How many SCSA Chemistry questions cover Science Inquiry Skills?

AusGrader has 91 SCSA Chemistry questions on Science Inquiry Skills, all with instant AI grading and detailed marking feedback.

WACE Chemistry — Science Inquiry Skills Questions & Answers

Q16

2021

VCAA

1 mark

Q16

1 mark

Which one of the following statements about IR spectroscopy is correct?

A

IR radiation changes the spin state of electrons.

B

Bond wave number is influenced only by bond strength.

C

An IR spectrum can be used to determine the purity of a sample.

D

In an IR spectrum, high transmittance corresponds to high absorption.

Reveal Answer

A

IR radiation changes the spin state of electrons.

IR radiation causes changes in the vibrational states of molecules, not the spin states of electrons (which is associated with Electron Paramagnetic Resonance spectroscopy).

B

Bond wave number is influenced only by bond strength.

The wavenumber of a bond's vibration depends on both the bond strength (force constant) and the reduced mass of the atoms involved, as described by Hooke's Law.

C

An IR spectrum can be used to determine the purity of a sample.

Correct Answer

An IR spectrum can reveal the presence of impurities if unexpected absorption peaks appear that do not belong to the pure compound.

D

In an IR spectrum, high transmittance corresponds to high absorption.

Transmittance and absorbance are inversely related; high transmittance means that most of the light passed through the sample, indicating low absorption.

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

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 $2112 \text{ MJ} / 33.67 \text{ MJ/L} = 62.7 \text{ L}$ .

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.

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.

Q25

2024

VCAA

1 mark

Q25

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.

Which of the following shows the chemicals in order of their retention times, from lowest to highest?

A

hexane, hexan-2-one, hexan-1-ol

B

hexane, hexan-1-ol, hexan-2-one

C

hexan-2-one, hexan-1-ol, hexane

D

hexan-1-ol, hexan-2-one, hexane

Reveal Answer

A

hexane, hexan-2-one, hexan-1-ol

This order represents the highest to lowest retention time, which would be the correct order if the stationary phase were polar (normal-phase HPLC).

B

hexane, hexan-1-ol, hexan-2-one

Hexane is the most non-polar compound, meaning it will interact most strongly with the non-polar stationary phase and have the highest, not lowest, retention time.

C

hexan-2-one, hexan-1-ol, hexane

While hexane correctly has the highest retention time, hexan-1-ol is more polar than hexan-2-one due to its ability to form hydrogen bonds, so hexan-1-ol will elute before hexan-2-one.

D

hexan-1-ol, hexan-2-one, hexane

Correct Answer

In reverse-phase HPLC (non-polar stationary phase, polar mobile phase), the most polar compound has the lowest retention time. Hexan-1-ol (most polar) elutes first, followed by hexan-2-one, and the non-polar hexane elutes last.

Q1

2022

VCAA

1 mark

Q1

1 mark

Scientific posters communicate the findings of scientific investigations.
Which section of a scientific poster should explain the reason for undertaking an investigation?

A

discussion

B

conclusion

C

introduction

D

methodology

Reveal Answer

A

discussion

The discussion section is used to interpret the results, explain their significance, and compare them to existing literature, not to state the initial reason for the study.

B

conclusion

The conclusion summarizes the main findings of the investigation and their broader implications, rather than explaining why the study was started.

C

introduction

Correct Answer

The introduction provides the background information, context, and the specific rationale or reason for undertaking the investigation.

D

methodology

The methodology section details the procedures, materials, and techniques used to conduct the investigation, not the reason for doing it.

Q6

2023

VCAA

1 mark

Q6

1 mark

Consider the following statements.

I. HPLC is a qualitative process.
II. HPLC is a quantitative process.
III. Triplets give information about molecule structure.

Which of the above statements apply to high-performance liquid chromatography (HPLC)?

