What Is Working Scientifically in HSC Physics and Chemistry?

A trolley travels 2.0 metres down a ramp in 1.5 seconds. You know the distance and the time, so can you calculate its acceleration? Not necessarily. You don’t know whether the trolley started from rest or received a push, and you haven’t established that its acceleration was constant. A calculation could give you a perfectly neat number based on assumptions you never tested.

This is the problem that Working scientifically solves in HSC Physics and Chemistry. It’s the set of skills you use to ask testable questions, plan investigations, collect and analyse evidence, solve problems, and justify conclusions. It’s not a separate chapter about laboratory equipment. It’s how you decide whether your scientific knowledge actually explains what you’ve observed.

01What are the seven Working scientifically skills?

The NSW Education Standards Authority (NESA) identifies the same seven Working scientifically skill areas in the 2025 Physics and Chemistry syllabuses.

They describe different parts of scientific inquiry, but you use them together.

These are the shared skill areas in the official Physics 11-12 Syllabus (2025) and Chemistry 11-12 Syllabus (2025).

You don’t necessarily use them in a fixed order. Analysing unexpected data might reveal a problem with your method, sending you back to plan a better investigation.

How the skills connect to course knowledge

Consider two investigations.

In Physics, you could investigate how changing the voltage across a resistor affects the electric current through it.

In Chemistry, you could investigate how changing hydrochloric acid concentration affects the rate at which magnesium produces hydrogen gas.

Both investigations use the same scientific skills. The knowledge needed to make sense of them is different.

In Physics, your prediction comes from the relationship between voltage, current, and resistance.

In Chemistry, your prediction comes from collision theory and the way reactant concentration influences reaction rate.

Scientific knowledge tells you what to expect. Working scientifically tells you how to test that expectation, whether the evidence supports it, and what limitations remain.

Knowing a formula isn’t enough if you use it on measurements that don’t represent the quantity you think they do.

02A fair test starts with one meaningful change

Imagine trying to find out whether higher acid concentration makes magnesium react faster.

You prepare two flasks. The first contains dilute acid and a short magnesium ribbon. The second contains concentrated acid and a much longer ribbon.

The second flask produces hydrogen faster.

What caused the difference?

You might immediately credit the higher acid concentration. But the longer magnesium ribbon could also have more exposed surface area, providing more places for reactions to occur.

You’ve changed two things at once. Your observations cannot isolate the effect of concentration.

A better investigation changes one chosen factor while keeping other relevant conditions constant.

Three types of variables help you plan this:

• Independent variable: What you deliberately change, such as acid concentration.
• Dependent variable: What you measure in response, such as the volume of hydrogen collected after 20 seconds.
• Controlled variables: Other relevant factors you keep constant, such as magnesium mass, exposed surface area, acid volume, and temperature.

The dependent variable must genuinely tell you something about the question.

For instance, measuring hydrogen after 20 seconds gives information about gas production over that interval. It does not directly measure the exact reaction rate at the twentieth second.

That’s the difference between obtaining a useful measurement and claiming more than it can establish.

A prediction needs conditions

Suppose you double the voltage across a resistor.

Would you expect the current to double?

For an ohmic resistor at constant temperature, yes. Ohm’s law states:

\[ V=IR \]

Here:

• \(V\) is the potential difference across the resistor, measured in volts (V).
• \(I\) is the current through the resistor, measured in amperes (A).
• \(R\) is the electrical resistance, measured in ohms (\(\Omega\)).

If \(R\) remains constant, doubling \(V\) doubles \(I\).

But increasing the voltage may heat the resistor. Its resistance might change with temperature, so doubling the voltage may no longer double the current.

The first model is useful, but it has a condition: resistance must remain constant.

A strong scientific prediction includes the relevant condition rather than silently assuming it.

04Why reliable results can still be wrong

Students often treat reliability, validity, accuracy, and precision as interchangeable. They’re not.

Imagine a balance that reads 0.50 g too high.

You place the same object on it five times. Every reading is almost identical.

The measurements are highly consistent, but they’re systematically biased. Repeating the measurement doesn’t remove the calibration error.

It’s useful to distinguish two broad sources of measurement error.

Random errors cause readings to fluctuate unpredictably. Repeated measurements and appropriate averaging can reduce their influence on a calculated mean.

