Skill Align SACE Stage 2 Chemistry - Free Online Pack 0 Booklet 2
An independently authored SACE Chemistry examination booklet with worked answers, atomic marking guidance and diagnostic review prompts.
- Paper
- Booklet 2
- Reading
- No separate reading time
- Writing
- Approximately 65 minutes
- Assessment
- 61 marks
Approved calculators may be used. A Skill Align-formatted data sheet and periodic table based on the current SACE Chemistry reference material are supplied in this booklet.
Booklet 2 - Questions 4 to 7
Answer every question and every part. Show working for calculations, include units and use the supplied chemical evidence where required.
Question 4
14 marksAn algal oil is converted to a diol, HO-R-OH, where R is the algal-derived carbon skeleton between the two hydroxyl groups. The diol reacts with butanedioic acid, HOOC-CH₂-CH₂-COOH, to form a polyester. The repeat unit contains ester links. A 12.0 g batch yields 8.64 g purified polymer; the theoretical mass is 10.8 g.
Question 5
15 marksA grape juice contains 180 g L⁻¹ glucose. Yeast converts glucose according to C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g). A 0.500 L batch consumes 75.0% of its glucose. M(glucose) = 180.16 g mol⁻¹.
Question 6
16 marksA flow cell uses V²⁺/V³⁺ at the negative electrode and VO₂⁺/VO²⁺ in acidic solution at the positive electrode during discharge. The reduction half-equation is VO₂⁺ + 2H⁺ + e⁻ → VO²⁺ + H₂O. A current of 2.50 A flows for 38.6 min; use F = 96485 C mol⁻¹.
Question 7
16 marksA wastewater plant precipitates struvite, MgNH₄PO₄·6H₂O, from a stream containing 0.0200 mol phosphate. The trial recovers 85.0% of the phosphate. One mole of struvite contains one mole of phosphorus; M(struvite) = 245.4 g mol⁻¹. The recovered struvite is intended for use as a land-applied fertiliser if it meets the required purity, contaminant and agronomic-quality specifications.
Worked Solutions And Marking Guide
General marking principles
- Award each listed raw mark independently when the required chemical evidence is demonstrated.
- Accept chemically correct equivalent equations, representations, working and wording.
- A conclusion or Science as a Human Endeavour judgement must be supported by relevant evidence and acknowledge material limitations or trade-offs.
Question 4
(a) - From the supplied HO-R-OH, a suitable repeat unit is [-O-R-O-C(=O)-CH₂-CH₂-C(=O)-]ₙ, where R is the defined algal-derived carbon skeleton and ester links occur in the backbone. - The single-bond ester oxygen comes from a hydroxyl group of the algal-derived diol. - The carbonyl oxygen remains with the carboxylic-acid-derived acyl group. - For each condensation linkage, the acid's hydroxyl oxygen and an alcohol hydroxyl hydrogen leave together as H₂O.
(b)(i) - A diol has two hydroxyl functional groups. - Butanedioic acid has two carboxyl functional groups.
(b)(ii) - Condensation between -OH and -COOH forms an ester link. - Water is eliminated for each ester linkage formed.
(c) - Yield ratio = 8.64/10.8 = 0.800. - Percentage yield = 0.800 × 100 = 80.0%.
(d)(i) - The 80.0% result establishes the purified polymer mass relative to the stated theoretical mass. - The isolated yield alone does not establish environmental superiority or complete reaction.
(d)(ii) - Compare equal-thickness algal and conventional polyester samples under the same temperature, light and exposure time. - Measure strength retention, mass change or degradation products after the controlled exposure.
Detailed marking criteria
Part a (1 mark)
Awards one mark for establishing this question-specific chemical point: From the supplied HO-R-OH, a suitable repeat unit is [-O-R-O-C(=O)-CH₂-CH₂-C(=O)-]ₙ, where R is the defined algal-derived carbon skeleton and ester links occur in the backbone.
Part a (1 mark)
Awards one mark for establishing this question-specific chemical point: The single-bond ester oxygen comes from a hydroxyl group of the algal-derived diol.
Part a (1 mark)
Awards one mark for establishing this question-specific chemical point: The carbonyl oxygen remains with the carboxylic-acid-derived acyl group.
Part a (1 mark)
Awards one mark for establishing this question-specific chemical point: For each condensation linkage, the acid's hydroxyl oxygen and an alcohol hydroxyl hydrogen leave together as H₂O.
Part (b)(i) (1 mark)
Awards one mark for establishing this question-specific chemical point: A diol has two hydroxyl functional groups.