A

I only

B

II only

C

I and II only

D

I, II and III

Reveal Answer

A

I only

While HPLC can be used qualitatively to identify substances based on retention time, this option is incomplete because HPLC is also a quantitative technique.

B

II only

While HPLC can be used quantitatively to determine concentration based on peak area, this option is incomplete because HPLC is also a qualitative technique.

C

I and II only

Correct Answer

HPLC is both a qualitative process (identifying components via retention time) and a quantitative process (determining concentration via peak area).

D

I, II and III

Statement III is incorrect because triplets refer to splitting patterns found in Nuclear Magnetic Resonance (NMR) spectroscopy, not HPLC.

Q19

2021

VCAA

1 mark

Q19

1 mark

A food chemist conducted an experiment in a bomb calorimeter to determine the energy content, in joules per gram, of a muesli bar. A 3.95 g sample of the muesli bar was combusted in the calorimeter and the temperature of the water rose by 16.7 °C. The calibration factor of the calorimeter was previously determined to be $4780 \text{ J } ^\circ\text{C}^{-1}$ .

The energy content of the muesli bar is

A

$3.51 \times 10^5 \text{ J g}^{-1}$

B

$2.02 \times 10^4 \text{ J g}^{-1}$

C

$1.13 \times 10^3 \text{ J g}^{-1}$

D

$7.25 \times 10 \text{ J g}^{-1}$

Reveal Answer

A

$3.51 \times 10^5 \text{ J g}^{-1}$

This value is incorrect and likely results from improperly multiplying the total energy by the mass or another calculation error, rather than dividing the total energy by the mass.

B

$2.02 \times 10^4 \text{ J g}^{-1}$

Correct Answer

The total energy released is the calibration factor multiplied by the temperature change ( $4780 \times 16.7 = 79826 \text{ J}$ ). Dividing this total energy by the mass of the sample ( $3.95 \text{ g}$ ) gives the correct energy content per gram ( $2.02 \times 10^4 \text{ J g}^{-1}$ ).

C

$1.13 \times 10^3 \text{ J g}^{-1}$

This incorrect value is obtained by multiplying the calibration factor by the mass and dividing by the temperature change, which does not correctly calculate energy content.

D

$7.25 \times 10 \text{ J g}^{-1}$

This incorrect value is obtained by dividing the calibration factor by both the temperature change and the mass, which is a misapplication of the calorimeter formula.

Q24

2022

VCAA

1 mark

Q24

1 mark

A high-performance liquid chromatography (HPLC) instrument is set up with a polar mobile phase and a non-polar stationary phase. Three amino acids – leucine, Leu, alanine, Ala, and asparagine, Asn – are added to the mobile phase and are run through the HPLC.

The order of the retention times, from shortest to longest, for these three amino acids is

A

Leu, Ala, Asn

B

Leu, Asn, Ala

C

Ala, Asn, Leu

D

Asn, Ala, Leu

Reveal Answer

A

Leu, Ala, Asn

This order represents the longest to shortest retention times. Leucine is the most non-polar and would interact most strongly with the non-polar stationary phase, resulting in the longest retention time.

B

Leu, Asn, Ala

Leucine is the most non-polar of the three amino acids and will have the longest retention time, not the shortest.

C

Ala, Asn, Leu

While leucine correctly has the longest retention time, asparagine is more polar than alanine due to its amide group and will elute before it.

D

Asn, Ala, Leu

Correct Answer

In reverse-phase HPLC (polar mobile phase, non-polar stationary phase), the most polar compound elutes first. Asparagine is the most polar, followed by the slightly non-polar alanine, and finally the most non-polar leucine.

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.

Q6

2020

VCAA

8 marks

Q6

Methane gas, $\text{CH}_4$ , can be captured from the breakdown of waste in landfills. $\text{CH}_4$ is also a primary component of natural gas. $\text{CH}_4$ can be used to produce energy through combustion.

Q6a

1 mark

Write the equation for the incomplete combustion of $\text{CH}_4$ to produce carbon monoxide, CO.

