How many NESA Chemistry questions cover Processing Data and Information?

AusGrader has 133 NESA Chemistry questions on Processing Data and Information, all with instant AI grading and detailed marking feedback.

HSC Chemistry — Processing Data and Information Questions & Answers

Q23

2020

VCAA

1 mark

Q23

1 mark

Use the following information to answer the question.

A solution of citric acid, $\text{C}_3\text{H}_5\text{O(COOH)}_3$ , was analysed by titration.
25.0 mL aliquots of the $\text{C}_3\text{H}_5\text{O(COOH)}_3$ solution were titrated against a standardised solution of 0.0250 M sodium hydroxide, NaOH. Phenolphthalein indicator was used and the average titre was found to be 24.0 mL.

Based on the titration, the concentration of $\text{C}_3\text{H}_5\text{O(COOH)}_3$ in the solution was

A

$8.0 \times 10^{-3} \text{ M}$

B

$8.7 \times 10^{-3} \text{ M}$

C

$2.6 \times 10^{-2} \text{ M}$

D

$7.2 \times 10^{-2} \text{ M}$

Reveal Answer

A

$8.0 \times 10^{-3} \text{ M}$

Correct Answer

Citric acid is triprotic, reacting with NaOH in a 1:3 ratio. The moles of NaOH used is $0.0250 \times 0.0240 = 6.0 \times 10^{-4}$ mol, so the moles of acid is $2.0 \times 10^{-4}$ mol. Dividing by the aliquot volume (0.0250 L) gives $8.0 \times 10^{-3}$ M.

B

$8.7 \times 10^{-3} \text{ M}$

This incorrect answer is obtained by mistakenly swapping the volumes of the acid and base in the concentration calculation.

C

$2.6 \times 10^{-2} \text{ M}$

This error occurs if the volumes of the acid and base are swapped and the 1:3 stoichiometric ratio between citric acid and NaOH is ignored.

D

$7.2 \times 10^{-2} \text{ M}$

This incorrect result comes from multiplying the moles of NaOH by 3 instead of dividing by 3 to find the moles of the triprotic citric acid.

Q5

2022

VCAA

1 mark

Q5

1 mark

Scientists often repeat trials of an experiment using the same experimental method and the same equipment.

Which one attribute of experimental data will be improved when there is an increase in the number of times that a trial is repeated?

A

bias

B

validity

C

accuracy

D

reliability

Reveal Answer

A

bias

Bias is a systematic error that consistently skews results in one direction. Repeating trials with the same method and equipment will simply reproduce the bias, not improve it.

B

validity

Validity refers to how well an experiment measures what it actually intends to measure. Repeating the exact same procedure does not change or improve its validity.

C

accuracy

Accuracy is how close a measurement is to the true or accepted value. If the equipment is miscalibrated, repeating trials will not make the average result any closer to the true value.

D

reliability

Correct Answer

Reliability refers to the consistency of experimental results. Increasing the number of trials reduces the impact of random errors and outliers, thereby improving the reliability of the data.

Q39

2025

SCSA

14 marks

Q39

14 marks

Freon-11 is a colourless chlorofluorocarbon that boils at 23.77 °C. Prior to the knowledge of the ozone-depleting potential of chlorofluorocarbons (CFCs) and other possible harmful effects on the environment, it was used as a refrigerant.

The following data was used to determine that Freon-11 is trichlorofluoromethane, with a molecular formula of $CCl_3F$ .

A Freon-11 sample of 4.121 g was combusted in excess oxygen. All the carbon in the compound was converted to carbon dioxide and in a separate process, all its chlorine was converted into hydrochloric acid. The carbon dioxide produced had a mass of 1.320 g and the hydrochloric acid formed, required 85.70 mL of 1.050 mol L $^{-1}$ of ammonia solution for complete neutralisation.
Another sample of the Freon-11 with a mass of 3.721 g occupied a volume of 0.6068 L at a pressure of 120.00 kPa and temperature of 50.6 °C.

Using the same data, use calculations and reasoning to demonstrate that this is the correct molecular formula.

