Band 4

CHEMISTRY, M5 2025 HSC 35

A purple solution at 25°C contains a mixture of two different cobalt$\text{(II)}$ complexes which are at equilibrium.

$\underset{\text{(blue)}}{\ce{CoCl4^{2-}(aq)}} \ce{+ 6H2O(l)} \rightleftharpoons
\underset{\text{(pink)}}{\ce{Co(H2O)6^{2+}(aq)}} \ce{+ 4Cl^{-}(aq)}$

The results of heating and cooling a sample of this solution are given in the table.

\begin{array}{|l|c|c|}
\hline\rule{0pt}{2.5ex} \textit{Temperature} \ \text{(°C)} \rule[-1ex]{0pt}{0pt}& 80 & 0 \\
\hline \rule{0pt}{2.5ex}\textit{Colour of solution} \rule[-1ex]{0pt}{0pt} & \quad \text{blue} \quad & \quad \text{pink} \quad \\
\hline
\end{array}

The energy profile diagram for this reaction is shown.

How do collision theory and Le Chatelier's principle account for the colour change to pink when the solution is cooled? Refer to the energy profile diagram in your answer. (5 marks)

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Energy Profile Diagram

The energy profile diagram shows that the forward reaction is exothermic, as the products have lower enthalpy than the reactants. The forward activation energy $(\text{E}_{a1})$ is smaller than the reverse activation energy $(\text{E}_{a2})$.

Le Chatelier’s Principle

Le Chatelier’s Principle says that if something disrupts a system in dynamic equilibrium, the system will shift in a way that works against that change.
Cooling the solution from 80 °C to 0 °C causes the system to favour the forward exothermic reaction, producing heat in response to the lowered temperature.
This shifts the equilibrium to the right, increasing $\left[ \ce{Co(H2O)6^{2+}}\right]$ and turning the solution pink.

Collision Theory

Collision theory states that for a reaction to occur, particles must collide with sufficient energy (above the activation energy) and correct orientation.
When the solution is cooled, the average kinetic energy of all particles decreases, reducing both the collision frequency and the proportion of successful collisions.
In this way, both forward and reverse reaction rates decrease.
However, the reverse reaction has a higher activation energy $(\text{E}_{a2})$ than the forward reaction $(\text{E}_{a1})$, as shown in the energy profile diagram.
Cooling causes a greater proportion of particles to fall below $(\text{E}_{a2})$ than $(\text{E}_{a1})$, so the reverse reaction rate decreases more significantly than the forward rate.
This causes a net shift toward products, increasing $\left[ \ce{Co(H2O)6^{2+}}\right]$ and changing the solution colour to pink.

Show Worked Solution

Energy Profile Diagram

The energy profile diagram shows that the forward reaction is exothermic, as the products have lower enthalpy than the reactants. The forward activation energy $(\text{E}_{a1})$ is smaller than the reverse activation energy $(\text{E}_{a2})$.

Le Chatelier’s Principle

Le Chatelier’s Principle says that if something disrupts a system in dynamic equilibrium, the system will shift in a way that works against that change.
Cooling the solution from 80 °C to 0 °C causes the system to favour the forward exothermic reaction, producing heat in response to the lowered temperature.
This shifts the equilibrium to the right, increasing $\left[ \ce{Co(H2O)6^{2+}}\right]$ and turning the solution pink.

Collision Theory

Collision theory states that for a reaction to occur, particles must collide with sufficient energy (above the activation energy) and correct orientation.
When the solution is cooled, the average kinetic energy of all particles decreases, reducing both the collision frequency and the proportion of successful collisions.
In this way, both forward and reverse reaction rates decrease.
However, the reverse reaction has a higher activation energy $(\text{E}_{a2})$ than the forward reaction $(\text{E}_{a1})$, as shown in the energy profile diagram.
Cooling causes a greater proportion of particles to fall below $(\text{E}_{a2})$ than $(\text{E}_{a1})$, so the reverse reaction rate decreases more significantly than the forward rate.
This causes a net shift toward products, increasing $\left[ \ce{Co(H2O)6^{2+}}\right]$ and changing the solution colour to pink.

CHEMISTRY, M6 2025 HSC 34

A 0.010 L aliquot of an acid was titrated with 0.10 mol L$^{-1} \ \ce{NaOH}$, resulting in the following titration curve.

Calculate the $K_a$ for the acid used. (3 marks)
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The concentration of the $\ce{NaOH}$ was 0.10 mol L$^{-1}$.
Explain why the pH of the final solution never reached 13. (2 marks)
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a. $\text{Strategy 1:}$

$\text{From the shape of the titration curve, the acid was weak.}$

$\text{Equivalence point}\ \ \Rightarrow \ \ \ce{NaOH}\ \text{added}\ = 0.24\ \text{L} $

$\text{Halfway to the equivalent point = 0.012 L}, \ce{[ HA ]=\left[ A^{-}\right]}$

$\text{Here the pH} \approx 4.4 , \text{or}\ \ce{\left[H^{+}\right] is 4.0 \times 10^{-5}}$.

$K_a=10^{-\text{pH}}=\ce{\left[H+\right]} \times \dfrac{\ce{\left[A^{-}\right]}}{\ce{[HA]}}$

$\text{At this pH,} \ \ce{\left[A^{-}\right]=[HA] so} \ K_a=4.0 \times 10^{-5}$.

$\text{Strategy 2:}$

$\text{Equivalence point is at} \ \ce{0.024 L NaOH added}$.

$\text{Shape of curve shows acid is monoprotic.}$

$\ce{[HA] \times 0.010=0.1 \times 0.024}$

$\ce{[HA]=0.24 mol L^{-1}}$

$\text{pH at start is approximately 2.5}$

$\text{So,} \ \ce{\left[H+\right]=3.16 \times 10^{-3}}$

$K_a=\dfrac{\ce{\left[H^{+}\right]\left[A^{-}\right]}}{\ce{[HA]}} \quad \ce{\left[A^{-}\right]=\left[H^{+}\right]}$

$\ce{[HA]=0.24-3.16 \times 10^{-3}=0.237}$

$K_a=\dfrac{\left(3.16 \times 10^{-3}\right)^2}{0.237}=4.2 \times 10^{-5}$

b. pH of the final solution < 13:

Some of the hydroxide was neutralised by the acid.
The 10 mL of acid also diluted the NaOH.
So the $\ce{NaOH}$ concentration of the mixture will be less than 0.1 mol L$^{-1}$ and the pH will be less than 13.

Show Worked Solution

a. $\text{Strategy 1:}$

$\text{From the shape of the titration curve, the acid was weak.}$

$\text{Equivalence point}\ \ \Rightarrow \ \ \ce{NaOH}\ \text{added}\ = 0.24\ \text{L} $

$\text{Halfway to the equivalent point = 0.012 L}, \ce{[ HA ]=\left[ A^{-}\right]}$

$\text{Here the pH} \approx 4.4 , \text{or}\ \ce{\left[H^{+}\right] is 4.0 \times 10^{-5}}$.

$K_a=10^{-\text{pH}}=\ce{\left[H+\right]} \times \dfrac{\ce{\left[A^{-}\right]}}{\ce{[HA]}}$

$\text{At this pH,} \ \ce{\left[A^{-}\right]=[HA] so} \ K_a=4.0 \times 10^{-5}$.

$\text{Strategy 2:}$

$\text{Equivalence point is at} \ \ce{0.024 L NaOH added}$.

$\text{Shape of curve shows acid is monoprotic.}$

$\ce{[HA] \times 0.010=0.1 \times 0.024}$

$\ce{[HA]=0.24 mol L^{-1}}$

$\text{pH at start is approximately 2.5}$

$\text{So,} \ \ce{\left[H+\right]=3.16 \times 10^{-3}}$

$K_a=\dfrac{\ce{\left[H^{+}\right]\left[A^{-}\right]}}{\ce{[HA]}} \quad \ce{\left[A^{-}\right]=\left[H^{+}\right]}$

$\ce{[HA]=0.24-3.16 \times 10^{-3}=0.237}$

$K_a=\dfrac{\left(3.16 \times 10^{-3}\right)^2}{0.237}=4.2 \times 10^{-5}$

b. pH of the final solution < 13:

Some of the hydroxide was neutralised by the acid.
The 10 mL of acid also diluted the NaOH.
So the $\ce{NaOH}$ concentration of the mixture will be less than 0.1 mol L$^{-1}$ and the pH will be less than 13.

CHEMISTRY, M6 2025 HSC 33

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.

\begin{array}{|l|c|c|c|}
\hline \rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}& \ \ \textit{Brand X} \ \ & \ \ \textit{Brand Y} \ \ & \ \ \textit{Brand Z} \ \ \\
\hline \rule{0pt}{2.5ex}\ce{CaCO3(\%)} \rule[-1ex]{0pt}{0pt}& 85.5 & 83.9 & 82.4 \\
\hline
\end{array}

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 mol L$^{-1} \ \ce{HCl(aq)}$ 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 mol L$^{-1} \ \ce{KOH}$ .

The results of the titrations are recorded.

\begin{array}{|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Burette volume}\text{(mL)} \rule[-1ex]{0pt}{0pt}& \textit{Trial 1} & \textit{Trial 2} & \textit{Trial 3} & \textit{Trial 4} \\
\hline
\rule{0pt}{2.5ex}\text{Final} \rule[-1ex]{0pt}{0pt}& 7.80 & 14.90 & 22.10 & 29.25 \\
\hline
\rule{0pt}{2.5ex}\text{Initial} \rule[-1ex]{0pt}{0pt}& 0.00 & 7.80 & 14.90 & 22.10 \\
\hline
\rule{0pt}{2.5ex}\text{Total used} \rule[-1ex]{0pt}{0pt}& 7.80 & 7.10 & 7.20 & 7.15 \\
\hline
\end{array}

Determine the brand of the chalk sample. Include a relevant chemical equation in your answer. (7 marks)

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$\text{Exclude the outlier (Trial 1):}$

$\text{Average volume} \ \ce{(KOH)} =\dfrac{7.10+7.20+7.15}{3}=0.00715 \ \text{L}$

$\ce{HCl(aq) + KOH(aq) \rightarrow KCl(aq) + H2O(l)}$

$\text{moles} \ \ce{KOH=0.10 \times 0.00715=0.000715 mol }$

$\text{Ratio}\ \ \ce{HCl:KOH=1: 1}$

$\ce{0.000715 mol HCl} \ \text{for each sample}$

$\ce{0.000715 \times 5=0.003575 mol}\ \text{total in sampled solution}$

$\text{Initial} \ \ \ce{n(HCl)=0.550 \times 0.1000=0.0550 mol}$

$\ce{n(HCl)}\ \text{that reacted with} \ \ce{CaCO3=0.0550-0.003575=0.051425 mol}$