Part (b)(i) (1 mark)
Awards one mark for establishing this question-specific chemical point: Butanedioic acid has two carboxyl functional groups.
Part (b)(ii) (1 mark)
Awards one mark for establishing this question-specific chemical point: Condensation between -OH and -COOH forms an ester link.
Part (b)(ii) (1 mark)
Awards one mark for establishing this question-specific chemical point: Water is eliminated for each ester linkage formed.
Part c (1 mark)
Awards one mark for establishing this question-specific chemical point: Yield ratio = 8.64/10.8 = 0.800.
Part c (1 mark)
Awards one mark for establishing this question-specific chemical point: Percentage yield = 0.800 × 100 = 80.0%.
Part (d)(i) (1 mark)
Awards one mark for establishing this question-specific chemical point: The 80.0% result establishes the purified polymer mass relative to the stated theoretical mass.
Part (d)(i) (1 mark)
Awards one mark for establishing this question-specific chemical point: The isolated yield alone does not establish environmental superiority or complete reaction.
Part (d)(ii) (1 mark)
Awards one mark for establishing this question-specific chemical point: Compare equal-thickness algal and conventional polyester samples under the same temperature, light and exposure time.
Part (d)(ii) (1 mark)
Awards one mark for establishing this question-specific chemical point: Measure strength retention, mass change or degradation products after the controlled exposure.
Question 5
(a)(i) - Glucose is a monosaccharide containing multiple hydroxyl groups. - The supplied equation gives a one-to-two glucose-to-ethanol mole ratio.
(a)(ii) - Fermentation is enzyme-catalysed under anaerobic conditions. - C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g) is balanced for carbon, hydrogen and oxygen atoms.
(b) - Initial glucose mass = 180 × 0.500 = 90.0 g. - Consumed glucose mass = 0.750 × 90.0 = 67.5 g. - n(glucose) = 67.5/180.16 = 0.3747 mol. - n(ethanol) = 2 × 0.3747 = 0.749 mol. - Using M(ethanol) = 46.07 gives 34.5 g ethanol.
(c) - The required reaction representation is C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g). - The carbohydrate glucose is converted into the alcohol ethanol and carbon dioxide while conserving six carbon atoms. - Fermentation includes coupled oxidation and reduction steps mediated by yeast enzymes.
(d)(i) - Residual glucose would support incomplete conversion of the supplied glucose rather than complete fermentation. - Ethanol lost during collection or carbon diverted to other products could lower recovered ethanol even if glucose was consumed, so recovery alone cannot select a cause.
(d)(ii) - Run replicate batches at fixed temperature, pH and yeast loading, measure ethanol concentration by gas chromatography or a validated density method, and compare measured yield with residual glucose and carbon-dioxide data.
Detailed marking criteria
Part (a)(i) (1 mark)
Awards one mark for establishing this question-specific chemical point: Glucose is a monosaccharide containing multiple hydroxyl groups.
Part (a)(i) (1 mark)
Awards one mark for establishing this question-specific chemical point: The supplied equation gives a one-to-two glucose-to-ethanol mole ratio.
Part (a)(ii) (1 mark)
Awards one mark for establishing this question-specific chemical point: Fermentation is enzyme-catalysed under anaerobic conditions.
Part (a)(ii) (1 mark)
Awards one mark for establishing this question-specific chemical point: C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g) is balanced for carbon, hydrogen and oxygen atoms.
Part b (1 mark)
Awards one mark for establishing this question-specific chemical point: Initial glucose mass = 180 × 0.500 = 90.0 g.
Part b (1 mark)
Awards one mark for establishing this question-specific chemical point: Consumed glucose mass = 0.750 × 90.0 = 67.5 g.
Part b (1 mark)
Awards one mark for establishing this question-specific chemical point: n(glucose) = 67.5/180.16 = 0.3747 mol.
Part b (1 mark)
Awards one mark for establishing this question-specific chemical point: n(ethanol) = 2 × 0.3747 = 0.749 mol.
Part b (1 mark)
Awards one mark for establishing this question-specific chemical point: Using M(ethanol) = 46.07 gives 34.5 g ethanol.
Part c (1 mark)
Awards one mark for establishing this question-specific chemical point: The required reaction representation is C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g).
Part c (1 mark)
Awards one mark for establishing this question-specific chemical point: The carbohydrate glucose is converted into the alcohol ethanol and carbon dioxide while conserving six carbon atoms.
Part c (1 mark)
Awards one mark for establishing this question-specific chemical point: Fermentation includes coupled oxidation and reduction steps mediated by yeast enzymes.