Systematic errors shift measurements consistently in one direction or otherwise distort them in a predictable way. They usually require a correction to the method, calibration, or equipment.

And an invalid comparison can remain invalid even when every instrument is accurate.

For example, if you change both acid concentration and magnesium surface area, a perfect gas syringe still cannot tell you which change caused the difference.

What to do when the data look unusual

Before rejecting a result or forcing it to fit your prediction, consider the actual measurement process.

Was the equipment calibrated? Were relevant variables controlled? Does the proposed relationship make sense with the known science? Is the unusual result reproducible?

An outlier is a result that lies noticeably away from the main pattern. It deserves investigation, not automatic deletion.

Sometimes the unusual result identifies a flawed measurement. Sometimes it reveals a limitation in the model.

Distinguishing those explanations is part of scientific analysis.

06How is Working scientifically assessed in the HSC?

Working scientifically isn’t just about practical exams or writing laboratory reports.

Under the 2025 replacement syllabuses, scientific investigations, data interpretation, mathematical problem-solving, and explanations using evidence are integrated with course content.

Which syllabus applies to your HSC year?

As at 8 November 2026, Physics and Chemistry have different implementation schedules.

This means:

• Students sitting the HSC in 2026 or 2027 study the 2017 Physics and Chemistry syllabuses.
• Students sitting the HSC in 2028 study the 2025 Physics syllabus but the 2017 Chemistry syllabus.
• Students sitting the HSC in 2029 are the first cohort to use both replacement syllabuses.

The familiar Module 1-8 labels belong to the 2017 courses. The 2025 syllabuses organise content into named focus areas instead.

For example, the 2025 Physics course includes Fundamentals of mechanics, Waves, and Electricity and magnetism in Year 11. The 2025 Chemistry course includes Properties and structure of matter, Quantitative chemistry, and Chemical reactions in Year 11.

Working scientifically applies throughout these focus areas. It isn’t an additional content topic to study afterwards.

What does the 60% Working scientifically weighting mean?

For both 2025 replacement courses, NESA specifies these school-based assessment component weightings in Year 11 and Year 12:

These are mandatory component weightings for the school-based assessment programs.

They do not mean that 60% of every test must involve laboratory work, or that 60% of the external HSC examination is a separate practical-skills section.

The new written HSC examinations integrate Working scientifically questions throughout the paper. A question might provide experimental results and ask you to calculate a quantity, evaluate a method, and explain the outcome using physics or chemistry.

The official requirements are available in the Physics school-based assessment guidance and Chemistry school-based assessment guidance.

What about depth studies and practical tasks?

The 2025 Physics and Chemistry syllabuses each require 10 hours of depth studies in Year 11 and 10 hours in Year 12, with at least one depth study in each year.

Depth studies are targeted investigations that use Working scientifically skills to explore course content more deeply.

Scientific investigations can involve laboratory experiments, simulations, fieldwork, or analysis of secondary-source information. Working scientifically doesn’t require every investigation to use chemicals or physical apparatus.

NESA sets the syllabus requirements, but schools have discretion over their particular investigations and assessment tasks within the applicable rules.

Under the NESA ACE assessment rules, updated in September 2026, the relevant cohorts are generally limited to one formal take-home assessment task per Preliminary or HSC course, worth no more than 15%, subject to specified course exemptions.

Your school’s assessment schedule tells you which outcomes each task assesses, when it occurs, and how much it contributes.

How should you actually prepare?

Treat practical skills, mathematical skills, and scientific explanation as connected habits.

When practising an investigation question, identify what was deliberately changed and what was measured. Then ask whether the design fairly isolates the intended relationship.

For a calculation, keep track of the physical quantity represented by each number. Convert units before substituting, show the relationship being used, and check whether the result makes physical or chemical sense.

If prefixes and conversions are slowing you down, tutorgum’s SI Prefix Conversions for HSC Physics and Chemistry explains that part separately.

For a graph, don’t stop after identifying that it rises or falls. Explain what its gradient, intercept, curvature, or unusual points mean in the context of the question.

For a longer response, use a simple reasoning structure:

Claim → evidence → scientific explanation → limitation, where relevant.

For example:

“The higher acid concentration produced a greater mean hydrogen volume after 20 seconds, increasing from 12.0 mL to 20.0 mL. This is consistent with a greater frequency of effective collisions at the magnesium surface. However, the results do not establish that reaction rate is directly proportional to acid concentration.”