Reveal Answer

2CH4(g) + 3O2(g) → 2CO(g) + 4H2O(l) or
CH4(g) + 1.5O2(g) → CO(g) + 2H2O(l)

Marking Criteria

Descriptor	Marks
Correct response.	1

Q6b

2 marks

If 20.0 g of $\text{CH}_4$ is kept in a 5.0 L sealed container at $25 ^\circ\text{C}$ , what would be the pressure in the container?

Reveal Answer

$n(\text{CH}_4) = 20.0 / 16.0 = 1.25 \text{ mol} *$
$\text{Pressure}(\text{CH}_4) = nRT / V = 1.25 \times 8.31 \times 298 / 5.0 = 6.2 \times 10^2 \text{ kPa} *$

Marking Criteria

Descriptor	Marks
Calculates correct amount of CH4 (1.25 mol).	1
Calculates correct pressure of CH4 (6.2 x 10^2 kPa).	1

Q6c

3 marks

A Bunsen burner is used to heat a beaker containing 350.0 g of water. Complete combustion of 0.485 g of $\text{CH}_4$ raises the temperature of the water from $20 ^\circ\text{C}$ to $32.3 ^\circ\text{C}$ .

Calculate the percentage of the Bunsen burner's energy that is lost to the environment.

Reveal Answer

$\text{Energy from CH}_4 = 0.485 \text{ g} \times 55.6 \text{ kJ g}^{-1} = 27.0 \text{ kJ}$

$\text{Energy absorbed by water} = 4.18 \text{ J g}^{-1} \text{ K}^{-1} \times 350.0 \times (32.3 - 20.0)$
$= 1.80 \times 10^4 \text{ J}$
$= 18.0 \text{ kJ}$

$\text{Energy lost to environment} = 27.0 - 18.0 = 9.0 \text{ kJ}$

$\% \text{ energy lost} = (9.0 / 27.0) \times 100 = 33.3 \%$

Marking Criteria

Descriptor	Marks
Calculating the energy from CH4.	1
Calculating the energy absorbed by water.	1
Working out the percentage of energy loss.	1

Q6d

2 marks

Compare the environmental impact of $\text{CH}_4$ obtained from landfill to the environmental impact of $\text{CH}_4$ obtained from natural gas.

Reveal Answer

Similarity – methane from both sources

Both produce atmospheric carbon dioxide through combustion.
Methane from both sources contains small amounts of nitrogen and sulfur; combustion of natural gas leads to the formation of acidic oxides such as SOx and NOx.

Difference – landfill versus natural gas

Methane from landfill can be produced renewably, whereas methane from natural gas releases stored carbon.
Methane from landfill is more carbon neutral, methane from natural gas increases atmospheric CO2 levels.
Obtaining methane from natural gas via fracking causes additional significant environmental damage, whereas when obtaining methane from a landfill the damage has already been done in the formation of the landfill.
Landfill gases contain less methane and release more CO2 (for the same amount of energy generated), natural gas contains more methane and releases comparatively less CO2.
Methane captured from landfill and used as a source on energy may have a positive impact as it is a more potent greenhouse gas than CO2.
CH4 from landfill is more easily collected compared to fracking/sourcing methane from fossil fuels.

Marking Criteria

Descriptor	Marks
1 mark for each valid comparison point (any 2 of): Both produce atmospheric carbon dioxide through combustion; Methane from both sources contains small amounts of nitrogen and sulfur; Methane from landfill can be produced renewably, whereas methane from natural gas releases stored carbon; Methane from landfill is more carbon neutral; Obtaining methane from natural gas via fracking causes additional significant environmental damage; Landfill gases contain less methane and release more CO2; Methane captured from landfill may have a positive impact as it is a more potent greenhouse gas than CO2; CH4 from landfill is more easily collected.	2

Descriptor

Marks

1 mark for each valid comparison point (any 2 of): Both produce atmospheric carbon dioxide through combustion; Methane from both sources contains small amounts of nitrogen and sulfur; Methane from landfill can be produced renewably, whereas methane from natural gas releases stored carbon; Methane from landfill is more carbon neutral; Obtaining methane from natural gas via fracking causes additional significant environmental damage; Landfill gases contain less methane and release more CO2; Methane captured from landfill may have a positive impact as it is a more potent greenhouse gas than CO2; CH4 from landfill is more easily collected.

