Skill Align

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

A calculator may be used. The SACE Chemistry data sheet and periodic table are reproduced as source-controlled reference pages 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 marks

Stimulus

An algal oil is converted to a diol and then reacted with butanedioic acid 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.

Use the supplied algal polyester route evidence to answer all parts of this question.

(a) 3 marks

Write or draw the structural or reaction representation relevant to the algal polyester route. Identify the functional-group or biomolecular change and explain the resulting chemical behaviour.

(b)(i) 1 mark

State the principal measured pattern in the supplied algal polyester route data and identify the measurements used.

(b)(ii) 2 marks

Interpret what those observations establish about the algal polyester route, including any stated threshold, control, uncertainty, or chemical basis.

(c) 4 marks

Calculate the percentage yield of purified polyester from the actual and theoretical masses.

(d)(i) 2 marks

State what the supplied result establishes about the algal polyester route, then identify one limitation of using that result alone to infer reaction completeness, composition, or product identity.

(d)(ii) 2 marks

Propose a specific analytical measurement or controlled trial for the algal polyester route, and explain how its result would test the stated chemical interpretation.

Answer

(a) - A suitable repeat-unit representation is [-O-R-O-C(=O)-CH₂-CH₂-C(=O)-]ₙ, where R is the algal-derived diol residue. - The displayed -O-C(=O)- groups are the ester linkages in the polymer backbone. - The carboxyl carbon remains bonded to the carbonyl oxygen, while the diol oxygen becomes the single-bond ester oxygen.

(b)(i) - A diol has two hydroxyl functional groups.

(b)(ii) - Butanedioic acid has two carboxyl functional groups. - Condensation between -OH and -COOH forms an ester link.

(c) - Percentage yield = 8.64/10.8 × 100 = 80.0%. - The purified mass is used as the actual yield. - The theoretical mass is the denominator in percentage yield. - Both masses use grams, so units cancel in the ratio.

(d)(i) - The 80.0% isolated yield alone does not establish environmental superiority. - Algal feedstock may reduce fossil carbon demand but cultivation uses energy and nutrients.

(d)(ii) - Compare cradle-to-grave impacts per kilogram of usable polymer. - Measure durability, recyclability and biodegradation under relevant conditions.

Question 5

15 marks

Stimulus

A 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.

Use the supplied grape-sugar fermentation evidence to answer all parts of this question.

(a) 4 marks

Write or draw the structural or reaction representation relevant to the grape-sugar fermentation. Identify the functional-group or biomolecular change and explain the resulting chemical behaviour.

(b)(i) 1 mark

State one structural feature of glucose and the mole ratio between glucose and ethanol in the supplied fermentation equation.

(b)(ii) 1 mark

Explain how the functional groups, enzyme-catalysed conditions and balanced equation describe glucose fermentation.

(c) 5 marks

Calculate the amount, in moles, and mass, in grams, of ethanol formed when 75.0% of the glucose in the batch is consumed.

(d)(i) 2 marks

State what the supplied result establishes about the grape-sugar fermentation, then identify one limitation of using that result alone to infer reaction completeness, composition, or product identity.

(d)(ii) 2 marks

Propose a specific analytical measurement or controlled trial for the grape-sugar fermentation, and explain how its result would test the stated chemical interpretation.

Answer

(a) - 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. - Enzymes lower activation energy without changing the displayed overall stoichiometry.

(b)(i) - Glucose is a monosaccharide containing multiple hydroxyl groups.

(b)(ii) - The aldehyde form and cyclic hemiacetal forms interconvert in solution.

(c) - 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.

(d)(i) - The calculation predicts a theoretical ethanol amount for consumed glucose. - Incomplete collection or evaporation would lower measured ethanol recovery.

(d)(ii) - Replicate batches distinguish process variability from a single anomalous run. - Monitor temperature and pH because both affect enzyme activity.

Question 6

16 marks

Stimulus

A 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.

Use the supplied vanadium flow-cell evidence and question-specific visual to answer all parts of this question.