That response explains more than simply writing “higher concentration means faster reaction”.

Does Working scientifically affect which subject you should choose?

Working scientifically is part of both subjects, not a separate HSC course or an admission requirement.

Subject selection is a different decision from assessment preparation.

The 2025 NESA syllabuses list no prerequisite for entering Year 11 Physics or Year 11 Chemistry, although schools may have their own subject-selection procedures or recommendations. Year 12 Physics and Chemistry each require the corresponding Year 11 course.

University entry requirements are separate again. For example, UNSW’s published 2027 domestic undergraduate guide lists Mathematics Extension 1 and Physics as assumed knowledge for its Bachelor of Engineering (Honours)/Bachelor of Engineering Science. That is not a statewide school prerequisite for HSC Physics.

Always check the particular university course and entry year, because requirements can change. A useful distinction is explained in the University of Sydney’s guide to prerequisites and assumed knowledge.

ATAR scaling is also separate. UAC’s ATAR calculation guidance explains how scaled marks contribute to the ATAR. Scaling depends on the performance of the relevant course cohorts and is not a fixed bonus guaranteed by choosing Physics or Chemistry.

Your interest in the subject, willingness to practise its particular skills, and future study plans are more useful starting points than assuming one science automatically delivers a higher ATAR.

07Questions and solutions

The questions below move from identifying a fair test to evaluating competing scientific explanations. The measurements and scenarios are hypothetical and are provided so each problem can be solved without conducting an experiment.

Question 1

A student investigates how the type of surface affects the rebound height of a basketball.

They drop the same basketball from a height of 1.50 m onto a wooden floor and a carpeted floor, without giving the ball an initial push or spin.

The ball rebounds to 0.72 m on wood and 0.30 m on carpet.

Identify the independent variable, the dependent variable, and two controlled variables. Explain whether one trial on each surface is sufficient to establish a reliable general conclusion.

Solution 1

The independent variable is the surface type, the dependent variable is the rebound height, and one trial per surface is insufficient to establish a reliable general conclusion.

The independent variable is deliberately changed between wood and carpet. The dependent variable is the maximum height reached after the first bounce.

The ball and the initial drop height must be controlled. The release technique, including the absence of initial spin, should also be kept consistent.

The single measurements show a higher rebound on wood in these trials:

\[ 0.72\,\mathrm{m}>0.30\,\mathrm{m} \]

However, one bounce on each surface provides no information about how much the measurements vary between trials.

The student should repeat the drops, measure the rebound heights consistently, and compare the mean values and their variation.

The tempting conclusion is that wood must always produce a higher rebound. The evidence supports that observation for these trials, but a stronger general claim needs repeated, controlled measurements.

Question 2

A student uses a burette containing hydrochloric acid labelled \(0.1000\,\mathrm{mol\,L^{-1}}\) to determine the concentration of sodium hydroxide.

Exactly 25.00 mL of sodium hydroxide is pipetted into a conical flask for each titration.

The neutralisation reaction is:

\[ \ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)} \]

Three concordant titres are recorded:

After the experiment, the student remembers that the burette was rinsed with distilled water before the acid was added. Some water remained inside, uniformly diluting the acid in the burette. The student did not rinse the burette with the standard acid.

Assume the indicator endpoint was correctly identified and the diluted acid concentration remained uniform throughout the trials.

Calculate the apparent sodium hydroxide concentration using the labelled acid concentration. Determine the direction of the systematic error.

Explain why the concordant titres do not remove the problem, and why adding some distilled water to the conical flask after pipetting the sodium hydroxide would not produce the same error.

Solution 2

The apparent sodium hydroxide concentration is \(0.1203\,\mathrm{mol\,L^{-1}}\), but this overestimates its true concentration. The acid was diluted inside the burette, and consistent titres cannot correct an incorrect standard concentration.

First, calculate the apparent concentration.

The mean titre is:

\[ \begin{aligned} \bar V&=\frac{30.10+30.05+30.10}{3}\\ &=30.083\,\mathrm{mL}\\ &=0.030083\,\mathrm{L} \end{aligned} \]

Let \(n\) be amount of substance in moles, \(c\) be concentration in \(\mathrm{mol\,L^{-1}}\), and \(V\) be solution volume in litres.