2

Q1

2021

VCAA

9 marks

Q1

Digesters use bacteria to convert organic waste into biogas, which contains mainly methane, $\text{CH}_4$ . Biogas can be used as a source of energy.

Q1b

A digester processed 1 kg of organic waste to produce 496.0 L of biogas at standard laboratory conditions (SLC). The biogas contained 60.0% $\text{CH}_4$ .

Q1c

Biogas was combusted to release $1.63 \times 10^3 \text{ kJ}$ of energy. This energy was used to heat 100 kg of water in a tank. The initial temperature of the water was 25.0 °C.

Q1a

1 mark

Both biogas and coal seam gas contain $\text{CH}_4$ as their main component.

Why is biogas considered a renewable energy source but coal seam gas is not?

Reveal Answer

Biogas is considered renewable because its production-and-use cycle is continuous so that it is constantly replenished whereas coal seam gas is used at a faster rate than it can be replenished.*

Marking Criteria

Descriptor	Marks
Provides a direct comparison of both biogas and coal seam gas indicating the period of time used to produce these materials	1

Q1b (i)

2 marks

Write the thermochemical equation for the complete combustion of $\text{CH}_4$ at SLC.

Reveal Answer

$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$ * $\Delta H = -890\ \mathrm{kJ\ mol^{-1}}$

Marking Criteria

Descriptor	Marks
Writes a correctly balanced chemical equation with associated states	1
Provides a molar enthalpy of combustion with a negative sign that matches the equation written	1

Q1b (ii)

3 marks

Calculate the amount of energy that could be produced by $\text{CH}_4$ from 1 kg of organic waste.

Reveal Answer

$n(\text{biogas}) = 496.0 / 24.8$

$= 20.0\ \mathrm{mol}$ *

$n(CH_4) = 0.60 \times 20.0$

$= 12.0\ \mathrm{mol}$ *

Energy $= 12.0\ \mathrm{mol} \times 890\ \mathrm{kJ\ mol^{-1}}$

$= 1.07 \times 10^4\ \mathrm{kJ}\ (10680\ \mathrm{kJ})\ \text{or}\ (10.7\ \mathrm{MJ})$ *

Marking Criteria

Descriptor	Marks
Calculates $n(\text{biogas})$ correctly	1
Calculates $n(\text{CH}_4)$ correctly	1
Calculates the energy produced correctly	1

Q1c (i)

2 marks

What is the maximum temperature that the water in the tank could reach?

Reveal Answer

Energy $= 4.18 \times m(H_2O) \times \Delta T$

$1.63 \times 10^3\ (\times 10^3) = 4.18 \times 100\ (\times 10^3) \times \Delta T$

$\Delta T = 1.63 \times 10^3 / (4.18 \times 100)$

$= 3.90\ ^\circ C$ *

$T_{\max} = 25.0 + 3.90$

$= 28.9\ ^\circ C\ (\text{or }302\ \mathrm{K})$ *

Marking Criteria

Descriptor	Marks
Calculates $\Delta T$ correctly	1
Calculates $T_{\max}$ correctly to three significant figures	1

Q1c (ii)

1 mark

State why this temperature may not be reached.