Reveal Answer

Carbon
$n(C) = n(CO_2) = 1.320/44.01 = 0.02999$ mol
$m(C) = 0.02299 \times 12.01 = 0.3602$ g

Chlorine
$n(Cl) = n(HCl) = n(NH_3) = 1.050 \times 0.08570 = 0.08999$ mol
$m(Cl) = 0.08999 \times 35.45 = 3.1900$ g

Fluorine
$m(F) = 4.121 - (0.3602 + 3.1900) = 0.5708$ g
$n(F) = 0.5708 / 19.00 = 0.03004$ mol

Mole Ratio
C: 0.02999
Cl: 0.08999
F: 0.03004

Simplify
Divide by 0.02999
C: 1
Cl: 3.00
F: 1.002

Empirical Formula
$CCl_3F$

Molecular Formula
Empirical formula mass (EFM) = 137.36 amu/g mol $^{-1}$
$n = PV/RT = (120.0 \times 0.6068)/(8.314 \times 323.75) = 0.027052476$ mol
Molecular formula mass (MFM) = $m/n = 3.721/0.0276503204 = 137.54$ g mol $^{-1}$
Molecular formula = MFM/EFM $\times$ Empirical formula = $137.54/137.36 \times CCl_3F = 1.0013 \times CCl_3F$

Molecular formula = Empirical formula = $CCl_3F$

Marking Criteria

Descriptor	Marks
n(C) calculation	1
m(C) calculation	1
n(Cl) calculation	1
m(Cl) calculation	1
m(F) calculation	1
n(F) calculation	1
Mole Ratio setup	1
Simplify ratio	1
Empirical Formula	1
Empirical formula mass (EFM)	1
n = PV/RT calculation	1
Molecular formula mass (MFM) calculation	1
Molecular formula calculation	1
Molecular formula = Empirical formula statement	1

Q39

2020

SCSA

12 marks

Q39

Fluorescent lights are glass tubes which are coated on the inside with rare earth metal phosphates (such as cerium, lanthanum and terbium phosphates) that provide light. Cerium, lanthanum and terbium are expensive, so are recovered once the fluorescent light is no longer functional.

The key steps in one method proposed for recovery of these rare earth metals are summarised below:

Step 1: Physical separation of the rare earth metal phosphates from the glass and any metallic components. This gives an impure powder consisting of cerium, lanthanum and terbium phosphates.
Step 2: Add excess solid sodium carbonate to the powder and heat, completely converting each rare earth metal phosphate to its corresponding oxide, as shown by the following balanced equations:
$2 \text{ LaPO}_4\text{(s)} + 3 \text{ Na}_2\text{CO}_3\text{(s)} \rightarrow \text{La}_2\text{O}_3\text{(s)} + 2 \text{ Na}_3\text{PO}_4\text{(s)} + 3 \text{ CO}_2\text{(g)}$
$4 \text{ CePO}_4\text{(s)} + 6 \text{ Na}_2\text{CO}_3\text{(s)} + \text{O}_2\text{(g)} \rightarrow 4 \text{ CeO}_2\text{(s)} + 4 \text{ Na}_3\text{PO}_4\text{(s)} + 6 \text{ CO}_2\text{(g)}$
$2 \text{ TbPO}_4\text{(s)} + 3 \text{ Na}_2\text{CO}_3\text{(s)} \rightarrow \text{Tb}_2\text{O}_3\text{(s)} + 2 \text{ Na}_3\text{PO}_4\text{(s)} + 3 \text{ CO}_2\text{(g)}$
Step 3: Wash the product from Step 2 with water.
Step 4: Add hydrochloric acid to the washed product from Step 3 to leach (dissolve) only the rare earth metal oxides.
Step 5: Use solvent extraction to separate the different rare earth metals from each other and create separate solutions of each of them.
Step 6: Add oxalic acid to the separated solutions to precipitate the rare earth metal ions as oxalate salts.
Step 7: Heat the oxalate salts to recover the rare earth metals as pure oxides, namely $\text{La}_2\text{O}_3$ , $\text{Tb}_4\text{O}_7$ and $\text{CeO}_2$ .

A chemist used the above procedure to determine the percentage by mass of lanthanum, terbium and cerium in some fluorescent lights and, after completing Step 1, had recovered 1.20 kg of the coating chemicals.

Q39b

The mass of the solid sent from Step 3 to Step 4 was 1.16 kg. This solid was leached with 6.00 mol L $^{-1}$ HCl at a solid to liquid ratio of 150 g per litre. Analysis of the solution at the end of leaching showed that it contained lanthanum, terbium and cerium, with its lanthanum concentration being $8.65 \times 10^{-3} \text{ mol L}^{-1}$ .

Q39a

3 marks

At the completion of Step 2, the mass of the mixture had decreased by 11.3 g. Calculate the mass of sodium carbonate that reacted with the rare earth metal phosphates.

