 

Note: 



marks. 

(ii) C6H6 (l) + 15 /2 O2(g) → 6CO2(g) + 3H2O(l) ∆H = - 3280 kJ 1 mark for balancing 

1 mark for states 

1 mark for ΔH 

(iii) (1) (A) Q (B) P 

(iii) (2) 

Enthalpy 

reactants 

products 

Reaction pathway 

Ea without 

catalyst 

(iii) (3) Effect: 

• catalyst increases rate of reaction. 

Explanation: [Any one:] 

• provides alternative pathway of lower activation energy 

• hence more reactant particles have required Ea on collision 

• hence more successful collisions per unit time 

Note: 

Diagram should be 

clearly labelled. 

Note: 



marks.

© SASTA 2009 2008 Exam Solutions 205 


(a) (i) 

(ii) 

1,2,3 – propanetriol or 

propane – 1,2,3 – triol 

(iii) (1) increased rate of reaction 

(iii) (2) Catalyst 

Increased temperature 

(iii) (3) (A) [Any one:] 

Increased melting temperature 

More solid 

(iii) (3) (B) Becomes more saturated / less unsaturated 

(b) (i) Time taken for component to pass through the column.. 

(c) (i) 

(ii) Identification: 

Triglyceride 2. 

(ii) soap 

Explanation – 2 points: 

• longer retention time/held for longer time on column 

• more strongly adsorbed/attracted/bound to non-polar stationary 

phase 

[Any three of:] 

• tail is non-polar 

• thus tail dissolves in non-polar oil 

• ionic / carboxylate head attracted to polar H2O molecules 

• forms micelles with oil droplet in centre 

Note: 

propan- acceptable. 

Eg. propan – 1,2,3 – triol 

Note: 

Take care when drawing 

or copying structures. 

Be careful with the 

positioning of the bonds. 

Note: 

Needs two points. 

Note: 

Must have three distinct 


marks. 

Note: 

Annotations on the 

diagram acceptable as 

long as referred to the 

answer.



(a) (i) 

(ii) 

(b) (i) 

(b) (ii) (1) 

Mg is more active metal than chromium or 

Mg higher in activity series than Cr 

Mg reduces Cr 3+ ions or 

Mg displaces Cr 3+ ions from solution. 

3Mg + 2Cr 3+ → 3Mg 2+ + 2Cr 

3CrO4 2− + 2Fe 3+ → Fe2(CrO4)3 

pOH = - log[OH − ] 

= 5.678 

pH = 14 ─ pOH 

= 8.3 

(2) • As pH increases [H+ ] decreases. 

• System adjusts to oppose the change and increase [H + ]. 

• Thus equilibrium shifts to the left/back reaction is favoured. 

Note: 

Needs two clear points. 

Should refer to metal 

activity series given. 

Care with terminology. 

Note: 

1 mark for correct species 

1 mark for balancing 

Note: 



Note: 

Should show full working 

for calculations. 

Note: 


clear points for full marks. 

(c) (i) amphoteric Note: 

Not amphiprotic 

(ii) 

Cr2O3 + 2OH − → 2CrO2 − + H2O 

(iii) Non metallic (as it is reacting with OH − ) 

Note: 


1 mark for balancing



(a) (i) Nitrogen or N2 Note: 

Name or formula acceptable 

(ii) (1) trigonal pyramid 

(2) (A) unbonded electron pair on nitrogen atom. 

(B) 

(b) (i) anaerobic 

(c) 

(ii) (1) N is a plant nutrient 

(2) [Any two of:] 

• Surface algal growth blocks sunlight from lower algae. 

• Plants cannot undergo photosynthesis. 

• Thus plants do not produce O2. 

• Aerobic decay of dead algae uses up oxygen. 