Reveal Answer

For example:

loss of heat/energy to the atmosphere
heat/energy loss in the combustion chamber
heat/energy loss since the tank material also is heated
heat/energy loss from the piping
faulty insulation

Marking Criteria

Descriptor	Marks
Provides any logical reason involving incomplete transfer of heat/energy to the water (e.g., loss of heat to the atmosphere, heat loss in the combustion chamber, heat loss to the tank material, heat loss from piping, or faulty insulation)	1

Q4

2022

VCAA

12 marks

Q4a

Collagen is found in muscles, joints and skin. It is the most abundant protein in the body. Gelatin can be made by boiling animal collagen in water for many hours. Collagen and gelatin are made from the same 18 amino acids. Gelatin is more easily metabolised by the body than collagen.

Q4c

Researchers have identified pathways that will enable production of the biofuel 2-methylpropan-1-ol from proteins. 2-methylpropan-1-ol can be used in petrol engines. 2-methylpropan-1-ol has a heat of combustion of $36.1 \text{ kJ g}^{-1}$ .

Q4a (i)

1 mark

Gelatin is produced when collagen is broken into smaller molecules.

Name the chemical reaction that produces gelatin from collagen.

Reveal Answer

Hydrolysis

Marking Criteria

Descriptor	Marks
Identifies hydrolysis as the chemical reaction that produces gelatin from collagen.	1

Q4a (ii)

3 marks

Explain how consuming gelatin can be useful in increasing collagen levels in the body. In your answer, identify any chemical processes involved.

Reveal Answer

Marking Criteria

Descriptor	Marks
Indicates that gelatin contains essential amino acids that must be included in the diet for collagen synthesis to occur.	1
Identifies that consumed gelatin is broken down via hydrolysis into its constituent amino acids.	1
Explains that the body forms required collagen through the condensation of these amino acids to form a polypeptide.	1

Q4b (i)

2 marks

Vitamin C is required for the production of collagen in the body.

What is the source of the vitamin C present in the body? Justify your answer.

Reveal Answer

Marking Criteria

Descriptor	Marks
States that vitamin C is sourced through diet or food.	1
Recognises that vitamin C is an essential vitamin and cannot be synthesised in the body.	1

Q4b (ii)

1 mark

Vitamin C is added to some foods to help prevent spoilage.

State how vitamin C can slow the rate of oxidative rancidity in some foods.

Reveal Answer

Marking Criteria

Descriptor	Marks
States that vitamin C is preferentially oxidised, slowing the rate of oxidative rancidity.	1

Q4c (i)

2 marks

Compare the energy content of octane and 2-methylpropan-1-ol. Explain the difference.

Reveal Answer

Marking Criteria

Descriptor	Marks
States that the energy content of octane ( $47.9 \text{ kJ g}^{-1}$ ) is higher than the energy content of 2-methylpropan-1-ol ( $36.1 \text{ kJ g}^{-1}$ ).	1
Explains that 2-methylpropan-1-ol contains oxygen and is therefore already partially oxidised.	1

Q4c (ii)

3 marks

A small fuel burner containing $2.36 \text{ g}$ of 2-methylpropan-1-ol was placed directly underneath a beaker containing $500.0 \text{ g}$ of water at standard laboratory conditions (SLC).

Calculate the maximum temperature that the water could reach if the contents of the fuel burner underwent complete combustion.

Reveal Answer

$\text{Energy released} = 2.36 \times 36.1 = 85.2 \text{ kJ}$
$E = mc\Delta T \text{, therefore } \Delta T = E / mc$
$= \frac{85.2 \times 10^3}{500 \times 4.18}$
$= 40.8 ^\circ\text{C}$
$\text{Max. Temp} = 25.0 + 40.8$
$= 65.8 / 66 ^\circ\text{C } (339 \text{ K})$

Marking Criteria

Descriptor	Marks
Calculates the correct energy released ( $85.2 \text{ kJ}$ ).	1
Calculates the correct change in temperature, $\Delta T$ ( $40.8 ^\circ\text{C}$ ).	1
Calculates the correct final temperature ( $65.8 ^\circ\text{C}$ or $66 ^\circ\text{C}$ or $339 \text{ K}$ ).	1

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