Part (d)(i) (1 mark)
Awards one mark for establishing this question-specific chemical point: Residual glucose would support incomplete conversion of the supplied glucose rather than complete fermentation.
Part (d)(i) (1 mark)
Awards one mark for establishing this question-specific chemical point: Ethanol lost during collection or carbon diverted to other products could lower recovered ethanol even if glucose was consumed, so recovery alone cannot select a cause.
Part (d)(ii) (1 mark)
Awards one mark for establishing this question-specific chemical point: Run replicate batches at fixed temperature, pH and yeast loading, measure ethanol concentration by gas chromatography or a validated density method, and compare measured yield with residual glucose and carbon-dioxide data.
Question 6
(a)(i) - The oxidation half-equation is V²⁺ → V³⁺ + e⁻.
(a)(ii) - The reduction half-equation consumes two H⁺ per electron. - Adding the half-equations conserves charge and electrons. - Electron flow is from the V²⁺ half-cell to the VO₂⁺ half-cell.
(b) - V²⁺ is oxidised to V³⁺ at the negative electrode during discharge. - VO₂⁺ is reduced to VO²⁺ at the positive electrode during discharge. - One mole of electrons is released per mole of V²⁺ oxidised and one mole of electrons is consumed per mole of VO₂⁺ reduced. - Electrons flow from the negative electrode through the external circuit to the positive electrode.
(c) - Charge Q = It = 2.50 × (38.6 × 60) = 5790 C. - n(e⁻) = 5790/96485 = 0.0600 mol. - The one-electron half-reaction gives 0.0600 mol VO₂⁺ reduced. - Time must be converted from minutes to seconds.
(d) - Before and after the same discharge, measure each tank's volume, total vanadium concentration and oxidation-state distribution separately to establish compartment-specific species inventories. - A loss of active vanadium from one compartment accompanied by a gain or misplaced vanadium species in the other supports membrane crossover. - Usable V²⁺ and VO₂⁺ remaining in their correct compartments at the discharge cutoff, without a corresponding transfer between tanks, support incomplete utilisation. - Membrane repair or replacement follows a crossover diagnosis; residual usable reactants instead support testing flow, mixing or discharge-cutoff settings.
Detailed marking criteria
Part (a)(i) (1 mark)
Awards one mark for establishing this question-specific chemical point: The oxidation half-equation is V²⁺ → V³⁺ + e⁻.
Part (a)(ii) (1 mark)
Awards one mark for establishing this question-specific chemical point: The reduction half-equation consumes two H⁺ per electron.
Part (a)(ii) (1 mark)
Awards one mark for establishing this question-specific chemical point: Adding the half-equations conserves charge and electrons.
Part (a)(ii) (1 mark)
Awards one mark for establishing this question-specific chemical point: Electron flow is from the V²⁺ half-cell to the VO₂⁺ half-cell.
Part b (1 mark)
Awards one mark for establishing this question-specific chemical point: V²⁺ is oxidised to V³⁺ at the negative electrode during discharge.
Part b (1 mark)
Awards one mark for establishing this question-specific chemical point: VO₂⁺ is reduced to VO²⁺ at the positive electrode during discharge.
Part b (1 mark)
Awards one mark for establishing this question-specific chemical point: One mole of electrons is released per mole of V²⁺ oxidised and one mole of electrons is consumed per mole of VO₂⁺ reduced.
Part b (1 mark)
Awards one mark for establishing this question-specific chemical point: Electrons flow from the negative electrode through the external circuit to the positive electrode.
Part c (1 mark)
Awards one mark for establishing this question-specific chemical point: Charge Q = It = 2.50 × (38.6 × 60) = 5790 C.
Part c (1 mark)
Awards one mark for establishing this question-specific chemical point: n(e⁻) = 5790/96485 = 0.0600 mol.
Part c (1 mark)
Awards one mark for establishing this question-specific chemical point: The one-electron half-reaction gives 0.0600 mol VO₂⁺ reduced.
Part c (1 mark)
Awards one mark for establishing this question-specific chemical point: Time must be converted from minutes to seconds.
Part d (1 mark)
Awards one mark for establishing this question-specific chemical point: Before and after the same discharge, measure each tank's volume, total vanadium concentration and oxidation-state distribution separately to establish compartment-specific species inventories.
Part d (1 mark)
Awards one mark for establishing this question-specific chemical point: A loss of active vanadium from one compartment accompanied by a gain or misplaced vanadium species in the other supports membrane crossover.
Part d (1 mark)
Awards one mark for establishing this question-specific chemical point: Usable V²⁺ and VO₂⁺ remaining in their correct compartments at the discharge cutoff, without a corresponding transfer between tanks, support incomplete utilisation.