$\ce{2HCl(aq) + CaCO3(s) \rightarrow CaCl2(aq) + H2O(l) + CO2(g)}$

$\text{Ratio}\ \ \ce{HCl:CaCO3=2:1}$

$\ce{n(CaCO3)}=\dfrac{0.051425}{2}=0.0257125\ \text{mol}$

$\ce{MM(CaCO3)}=40.08+12.01+3 \times 16=100.09$

$\text{Mass} \ \ce{CaCO3} =0.0257125 \times 100.09=2.5735641\ \text{g}$

$\% \ce{CaCO3}=\dfrac{2.5735641}{3.00} \times 100=85.7854 \% \approx 85.8 \%$

$\text{Chalk sample has to be Brand X.}$

Show Worked Solution

$\text{Exclude the outlier (Trial 1):}$

$\text{Average volume} \ \ce{(KOH)} =\dfrac{7.10+7.20+7.15}{3}=0.00715 \ \text{L}$

$\ce{HCl(aq) + KOH(aq) \rightarrow KCl(aq) + H2O(l)}$

$\text{moles} \ \ce{KOH=0.10 \times 0.00715=0.000715 mol }$

$\text{Ratio}\ \ \ce{HCl:KOH=1: 1}$

$\ce{0.000715 mol HCl} \ \text{for each sample}$

$\ce{0.000715 \times 5=0.003575 mol}\ \text{total in sampled solution}$

$\text{Calculate}\ \ce{HCl}\ \text{that reacted with}\ \ce{CaCO3}:$

$\text{Initial} \ \ \ce{n(HCl)=0.550 \times 0.1000=0.0550 mol}$

$\ce{n(HCl)}\ \text{that reacted with} \ \ce{CaCO3=0.0550-0.003575=0.051425 mol}$

$\ce{2HCl(aq) + CaCO3(s) \rightarrow CaCl2(aq) + H2O(l) + CO2(g)}$

$\text{Ratio}\ \ \ce{HCl:CaCO3=2:1}$

$\ce{n(CaCO3)}=\dfrac{0.051425}{2}=0.0257125\ \text{mol}$

$\ce{MM(CaCO3)}=40.08+12.01+3 \times 16=100.09$

$\text{Mass} \ \ce{CaCO3} =0.0257125 \times 100.09=2.5735641\ \text{g}$

$\% \ce{CaCO3}=\dfrac{2.5735641}{3.00} \times 100=85.7854 \% \approx 85.8 \%$

$\text{Chalk sample has to be Brand X.}$

CHEMISTRY, M5 2025 HSC 32

The following three solids were added together to 1 litre of water:

$\ce{0.006\ \text{mol}\ Mg(NO3)2}$
$\ce{0.010\ \text{mol}\ NaOH}$
$\ce{0.002\ \text{mol}\ Na2CO3}$.

Which precipitate(s), if any, will form? Justify your answer with appropriate calculations. (5 marks)

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All sodium and nitrate salts are soluble $\Rightarrow$ possible precipitates are $\ce{Mg(OH)2}$ and $\ce{MgCO3}$.

$\ce{\left[Mg^{2+}\right]=6 \times 10^{-3} \quad\left[ OH^{-}\right]=1 \times 10^{-2} \quad\left[ CO3^{2-}\right]=2 \times 10^{-3}}$

$\ce{\left[Mg^{2+}\right]\left[ OH^{-}\right]^2=6 \times 10^{-7} \quad \quad \quad \ \ \ K_{\textit{sp}}=5.61 \times 10^{-12}}$

$\ce{\left[ Mg^{2+}\right]\left[ CO3^{2-}\right]=1.2 \times 10^{-5} \quad \quad K_{\textit{sp}}=6.82 \times 10^{-6}}$

$\ce{Mg(OH)2}$ is a lot less soluble than $\ce{MgCO3}$ and will precipitate preferentially.

$\ce{\left[Mg^{2+}\right]\left[OH^{-}\right]^2 > K_{\textit{sp}}}$

$\Rightarrow \ce{Mg(OH)2}$ will precipitate.

$\ce{Mg^{2+}(aq) + 2OH-(aq) \rightarrow Mg(OH)2(s)}$

Since $K_{\textit{sp}}$ is small, assume reaction goes to completion.

$\ce{n(Mg^{2+})=0.006\ mol, n(OH^{-})=0.010\ mol}$

Using stoichiometric ratio $(1:2)$

$0.006 > \dfrac{0.010}{2}=0.005\ \ \Rightarrow \ce{Mg^{2+}}$ is in excess.

Concentration drops to: $6 \times 10^{-3}-5 \times 10^{-3}=1 \times 10^{-3}$.

Check if $\ce{MgCO3}$ will precipitate:

$\ce{\left[Mg^{2+}\right]\left[CO3^{2-}\right]}$ becomes $2 \times 10^{-6} <K_{\textit{sp}}$.

$\ce{\Rightarrow\ MgCO3}$ won’t precipitate.

Show Worked Solution

All sodium and nitrate salts are soluble $\Rightarrow$ possible precipitates are $\ce{Mg(OH)2}$ and $\ce{MgCO3}$.

$\ce{\left[Mg^{2+}\right]=6 \times 10^{-3} \quad\left[ OH^{-}\right]=1 \times 10^{-2} \quad\left[ CO3^{2-}\right]=2 \times 10^{-3}}$

$\ce{\left[Mg^{2+}\right]\left[ OH^{-}\right]^2=6 \times 10^{-7} \quad \quad \quad \ \ \ K_{\textit{sp}}=5.61 \times 10^{-12}}$

$\ce{\left[ Mg^{2+}\right]\left[ CO3^{2-}\right]=1.2 \times 10^{-5} \quad \quad K_{\textit{sp}}=6.82 \times 10^{-6}}$

$\ce{Mg(OH)2}$ is a lot less soluble than $\ce{MgCO3}$ and will precipitate preferentially.

$\ce{\left[Mg^{2+}\right]\left[OH^{-}\right]^2 > K_{\textit{sp}}}$

$\Rightarrow \ce{Mg(OH)2}$ will precipitate.

$\ce{Mg^{2+}(aq) + 2OH-(aq) \rightarrow Mg(OH)2(s)}$

Since $K_{\textit{sp}}$ is small, assume reaction goes to completion.

$\ce{n(Mg^{2+})=0.006\ mol, n(OH^{-})=0.010\ mol}$

Using stoichiometric ratio $(1:2)$

$0.006 > \dfrac{0.010}{2}=0.005\ \ \Rightarrow \ce{Mg^{2+}}$ is in excess.

Concentration drops to: $6 \times 10^{-3}-5 \times 10^{-3}=1 \times 10^{-3}$.

Check if $\ce{MgCO3}$ will precipitate:

$\ce{\left[Mg^{2+}\right]\left[CO3^{2-}\right]}$ becomes $2 \times 10^{-6} <K_{\textit{sp}}$.

$\ce{\Rightarrow\ MgCO3}$ won’t precipitate.

Algebra, STD2 EQ-Bank 11

The amount of water ($W$) in litres used by a garden irrigation system varies directly with the time ($t$) in minutes that the system operates.

This relationship is modelled by the formula $W=kt$, where $k$ is a constant.

The irrigation system uses 96 litres of water when it operates for 24 minutes.

Show that the value of $k$ is 4. (1 mark)
--- 3 WORK AREA LINES (style=lined) ---
The water tank for the irrigation system contains 650 litres of water. Calculate how many minutes the irrigation system can operate before the tank is empty. (2 marks)
--- 5 WORK AREA LINES (style=lined) ---

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a. $W=kt$

$\text{When } W = 96 \text{ and } t = 24:$

$96$	$=k \times 24$
$k$	$=\dfrac{96}{24}$
$k$	$=4\ \text{(as required)}$

b. $162.5\ \text{minutes}$

Show Worked Solution

a. $W=kt$

$\text{When } W = 96 \text{ and } t = 24:$

$96$	$=k \times 24$
$k$	$=\dfrac{96}{24}$
$k$	$=4\ \text{(as required)}$

b. $W = 4t$

$\text{When } W = 650:$

$650$	$=4\times t$
$t$	$ =\dfrac{650}{4}$
	$=162.5\ \text{minutes}$

$\therefore\ \text{The irrigation system can operate for } 162.5 \text{ minutes.}$

Algebra, STD2 EQ-Bank 10

The cost ($C$) of copper wire varies directly with the length ($L$) in metres of the wire.

This relationship is modelled by the formula $C = kL$, where $k$ is a constant.

A 250 metre roll of copper wire costs $87.50.

Show that the value of $k$ is 0.35. (1 mark)
--- 3 WORK AREA LINES (style=lined) ---
A builder has a budget of $140 for copper wire. Calculate the maximum length of wire that can be purchased. (2 marks)
--- 5 WORK AREA LINES (style=lined) ---

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a. $C=kL$

$\text{When } C = 87.50 \text{ and } L = 250:$

$87.50$	$=k \times 250$
$k$	$=\dfrac{87.50}{250}$
$k$	$=0.35\ \text{(as required)}$

b. $400\ \text{m}$

Show Worked Solution

a. $C=kL$

$\text{When } C = 87.50 \text{ and } L = 250:$

$87.50$	$=k \times 250$
$k$	$=\dfrac{87.50}{250}$
$k$	$=0.35\ \text{(as required)}$

b. $C = 0.35L$

$\text{When } C = 140:$

$140$	$=0.35\times L$
$L$	$ =\dfrac{140}{0.35}$
	$=400\ \text{m}$

$\therefore\ \text{The builder can purchase } 400 \text{ metres of wire.}$

Algebra, STD2 EQ-Bank 08 MC

The energy ($E$) required to heat water varies directly with the mass ($m$) of the water.

It takes 2100 joules of energy to heat 500 grams of water by 1°C.

Which equation represents the relationship between $E$ and $m$?

$E = 4.2m$
$E = 0.24m$
$m = 4.2E$
$m = 0.24E$

Show Answers Only

$A$

Show Worked Solution

$E \propto m$

$E=km$

$\text{When } E = 2100 \text{ and } m = 500:$

$2100$	$=k \times 500$
$k$	$=\dfrac{2100}{500}$
$k$	$=4.2$

$\therefore\ E=4.2m$

$\Rightarrow A$

Algebra, STD2 EQ-Bank 07 MC

The distance ($d$) a spring stretches varies directly with the force ($F$) applied to it.

When a force of 18 newtons is applied, the spring stretches 27 mm.

What is the value of the constant of variation ($k$)?