Reveal Answer

n(CO₂) = 11.3/44.01 = 0.257 mol

n(CO₂) = n(C) = n(Na₂CO₃) = 0.257 mol

m(Na₂CO₃) = 0.257 × 105.99 = 27.2 g

Marking Criteria

Descriptor	Marks
n(CO₂) = 11.3/44.01 = 0.257 mol	1
n(CO₂) = n(C) = n(Na₂CO₃) = 0.257 mol	1
m(Na₂CO₃) = 0.257 × 105.99 = 27.2 g	1

Q39b

5 marks

Calculate the percentage, by mass, of lanthanum in the fluorescent light coating chemical, given that the leaching efficiency for lanthanum was 86%.

Note that the balanced equation for the leaching of lanthanum with hydrochloric acid is:
$\text{La}_2\text{O}_3\text{(s)} + 6 \text{ HCl(aq)} \rightarrow 2 \text{ LaCl}_3\text{(aq)} + 3 \text{ H}_2\text{O(l)}$

Reveal Answer

Volume of HCl used: 1160/150 = 7.73 L

n(La) = cV = 8.65 × 10⁻³ × 7.73 = 0.0669 mol La in solution

Taking into account the leaching efficiency,
n(La in the solid that was leached) = 0.0669/0.86 = 0.0778

m(La in the solid that was leached) = 0.0778 × 138.9 = 10.8 g

% La in the coating chemical = (10.8/1200) × 100 = 0.900%

Marking Criteria

Descriptor	Marks
Volume of HCl used: 1160/150 = 7.73 L	1
n(La) = cV = 8.65 × 10⁻³ × 7.73 = 0.0669 mol La in solution	1
Taking into account the leaching efficiency, n(La in the solid that was leached) = 0.0669/0.86 = 0.0778	1
m(La in the solid that was leached) = 0.0778 × 138.9 = 10.8 g	1
% La in the coating chemical = (10.8/1200) × 100 = 0.900%	1

Q39c

4 marks

Analysis of the cerium-containing solution produced in Step 5 showed that its cerium concentration was 0.146 mol L $^{-1}$ . This solution, which had a volume of 424 mL, was added to 110 mL of aqueous 1.15 mol L $^{-1}$ oxalic acid during Step 6, resulting in the precipitation of cerium oxalate, $\text{Ce(C}_2\text{O}_4\text{)}_2$ . The balanced equation for this reaction is:

$\text{CeCl}_4\text{(aq)} + 2 \text{ H}_2\text{C}_2\text{O}_4\text{(aq)} \rightarrow \text{Ce(C}_2\text{O}_4\text{)}_2\text{(s)} + 4 \text{ HCl(aq)}$

Did the chemist add enough oxalic acid solution to precipitate all of the cerium? Use calculations to support your answer.

Reveal Answer

n(cerium) = 0.146 × 0.424 = 0.0619 mol

n(oxalic acid needed to react with cerium) = 2 × 0.0619 = 0.124 mol

n(oxalic acid available) = 0.110 × 1.15 = 0.127 mol

comparison of the moles of oxalic acid shows that enough oxalic acid was added

Marking Criteria

Descriptor	Marks
n(cerium) = 0.146 × 0.424 = 0.0619 mol	1
n(oxalic acid needed to react with cerium) = 2 × 0.0619 = 0.124 mol	1
n(oxalic acid available) = 0.110 × 1.15 = 0.127 mol	1
comparison of the moles of oxalic acid shows that enough oxalic acid was added	1

Q17

2023

VCAA

1 mark

Q17

1 mark

Lignite is a type of brown coal. When lignite is completely combusted in a power station, 19.0 MJ/tonne of energy is released. The efficiency of the power station is 39%.

What mass of lignite is required to produce 42.0 MJ of usable energy in the power station?

A

0.862 tonnes

B

1.16 tonnes

C

2.21 tonnes

D

5.67 tonnes

Reveal Answer

A

0.862 tonnes

This incorrect answer is obtained by calculating $42.0 / (19.0 / 0.39)$ , which incorrectly divides the energy density by the efficiency instead of multiplying them to find the usable energy per tonne.

B

1.16 tonnes

This incorrect answer is obtained by calculating $19.0 / (42.0 \times 0.39)$ , which incorrectly divides the energy density by the required energy instead of dividing the required energy by the usable energy per tonne.

C

2.21 tonnes

This incorrect answer is obtained by calculating $42.0 / 19.0$ , which represents the mass needed if the power station were 100% efficient, completely ignoring the 39% efficiency factor.