Possible equation [Any one of:]: 

• 2NO + O2 → 2NO2 

• NO2 → NO + O 

• O + O2 → O3 

Increase in ozone [At least one point]: 

• NO reacts with atmospheric O2 to form NO2 

• NO2 dissociates in the presence of sunlight 

• NO2 dissociates to form O atoms 

• O atoms react with O2 to form O3 

• increased number of plants results in more NO released and 

hence an increase in O3 levels. 

Increase undesirable [At least one point]: 

• build-up of O3 in lower atmosphere leads to the formation of 

photochemical smog 

• O3 reacts with hydrocarbons to form a range of toxic 

chemicals (eg aldehydes, ketones, PAN) 

• O3 damages plants/reduces rate of photosynthesis/stunts 

plant growth 

• O3 causes rubber/plastics/paints to deteriorate 

• O3 damages skin 

• O3 is a respiratory irritant. 

Note: 

⊕ needs to be next to the 

nitrogen atom in the first 

diagram. 

Note: 

Must have two distinct clear 

points for full marks. 

1. Be careful to answer the 

question ie 

increase in O3 and 

increase undesirable. 

2. Plan your answer on 

scrap paper before 

writing. 

3. Marks are allocated for 

communication skills 

including grammar, 

spelling and writing in 

sentences. 

4. Try for six well 

presented points 

5. Equations do not need to 

be balanced but must 

have correct formulae 

Do not write your answer 

in dot point form unless 

you are running out of 

time and it is the only way 

you are going to get an 

answer down.



(a) (i) C5H8 Note: 

Must be molecular 

formula 

(b) 

(c) (i) 

(ii) [Any one of:] 

• Contains a carbon-carbon double bond or 

• contains an alkene group 

[Any one of:] 

• monosaccharides are renewable 

• more petroleum available to use as chemical feedstock 

• carotene molecule is non-polar 

• carotene molecule is very large compared to water 

• H2O molecule is polar 

• non-polar molecules not attracted to / cannot form 

H-bonds with polar H2O molecules 

(ii) (1) initial colour final colour 

β-carotene brown colourless/faded brown 

retinal brown colourless/faded brown 

(2) [Any one of:] 

• both contain C=C group / alkene group / are unsaturated 

• both undergo an addition reaction 

(iii) (1) β-carotene no change/no silver mirror formed 

Retinal silver mirror forms 

Note: 



marks. 

(2) aldehyde Note: 

Must be name; -CHO 

formula not acceptable. 

(3) oxidation 

(4) 

Note: 




positioning of the bonds.



(a) (i) chemical → electrical 

(b) (i) 

(ii) (1) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 Note: 

correct convention 

(s, p, d etc and 

superscripts) 

(iii) (1) 

(2) +5 Note: 

Sign before digit with 

oxidation number 

(3) [Any two of:] 

• oxidation no. of V decreases 

• electrons are gained 

• reduction occurs 

(4) (arrow marked on the diagram) Note: 

Arrow must be in / 

near external circuit. 

electrolytic cell 

(2) V2+ or 

vanadium(II) ion 

• zeolite consists of a continuous network of 

strong/primary/ionic/covalent bonds 

• a large amount of energy is needed to overcome these strong 

bonds 

(ii) (1) [Al2Si10O24] 2− 

(2) 


• Al is (covalently) bonded within the lattice network. 

• Al has substituted for silicon within the lattice network. 

• Na and Ca ions are adsorbed/stuck/attracted to the surface of 

the zeolite; Al is not. 

Note: 

Not vanadium ion as 

not specific enough 

Note: 

Must be adsorbed 

not absorbed.