$\dfrac{2}{5}$
$\dfrac{5}{2}$
$\dfrac{2}{3}$
$\dfrac{3}{2}$

Show Answers Only

$D$

Show Worked Solution

$d \propto F$

$d=kF$

$\text{When } d = 27 \text{ and } F = 18:$

$27$	$=k \times 18$
$k$	$=\dfrac{27}{18}$
$k$	$=\dfrac{3}{2}$

$\Rightarrow D$

PHYSICS, M8 2025 HSC 15 MC

Two stars, $X$ and $Y$, are identified on the Hertzsprung-Russell diagram.

In what way are these two stars different?

$X$ has a higher luminosity than $Y$.
$X$ is a red star, and $Y$ is a blue star.
$X$ has a lower core temperature than $Y$.
$X$ has a higher surface temperature than $Y$.

Show Answers Only

$C$

Show Worked Solution

$C$ is correct: The $x$-axis shows surface temperature, not core temperature. $Y$ is a giant and $X$ is a main-sequence star; giants have higher core temperatures due to heavier-element fusion, so this statement is correct.

Other options:

$A$ is incorrect. Luminosity increases up the $y$-axis; since $X$ is below $Y$, it is less luminous.
$B$ is incorrect. Colour depends on surface temperature ($x$-axis). Since $X$ and $Y$ are located above the same $x$-value, they have the same temperature and colour.
$D$ is incorrect. $X$ and $Y$ share the same $x$-position, so they have the same surface temperature.

PHYSICS, M6 2025 HSC 14 MC

A proton having a velocity of $1 \times 10^6 \ \text{m s}^{-1}$ enters a uniform field with a trajectory that is initially perpendicular to the field.

Which row in the table correctly identifies the field, and its effect on the kinetic energy of the proton?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ & \\
\ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ \textit{Type of field}\ \ & \textit{Effect on} \\
\textit{}\rule[-1ex]{0pt}{0pt}& \textit{kinetic energy } \\
\hline
\rule{0pt}{2.5ex}\text{Electric}\rule[-1ex]{0pt}{0pt}&\text{Decreases}\\
\hline
\rule{0pt}{2.5ex}\text{Electric}\rule[-1ex]{0pt}{0pt}& \text{Increases}\\
\hline
\rule{0pt}{2.5ex}\text{Magnetic}\rule[-1ex]{0pt}{0pt}& \text{Decreases} \\
\hline
\rule{0pt}{2.5ex}\text{Magnetic}\rule[-1ex]{0pt}{0pt}& \text{Increases} \\
\hline
\end{array}
\end{align*}

Show Answers Only

$B$

Show Worked Solution

In a magnetic field, the magnetic force is always perpendicular to the velocity, so it changes the direction of motion but does no work (i.e. there is no change in kinetic energy – eliminate $C$ and $D$).
Since the proton enters perpendicular to the electric field, the electric force acts either upward or downward (depending on field direction), giving the proton an acceleration parallel to the field.
This acceleration adds a new velocity component, increasing the proton’s overall speed and kinetic energy.

$\Rightarrow B$

PHYSICS, M5 2025 HSC 13 MC

A mass is attached by a light, inextensible string of length $l$ to a vertical rigid support.

The mass rotates with uniform speed, $v$, in a horizontal circle as shown.

What is the acceleration of the mass?

$g$
$\dfrac{v^2}{l}$
$\dfrac{v^2}{l\, \sin \theta}$
$\dfrac{v^2}{l\, \cos \theta}$

Show Answers Only

$C$

Show Worked Solution

Mass moves in a horizontal circle $\Rightarrow\ \ F_c=\dfrac{mv^2}{r} $
Radius of circle $(r) = l\,\sin\,\theta$
$ma=\dfrac{mv^2}{l\,\sin\,\theta}\ \ \Rightarrow \ \ a=\dfrac{v^2}{l\,\sin\,\theta} $

$\Rightarrow C$

PHYSICS, M7 2025 HSC 9 MC

The diagram shows the absorption spectra of two different stars in the Milky Way galaxy.

Based on the information in the diagram, which of the following statements about the two stars is true?

Star $A$ has a lower density than Star $B$.
Star $A$ has a greater rotational velocity than Star $B$.
Star $A$ has a greater translational velocity than Star $B$.
Star $A$ and Star $B$ have different chemical compositions.

Show Answers Only

$A$

Show Worked Solution

Broader spectral lines occur if a star has a higher density or a higher rotational velocity.

$\Rightarrow A$

PHYSICS, M5 2025 HSC 8 MC

A projectile is launched from point $P$, with speed $u$, at angle $\theta$ to the horizontal. It lands at point $Q$.

The time of flight of the projectile is $t$.

Which row in the table best describes the time to reach maximum height and the speed of the projectile at $Q$?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ & \\
\ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{3.2ex}\textbf{A.}\rule[-1.5ex]{0pt}{0pt}\\
\rule{0pt}{3.2ex}\textbf{B.}\rule[-1.5ex]{0pt}{0pt}\\
\rule{0pt}{3.2ex}\textbf{C.}\rule[-1.5ex]{0pt}{0pt}\\
\rule{0pt}{3.2ex}\textbf{D.}\rule[-1.5ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Time to reach}& \textit{Speed of projectile} \\
\ \textit{maximum height}\ \rule[-1ex]{0pt}{0pt}& \textit{at Q} \\
\hline
\rule{0pt}{2.5ex}>\dfrac{t}{2}\rule[-1ex]{0pt}{0pt}&<u\\
\hline
\rule{0pt}{2.5ex}>\dfrac{t}{2}\rule[-1ex]{0pt}{0pt}& =u\\
\hline
\rule{0pt}{2.5ex}=\dfrac{t}{2}\rule[-1ex]{0pt}{0pt}& =u\\
\hline
\rule{0pt}{2.5ex}=\dfrac{t}{2}\rule[-1ex]{0pt}{0pt}& <u \\
\hline
\end{array}
\end{align*}

Show Answers Only

$A$

Show Worked Solution

Max height occurs at $\frac{t}{2}$ only if the projectile lands at the same height as it is launched. Here, since point $Q$ is higher than point $P$, time to reach max > $\frac{t}{2}$ .
Since the projectile gains GPE at $Q$ and by the Law of Conservation of Energy, it must have a lower speed at $Q$.

$\Rightarrow A$

PHYSICS, M8 2025 HSC 30

A beam of electrons travelling at $4 \times 10^3 \ \text{m s}^{-1}$ exits an electron gun and is directed toward two narrow slits with a separation, $d$, of 1 $\mu\text{m}$. The resulting interference pattern is detected on a screen 50 cm from the slits.

Show that the wavelength of the electrons in this experiment is 182 nm. (2 marks)
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An interference fringe occurs on the screen where constructive interference takes place.

Determine the distance between the central interference fringe $A$ and the centre of the next bright fringe $B$. (2 marks)

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Determine the potential difference acting in the electron gun to accelerate the electrons in the beam from rest to $4 \times 10^3 \ \text{m s}^{-1}$. (2 marks)

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a. $\text {Using} \ \ \lambda=\dfrac{h}{m v}:$

$\lambda=\dfrac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 4 \times 10^3}=1.82 \times 10^{-7} \ \text{m}=182 \ \text{nm}$

b. $x=9.1 \ \text{cm}$

c. $V=4.5 \times 10^{-5} \ \text{V}$

Show Worked Solution

a. $\text {Using} \ \ \lambda=\dfrac{h}{m v}:$

$\lambda=\dfrac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 4 \times 10^3}=1.82 \times 10^{-7} \ \text{m}=182 \ \text{nm}$

b. $\text {Using}\ \ x=\dfrac{\lambda m L}{d}$

$\text{where} \ \ x=\text{distance between middle of adjacent bright spots}$

$x=\dfrac{182 \times 10^{-9} \times 1 \times 0.5}{1 \times 10^{-6}}=0.091 \ \text{m}=9.1 \ \text{cm}$

c. $\text{Work done by field}=\Delta K=K_f-K_i$

$qV$	$=\dfrac{1}{2} m v^2-0$
$V$	$=\dfrac{m v^2}{2 q}=\dfrac{9.109 \times 10^{-31} \times\left(4 \times 10^3\right)^2}{2 \times 1.602 \times 10^{-19}}=4.5 \times 10^{-5} \ \text{V}$

PHYSICS, M8 2025 HSC 33

Analyse the role of experimental evidence and theoretical ideas in developing the Standard Model of matter. (6 marks)

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Overview Statement

The Standard Model’s development depends on the cyclical relationship between experimental evidence and theoretical predictions.
These components interact with each other, where theory guides experiments and results validate or refine theory.

Particle Discovery

Theoretical predictions lead to targeted experimental searches for specific particles.
The Higgs Boson was theoretically proposed decades before its experimental discovery to explain particle mass.
Cloud chambers discovered antimatter after theory predicted its existence.
Particle accelerators verified quarks existed by revealing the internal structure of protons and neutrons.
This pattern shows theory provides the framework while experiments confirm reality.
Consequently, successful verification enables confidence in theoretical models and guides further predictions.

Experimental Tools Driving Theoretical Refinement

High-energy particle accelerators create small wavelength ‘matter probes’ allowing high-resolution investigation of matter’s structure.
These experiments verified electroweak theory by demonstrating electromagnetic and weak nuclear forces result from the same underlying interaction.
Unexpected experimental results sometimes cause theoretical modifications or new predictions.
The significance is that increasingly powerful experimental tools reveal deeper layers of matter structure.

Implications and Synthesis

This reveals the Standard Model emerged from iterative cycles where theory and experiment continuously influence each other.
Neither component alone could have produced the model.
Together, they form a self-correcting system advancing our understanding of fundamental matter.

Show Worked Solution

Overview Statement

The Standard Model’s development depends on the cyclical relationship between experimental evidence and theoretical predictions.
These components interact with each other, where theory guides experiments and results validate or refine theory.

Particle Discovery

Theoretical predictions lead to targeted experimental searches for specific particles.
The Higgs Boson was theoretically proposed decades before its experimental discovery to explain particle mass.
Cloud chambers discovered antimatter after theory predicted its existence.
Particle accelerators verified quarks existed by revealing the internal structure of protons and neutrons.
This pattern shows theory provides the framework while experiments confirm reality.
Consequently, successful verification enables confidence in theoretical models and guides further predictions.

Experimental Tools Driving Theoretical Refinement

High-energy particle accelerators create small wavelength ‘matter probes’ allowing high-resolution investigation of matter’s structure.
These experiments verified electroweak theory by demonstrating electromagnetic and weak nuclear forces result from the same underlying interaction.
Unexpected experimental results sometimes cause theoretical modifications or new predictions.
The significance is that increasingly powerful experimental tools reveal deeper layers of matter structure.