D

5.67 tonnes

Correct Answer

The usable energy per tonne is $19.0 \times 0.39 = 7.41 \text{ MJ/tonne}$ . To produce 42.0 MJ of usable energy, the mass required is $42.0 / 7.41 = 5.67 \text{ tonnes}$ .

Q9

2024

VCAA

1 mark

Q9

1 mark

The overall reaction in an alkaline–ethanol fuel cell is shown below.

$\text{C}_2\text{H}_5\text{OH}(\text{l}) + 3\text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) + 3\text{H}_2\text{O}(\text{l})$

What amount of hydroxide ions, $\text{OH}^-$ , reacts with 1 mol of ethanol, $\text{C}_2\text{H}_5\text{OH}$ , at the negative electrode of the fuel cell?

A

4 mol

B

6 mol

C

8 mol

D

12 mol

Reveal Answer

A

4 mol

Incorrect. 4 moles of $\text{OH}^-$ is insufficient to balance the oxygen and hydrogen atoms in the oxidation half-reaction of ethanol to $\text{CO}_2$ .

B

6 mol

Incorrect. While ethanol contains 6 hydrogen atoms, balancing the full oxidation half-reaction to $\text{CO}_2$ and $\text{H}_2\text{O}$ requires 12 moles of $\text{OH}^-$ .

C

8 mol

Incorrect. 8 moles of $\text{OH}^-$ does not properly balance the mass and charge for the complete oxidation of ethanol in an alkaline medium.

D

12 mol

Correct Answer

Correct. The balanced half-equation at the negative electrode (anode) is $\text{C}_2\text{H}_5\text{OH} + 12\text{OH}^- \rightarrow 2\text{CO}_2 + 9\text{H}_2\text{O} + 12\text{e}^-$ , showing that 12 moles of $\text{OH}^-$ react with 1 mole of ethanol.

Q22

2020

VCAA

1 mark

Q22

1 mark

The combustion of which fuel provides the most energy per 100 g?

A

pentane ( $\text{M} = 72 \text{ g mol}^{-1}$ ), which releases $49\,097 \text{ MJ tonne}^{-1}$

B

nitromethane ( $\text{M} = 61 \text{ g mol}^{-1}$ ), which releases $11.63 \text{ kJ g}^{-1}$

C

butanol ( $\text{M} = 74 \text{ g mol}^{-1}$ ), which releases $2670 \text{ kJ mol}^{-1}$

D

ethyne ( $\text{M} = 26 \text{ g mol}^{-1}$ ), which releases $1300 \text{ kJ mol}^{-1}$

Reveal Answer

A

pentane ( $\text{M} = 72 \text{ g mol}^{-1}$ ), which releases $49\,097 \text{ MJ tonne}^{-1}$

Converting $49\,097 \text{ MJ tonne}^{-1}$ to $\text{kJ g}^{-1}$ gives $49.1 \text{ kJ g}^{-1}$ , which yields $4910 \text{ kJ}$ per $100 \text{ g}$ . This is slightly less than the energy provided by ethyne.

B

nitromethane ( $\text{M} = 61 \text{ g mol}^{-1}$ ), which releases $11.63 \text{ kJ g}^{-1}$

An energy release of $11.63 \text{ kJ g}^{-1}$ yields $1163 \text{ kJ}$ per $100 \text{ g}$ , which is the lowest energy output among the given choices.

C

butanol ( $\text{M} = 74 \text{ g mol}^{-1}$ ), which releases $2670 \text{ kJ mol}^{-1}$

Dividing $2670 \text{ kJ mol}^{-1}$ by the molar mass of $74 \text{ g mol}^{-1}$ gives $36.1 \text{ kJ g}^{-1}$ , resulting in $3610 \text{ kJ}$ per $100 \text{ g}$ .

D

ethyne ( $\text{M} = 26 \text{ g mol}^{-1}$ ), which releases $1300 \text{ kJ mol}^{-1}$

Correct Answer

Dividing $1300 \text{ kJ mol}^{-1}$ by the molar mass of $26 \text{ g mol}^{-1}$ gives $50 \text{ kJ g}^{-1}$ . This yields $5000 \text{ kJ}$ per $100 \text{ g}$ , which is the highest energy output of all the options.

Q23

2022

QCAA

Paper 1

4 marks

Q23

4 marks

Ibuprofen is manufactured using two different processes.