Using \(n=cV\) and the labelled acid concentration:

\[ \begin{aligned} n_\mathrm{HCl,app} &=c_\mathrm{HCl}V_\mathrm{HCl}\\ &=(0.1000)(0.030083)\\ &=0.0030083\,\mathrm{mol} \end{aligned} \]

The reaction has a 1:1 mole ratio between HCl and NaOH.

Therefore:

\[ \begin{aligned} c_\mathrm{NaOH,app} &=\frac{n_\mathrm{HCl,app}}{V_\mathrm{NaOH}}\\ &=\frac{0.0030083\,\mathrm{mol}} {0.02500\,\mathrm{L}}\\ &=0.1203\,\mathrm{mol\,L^{-1}} \end{aligned} \]

The numerical substitution for the acid uses concentration in \(\mathrm{mol\,L^{-1}}\) and volume in litres.

Next, consider the actual acid concentration.

The water remaining in the burette reduced the acid concentration below the labelled value.

A larger volume of this diluted acid was therefore required to neutralise the same amount of sodium hydroxide.

By calculating with \(0.1000\,\mathrm{mol\,L^{-1}}\) instead of the lower actual acid concentration, the student overestimated the number of moles of HCl delivered, and therefore the number of moles of NaOH.

The true sodium hydroxide concentration is lower than \(0.1203\,\mathrm{mol\,L^{-1}}\). Its exact value cannot be determined without knowing the actual acid concentration.

The concordant titres show consistency, not accuracy.

All three titrations used the same incorrectly prepared acid. A shared systematic error can produce very similar results.

Adding water to the conical flask is different.

Once exactly 25.00 mL of sodium hydroxide has been transferred, adding distilled water changes the solution’s concentration in the flask but not the number of moles of NaOH present.

The same number of moles of HCl is still required for neutralisation, assuming no solution is lost and the endpoint remains detectable.

The location of the added water changes the scientific result. Treating every dilution as an equivalent error misses that distinction.

Question 3

A student investigates a trolley travelling down a ramp.

A timing system records the trolley’s position relative to a marked starting line. The measurements are:

The timing and position measurements are assumed exact for this question.

Two students suggest competing models:

• Model A: The trolley started from rest and moved with constant acceleration.
• Model B: The trolley had a non-zero initial speed and moved with constant acceleration.

Determine whether Model A is consistent with both measurements.

For Model B, calculate the initial speed and acceleration.

Explain whether the measurements prove that acceleration was constant, and identify an additional measurement that would help test the model.

Solution 3

Model A is inconsistent with the exact measurements. Model B fits both with an initial speed of approximately \(0.333\,\mathrm{m\,s^{-1}}\) and an acceleration of \(1.33\,\mathrm{m\,s^{-2}}\). However, two position measurements do not prove constant acceleration.

For motion with constant acceleration:

\[ s=v_0t+\frac12at^2 \]

Here, \(s\) is displacement in metres, \(v_0\) is initial velocity in \(\mathrm{m\,s^{-1}}\), \(t\) is elapsed time in seconds, and \(a\) is acceleration in \(\mathrm{m\,s^{-2}}\).

The positive direction is down the ramp.

Test Model A.

Starting from rest means \(v_0=0\), so:

\[ s=\frac12at^2 \]

Using the first measurement:

\[ \begin{aligned} a&=\frac{2s}{t^2}\\ &=\frac{2(1.00\,\mathrm{m})} {(1.00\,\mathrm{s})^2}\\ &=2.00\,\mathrm{m\,s^{-2}} \end{aligned} \]

If that acceleration were constant, the displacement after 1.50 seconds would be:

\[ \begin{aligned} s&=\frac12at^2\\ &=\frac12(2.00)(1.50)^2\\ &=2.25\,\mathrm{m} \end{aligned} \]

The measured displacement is only 2.00 m.

Therefore, the trolley cannot have both started from rest and maintained a constant acceleration of \(2.00\,\mathrm{m\,s^{-2}}\), given exact measurements.

Test Model B.

Now allow a non-zero initial speed.