Implications and Synthesis

This reveals the Standard Model emerged from iterative cycles where theory and experiment continuously influence each other.
Neither component alone could have produced the model.
Together, they form a self-correcting system advancing our understanding of fundamental matter.

PHYSICS, M7 2025 HSC 32

Analyse the consequences of the theory of special relativity in relation to length, time and motion. Support your answer with reference to experimental evidence. (8 marks)

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Overview Statement

Special relativity predicts that time dilation, length contraction and relativistic momentum arise from the principle that the speed of light is constant for all observers.
These effects change how time, distance and motion are measured and each consequence is supported by experimental evidence.

Time–Length Relationship in Muon Observations

Muon-decay experiments show how time dilation and length contraction depend on relative motion.
Muons created high in the atmosphere have such short lifetimes that, in classical physics, they should decay long before reaching Earth’s surface. Yet far more muons are detected at ground level than predicted.
In Earth’s frame of reference, the muons experience time dilation, so they “live longer” and travel further.
In the muon’s frame of reference, the atmosphere is length-contracted according to the equation $l=l_0 \sqrt{\left(1-\dfrac{v^{2}}{c^{2}}\right)}$, so the distance they travel is much shorter.
Both viewpoints are valid within their own reference frames, showing that time and length are not absolute but depend on relative motion.

Momentum–Energy Relationship in Particle Accelerators

Relativistic momentum explains how objects behave as they approach light speed.
Particle accelerators show that enormous increases in energy produce only small increases in speed at high velocities.
Particles act as though their mass increases, so each additional acceleration requires disproportionately more energy.
This makes it impossible for any object with mass to reach the speed of light, since doing so requires infinite energy.
Thus, relativistic momentum preserves light speed as a universal limit.

Implications and Synthesis

The consequences of these observations reveal that space and time form a single, interconnected framework rather than separate absolute quantities.
At high velocities, motion fundamentally alters measurements of time, distance and momentum.
Together, these consequences confirm that classical physics fails at relativistic speeds and that special relativity accurately describes the behaviour of fast-moving objects.

Show Worked Solution

Overview Statement

Special relativity predicts that time dilation, length contraction and relativistic momentum arise from the principle that the speed of light is constant for all observers.
These effects change how time, distance and motion are measured and each consequence is supported by experimental evidence.

Time–Length Relationship in Muon Observations

Muon-decay experiments show how time dilation and length contraction depend on relative motion.
Muons created high in the atmosphere have such short lifetimes that, in classical physics, they should decay long before reaching Earth’s surface. Yet far more muons are detected at ground level than predicted.
In Earth’s frame of reference, the muons experience time dilation, so they “live longer” and travel further.
In the muon’s frame of reference, the atmosphere is length-contracted according to the equation $l=l_0 \sqrt{\left(1-\dfrac{v^{2}}{c^{2}}\right)}$, so the distance they travel is much shorter.
Both viewpoints are valid within their own reference frames, showing that time and length are not absolute but depend on relative motion.

Momentum–Energy Relationship in Particle Accelerators

Relativistic momentum explains how objects behave as they approach light speed.
Particle accelerators show that enormous increases in energy produce only small increases in speed at high velocities.
Particles act as though their mass increases, so each additional acceleration requires disproportionately more energy.
This makes it impossible for any object with mass to reach the speed of light, since doing so requires infinite energy.
Thus, relativistic momentum preserves light speed as a universal limit.

Implications and Synthesis

The consequences of these observations reveal that space and time form a single, interconnected framework rather than separate absolute quantities.
At high velocities, motion fundamentally alters measurements of time, distance and momentum.
Together, these consequences confirm that classical physics fails at relativistic speeds and that special relativity accurately describes the behaviour of fast-moving objects.

PHYSICS, M8 2025 HSC 31

Experiments have been carried out by scientists to investigate cathode rays.

Assess the contribution of the results of these experiments in developing an understanding of the existence and properties of electrons. (5 marks)

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Judgment Statement

The cathode ray experiments were highly valuable in establishing both the existence and properties of electrons through definitive experimental evidence and quantitative measurements.

Demonstrating Particle Nature

Electric field deflection experiments produced significant results by proving cathode rays were negatively charged particles rather than electromagnetic radiation.
This was highly effective because it eliminated the competing theory that cathode rays were electromagnetic waves, since waves are not deflected by electric fields.
The consistent deflection pattern across many experiments provided strong evidence for the particle nature of electrons.

Quantifying Electron Properties

By adjusting electric and magnetic field strengths within experiments, the charge-to-mass ratio of the electron was determined.
This measurement proved highly effective as it provided the first quantitative property of electrons.
The e/m ratio demonstrated considerable value by revealing electrons were much lighter than atoms, indicating subatomic particles existed.

Overall Assessment

Assessment reveals these experiments achieved major significance in atomic theory development.
The combined results produced measurable, reproducible data that definitively established electrons as fundamental charged particles with specific properties.
Overall, these contributions proved essential for understanding atomic structure.

PHYSICS, M6 2025 HSC 6 MC

In the diagram, an electron enters the shaded region where it is subjected to an external magnetic field that causes it to move in a circular arc, as indicated by the dotted line.

Which magnetic field could produce the motion of the electron shown?

Show Answers Only

$A$

Show Worked Solution

The electron curves in a circular path, so a magnetic force must act at right angles to its velocity.
When it first enters the shaded region, the force on the electron is upward.
For a positive charge, the right-hand palm rule $(F = qvB)$ would give a magnetic field into or out of the page depending on the direction of the force.
Since an electron is negatively charged, the actual direction of the magnetic field is the opposite of what the right-hand rule predicts.
Reversing the direction gives a magnetic field out of the page as it is the only field direction that produces the observed upward force on a negative charge.

$\Rightarrow A$

PHYSICS, M5 2025 HSC 5 MC

A planet orbits a star in an elliptical orbit as shown.

At which point in its orbit is the planet's kinetic energy increasing?

$W$
$X$
$Y$
$Z$

Show Answers Only

$D$

Show Worked Solution

As the planet gets closer to the sun in its orbit, its gravitational potential energy decreases.
The Law of Conservation of Energy means that as the planet’s potential energy decreases, its kinetic energy increases.
This occurs at point $Z$, noting that the planet is moving perpendicular to the sun at $W$ and $Y$.

$\Rightarrow D$

PHYSICS, M8 2025 HSC 4 MC

Which row in the table identifies the particle with the shortest de Broglie wavelength?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \textit{Particle}\quad\rule[-1ex]{0pt}{0pt}& \quad\textit{Velocity}\quad \\
\hline
\rule{0pt}{2.5ex}\text{Electron}\rule[-1ex]{0pt}{0pt}&0.1 c\\
\hline
\rule{0pt}{2.5ex}\text{Electron}\rule[-1ex]{0pt}{0pt}& 0.9 c\\
\hline
\rule{0pt}{2.5ex}\text{Proton}\rule[-1ex]{0pt}{0pt}& 0.1 c \\
\hline
\rule{0pt}{2.5ex}\text{Proton}\rule[-1ex]{0pt}{0pt}& 0.9 c \\
\hline
\end{array}
\end{align*}

Show Answers Only

$D$

Show Worked Solution

Using $\lambda = \dfrac{h}{p}$, the large momentum $(p = mv)$ of a fast, heavy particle makes $\lambda$ smallest.
The proton at 0.9$c$ is the heaviest and fastest option. Since this has the greatest momentum, it has the shortest de Broglie wavelength.

$\Rightarrow D$

PHYSICS, M5 2025 HSC 29

A mass moves around a vertical circular path of radius $r$, in Earth's gravitational field, without loss of mechanical energy. A string of length $r$ maintains the circular motion of the mass.

When the mass is at its highest point $B$, the tension in the string is zero.

Show that the speed of the mass at the highest point, $B$, is given by $v=\sqrt{r g}$. (2 marks)

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Compare the speed of the mass at point $A$ to that at point $B$. Support your answer using appropriate mathematical relationships. (3 marks)

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a. $\text {At point} \ B \ \Rightarrow \ F_c=F_{\text {net}}$

$\dfrac{mv_B^2}{r}$	$=T+mg$
$v_B^2$	$=rg \ \ (T=0)$
$v_B$	$=\sqrt{r g}$

b. $\text {Total} \ ME=E_k+GPE$

$\text{At point} \ B :$

$\text {Total} \ ME$	$=\dfrac{1}{2} m v_B^2+mg(2r)$
	$=\dfrac{1}{2} m \times r g+2 mrg$
	$=\dfrac{5}{2} m r g$

$\text{At point} \ A :$

$\text {Total} \ ME=\dfrac{1}{2} mv_A^2+mrg$

$\text{Since \(ME$ is conserved:}\)

$\dfrac{5}{2} mrg$	$=\dfrac{1}{2} m v_A^2+m r g$
$\dfrac{1}{2} mv_A^2$	$=\dfrac{3}{2} m r g$
$v_A^2$	$=3 rg$
$v_A$	$=\sqrt{3rg}=\sqrt{3} \times v_B$

Show Worked Solution

a. $\text {At point} \ B \ \Rightarrow \ F_c=F_{\text {net}}$

$\dfrac{mv_B^2}{r}$	$=T+mg$
$v_B^2$	$=rg \ \ (T=0)$
$v_B$	$=\sqrt{r g}$

b. $\text {Total} \ ME=E_k+GPE$

$\text{At point} \ B :$

$\text {Total} \ ME$	$=\dfrac{1}{2} m v_B^2+mg(2r)$
	$=\dfrac{1}{2} m \times r g+2 mrg$
	$=\dfrac{5}{2} m r g$

$\text{At point} \ A :$

$\text {Total} \ ME=\dfrac{1}{2} mv_A^2+mrg$

$\text{Since \(ME$ is conserved:}\)

$\dfrac{5}{2} mrg$	$=\dfrac{1}{2} m v_A^2+m r g$
$\dfrac{1}{2} mv_A^2$	$=\dfrac{3}{2} m r g$
$v_A^2$	$=3 rg$
$v_A$	$=\sqrt{3rg}=\sqrt{3} \times v_B$

PHYSICS, M6 2025 HSC 26

The starting position of a simple AC generator is shown. It consists of a single rectangular loop of wire in a uniform magnetic field of 0.5 T. This loop is connected to two slip rings and the slip rings are connected via brushes to a voltmeter.

The loop is rotated at a constant rate through an angle of 90 degrees from the starting position in the direction indicated, in 0.1 seconds.
Calculate the magnitude of the average emf generated during this rotation. (2 marks)

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The same coil was then rotated at 10 revolutions per second from the starting position. The voltage varies with time, as shown in the graph.