Diagram Preview

(a)(i) 1 mark

Write or draw the chemical equation, structural formula, half-equation, equilibrium expression, or process representation relevant to the vanadium flow-cell.

(a)(ii) 3 marks

Use that representation to explain the chemical behaviour and measured outcome in the vanadium flow-cell.

(b) 4 marks

Identify the oxidation and reduction processes, electron stoichiometry, electrode changes, and charge-transfer evidence in the supplied vanadium flow-cell context.

(c) 4 marks

Calculate the charge transferred and the amount, in moles, of VO₂⁺ reduced during the 38.6-minute discharge.

(d) 4 marks

Assess the predicted vanadium flow-cell outcome. Identify one electrochemical assumption or side reaction and state a measurement that would test current efficiency, product amount, or cell performance.

Answer

(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. - Electrons leave the negative electrode through the external circuit. - VO₂⁺ is reduced to VO²⁺ at the positive electrode. - The membrane permits ionic charge balance while limiting bulk mixing.

(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) - The charge calculation assumes 100% current efficiency. - Side reactions such as hydrogen evolution would reduce vanadium conversion. - Measure vanadium oxidation states before and after charging to test efficiency. - Using the same element on both sides reduces permanent cross-contamination effects.

Question 7

16 marks

Stimulus

A 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.

Use the supplied phosphorus-recovery trial evidence to answer all parts of this question.

(a) 4 marks

Explain how phosphate is transferred into struvite, identify the one-to-one phosphate-to-product relationship, and state the chemical or operational factors that affect recovery.

(b) 4 marks

Explain how magnesium availability, phosphate speciation and pH control struvite precipitation, then outline how the solid is separated and checked for purity. No unsupported net-ionic equation is required.

(c) 4 marks

Calculate the mass of struvite recovered and the amount, in moles, of phosphate left unrecovered.

(d) 4 marks

Make a justified resource or treatment choice for the phosphorus-recovery trial using the stated functional outcome, then evaluate one omitted lifecycle, quality, or field-performance factor.

Answer

(a) - Struvite precipitation transfers dissolved phosphate into a solid product. - The one-to-one formula links phosphate moles to struvite moles. - Recovery reduces phosphate discharge that can promote eutrophication. - Magnesium availability, pH and competing ions affect precipitation.

(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.

(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) - The relevant functional unit is plant-available phosphorus delivered to soil. - Benefits include nutrient recovery and reduced eutrophication risk. - Costs include magnesium reagent, aeration, pumping and dewatering energy. - Heavy metals or pathogens must be measured before land application.

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

(b)(i) - A diol has two hydroxyl functional groups.

(b)(ii) - Butanedioic acid has two carboxyl functional groups. - Condensation between -OH and -COOH forms an ester link.

(d)(i) - The 80.0% isolated yield alone does not establish environmental superiority. - Algal feedstock may reduce fossil carbon demand but cultivation uses energy and nutrients.

(d)(ii) - Compare cradle-to-grave impacts per kilogram of usable polymer. - Measure durability, recyclability and biodegradation under relevant conditions.

Detailed marking criteria

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: A suitable repeat-unit representation is [-O-R-O-C(=O)-CH₂-CH₂-C(=O)-]ₙ, where R is the algal-derived diol residue.

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: The displayed -O-C(=O)- groups are the ester linkages in the polymer backbone.

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: The carboxyl carbon remains bonded to the carbonyl oxygen, while the diol oxygen becomes the single-bond ester oxygen.

Part (b)(i) (1 mark)

Awards one mark for establishing this question-specific chemical point: A diol has two hydroxyl functional groups.

Part (b)(ii) (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 c (1 mark)

Awards one mark for establishing this question-specific chemical point: Percentage yield = 8.64/10.8 × 100 = 80.0%.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: The purified mass is used as the actual yield.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: The theoretical mass is the denominator in percentage yield.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: Both masses use grams, so units cancel in the ratio.