Substitute both measurements into the displacement equation. Using metres and seconds throughout, the numerical equations are:

\[ \begin{aligned} 1.00&=1.00v_0+0.50a\\ 2.00&=1.50v_0+1.125a \end{aligned} \]

From the first equation:

\[ v_0=1.00-0.50a \]

Substitute into the second:

\[ \begin{aligned} 2.00&=1.50(1.00-0.50a)+1.125a\\ 2.00&=1.50+0.375a\\ a&=\frac{0.50}{0.375}\\ &=1.33\,\mathrm{m\,s^{-2}} \end{aligned} \]

Then:

\[ \begin{aligned} v_0&=1.00-0.50(1.333\ldots)\\ &=0.333\,\mathrm{m\,s^{-1}} \end{aligned} \]

These values reproduce both measured positions.

But fitting two measurements doesn’t prove the model.

Constant acceleration with a non-zero starting speed is one possible explanation. A changing acceleration could also produce the same two recorded positions.

The student could measure the trolley’s velocity as it crosses the starting line and take additional position-time measurements to see whether a single constant-acceleration model continues to fit.

The tempting mistake is to treat a mathematically successful model as uniquely proven. A model that fits limited evidence may still have competitors.

Question 4

A wheeled trolley moves along a horizontal track while a force sensor applies a constant horizontal pulling force of 4.0 N.

The experiment is repeated with additional mass placed securely on the trolley.

The wheels, track, and pulling direction remain unchanged. Assume the measurements are accurate, air resistance is negligible, and the only opposing horizontal force is resistance from the trolley-track interaction.

One student argues that doubling the mass must have doubled the resistive force, explaining why acceleration decreased.

Another argues that the resistive force was unchanged.

Use Newton’s second law to evaluate both explanations.

Then predict the acceleration of the 2.0 kg trolley if the applied pulling force increases to 8.0 N. For this final part, assume the resistive force remains unchanged over the speeds involved.

Solution 4

The resistive force is \(1.0\,\mathrm{N}\) in both trials. The decrease in acceleration is explained by the increase in inertial mass, not by an increase in resistance. Under the final stated assumption, an 8.0 N pull produces an acceleration of \(3.5\,\mathrm{m\,s^{-2}}\).

Find the resistive force in each trial.

Newton’s second law states:

\[ F_\mathrm{net}=ma \]

Here, \(F_\mathrm{net}\) is the net force in newtons, \(m\) is mass in kilograms, and \(a\) is acceleration in \(\mathrm{m\,s^{-2}}\).

For the horizontal trolley:

\[ F_\mathrm{pull}-F_\mathrm{res}=ma \]

The quantities \(F_\mathrm{pull}\) and \(F_\mathrm{res}\) are the magnitudes of the applied pulling force and opposing resistive force, respectively.

For the 1.0 kg trolley:

\[ \begin{aligned} F_\mathrm{res} &=F_\mathrm{pull}-ma\\ &=4.0-(1.0)(3.0)\\ &=1.0\,\mathrm{N} \end{aligned} \]

For the 2.0 kg trolley:

\[ \begin{aligned} F_\mathrm{res} &=4.0-(2.0)(1.50)\\ &=1.0\,\mathrm{N} \end{aligned} \]

The units in these substitutions are newtons, since \(\mathrm{kg\,m\,s^{-2}}\) is equivalent to a newton.

Explain the falling acceleration.

The net horizontal force is 3.0 N in both trials.

When the mass doubles from 1.0 kg to 2.0 kg, the same net force produces half the acceleration.

The first student’s interpretation is tempting because a heavier object can experience greater contact resistance in some circumstances. But the acceleration data do not support that explanation here.

A decrease in acceleration does not, by itself, demonstrate an increase in opposing force. Mass also affects acceleration.

Now double the applied pulling force.

For the 2.0 kg trolley, with the stated constant resistance:

\[ \begin{aligned} F_\mathrm{net} &=8.0-1.0\\ &=7.0\,\mathrm{N}\\ a&=\frac{7.0\,\mathrm{N}}{2.0\,\mathrm{kg}}\\ &=3.5\,\mathrm{m\,s^{-2}} \end{aligned} \]

This is greater than twice the previous acceleration of \(1.50\,\mathrm{m\,s^{-2}}\).

Why? Doubling the applied force does not double the net force when a constant opposing force is present.

These two trials support approximately constant resistance under the stated conditions. They do not establish that resistance can never vary with load, speed, or surface conditions.

For the broader distinction between contact forces and simplified friction formulas, see tutorgum’s Why Friction Is Not Always μmg: HSC Physics Explained.