(a) (i) 

(ii) 

(b) (i) 

[Any three of:] 

• Al 3+ very small and highly charged/high charge density 

• attract negatively charged clay particles (& other suspended 

matter) 

• Al 3+ forms Al(OH)3 gel trapping suspended matter 

• large floc/aggregate/clump forms and settles to the bottom 


• clearer 

• improves clarity 

• floc settles to the bottom 

• can be filtered 

(iii) (1) Al2O3 + 6H + → 2Al 3+ + 3H2O 


• H + exchanges with/displaces Al 3+ on clay surfaces 

• cation exchange occurs 

• HClO is an oxidiser 

• kills bacteria/sterilises the water 

(ii) (1) 2Cl − → Cl2 + 2e − 

(2) + electrode 

(3) Explanation: 

• less energy needed to reduce H2O than Na + 

What produced: 

• H2O is reduced at the cathode / H2 produced at the cathode 


• molten NaCl (or another sodium ionic compound) 

• molten sodium compound 

Note: 

Must have three 

distinct clear points 

for full marks. 

Note: 


Note: 


Note: 

Needs molten. 

Not molten salt (all 

ionic compounds are 

salts)



(a) 

(b) (i) 

(ii) (1) [Any two of:] 

• Enzyme chain gains energy. 

• Hydrogen bonds/ion-dipole and dispersion forces are 

overcome. 

• The 3-D shape of the enzyme will change. 

Must mention: 

• If the structure of the enzyme changes then it is unable to 

carry out its biological function. 

(ii) (2)(A) CO2 + H2O → H2CO3 

(B) 

CO2 + H2O H2CO3 

Acidic or 

low pH 

(c) (ii) (1) 0.21 – 0.23 

(ii) 

(2) ppb = µg/L 

no. µg Cd = 3.5 x 0.010 

(3) ppb = µg/kg 

= 0.035 µg 

= = 0.035/0.00045 = 77.7777 

= 78 ppb (2 sf) 

[Any two of:] 

• Cd lamp is used 

• wavelength used is absorbed only by Cd 

• Na absorbs a different wavelength/does not absorb this 

wavelength 

Note: 




positioning of the bonds. 

Note: 

Section of molecule 

therefore open at least 

one end or both. 

Note: 



marks. 

Note: 

1 mark for correct 

species 


Note: 

Should show full 

working for calculations. 

Note: 

Should show full 

working for calculations. 

Note: 



marks. 

(iii) Has non zero absorbance for zero concentration of cadmium. Note: 

Not does not start at zero



(a) NO (Regenerated in Step 4 and used in Step 2) Note: 

Name or formula. 

(b) Reason: Advantage: 

More product produced Thus more cost effective. 

per unit time. 

or 

Increased rate at lower Thus save on energy. 

temperatures. 

(c) 

(d) (i) 

(e) 

Effect: 

• Increases yield of NO2. 

Explanation [Any two of]: 

• System adjusts to oppose the high pressure. 

• Thus favours reaction that forms fewer moles of gas 

particles. 

• Favours forward reaction/equilibrium shifts to the right. 

Kc = 

(ii) (1) 2NO2(g) N2O4(g) 

(2) 

(3) 

initial moles 0.132 0 

attaining eq m 0.0800 0.0400 

at equilibrium (0.132 – 0.0800) 0.0400 

Kc = 

= 

= 

= 14.79 

= 0.0520 

Reaction: Exothermic 

Explanation [Any two of]: 

• Kc decreases as temperature increases 

• there are less products and more reactants 

• the reverse reaction is endothermic 

Fe + 2H + → H2 + Fe 2+ or 

2Fe + 6H + → 3H2 + 2Fe 3+ 

Note: 

Advantage must 

correspond with reason 

Note: 



Note: 

Should show full working 

for calculations. 

Note: 



Note: 

Must be ionic equation 


1 mark for balancing



(a) (i) 6CO2 + 24H + + 24e − → C6H12O6 + 6H2O Note: 



(b) (i) 

(ii) n(CO2) = 

n(C6H12O6) = x n(CO2) 

= x 

Energy required = 2820 x x 

= 10 679 kJ 

Note: 

Should show full working for 

calculations. 

(iii) (1) C6H12O6 → 2C2H5OH + 2CO2 Note: 



(2) • Yeasts provide enzymes. 