Part (d)(i) (1 mark)

Awards one mark for establishing this question-specific chemical point: The 80.0% isolated yield alone does not establish environmental superiority.

Part (d)(i) (1 mark)

Awards one mark for establishing this question-specific chemical point: Algal feedstock may reduce fossil carbon demand but cultivation uses energy and nutrients.

Part (d)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: Compare cradle-to-grave impacts per kilogram of usable polymer.

Part (d)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: Measure durability, recyclability and biodegradation under relevant conditions.

Question 5

(b)(i) - Glucose is a monosaccharide containing multiple hydroxyl groups.

(b)(ii) - The aldehyde form and cyclic hemiacetal forms interconvert in solution.

(d)(i) - The calculation predicts a theoretical ethanol amount for consumed glucose. - Incomplete collection or evaporation would lower measured ethanol recovery.

(d)(ii) - Replicate batches distinguish process variability from a single anomalous run. - Monitor temperature and pH because both affect enzyme activity.

Detailed marking criteria

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: The required reaction representation is C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g).

Part a (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 a (1 mark)

Awards one mark for establishing this question-specific chemical point: Fermentation includes coupled oxidation and reduction steps mediated by yeast enzymes.

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: Enzymes lower activation energy without changing the displayed overall stoichiometry.

Part (b)(i) (1 mark)

Awards one mark for establishing this question-specific chemical point: Glucose is a monosaccharide containing multiple hydroxyl groups.

Part (b)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: The aldehyde form and cyclic hemiacetal forms interconvert in solution.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: Initial glucose mass = 180 × 0.500 = 90.0 g.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: Consumed glucose mass = 0.750 × 90.0 = 67.5 g.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: n(glucose) = 67.5/180.16 = 0.3747 mol.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: n(ethanol) = 2 × 0.3747 = 0.749 mol.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: Using M(ethanol) = 46.07 gives 34.5 g ethanol.

Part (d)(i) (1 mark)

Awards one mark for establishing this question-specific chemical point: The calculation predicts a theoretical ethanol amount for consumed glucose.

Part (d)(i) (1 mark)

Awards one mark for establishing this question-specific chemical point: Incomplete collection or evaporation would lower measured ethanol recovery.

Part (d)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: Replicate batches distinguish process variability from a single anomalous run.

Part (d)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: Monitor temperature and pH because both affect enzyme activity.

Question 6

(a)(i) - The oxidation half-equation is V²⁺ → V³⁺ + e⁻.

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: Electrons leave the negative electrode through the external circuit.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: VO₂⁺ is reduced to VO²⁺ at the positive electrode.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: The membrane permits ionic charge balance while limiting bulk mixing.

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: The charge calculation assumes 100% current efficiency.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Side reactions such as hydrogen evolution would reduce vanadium conversion.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Measure vanadium oxidation states before and after charging to test efficiency.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Using the same element on both sides reduces permanent cross-contamination effects.

Question 7

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: Recovery reduces phosphate discharge that can promote eutrophication.

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: Magnesium availability, pH and competing ions affect precipitation.

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 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: The relevant functional unit is plant-available phosphorus delivered to soil.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Benefits include nutrient recovery and reduced eutrophication risk.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Costs include magnesium reagent, aeration, pumping and dewatering energy.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Heavy metals or pathogens must be measured before land application.

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.

SACE Chemistry Stage 2 — Booklet 2 — Free Online Pack 0

Exam-pack paper structure

Read SACE Chemistry Stage 2 — Booklet 2 online

Skill Align SACE Stage 2 Chemistry - Free Online Pack 0 Booklet 2

Booklet 2 - Questions 4 to 7

Question 4

Question 5

Question 6

Question 7

Worked Solutions And Marking Guide

General marking principles

Question 4

Detailed marking criteria

Question 5

Detailed marking criteria

Question 6

Detailed marking criteria

Question 7

Detailed marking criteria

Diagnostic Checklist

What is included

Independent practice resource

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