On the same axes, sketch a graph that shows the variation of voltage with time if the rotational speed is 20 revolutions per second in the opposite direction, beginning at the original starting position. (3 marks)

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a. $\text{Using}\ \ \phi=BA \ \ \text{and} \ \ \varepsilon=\dfrac{\Delta \phi}{\Delta t}$

$\varepsilon=\dfrac{\Delta(B A)}{\Delta t}=\dfrac{B \Delta A}{\Delta t}=\dfrac{0.5 \times 0.4 \times 0.3}{0.1}=0.6\ \text{V}$

Show Worked Solution

a. $\text{Using}\ \ \phi=BA \ \ \text{and} \ \ \varepsilon=\dfrac{\Delta \phi}{\Delta t}$

$\varepsilon=\dfrac{\Delta(B A)}{\Delta t}=\dfrac{B \Delta A}{\Delta t}=\dfrac{0.5 \times 0.4 \times 0.3}{0.1}=0.6\ \text{V}$

PHYSICS, M7 2025 HSC 25

A student conducts an experiment to determine the work function of potassium.

The following diagram depicts the experimental setup used, where light of varying frequency is incident on a potassium electrode inside an evacuated tube.

For each frequency of light tested, the voltage in the circuit is varied, and the minimum voltage (called the stopping voltage) required to bring the current in the circuit to zero is recorded.

\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ \textit{Frequency of incident light} \ \ & \quad \quad \quad \textit{Stopping voltage} \quad \quad \quad \\
\left(\times 10^{15} \ \text{Hz}\right) \rule[-1ex]{0pt}{0pt}& \text{(V)}\\
\hline
\rule{0pt}{2.5ex}0.9 \rule[-1ex]{0pt}{0pt}& 1.5 \\
\hline
\rule{0pt}{2.5ex}1.1 \rule[-1ex]{0pt}{0pt}& 2.0 \\
\hline
\rule{0pt}{2.5ex}1.2 \rule[-1ex]{0pt}{0pt}& 2.5 \\
\hline
\rule{0pt}{2.5ex}1.3 \rule[-1ex]{0pt}{0pt}& 3.0 \\
\hline
\rule{0pt}{2.5ex}1.4 \rule[-1ex]{0pt}{0pt}& 3.5 \\
\hline
\rule{0pt}{2.5ex}1.5 \rule[-1ex]{0pt}{0pt}& 4.0 \\
\hline
\end{array}

Construct an appropriate graph using the data provided, and from this, determine the threshold frequency of potassium. (3 marks)

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Using the particle model of light, explain the features shown in the experimental results. (3 marks)

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Show Answers Only

b. Features shown in experiment results:

Light is made of discrete energy packets called photons.
A photon’s energy is directly proportional to its frequency: $E = hf$.
When photons hit a metal surface, they transfer their energy to electrons.
If the photon energy is greater than the metal’s work function $(\phi)$, the threshold frequency is exceeded and electrons are ejected with kinetic energy equal to the excess energy according to the equation $KE = hf-\phi$.
Increasing the frequency increases the photon energy which results in ejected electrons with greater kinetic energy.
A larger stopping voltage is therefore needed to halt these more energetic electrons.

Show Worked Solution

b. Features shown in experiment results:

Light is made of discrete energy packets called photons.
A photon’s energy is directly proportional to its frequency: $E = hf$.
When photons hit a metal surface, they transfer their energy to electrons.
If the photon energy is greater than the metal’s work function $(\phi)$, the threshold frequency is exceeded and electrons are ejected with kinetic energy equal to the excess energy according to the equation $KE = hf-\phi$.
Increasing the frequency increases the photon energy which results in ejected electrons with greater kinetic energy.
A larger stopping voltage is therefore needed to halt these more energetic electrons.

PHYSICS, M8 2025 HSC 27

Outline TWO ways in which Schrödinger’s model of electron behaviour is different from electron behaviour in the atomic models of Rutherford and Bohr. (3 marks)

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Answers could include two of the following:

Electron location

Bohr/Rutherford: Electrons move in fixed paths (or orbits) around the nucleus.
Schrödinger: Electrons exist in orbitals, which are regions where they are likely to be found.

Nature of the electron

Bohr/Rutherford: Electrons are treated mainly as particles.
Schrödinger: Electrons behave as waves, following de Broglie’s wave ideas.

Certainty vs probability (extra option)

Bohr/Rutherford: The position of an electron can be predicted exactly in its orbit.
Schrödinger: Only the probability of an electron’s position can be known, not its exact location.

Show Worked Solution

Answers could include two of the following:

Electron location

Bohr/Rutherford: Electrons move in fixed paths (or orbits) around the nucleus.
Schrödinger: Electrons exist in orbitals, which are regions where they are likely to be found.

Nature of the electron

Bohr/Rutherford: Electrons are treated mainly as particles.
Schrödinger: Electrons behave as waves, following de Broglie’s wave ideas.

Certainty vs probability (extra option)

Bohr/Rutherford: The position of an electron can be predicted exactly in its orbit.
Schrödinger: Only the probability of an electron’s position can be known, not its exact location.

PHYSICS, M5 2025 HSC 24

Two satellites, $A$ and $B$, are in stable circular orbits around the Earth. The radius of satellite $A$'s orbit is three times that of satellite $B$'s orbit. Both satellites have the same kinetic energy.

Show that the mass of $A$ is three times the mass of $B$. (3 marks)

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$\text{Show}\ \ m_A=3 \times m_B$

$\text{Since \(A$ and $B$ are in stable orbits:}\)

$F_C$	$=F_G$
$\dfrac{m v^2}{r}$	$=\dfrac{G Mm}{r}$
$\dfrac{1}{2} m v^2$	$=\dfrac{G M m}{2 r}$

$\text{Since}\ \ K=\dfrac{1}{2} m v^2 \ \ \text{and} \ \ K_A=K_B \ \text{(given)}:$

$\dfrac{GMm_{A}}{2r_A}$	$=\dfrac{GMm_{B}}{2r_B}$
$\dfrac{m_A}{r_A}$	$=\dfrac{m_B}{r_B}$
$\dfrac{m_A}{3r_B}$	$=\dfrac{m_B}{r_B}\left(r_A=3 \times r_B\right)$
$m_A$	$=3 \times m_B \ \text{… as required}$

Show Worked Solution

$\text{Show}\ \ m_A=3 \times m_B$

$\text{Since \(A$ and $B$ are in stable orbits:}\)

$F_C$	$=F_G$
$\dfrac{m v^2}{r}$	$=\dfrac{G Mm}{r}$
$\dfrac{1}{2} m v^2$	$=\dfrac{G M m}{2 r}$

$\text{Since}\ \ K=\dfrac{1}{2} m v^2 \ \ \text{and} \ \ K_A=K_B \ \text{(given)}:$

$\dfrac{GMm_{A}}{2r_A}$	$=\dfrac{GMm_{B}}{2r_B}$
$\dfrac{m_A}{r_A}$	$=\dfrac{m_B}{r_B}$
$\dfrac{m_A}{3r_B}$	$=\dfrac{m_B}{r_B}\left(r_A=3 \times r_B\right)$
$m_A$	$=3 \times m_B \ \text{… as required}$

PHYSICS, M6 2025 HSC 23

A wire loop is carrying a current of 2 A from $A$ to $B$ as shown. The length of wire within the magnetic field is 5 cm. The loop is free to pivot around the axis. The magnetic field is of magnitude $3 \times 10^{-2}\ \text{T}$ and at right angles to the wire.

Determine the torque produced on the wire loop due to the motor effect. (3 marks)

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Both the current and the magnetic field were changed, and the torque was observed to be in the same direction but twice the magnitude.
What changes to the magnitude of BOTH the current and the magnetic field are required to produce this result? (2 marks)

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a. $\tau=6 \times 10^{-4} \ \text{Nm clockwise from side} \ B$

b. $\text{Torque is doubled in same direction (given)}$

$\text{Using}\ \ \tau=BILd:$

$\text{If current is halved}\ \ \left(I \rightarrow \frac{1}{2} I\right) \ \ \text{and the magnetic field is}$

$\text {increased by a factor of} \ 4 \ (B \rightarrow 4B),\ \text{torque is doubled.}$

$\tau_{\text{new}}=4B \times \frac{1}{2} I \times Ld=2 \times BILd=2 \tau_{\text {orig }}$

Show Worked Solution

a. $\text{Convert units:}\ \ 5 \ \text{cm} \ \Rightarrow \ \dfrac{5}{100}=0.05 \ \text{m}$

$F=BIL=3 \times 10^{-2} \times 2 \times 0.05=0.003 \ \text{N}$

$\text{Convert units:}\ \ 20\ \text{cm} \ \Rightarrow \ \ 0.2 \ \text{m}$

$\tau$	$=F d$
	$=0.003 \times 0.2$
	$=6 \times 10^{-4} \ \text{Nm clockwise from side} \ B$

b. $\text{Torque is doubled in same direction (given)}$

$\text{Using}\ \ \tau=BILd:$

$\text{If current is halved}\ \ \left(I \rightarrow \frac{1}{2} I\right) \ \ \text{and the magnetic field is}$

$\text {increased by a factor of} \ 4 \ (B \rightarrow 4B),\ \text{torque is doubled.}$

$\tau_{\text{new}}=4B \times \frac{1}{2} I \times Ld=2 \times BILd=2 \tau_{\text {orig }}$

Financial Maths, STD2 EQ-Bank 02

Kylie is a financial broker who earns a salary of $\$93\,600$ per annum.
She has 35% of her salary deducted for tax.
Show that her net weekly pay is $1170. (1 marks)
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Ben, Kylie's partner, works as a carpenter.
He works 36 hours per week at a normal rate of $40 per hour and averages 6 hours overtime at time-and-a-half.
Show that his average weekly pay before tax is $1800. (2 marks)
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Kylie and Ben drew up the weekly budget below.
They need to save $400 per week for an overseas holiday and also want to continue to save for a home.
Ben has 25% of his gross wage deducted for taxation.
Comment on the budget as it appears in the table, indicating where they may have made an error, and suggest some practical ways to make the budget work for them. (3 marks)
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Show Answers Only

a.	$\text{Kylie’s net salary}$	$=\dfrac{$93\,600\times 0.65}{52}$
		$=$1170$

b.	$\text{Ben’s gross pay}$	$=$40\times (36+1.5\times 6)$
		$=$1800$

c.	$\text{Ben’s net pay}$	$=1800\times 0.75$
		$=$1350$
	$\text{Total net income}$	$=1170+1350=$2520$
	$\text{Total expenses}$	$=1000+360+210+250+250+400+500$
		$=$2720$

$\text{They have mistakenly used Ben’s gross salary in their}$

$\text{calculations which will leave them short of their savings}$

$\text{target of }$500\ \text{per week by } $200. $

$\text{Option 1. Reduce savings by }$200\ \text{per week.}$

$\text{Option 2. Reduce entertainment by by }$100 $

$\text{per week and reduce savings by }$100\ \text{per week.}$

Show Worked Solution

a.	$\text{Kylie’s net salary}$	$=\dfrac{93\,600\times 0.65}{52}$
		$=$1170$

b.	$\text{Ben’s gross pay}$	$=40\times (36+1.5\times 6)$
		$=$1800$

c.	$\text{Ben’s net pay}$	$=1800\times 0.75$
		$=$1350$
	$\text{Total net income}$	$=1170+1350=$2520$
	$\text{Total expenses}$	$=1000+360+210+250+250+400+500$
		$=$2720$

$\text{They have mistakenly used Ben’s gross salary in their}$

$\text{calculations which will leave them short of their savings}$

$\text{target of }$500\ \text{per week by } $200. $

$\text{Option 1. Reduce savings by }$200\ \text{per week.}$

$\text{Option 2. Reduce entertainment by by }$100 $

$\text{per week and reduce savings by }$100\ \text{per week.}$

BIOLOGY, M7 2025 HSC 21

The diagram shows components of the innate immune system in humans.