Part d (1 mark)
Awards one mark for establishing this question-specific chemical point: Membrane repair or replacement follows a crossover diagnosis; residual usable reactants instead support testing flow, mixing or discharge-cutoff settings.
Question 7
(a) - Struvite precipitation transfers dissolved phosphate into a solid product. - The one-to-one formula links phosphate moles to struvite moles. - Magnesium availability, pH, phosphate speciation, mixing and competing ions affect struvite recovery.
(b) - Precipitation occurs when the ionic product exceeds the solubility threshold. - Increasing suitable magnesium ion concentration can promote nucleation. - Controlled pH changes phosphate speciation and struvite formation. - Filtration separates the solid from treated water. - Wash the recovered solid and assay its phosphorus, magnesium and contaminant content before claiming struvite purity.
(c) - Recovered phosphate = 0.850 × 0.0200 = 0.0170 mol. - The stoichiometric struvite amount is also 0.0170 mol. - m(struvite) = 0.0170 × 245.4 = 4.17 g. - Unrecovered phosphate = 0.0200-0.0170 = 0.00300 mol.
(d) - Adopt struvite recovery only when the separated solid meets phosphorus-availability, contaminant and purity requirements for land application. - Use plant-available phosphorus delivered to soil as the functional unit. - Measure heavy metals, pathogens, crop uptake and runoff in a field trial. - Include magnesium reagent, aeration, pumping and dewatering in the lifecycle comparison.
Detailed marking criteria
Part a (1 mark)
Awards one mark for establishing this question-specific chemical point: Struvite precipitation transfers dissolved phosphate into a solid product.
Part a (1 mark)
Awards one mark for establishing this question-specific chemical point: The one-to-one formula links phosphate moles to struvite moles.
Part a (1 mark)
Awards one mark for establishing this question-specific chemical point: Magnesium availability, pH, phosphate speciation, mixing and competing ions affect struvite recovery.
Part b (1 mark)
Awards one mark for establishing this question-specific chemical point: Precipitation occurs when the ionic product exceeds the solubility threshold.
Part b (1 mark)
Awards one mark for establishing this question-specific chemical point: Increasing suitable magnesium ion concentration can promote nucleation.
Part b (1 mark)
Awards one mark for establishing this question-specific chemical point: Controlled pH changes phosphate speciation and struvite formation.
Part b (1 mark)
Awards one mark for establishing this question-specific chemical point: Filtration separates the solid from treated water.
Part b (1 mark)
Awards one mark for establishing this question-specific chemical point: Wash the recovered solid and assay its phosphorus, magnesium and contaminant content before claiming struvite purity.
Part c (1 mark)
Awards one mark for establishing this question-specific chemical point: Recovered phosphate = 0.850 × 0.0200 = 0.0170 mol.
Part c (1 mark)
Awards one mark for establishing this question-specific chemical point: The stoichiometric struvite amount is also 0.0170 mol.
Part c (1 mark)
Awards one mark for establishing this question-specific chemical point: m(struvite) = 0.0170 × 245.4 = 4.17 g.
Part c (1 mark)
Awards one mark for establishing this question-specific chemical point: Unrecovered phosphate = 0.0200-0.0170 = 0.00300 mol.
Part d (1 mark)
Awards one mark for establishing this question-specific chemical point: Adopt struvite recovery only when the separated solid meets phosphorus-availability, contaminant and purity requirements for land application.
Part d (1 mark)
Awards one mark for establishing this question-specific chemical point: Use plant-available phosphorus delivered to soil as the functional unit.
Part d (1 mark)
Awards one mark for establishing this question-specific chemical point: Measure heavy metals, pathogens, crop uptake and runoff in a field trial.
Part d (1 mark)
Awards one mark for establishing this question-specific chemical point: Include magnesium reagent, aeration, pumping and dewatering in the lifecycle comparison.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Topic 3 - Biological chemistry and biomolecules | Q5 | 15 | ___ | Review carbohydrates, triglycerides, amino acids, proteins, hydrolysis, condensation and biological structure-function links. |
| Topic 3 - Organic structure and reactions | Q4 | 14 | ___ | Review functional groups, naming, structure-property relationships, reactions, synthesis and polymer chemistry. |
| Topic 4 - Electrochemistry and redox applications | Q6 | 16 | ___ | Review oxidation states, half-equations, cell conventions, electron flow, electrolysis and redox applications. |
| Topic 4 - Materials, resources and sustainability | Q7 | 16 | ___ | Review energy, water, soil, materials, resource recovery, functional units and evidence-based sustainability judgements. |