• Enzymes increase the rate of reaction. 

(3) CO2 released is balanced by CO2 absorbed during 

photosynthesis. 

(ii) (1) ester 

[Any two of:] 

• HCO3 − is basic 

• neutralises H + 

• removes/lowers concentration of H + 

Note: 



Note: 



(2) (A) CH3OH or methanol Note: 

Name or formula 

(B) • carboxylate ion forms (actually 2 carboxylate ions) 

• ion-dipole forces between ion and water 

• weaker dipole-dipole forces between chlorophyll 

molecule and water. 

Note: 

Must have three distinct clear 

points for full marks.



(a) (i) d-block 

sig fig 3 sig figs must be used for answer in (a) 

(ii) n = C.V 

= 0.0100 x 0.250 

= 0.00250 

m = n.M 

= 0.00250 x 433.026 

= 1.082565 

= 1.08 g (3 sf) 

(iii) (1) n = C.V 

= 0.0100 x 0.01786 

= 1.786 x 10 −4 

= 1.79 x 10 −4 (3 sf) 

(2) n(F − ) = 3n(La 3+ ) 

= 3 x 1.786 x 10 −4 

= 5.358 x 10 −4 

= 5.36 x 10 −4 (3 sf) 

(3) m = n.m 

= 5.358 x 10 −4 x 41.99 

= 0.022498…. 

= 0.0225 g 

= 22.5 mg (3 sf) 

(4) % = x 100 

(5) Increased. 

= 4.4996 

= 4.50% (3 sf) 

(b) mg needed = 5 x 0.95 

(c) (i) 

∴ no. tablets needed = = 3.96 

Thus need 4 tablets. 

That the solubility of F − / % of F − soluble in water 

decreases as water hardness increases. 

(ii) Data from graph [Any one]: 

• water hardness increases from 0 – 200 

• F − changes 100 – 90% 

Comment[Any one]: 

• most F − still available 

• minor reduction 

Note: 

Full working for 

calculations should be 

shown. 

Units not necessary as 

stated in question 

Note: 

Must use data from graph 

and comment for full marks



(a) (i) (1) Condensation 

(b) 

(c) (i) 

(2) 1 mark for any one of: 

• small molecule/H2O eliminated/lost during reaction 

• mass of polymer chain less than total mass of 

monomer used 

• reaction between carboxylic acid and amine 

(ii) (1) (1) amide 

(2) 

(3) 1 mark for any one of: 

• small molecule/H2O not eliminated during reaction 

• mass of polymer chain = total mass of monomer used 

1,6 – hexanedioic/hexandioic acid or 

hexanedioic/hexandioic acid 

1 mark for any one of: 

• the polymers consist of chains of different length/not 

all chains are the same length 

• some monomer remains 

• contaminants or additives may be present 

(ii) 1 mark for: 

• Stanyl softens at higher temperature than other two 

polymers. 

1 mark for: 

• PP non-polar 

• dispersion forces only between PP chains 

• little energy needed to overcome weak interactions 

between PP chains 

• Nylon 6/Stanyl contain polar amide group 

• hydrogen bonding between Nylon 6/Stanyl chains 

• stronger forces between/higher temperatures needed 

to overcome forces between Nylon 6/Stanyl chains 

than PP chains 

• shorter non-polar segments in Stanyl chains than 

Nylon 6 chains 

• more amide groups/hydrogen bonds per unit length of 

polymer chain in Stanyl than Nylon 6 

1. Be careful to answer 

the question ie 

Which polymer 

2. Plan your answer on 

scrap paper before 

writing. 

3. Marks are allocated for 

communication skills 

including grammar, 

spelling and writing in 

sentences. 

4. Try for six well 

presented points 

Do not write your answer 

in dot point form unless 

you are running out of 

time and it is the only way 

you are going to get an 

answer down.

Suggested Stage 2 Chemistry 2008 SACE Board of SA Exam ...

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?