Process	Number of reagents used	Reagents	Reagents	Ibuprofen	Ibuprofen	Waste products	Waste products
		Atoms	$\text{M}_\text{r}$	Atoms	$\text{M}_\text{r}$	Atoms	$\text{M}_\text{r}$
1	7	$\text{C}_{20}\text{H}_{42}\text{NO}_{10}\text{ClNa}$	514.5	$\text{C}_{13}\text{H}_{18}\text{O}_2$	206.0	$\text{C}_7\text{H}_{24}\text{NO}_8\text{ClNa}$	308.5
2	4	$\text{C}_{15}\text{H}_{22}\text{O}_4$	266.0	$\text{C}_{13}\text{H}_{18}\text{O}_2$	206.0	$\text{C}_2\text{H}_4\text{O}_2$	60.0

Calculate the atom economy for each process and draw conclusions about the economic and environmental impact of each process.

Reveal Answer

Process 1: atom economy = $206.0/514.5 \times 100 = 40.04\%$
Process 2: atom economy = $206.0/266.0 \times 100 = 77.44\%$
Process 2 has 37.4% better atom economy than process 1

Economic impact: Process 2 has a better atom economy than process 1 (fewer reagents are required).
Environmental impact: Process 2 is greener than process 1 because fewer waste products (atoms) are produced.

Marking Criteria

Descriptor	Marks
Calculates atom economy for Process 1 as 40%	1
Calculates atom economy for Process 2 as 77%	1
Concludes process 2 is cheaper as fewer reagent atoms are required	1
Concludes process 2 is greener as fewer waste atoms are produced	1

Q26

2024

VCAA

1 mark

Q26

1 mark

Use the following information to answer the question.

A chemist runs a mixture of hexane, hexan-1-ol and hexan-2-one through a high-performance liquid chromatography (HPLC) column using a polar mobile phase and a non-polar stationary phase.

The chemist wants to determine the concentration of hexane in the mixture.

Which one of the following will provide information to allow the hexane concentration to be accurately calculated?

A

running a series of known concentrations of hexane through the HPLC column under the same conditions

B

running the HPLC experiment using a non-polar mobile phase and a polar stationary phase

C

using published retention times and peak sizes of standard hexane chromatographs

D

reducing the HPLC column temperature to achieve better separation of the compounds

Reveal Answer

A

running a series of known concentrations of hexane through the HPLC column under the same conditions

Correct Answer

To accurately determine concentration, a calibration curve must be constructed by running standard solutions of known concentrations under the exact same experimental conditions to compare peak areas.

B

running the HPLC experiment using a non-polar mobile phase and a polar stationary phase

Changing the polarity of the mobile and stationary phases alters the separation method (switching to normal phase chromatography) but does not provide the quantitative reference data needed to calculate concentration.

C

using published retention times and peak sizes of standard hexane chromatographs

Retention times and peak sizes are highly dependent on the specific instrument, column age, and exact experimental conditions, so published data cannot be reliably used for quantitative analysis.

D

reducing the HPLC column temperature to achieve better separation of the compounds

While reducing the temperature might improve the separation (resolution) of the peaks, it does not provide the reference standards required to calculate the actual concentration of the compound.

Q22

2021

VCAA

1 mark

Q22

1 mark

1 L of octane has a mass of 703 g at SLC. The efficiency of the reaction when octane undergoes combustion in the petrol engine of a car is 25.0%.

What volume of octane stored in a petrol tank at SLC is required to produce 528 MJ of usable energy in a combustion engine?

A

3.92 L

B

11.8 L

C

15.7 L

D

62.7 L

Reveal Answer

A

3.92 L

This result comes from incorrectly multiplying the theoretical volume by the efficiency ( $15.7 \text{ L} \times 0.25$ ), rather than dividing by it to account for the extra fuel needed due to energy loss.

B

11.8 L

This value is obtained by incorrectly multiplying the theoretical volume by $0.75$ (or $100\% - 25\%$ ), which is an incorrect application of the efficiency percentage.

C

15.7 L

This is the volume of octane required if the engine were $100\%$ efficient ( $528 \text{ MJ} / 33.67 \text{ MJ/L}$ ). It fails to account for the $25.0\%$ efficiency of the engine, which requires more fuel to be burned.

D

62.7 L

Correct Answer

The total energy needed is $528 \text{ MJ} / 0.25 = 2112 \text{ MJ}$ . With an energy density of $33.67 \text{ MJ/L}$ (calculated from octane's heat of combustion of $5460 \text{ kJ/mol}$ , molar mass of $114.0 \text{ g/mol}$ , and density of $703 \text{ g/L}$ ), the required volume is $2112 \text{ MJ} / 33.67 \text{ MJ/L} = 62.7 \text{ L}$ .

