How many NESA Chemistry questions cover Processing Data and Information?

AusGrader has 64 NESA Chemistry questions on Processing Data and Information, all with instant AI grading and detailed marking feedback.

NESA (HSC) Chemistry — Processing Data and Information Questions & Answers | AusGrader

Q13

2024

NESA

1 mark

Q13

1 mark

A fuel has these enthalpies of combustion: $-2057.8 \text{ kJ mol}^{-1}$ and $-48.9 \text{ kJ g}^{-1}$ .

Which of the following correctly identifies the fuel?

A

Ethanol ( $MM = 46.1 \text{ g mol}^{-1}$ )

B

Propane ( $MM = 44.1 \text{ g mol}^{-1}$ )

C

Propene ( $MM = 42.1 \text{ g mol}^{-1}$ )

D

Hydrogen ( $MM = 2.02 \text{ g mol}^{-1}$ )

Q14

2023

NESA

1 mark

Q14

1 mark

What volume of $0.540 \text{ mol L}^{-1}$ hydrochloric acid will react completely with $1.34 \text{ g}$ of sodium carbonate?

A

11.7 mL

B

23.4 mL

C

29.9 mL

D

46.8 mL

Correct Answer

Correct. $1.34 \text{ g}$ of $\text{Na}_2\text{CO}_3$ is $0.0126 \text{ mol}$ , which requires $0.0253 \text{ mol}$ of $\text{HCl}$ due to the 1:2 molar ratio. Dividing this by the concentration ( $0.540 \text{ mol L}^{-1}$ ) gives $0.0468 \text{ L}$ , or $46.8 \text{ mL}$ .

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.

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

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.

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

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.

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

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