State the role of TWO components that protect against infection. (2 marks)

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \quad \textit{Component} \quad \quad \rule[-1ex]{0pt}{0pt} & \quad \quad \textit{How it protects against infection}\quad \quad \rule[-1ex]{0pt}{0pt}\\
\hline
\ & \\
\ & \\
\ & \\
\ & \\
\ & \\
\hline
\ & \\
\ & \\
\ & \\
\ & \\
\ & \\
\hline
\end{array}

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Any TWO of the following components

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex}\textit{Component} \quad \quad \rule[-1ex]{0pt}{0pt} & \textit{How it protects against infection}\quad\rule[-1ex]{0pt}{0pt}\\
\hline
\text{Skin} & \text{Acts as a physical barrier}\\
\ & \text{preventing pathogen entry into} \\
\ & \text{tissue}\\
\hline
\text{Stomach} &\text{Destroys ingested pathogens} \\
\text{acid} & \text{through low pH chemical}\\
\ & \text{environment.}\\
\hline
\text{Mucus} &\text{Traps pathogens and prevents} \\
\text{lining} & \text{their entry into underlying}\\
\ & \text{tissues.}\\
\hline
\text{Nasal} &\text{Filters and traps airborne} \\
\text{hair} & \text{pathogens, preventing}\\
\ & \text{respiratory entry.}\\
\hline
\text{Tear glands} &\text{Produce lysozyme enzyme that} \\
\ & \text{destroys bacterial cell walls.}\\
\hline
\text{Urinary} &\text{Flushes pathogens from the} \\
\text{tract} & \text{urethra preventing infection.}\\
\hline
\end{array}

Show Worked Solution

Any TWO of the following components

BIOLOGY, M8 2025 HSC 24

The following flow chart represents the control of body temperature in humans.

Complete the flow chart to give an example of mechanism A and an example of mechanism B. (2 marks)
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Outline how mechanism B maintains homeostasis. (2 marks)
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a. Mechanism A (Decreases temperature):

Sweating/perspiration → Vasodilation

Mechanism B (Increases temperature):

Shivering → Vasoconstriction

b. Maintaining homeostasis

When body temperature drops below normal range, thermoreceptors detect the change.
The hypothalamus (control centre) activates mechanism B responses like shivering and vasoconstriction.
Shivering generates heat through muscle contractions whilst vasoconstriction reduces heat loss.
Body temperature increases back to normal range, restoring homeostasis.

Show Worked Solution

a. Mechanism A (Decreases temperature):

Sweating/perspiration → Vasodilation

Mechanism B (Increases temperature):

Shivering → Vasoconstriction

b. Maintaining homeostasis

When body temperature drops below normal range, thermoreceptors detect the change.
The hypothalamus (control centre) activates mechanism B responses like shivering and vasoconstriction.
Shivering generates heat through muscle contractions whilst vasoconstriction reduces heat loss.
Body temperature increases back to normal range, restoring homeostasis.

BIOLOGY, M5 2025 HSC 31

Congenital amegakaryocytic thrombocytopenia (CAMT) is a rare, inherited disorder where bone marrow no longer makes platelets that are important for clotting and preventing bleeding. The pedigree below shows the inheritance of CAMT in a family.

What type of inheritance is shown in the pedigree above? Justify your answer? (3 marks)

Type of Inheritance:

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A CAMT mutation was found to produce the following amino acid sequence:
1. Glutamine – Tyrosine – Isoleucine – Aspartic acid.
The same DNA fragment has been sequenced from an unaffected individual.
Template strand GTC ATA CAG CTG.
The following codon chart displays all the codons and corresponding amino acids. The chart translates mRNA sequences into amino acids.
Use the codon chart shown to explain the type of mutation which causes CAMT. (3 marks)

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a. Type of inheritance: Autosomal recessive

Both males and females are affected equally, ruling out sex-linked inheritance.
Two affected parents (10 and 11) produce only affected offspring (17, 18, 19). This is consistent with autosomal recessive inheritance (aa × aa = all aa).
The disorder skips generations. Unaffected carriers can pass on the recessive allele without expressing the phenotype.
Affected individual 5 and unaffected individual 6 produce affected child 13, confirming individual 6 is a heterozygous carrier.

b. Type of mutation causing CAMT

The template strand transcribes to mRNA CAG UAU GUC GAC, which translates to Glutamine-Tyrosine-Valine-Aspartic acid in unaffected individuals.
In CAMT, Isoleucine replaces Valine at position 3. This results from a single nucleotide substitution changing the codon from GUC to an Isoleucine codon.
This is a point mutation (missense mutation). This causes one amino acid replacement, which affects protein function and leads to impaired platelet production.

Show Worked Solution

a. Type of inheritance: Autosomal recessive

Both males and females are affected equally, ruling out sex-linked inheritance.
Two affected parents (10 and 11) produce only affected offspring (17, 18, 19). This is consistent with autosomal recessive inheritance (aa × aa = all aa).
The disorder skips generations. Unaffected carriers can pass on the recessive allele without expressing the phenotype.
Affected individual 5 and unaffected individual 6 produce affected child 13, confirming individual 6 is a heterozygous carrier.

b. Type of mutation causing CAMT

The template strand transcribes to mRNA CAG UAU GUC GAC, which translates to Glutamine-Tyrosine-Valine-Aspartic acid in unaffected individuals.
In CAMT, Isoleucine replaces Valine at position 3. This results from a single nucleotide substitution changing the codon from GUC to an Isoleucine codon.
This is a point mutation (missense mutation). This causes one amino acid replacement, which affects protein function and leads to impaired platelet production.

BIOLOGY, M6 2025 HSC 34

The following graph shows the changes in allele frequencies in two separate populations of the same species. Each line represents an introduced allele.

Explain why the fluctuations of the allele frequencies are more pronounced in the small population, compared to the larger population. (2 marks)

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Evaluate the effects of gene flow on the gene pools of the two populations. (4 marks)
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a. Genetic Drift and Population Size

Small populations are more susceptible to genetic drift because each individual represents a larger proportion of the gene pool.
Random events cause greater fluctuations, while larger populations buffer these changes, resulting in stable allele frequencies.

b. Evaluation Statement

Gene flow is highly effective for maintaining genetic diversity in the larger population but shows limited effectiveness in the smaller population.

Population Size Differences

The large population (2000) demonstrates strong effectiveness in maintaining stable allele frequencies around 0.5.
This occurs because gene flow introduces consistent genetic material that prevents random loss of alleles.
The population size allows introduced alleles to establish without being lost through drift.

Vulnerability in Small Populations

The small population (20) shows limited benefit from gene flow.
Despite introduction of new alleles, genetic drift overwhelms the stabilising effect.
Multiple alleles are lost completely, demonstrating that population size critically determines whether gene flow can maintain genetic diversity.

Final Evaluation

Overall, gene flow proves highly effective in large populations for maintaining diversity,.
However, it demonstrates insufficient effectiveness in small populations where stochastic processes dominate.

Show Worked Solution

a. Genetic Drift and Population Size

Small populations are more susceptible to genetic drift because each individual represents a larger proportion of the gene pool.
Random events cause greater fluctuations, while larger populations buffer these changes, resulting in stable allele frequencies.

b. Evaluation Statement

Gene flow is highly effective for maintaining genetic diversity in the larger population but shows limited effectiveness in the smaller population.

Population Size Differences

The large population (2000) demonstrates strong effectiveness in maintaining stable allele frequencies around 0.5.
This occurs because gene flow introduces consistent genetic material that prevents random loss of alleles.
The population size allows introduced alleles to establish without being lost through drift.

Vulnerability in Small Populations

The small population (20) shows limited benefit from gene flow.
Despite introduction of new alleles, genetic drift overwhelms the stabilising effect.
Multiple alleles are lost completely, demonstrating that population size critically determines whether gene flow can maintain genetic diversity.

Final Evaluation

Overall, gene flow proves highly effective in large populations for maintaining diversity,.
However, it demonstrates insufficient effectiveness in small populations where stochastic processes dominate.

BIOLOGY, M8 2025 HSC 26

The diagram shows the steps in LASIK (Laser-Assisted In Situ Keratomileusis) surgery.

Compare the LASIK technology shown with ONE other technology that can be used to treat a named visual disorder. (4 marks)

Visual disorder:

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Visual disorder: Myopia (short-sightedness)

Similarities:

Both LASIK surgery and corrective spectacles correct refractive errors by changing light focus.
Both technologies enable clear vision at distance for myopia patients.

Differences:

LASIK permanently reshapes the cornea using laser ablation to correct vision.
Spectacles use external convex or concave lenses to refract light without altering eye structure.
LASIK requires surgical procedure with recovery time whilst spectacles require no invasive procedure.
LASIK provides permanent correction whereas spectacles require continuous wear for vision correction.

Show Worked Solution

Visual disorder: Myopia (short-sightedness)

Similarities:

Both LASIK surgery and corrective spectacles correct refractive errors by changing light focus.
Both technologies enable clear vision at distance for myopia patients.

Differences:

LASIK permanently reshapes the cornea using laser ablation to correct vision.
Spectacles use external convex or concave lenses to refract light without altering eye structure.
LASIK requires surgical procedure with recovery time whilst spectacles require no invasive procedure.
LASIK provides permanent correction whereas spectacles require continuous wear for vision correction.

BIOLOGY, M7 2025 HSC 28

Alpha-gal syndrome (AGS) is a tick-borne allergy to red meat caused by tick bites. Alpha-gal is a sugar molecule found in most mammals but not humans, and can also be found in the saliva of ticks. The diagram shows how a tick bite might cause a person to develop an allergic reaction to red meat.