Q12

2021

VCAA

1 mark

Q12

1 mark

Butane, $\text{C}_4\text{H}_{10}$ , undergoes complete combustion according to the following equation.

2\text{C}_4\text{H}_{10}\text{(g)} + 13\text{O}_2\text{(g)} \rightarrow 8\text{CO}_2\text{(g)} + 10\text{H}_2\text{O(g)}

67.0 g of $\text{C}_4\text{H}_{10}$ released 3330 kJ of energy during complete combustion at standard laboratory conditions (SLC).

The mass of carbon dioxide, $\text{CO}_2$ , produced was

A

0.105 g

B

3.18 g

C

50.9 g

D

204 g

Reveal Answer

A

0.105 g

This value is obtained by incorrectly dividing the moles of $\text{CO}_2$ (4.62 mol) by its molar mass (44.0 g/mol) instead of multiplying.

B

3.18 g

This incorrect answer results from a series of calculation errors, likely involving dividing the mass of butane by the product of the molar masses of butane and carbon dioxide.

C

50.9 g

This mass is calculated by incorrectly assuming a 1:1 molar ratio between butane and carbon dioxide, rather than the correct 2:8 ratio given in the balanced equation.

D

204 g

Correct Answer

First, calculate the moles of butane: $67.0 \text{ g} / 58.0 \text{ g/mol} = 1.155 \text{ mol}$ . Using the 2:8 molar ratio from the equation, the moles of $\text{CO}_2$ produced is $1.155 \times 4 = 4.62 \text{ mol}$ . Finally, multiply by the molar mass of $\text{CO}_2$ (44.0 g/mol) to get $4.62 \times 44.0 = 203.3 \text{ g}$ , which rounds to 204 g.

Q27

2023

NESA

4 marks

Q27

4 marks

A student has been asked to produce 185 mL of ethanol ( $MM = 46.068 \text{ g mol}^{-1}$ ) by fermenting glucose using yeast, as shown in the equation.

C_6H_{12}O_6(aq) \rightarrow 2C_2H_5OH(aq) + 2CO_2(g)

Given that the density of ethanol is $0.789 \text{ g mL}^{-1}$ , calculate the volume of carbon dioxide gas produced at 310 K and 100 kPa.

Reveal Answer

Calculate mass of ethanol required from density

$\rho = \frac{m}{V}$

$m(\text{ethanol}) = 0.789\text{ g mL}^{-1} \times 185\text{ mL} = 146\text{ g}$

$n(\text{ethanol}) = \frac{m}{MM} = \frac{146\text{ g}}{46.068\text{ g mol}^{-1}} = 3.17\text{ mol}$

\begin{align*} V &= \frac{nRT}{P}\\ V &= \frac{3.17\text{ mol} \times 8.314\text{ J K}^{-1}\text{ mol}^{-1} \times 310\text{ K}}{100\text{ kPa}}\\ &= 81.7\text{ L} \end{align*}

Marking Criteria

Descriptor	Marks
Calculates the volume of carbon dioxide produced	4
Provides the main steps of the calculation	3
Provides some steps of the calculation	2
Provides some relevant information	1
None of the above	0

Q7

2024

VCAA

1 mark

Q7

1 mark

In a galvanic cell, copper(II) ions, $\text{Cu}^{2+}$ , are converted to copper metal, Cu.

What mass of Cu is deposited for each 0.50 mol of electrons transferred in the cell?

A

$1.6 \times 10 \text{ g}$

B

$3.2 \times 10 \text{ g}$

C

$6.4 \times 10 \text{ g}$

D

$1.3 \times 10^2 \text{ g}$

Reveal Answer

A

$1.6 \times 10 \text{ g}$

Correct Answer

The half-reaction $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ shows that 2 moles of electrons produce 1 mole of Cu. Therefore, 0.50 mol of electrons produces 0.25 mol of Cu, which has a mass of $0.25 \text{ mol} \times 63.55 \text{ g/mol} \approx 16 \text{ g}$ .

B

$3.2 \times 10 \text{ g}$

This answer incorrectly assumes a 1:1 ratio of electrons to copper, which would produce 0.50 mol of Cu ( $32 \text{ g}$ ).

C

$6.4 \times 10 \text{ g}$

This is the mass of 1.0 mole of Cu ( $64 \text{ g}$ ), which would require 2.0 moles of electrons to be transferred, not 0.50 moles.

D

$1.3 \times 10^2 \text{ g}$

This is the mass of 2.0 moles of Cu ( $127 \text{ g}$ ), which would require 4.0 moles of electrons to be transferred.