Question 5

Two students investigate the reaction between calcium carbonate and excess hydrochloric acid:

\[ \ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g)} \]

They use separate sealed reaction vessels, each connected to a gas-volume sensor.

Vessel A initially contains 0.50 g of pure calcium carbonate. Vessel B contains 1.00 g.

The acid concentration and volume are the same in both vessels, and acid remains in excess throughout each reaction.

Both reactions occur at the same constant temperature and pressure. Assume no gas leaks, no gas dissolves, no other gases are produced, and all calcium carbonate eventually reacts.

The sensors are different models. Neither has been calibrated in millilitres.

Each reports a numerical reading \(S\) in arbitrary units. The relationship between sensor reading and actual collected gas volume \(V\) is:

\[ S=kV+b \]

Here, \(k\) is a positive but unknown sensor gain and \(b\) is an unknown zero offset. Each sensor has its own values of \(k\) and \(b\), which remain constant during its experiment.

The recorded results are:

Use a molar mass of \(100\,\mathrm{g\,mol^{-1}}\) for calcium carbonate.

A student claims:

“Vessel B has the larger sensor reading after 30 seconds, so it must have reacted faster. And because Vessel A has a larger fraction of its final reading by that time, the two sensors contradict each other.”

Evaluate the claim by determining:

1. The fraction of each reaction completed after 30.0 seconds.
2. The actual amount of carbon dioxide formed in each vessel during that interval, and the average rate of formation in \(\mathrm{mol\,s^{-1}}\).
3. Which conclusions about reaction progress and rate are supported, and whether either vessel has been shown to have intrinsically faster reaction kinetics.
4. Whether the method for determining fraction reacted would still work if the sensors had an unknown, non-linear response to gas volume instead of the stated linear relationship.

Solution 5

Vessel A has completed a greater fraction of its reaction after 30 seconds (60% versus 50%), but Vessel B has produced more carbon dioxide in absolute terms (\(0.0050\,\mathrm{mol}\) versus \(0.0030\,\mathrm{mol}\)). Neither raw sensor readings nor these results alone establish which system has intrinsically faster reaction kinetics. The fraction calculation works because the sensors have a stable linear response; it would not generally work for an unknown non-linear response.

The critical distinction is between a measurement, the fraction of a reaction completed, and the actual amount of product formed.

First, remove the sensor offsets and gains.

The sensor model is:

\[ S=kV+b \]

Initially, no reaction gas has been collected, so the initial reading represents the offset \(b\).

At any time, subtracting the initial reading removes that offset:

\[ S_t-S_0=kV_t \]

Here, \(S_0\) is the initial sensor reading, \(S_t\) is the reading at time \(t\), and \(V_t\) is the gas volume produced by that time.

At complete reaction:

\[ S_\infty-S_0=kV_\infty \]

The subscript \(\infty\) denotes the final reading after the reaction has finished.

Divide the two relationships:

\[ \frac{S_t-S_0}{S_\infty-S_0} =\frac{V_t}{V_\infty} \]

The unknown gain \(k\) cancels.

Because temperature and pressure remain constant, collected gas volume is proportional to the amount of carbon dioxide formed.

Therefore, the ratio gives the fraction \(f\) of the eventual carbon dioxide yield produced by time \(t\):

\[ f=\frac{S_t-S_0}{S_\infty-S_0} \]

For Vessel A:

\[ \begin{aligned} f_A&=\frac{80-20}{120-20}\\ &=\frac{60}{100}\\ &=0.60 \end{aligned} \]

For Vessel B:

\[ \begin{aligned} f_B&=\frac{100-50}{150-50}\\ &=\frac{50}{100}\\ &=0.50 \end{aligned} \]

Thus, after 30.0 seconds, Vessel A has produced 60% of its total carbon dioxide yield, while Vessel B has produced 50%.

These fractions are valid even though neither sensor has a known zero offset or gain.

Second, determine the total possible carbon dioxide yield.

From the balanced equation, one mole of calcium carbonate produces one mole of carbon dioxide.

The amount of calcium carbonate is calculated using:

\[ n=\frac{m}{M} \]

Here, \(n\) is amount in moles, \(m\) is mass in grams, and \(M\) is molar mass in \(\mathrm{g\,mol^{-1}}\).