The flow chart shows the process of antibody production following exposure to alpha-gal.
Describe the role of X, Y and Z in the process of antibody production. (4 marks)

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An allergic reaction to alpha-gal sugar is similar to a secondary immune response.
Describe the features of antibody production shown in the graph. (2 marks)

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Explain the role of memory cells in the immune response. (3 marks)

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a. Antibody Production Process

X is a Helper T-cell that recognises the alpha-gal antigen presented by macrophages on MHC-II molecules.
Helper T-cells activate and coordinate the adaptive immune response through cytokine release.
Y is a B-cell that has receptors specific to the alpha-gal antigen.
B-cells are activated by Helper T-cells and undergo clonal expansion.
Some B-cells differentiate into memory cells for long-term immunity.
Z is a Plasma cell, which is a differentiated B-cell specialised for antibody production.
Plasma cells produce large quantities of antibodies specific to alpha-gal that circulate in the bloodstream.

b. Features of Antibody Production

Initial tick bite produces low antibody concentration with slow, gradual increase over time, representing primary immune response.
Subsequent meat consumption triggers rapid elevation to higher antibody concentration, demonstrating secondary immune response with accelerated, amplified production.

c. Role of Memory Cells

Memory cells are produced during primary exposure and remain in circulation for years, maintaining immunological memory.
Upon re-exposure, memory cells rapidly recognise the specific antigen, which triggers immediate clonal expansion.
This results in faster and stronger antibody production because memory cells bypass the initial activation phase. Hence, providing enhanced immune protection against subsequent infections.

Show Worked Solution

a. Antibody Production Process

X is a Helper T-cell that recognises the alpha-gal antigen presented by macrophages on MHC-II molecules.
Helper T-cells activate and coordinate the adaptive immune response through cytokine release.
Y is a B-cell that has receptors specific to the alpha-gal antigen.
B-cells are activated by Helper T-cells and undergo clonal expansion.
Some B-cells differentiate into memory cells for long-term immunity.
Z is a Plasma cell, which is a differentiated B-cell specialised for antibody production.
Plasma cells produce large quantities of antibodies specific to alpha-gal that circulate in the bloodstream.

b. Features of Antibody Production

Initial tick bite produces low antibody concentration with slow, gradual increase over time, representing primary immune response.
Subsequent meat consumption triggers rapid elevation to higher antibody concentration, demonstrating secondary immune response with accelerated, amplified production.

c. Role of Memory Cells

Memory cells are produced during primary exposure and remain in circulation for years, maintaining immunological memory.
Upon re-exposure, memory cells rapidly recognise the specific antigen, which triggers immediate clonal expansion.
This results in faster and stronger antibody production because memory cells bypass the initial activation phase. Hence, providing enhanced immune protection against subsequent infections.

PHYSICS, M6 2025 HSC 22

The diagram represents the parts of the AC system used to transfer energy from a power station to people's houses.

Describe the energy transformations that take place in the transformers, and in the transmission line. (4 marks)

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Transformers $A$ and $B$: Some electrical energy is always lost, so each transformer outputs less electrical energy than it receives.
Core losses in transformers: Part of the input electrical energy is transformed into heat in the iron core due to eddy currents and magnetic effects.
Resistive losses in transformers: Some electrical energy becomes heat in the coils because of their resistance. Small amounts may also become sound/vibration energy.
Transmission lines: Electrical energy is also transformed into heat in the wires due to their resistance.

Show Worked Solution

Transformers $A$ and $B$: Some electrical energy is always lost, so each transformer outputs less electrical energy than it receives.
Core losses in transformers: Part of the input electrical energy is transformed into heat in the iron core due to eddy currents and magnetic effects.
Resistive losses in transformers: Some electrical energy becomes heat in the coils because of their resistance. Small amounts may also become sound/vibration energy.
Transmission lines: Electrical energy is also transformed into heat in the wires due to their resistance.

Functions, 2ADV EQ-Bank 10

Express $y=x^2-4 x+6$ in the form $y=(x-a)^2+c$. (1 mark)

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Graph the parabola, labelling its vertex and $y$-intercept. (2 marks)

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a. $y=(x-2)^2+2$

b. $\text {Vertex}=(2,2)$

$y \text{-intercept}=(0,6)$

Show Worked Solution

a.	$y$	$=x^2-4 x+6$
		$=x^2-4 x+4+2$
		$=(x-2)^2+2$

b. $\text {Vertex}=(2,2)$

$y \text{-intercept}=(0,6)$

Functions, 2ADV EQ-Bank 09

Using the discriminant, or otherwise, justify why the graph of $f(x)=-x^2+2 x-2$ lies entirely below the $x$-axis. (2 marks)

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$\Delta=b^2-4 a c=2^2-4(-1)(-2)=-4$

$\text{Since \(\ \Delta<0, \ y=-x^2+2 x-2\ $ does not intersect the $x$-axis.}\)

$\text{Since \(\ a=-1<0, f(x)$ is an upside down parabola.}\)

$\Rightarrow f(x)\ \text{must lie entirely below} \ x\text{-axis.}$

Show Worked Solution

$\Delta=b^2-4 a c=2^2-4(-1)(-2)=-4$

$\text{Since \(\ \Delta<0, \ y=-x^2+2 x-2\ $ does not intersect the $x$-axis.}\)

$\text{Since \(\ a=-1<0, f(x)$ is an upside down parabola.}\)

$\Rightarrow f(x)\ \text{must lie entirely below} \ x\text{-axis.}$

Functions, 2ADV EQ-Bank 02

Find the equation of the line that passes through $(2,1)$ and $(-3,4)$. (2 marks)

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Determine whether $(7,-2)$ lies on the line. (1 mark)

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a. $y=\dfrac{4-1}{-3-2}=-\dfrac{3}{5}$

b. $\text {Substitute}\ (7,-2) \ \text{into equation:}$

$-2$	$=-\dfrac{3}{5} \times 7+\dfrac{11}{5}$
$-2$	$=-\dfrac{21}{5}+\dfrac{11}{5}$
$-2$	$=-2 \ \text{(correct)}$

$\therefore (7,-2) \text{ lies on line.}$

Show Worked Solution

a. $(2,1),(-3,4)$

$\text{Gradient}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{4-1}{-3-2}=-\dfrac{3}{5}$

$\text{Find equation with} \ \ m=-\dfrac{3}{5} \ \ \text{through}\ \ (2,1):$

$y-1$	$=-\dfrac{3}{5}(x-2)$
$y$	$=-\dfrac{3}{5} x+\dfrac{11}{5}$

b. $\text {Substitute}\ (7,-2)\ \text{into equation:}$

$-2$	$=-\dfrac{3}{5} \times 7+\dfrac{11}{5}$
$-2$	$=-\dfrac{21}{5}+\dfrac{11}{5}$
$-2$	$=-2 \ \text{(correct)}$

$\therefore (7,-2) \text{ lies on line.}$

Functions, 2ADV EQ-Bank 05

Simplify $\dfrac{p+1}{q-q^3} \ ÷ \ \dfrac{p^3+p^2}{q^2-q}$. (2 marks)

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$\dfrac{1}{p^2(1+q)}$

Show Worked Solution

$\dfrac{p+1}{q-q^3} \ ÷ \ \dfrac{p^3+p^2}{q^2-q}$	$=\dfrac{p+1}{q\left(1-q^2\right)} \times \dfrac{q\left(1-q\right)}{p^2(p+1)}$
	$=\dfrac{(1-q)}{(1+q)(1-q) p^2}$
	$=\dfrac{1}{p^2(1+q)}$

Functions, 2ADV EQ-Bank 9

The quantity $y$ varies inversely with the square of $x$.

When $x = 4, \ y = 10$.

Find the value of $y$ when $x = 8$. (2 marks)

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$y=\dfrac{5}{2}$

Show Worked Solution

$y \propto \dfrac{1}{x^2}\ \ \Rightarrow\ \ y= \dfrac{k}{x^2}$

$\text{When}\ \ x=4, \ y=10:$

$10= \dfrac{k}{4^2}\ \ \Rightarrow\ \ k=160$

$\text{Find}\ y\ \text{when}\ x=8:$

$y=\dfrac{160}{8^2}=\dfrac{5}{2}$

L&E, 2ADV EQ-Bank 5

The mass `M` kg of a baby pig at age `x` days is given by `M = A(1.1)^x` where `A` is a constant. The graph of this equation is shown.

What is the value of `A`? (1 mark)

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What is the daily growth rate of the pig’s mass? Write your answer as a percentage. (1 mark)

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i. `1.5\ text(kg)`

ii. `10text(%)`

Show Worked Solution

i. `text(When)\ x = 0:`

♦ Mean mark (i) 48%.
♦♦♦ Mean mark part (ii) 6%!

`1.5`	`= A(1.1)^0`
`:. A`	`= 1.5\ text(kg)`

ii. `text(Daily growth rate)\ = 0.1 = 10text(%)`

L&E, 2ADV EQ-Bank 4

The number of bacteria in a culture grows from 100 to 114 in one hour.

What is the percentage increase in the number of bacteria? (1 mark)

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The bacteria continue to grow according to the formula `n = 100(1.14)^t`, where `n` is the number of bacteria after `t` hours.

What is the number of bacteria after 15 hours? (1 mark)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Time in hours $(t)$} \rule[-1ex]{0pt}{0pt} & \;\; 0 \;\; & \;\; 5 \;\; & \;\; 10 \;\; & \;\; 15 \;\; \\
\hline
\rule{0pt}{2.5ex} \text{Number of bacteria ( $n$ )} \rule[-1ex]{0pt}{0pt} & \;\; 100 \;\; & \;\; 193 \;\; & \;\; 371 \;\; & \;\; ? \;\; \\
\hline
\end{array}

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Use the values of `n` from `t = 0` to `t = 15` to draw a graph of `n = 100(1.14)^t`.

Use about half a page for your graph and mark a scale on each axis. (2 marks)

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Using your graph or otherwise, estimate the time in hours for the number of bacteria to reach 300. (1 mark)

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i. `text(14%)`

ii. `714`

iii. `text(Proof)\ \ text{(See Worked Solutions)}`

iv. `text(8.4 hours)`

Show Worked Solution

i. `text(Percentage increase)`

COMMENT: Common ADV/STD2 content in new syllabus.