Q40

2024

SCSA

17 marks

Q40

Nickel mining is one of the largest mining industries in Western Australia. One process for extracting nickel from nickel oxide ore is the Mond process, which was developed at the end of the 19th Century.

Step 1
Nickel(II) oxide is reacted with hydrogen to produce nickel metal and water at 200 °C, according to the equation:

\text{NiO(s)} + \text{H}_2\text{(g)} \rightleftharpoons \text{Ni(s)} + \text{H}_2\text{O(g)}

The nickel produced is impure and must be further purified.

Step 2
The impure nickel is reacted with carbon monoxide to produce nickel carbonyl ( $\text{Ni(CO)}_4$ ).

\text{Ni(s)} + 4\text{CO(g)} \rightleftharpoons \text{Ni(CO)}_4\text{(g)}

The impurities in the nickel are left as solids and separated.

Step 3
The nickel carbonyl gas is then passed over a platinum catalyst, causing the compound to decompose.

\text{Ni(CO)}_4\text{(g)} \rightleftharpoons \text{Ni(s)} + 4\text{CO(g)}

Q40a

5 marks

If the nickel produced in Step 1 is 95.7% pure, calculate the mass of nickel(II) oxide that would be required to produce 2245 tonne of the impure nickel.

Reveal Answer

Convert the mass of nickel from tonnes to grams:
$m(\mathrm{Ni}) = 2245 \times (1 \times 10^{6}) = 2.245 \times 10^{9}\ \mathrm{g}$

Calculate the mass of nickel in the impure sample:
$m(\mathrm{Ni}) = (2.245 \times 10^{9}) \times 95.7/100 = 2.148 \times 10^{9}\ \mathrm{g}$

Calculate the $n(\mathrm{Ni})$
$n(\mathrm{Ni}) = (2.148 \times 10^{9})/58.69 = 3.661 \times 10^{7}\ \mathrm{mol}$

Calculate the $n(\mathrm{NiO})$
$n(\mathrm{Ni}) = n(\mathrm{NiO}) = 3.661 \times 10^{7}\ \mathrm{mol}$

Calculate the mass of $\mathrm{NiO}$ :
$m(\mathrm{NiO}) = (3.661 \times 10^{7}) \times 74.69 = 2.73 \times 10^{9}\ \mathrm{g}$ (2730 tonne)

Marking Criteria

Descriptor	Marks
Convert the mass of nickel from tonnes to grams	1
Calculate the mass of nickel in the impure sample	1
Calculate the n(Ni)	1
Calculate the n(NiO)	1
Calculate the mass of NiO	1

Q40b

4 marks

In Step 2, if the impure nickel mixture contains 521 tonne of pure nickel and produces 1025 tonne of $\text{Ni(CO)}_4$ , calculate the percentage yield of this reaction.

Reveal Answer

$n(\text{Ni}) = (5.21 \times 10^8) / 58.69 = 8.877 \times 10^6 \text{ mol}$ .

The theoretical $n(\text{Ni(CO)}_4) = n(\text{Ni}) = 8.877 \times 10^6 \text{ mol}$ .

The theoretical mass $= 170.73 \times (8.877 \times 10^6 \text{ mol}) = 1.516 \times 10^9 \text{ g}$ .

The percentage yield $= (1.025 \times 10^9) / (1.516 \times 10^9) \times 100 = 67.6\%$ .

Marking Criteria

Descriptor	Marks
Calculates $n(\text{Ni})$ as $8.877 \times 10^6 \text{ mol}$	1
Calculates the theoretical $n(\text{Ni(CO)}_4)$ as $8.877 \times 10^6 \text{ mol}$	1
Calculates the theoretical mass as $1.516 \times 10^9 \text{ g}$	1
Calculates the percentage yield as $67.6\%$	1

Q40c

8 marks

A 19.22 g sample of the impure nickel from Step 1 was reacted with 90.00 mL of a 5.02 mol L⁻¹ solution of nitric acid. This reaction can be represented by the following equation:

3\text{Ni(s)} + 8\text{HNO}_3\text{(aq)} \rightarrow 3\text{Ni(NO}_3\text{)}_2\text{(aq)} + 2\text{NO(g)} + 4\text{H}_2\text{O(l)}

Calculate the volume of nitrogen monoxide produced measured at standard temperature and pressure (STP).