For Vessel A:

\[ \begin{aligned} n_{A,\mathrm{final}} &=\frac{0.50\,\mathrm{g}} {100\,\mathrm{g\,mol^{-1}}}\\ &=0.0050\,\mathrm{mol} \end{aligned} \]

For Vessel B:

\[ \begin{aligned} n_{B,\mathrm{final}} &=\frac{1.00\,\mathrm{g}} {100\,\mathrm{g\,mol^{-1}}}\\ &=0.0100\,\mathrm{mol} \end{aligned} \]

The quantities \(n_{A,\mathrm{final}}\) and \(n_{B,\mathrm{final}}\) are the final amounts of carbon dioxide in each vessel.

Vessel B can eventually produce twice as much carbon dioxide because it started with twice as much calcium carbonate.

Third, calculate the amount formed after 30 seconds.

Multiply each final yield by its fraction completed:

\[ \begin{aligned} n_{A,30} &=0.60(0.0050)\\ &=0.0030\,\mathrm{mol}\\[4pt] n_{B,30} &=0.50(0.0100)\\ &=0.0050\,\mathrm{mol} \end{aligned} \]

Despite having completed a smaller fraction of its reaction, Vessel B has produced more carbon dioxide.

Now calculate the average rates of formation over the first 30.0 seconds:

\[ \bar r=\frac{\Delta n}{\Delta t} \]

Here, \(\bar r\) is average molar production rate in \(\mathrm{mol\,s^{-1}}\), \(\Delta n\) is amount of carbon dioxide formed, and \(\Delta t\) is elapsed time.

For Vessel A:

\[ \begin{aligned} \bar r_A &=\frac{0.0030\,\mathrm{mol}} {30.0\,\mathrm{s}}\\ &=1.0\times10^{-4}\, \mathrm{mol\,s^{-1}} \end{aligned} \]

For Vessel B:

\[ \begin{aligned} \bar r_B &=\frac{0.0050\,\mathrm{mol}} {30.0\,\mathrm{s}}\\ &=1.67\times10^{-4}\, \mathrm{mol\,s^{-1}} \end{aligned} \]

Vessel B has the greater average carbon dioxide production rate when expressed in moles per second.

Fourth, evaluate the student’s claim.

The raw sensor readings cannot be compared directly because the instruments have different unknown gains and offsets.

A reading of 100 on one sensor does not necessarily represent more gas than a reading of 80 on another.

Normalising each sensor’s readings against its own initial and final values removes those differences under the stated linear model.

But a larger fraction completed and a larger amount produced are not the same thing.

Vessel A has progressed further towards its own final yield. Vessel B has formed more product per second in absolute terms.

Neither result establishes an intrinsic difference in reaction kinetics.

The vessels began with different masses of calcium carbonate. Their total available reactant amounts differ, and their exposed reactive surface areas may also differ.

To isolate the effect of a particular variable, the experiment would need to control relevant factors such as carbonate amount, exposed surface area, acid conditions, and temperature, while changing only the factor under investigation.

The two time points also cannot establish the instantaneous rate at exactly 30.0 seconds.

Finally, test the boundary of the method.

The normalisation works because the sensor reading is linear in actual gas volume:

\[ S=kV+b \]

Subtracting the initial reading removes \(b\), and dividing by the total change removes \(k\).

But suppose a sensor instead followed an unknown non-linear relationship, such as:

\[ S=kV^2+b \]

Subtracting the initial reading and dividing by the final change would now give:

\[ \frac{S_t-S_0}{S_\infty-S_0} =\frac{V_t^2}{V_\infty^2} \]

That is not generally equal to the fraction of gas produced.

Knowing the initial and final readings would no longer be sufficient to determine the intermediate gas volume fraction without further information about the sensor’s response.

The sensors would need to be calibrated across the relevant volume range, or another suitable measurement method would be required.

This is the central scientific insight: even a clever calculation only supports a conclusion when the model connecting the instrument reading to the physical quantity is justified.

08The next step: testing competing explanations

Once you can recognise the difference between a measurement and a supported conclusion, the next useful skill is deciding between rival scientific models.

Suppose two explanations both fit the available data. Rather than choosing the one that sounds more familiar, identify a new observation on which they make different predictions.

Then design an investigation that can distinguish them.

That is how Working scientifically connects the equations and theories of HSC Physics and Chemistry to the evidence used to test them.