Reveal Answer

Calculate the number of moles of Ni:
$= 19.22 \times 0.957 / 58.69 = 0.3134\ \mathrm{mol}$

Calculate the number of moles nitric acid:
$n(\mathrm{HNO_3}) = 0.0900 \times 5.02 = 0.4518\ \mathrm{mol}$

Determine the limiting reagent:
$n(\mathrm{Ni}) = \frac{3}{8}\,n(\mathrm{HNO_3})$

From $0.4518\ \mathrm{mol}$ of $\mathrm{HNO_3}$ :
$n(\mathrm{Ni}) = \frac{3}{8}(0.4518) = 0.1694\ \mathrm{mol}$ is produced

$n(\mathrm{HNO_3\ required}) < n(\mathrm{HNO_3\ available})$
Therefore, $\mathrm{HNO_3}$ is limiting reagent.

Calculate the number of moles of NO:
$n(\mathrm{NO}) = \frac{2}{8}n(\mathrm{HNO_3}) = 0.4518/4 = 0.1129$

Calculate the volume of NO:
$V(\mathrm{NO}) = 22.71 \times 0.1129 = 2.57\ \mathrm{L}$

Marking Criteria

Descriptor	Marks
Calculate the number of moles of Ni	2
Calculate the number of moles nitric acid	1
Determine the limiting reagent	3
Calculate the number of moles of NO	1
Calculate the volume of NO	1

Q33

2025

NESA

7 marks

Q33

7 marks

Chalk is predominantly calcium carbonate. Different brands of chalk vary in their calcium carbonate composition.

The table shows the composition of three different brands of chalk.

	Brand X	Brand Y	Brand Z
$\text{CaCO}_3$ (%)	85.5	83.9	82.4

The following procedure was used to determine the calcium carbonate composition of a chalk sample.

A sample of chalk was crushed in a mortar and pestle.
A 3.00 g sample of the crushed chalk was placed in a conical flask.
100.0 mL of $0.550 \text{ mol L}^{-1} \text{ HCl}(aq)$ was added to the sample and left to react completely, resulting in a clear solution.
Four 20 mL aliquots of this mixture were then titrated with $0.10 \text{ mol L}^{-1} \text{ KOH}$ .

The results of the titrations are recorded.

Burette volume (mL)	Trial 1	Trial 2	Trial 3	Trial 4
Final	7.80	14.90	22.10	29.25
Initial	0.00	7.80	14.90	22.10
Total used	7.80	7.10	7.20	7.15

Determine the brand of the chalk sample. Include a relevant chemical equation in your answer.

Reveal Answer

Exclude the outlier.

Average volume KOH used excluding Trial 1 = $\frac{7.10 + 7.20 + 7.15}{3} = 0.00715$ L

HCl( $aq$ ) + KOH( $aq$ ) $\rightarrow$ KCl( $aq$ ) + H $_2$ O( $l$ )

moles KOH = $0.10 \times 0.00715 = 0.000715$ mol KOH

Ratio HCl : KOH = 1 : 1
$\therefore$ 0.000715 mol HCl for each sample
$\therefore$ $0.000715 \times 5 = 0.003575$ mol total in sampled solution

Initial $n$ HCl = $0.550 \times 0.1000 = 0.0550$ mol HCl

$n$ HCl that reacted with CaCO $_3$ = $0.0550 - 0.003575$ = 0.051425 mol HCl

2HCl( $aq$ ) + CaCO $_3$ ( $s$ ) $\rightarrow$ CaCl $_2$ ( $aq$ ) + H $_2$ O( $l$ ) + CO $_2$ ( $g$ )

Ratio HCl : CaCO $_3$ = 2 : 1
$\therefore$ $\frac{0.051425}{2} = 0.0257125$ mol CaCO $_3$

Mass CaCO $_3$ = $0.0257125 \times MM \text{ CaCO}_3 \, (100.09)$ = 2.5735641 g

% CaCO $_3$ = $\frac{2.5735641}{3.00} \times 100 = 85.7854\% \approx 85.8\%$

Chalk sample has to be Brand X.

Marking Criteria

Descriptor	Marks
Correctly identifies the brand AND Provides all correct calculations AND Provides a balanced chemical equation including states	7
Provides substantially correct calculations AND Provides a balanced chemical equation	6
Provides the main steps in the calculation AND Provides a balanced chemical equation	5
Provides the main steps in the calculation AND Provides a balanced chemical equation	4
Provides some steps in the calculation	3
Provides some steps in the calculation	2
Provides some relevant information	1
None of the above	0

NESA Chemistry Processing Data and Information

Frequently Asked Questions

Ready to practise NESA Chemistry?

NESA Chemistry Processing Data and Information

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Frequently Asked Questions

Ready to practise NESA Chemistry?