`= (114 -100)/100 xx 100`

`= 14text(%)`

ii. `n = 100(1.14)^t`

`text(When)\ \ t = 15,`

`n= 100(1.14)^15= 713.793\ …\ = 714\ \ \ text{(nearest whole)}`

iii.

iv. `text(Using the graph)`

`text(The number of bacteria reaches 300 after ~ 8.4 hours.)`

Functions, 2ADV EQ-Bank 6

Identify where the graph $f(x)=\dfrac{\abs{x}}{x}$ is not continuous. (1 mark)

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Sketch the graph of $f(x)$. (2 marks)

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a. $\text {Denominator} \neq 0$

$f(x)\ \text{is not continuous when} \ \ x=0$

b. $\text{If} \ \ x>0 \ \Rightarrow \ f(x)=\dfrac{x}{x}=1$

$\text{If} \ \ x<0 \ \Rightarrow \ f(x)=-\dfrac{x}{x}=-1$

Show Worked Solution

a. $\text {Denominator} \neq 0$

$f(x)\ \text{is not continuous when} \ \ x=0$

b. $\text{If} \ \ x>0 \ \Rightarrow \ f(x)=\dfrac{x}{x}=1$

$\text{If} \ \ x<0 \ \Rightarrow \ f(x)=-\dfrac{x}{x}=-1$

Functions, 2ADV EQ-Bank 5

Identify where the graph $f(x)=\dfrac{x^2-1}{x-1}$ is not continuous. (1 mark)

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Sketch the graph of $f(x)$. (2 marks)

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a. $f(x)=\dfrac{x^2-1}{x-1}=\dfrac{(x+1)(x-1)}{(x-1)}=x+1$

$\text{Since denominator} \neq 0$

$f(x) \ \ \text{is not continuous when} \ \ x=1.$

Show Worked Solution

a. $f(x)=\dfrac{x^2-1}{x-1}=\dfrac{(x+1)(x-1)}{(x-1)}=x+1$

$\text{Since denominator} \neq 0$

$f(x) \ \ \text{is not continuous when} \ \ x=1.$

Functions, 2ADV EQ-Bank 4

Consider the function $y=f(x)$ where

$f(x)= \begin{cases}x^2+6, & \text { for } x \leqslant 0 \\ 6, & \text { for } 0<x \leqslant 3 \\ 2^x, & \text { for } x>3\end{cases}$

Sketch $y=f(x)$ (3 marks)

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For what value of $x$ is $y=f(x)$ NOT continuous? (1 mark)

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b. $f(x)\ \ \text {is NOT continuous at}\ \ x=3.$

Show Worked Solution

b. $f(x)\ \ \text {is NOT continuous at}\ \ x=3.$

Measurement, STD2 EQ-Bank 06

The distance between planets is measured in Astronomical Units (AU).

$1\ \text{AU}\ \approx 1.496\times 10^8$ kilometres.

Given Venus is approximately 0.7 AU from Earth, calculate this distance in kilometres giving your answer in scientific notation, correct to 3 significant figures. (2 marks)

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$1.03\times 10^8\ \text{(3 sig. fig.)}$

Show Worked Solution

$\text{Distance}$	$=0.7\times \left(1.496\times 10^8\right)$
	$=102\,830\,000$
	$=1.0283\times 10^8$
	$=1.03\times 10^8\ \text{(3 sig. fig.)}$

Statistics, STD2 EQ-Bank 03 MC

For the data below, which is the correct five figure summary?

$12,\ \ 16, \ \ 4,\ \ 6, \ \ 4, \ \ 5, \ \ 22, \ \ 20, \ \ 12, \ 8\ $

$4,\ \ 5, \ \ 8,\ \ 16, \ \ 21$
$4,\ \ 5, \ \ 8,\ \ 16, \ \ 22$
$4,\ \ 5, \ \ 10,\ \ 16, \ \ 21$
$4,\ \ 5, \ \ 10,\ \ 16, \ \ 22$

Show Answers Only

$D$

Show Worked Solution

$\text{Ordered data set}\ \rightarrow\ \ 4,\ \ 4, \ \ 5,\ \ 6, \ \ 8, \ \ 12, \ \ 12, \ \ 16, \ \ 20, \ 22\ $

$\text{Five number Summary}$

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Minimum} \rule[-1ex]{0pt}{0pt} & 4\\
\hline
\rule{0pt}{2.5ex} \ Q_1 \rule[-1ex]{0pt}{0pt} & 5 \\
\hline
\rule{0pt}{2.5ex} \text{Median} \rule[-1ex]{0pt}{0pt} & \dfrac{8+12}{2}=10\\
\hline
\rule{0pt}{2.5ex} \ Q_3 \rule[-1ex]{0pt}{0pt} & 16 \\
\hline
\rule{0pt}{2.5ex} \text{Maximum} \rule[-1ex]{0pt}{0pt} & 22\\
\hline
\end{array}

$\Rightarrow D$

Statistics, STD2 EQ-Bank 02 MC

The results of a test are displayed in the box-and-whisker plot below.

Which of the following statements is false?

The median is 155
The range is 60
The interquartile range is 50
25% of the scores are below 150

Show Answers Only

$C$

Show Worked Solution

$\text{Five number Summary}$

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Minimum} \rule[-1ex]{0pt}{0pt} & 140\\
\hline
\rule{0pt}{2.5ex} \ Q_1 \rule[-1ex]{0pt}{0pt} & 150 \\
\hline
\rule{0pt}{2.5ex} \text{Median} \rule[-1ex]{0pt}{0pt} & 155\\
\hline
\rule{0pt}{2.5ex} \ Q_3 \rule[-1ex]{0pt}{0pt} & 190 \\
\hline
\rule{0pt}{2.5ex} \text{Maximum} \rule[-1ex]{0pt}{0pt} & 200\\
\hline
\end{array}

$\text{Median}\ =\ 155\ \checkmark$

$\text{Range}\ =\ 200-140=60\ \checkmark$

$\text{IQR}\ =\ 190-150=40\ \text{not}\ 50\ $X

$\text{Q1}\ =\ 150\ \therefore\ 25\%\ \text{of scores lie below 150}\ \checkmark$

$\Rightarrow C$

Measurement, STD2 EQ-Bank 04 MC

Kathmandu is 30$^{\circ}$ west of Perth. Using longitudinal distance, what is the time in Kathmandu when it is noon in Perth?

10:00 am
11:30 am
12:30 pm
2:00 pm

Show Answers Only

$A$

Show Worked Solution

$15^{\circ}\ =\text{1 hour time difference}$

$\text{Longitudinal distance}$	$=30^{\circ}$
$\therefore\ \text{Time Difference}$	$=\dfrac{30}{15}$
	$=2\ \text{hours}$

$\text{Time in Perth}\ =\ 12\ \text{pm}$

$\therefore\ \text{Time in Kathmandu}$	$ =\ 12\ \text{pm}\ -\ 2\ \text{hours}$
	$=\ 10:00\ \text{am}$

$\Rightarrow A$

Measurement, STD2 EQ-Bank 03

The table shows the approximate coordinates of two cities.

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{City} \rule[-1ex]{0pt}{0pt} & \textit{Latitude} \rule[-1ex]{0pt}{0pt} & \textit{Longitude}\\
\hline
\rule{0pt}{2.5ex} \text{Buenos Aires} \rule[-1ex]{0pt}{0pt} & 35^{\circ}\ \text{S} \rule[-1ex]{0pt}{0pt} & 60^{\circ}\ \text{W} \\
\hline
\rule{0pt}{2.5ex} \text{Adelaide} \rule[-1ex]{0pt}{0pt} & 35^{\circ}\ \text{S} \rule[-1ex]{0pt}{0pt} & 140^{\circ}\ \text{E} \\
\hline
\end{array}

What is the time difference between Adelaide and Buenos Aires? (2 marks)
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Roy lives in Adelaide and his cousin Juan lives in Buenos Aires. Roy wants to telephone Juan at 7 pm on Friday night, Buenos Aires time.
At what time, and on what day, should Roy make the call? (2 marks)
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a. $13\ \text{hours}\ 20\ \text{minutes}$

b. $8:20\ \text{am on Saturday}$

Show Worked Solution

a. $15^{\circ}\ =\text{1 hour time difference}$

$\text{Angular distance}$	$=60+140=200^{\circ}$
$\therefore\ \text{Time Difference}$	$=\dfrac{200}{15}$
	$=13.\dot{3}$
	$=13\ \text{hours}\ 20\ \text{minutes}$

b. $\text{Time in Buenos Aires}\ =\ 7\ \text{pm Friday night}$

$\therefore\ \text{Time in Adelaide}$	$ =\ 7\ \text{pm}\ +\ 13\ \text{hours}\ 20\ \text{minutes}$
	$=\ 8:20\ \text{am on Saturday}$

Statistics, STD2 EQ-Bank 01

A Physics class of 12 students is going on a 4 day excursion by bus.

The students are asked to each pack one bag for the trip. The bags are weighed, and the weights (in kg) are listed in order as follows:

$8,\ \ 9, \ \ 10,\ \ 10, \ \ 15, \ \ 18, \ \ 22, \ \ 25, \ \ 29, \ \ 35, \ \ 38, \ \ 41 $

Use the above data to produce a five number summary for the weights of the bags. (2 marks)
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Using your five number summary from part (a), calculate the interquartile range of the weights. (2 marks)
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a. $\text{Five number Summary}$

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Minimum} \rule[-1ex]{0pt}{0pt} & 8\\
\hline
\rule{0pt}{2.5ex} \ Q_1 \rule[-1ex]{0pt}{0pt} & 10 \\
\hline
\rule{0pt}{2.5ex} \text{Median} \rule[-1ex]{0pt}{0pt} & 20\\
\hline
\rule{0pt}{2.5ex} \ Q_3 \rule[-1ex]{0pt}{0pt} & 32 \\
\hline
\rule{0pt}{2.5ex} \text{Maximum} \rule[-1ex]{0pt}{0pt} & 41\\
\hline
\end{array}

b. $22$

Show Worked Solution

a. $\text{Five number Summary}$

b.	$\text{IQR}$	$=Q_3-Q_1$
		$=32-10=22$

Algebra, STD2 EQ-Bank 05

Jordan visits Italy on his holidays. He pays €180 (180 euros) for a pair of Italian leather boots.

How much is €180 in Australian dollars if AUD1 is worth €0.58? (2 marks)

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$$310.34$

Show Worked Solution

$0.58\ $€ $\ =\ \text{AUD 1}$

$\rightarrow\ 1\ $€$\ =$$\ \text{AUD }$$\dfrac{1}{0.58}$

$\rightarrow\ 180\ $€$=180\times\dfrac{1}{0.58}=310.344…$

$\therefore\ \text{Jordan’s boots cost }$310.34\ \text{in Australian dollars.}$

Algebra, STD2 EQ-Bank 04 MC

The graph shows the tax payable against taxable incomes up to $60 000 in a proposed tax system.

How much of each dollar earned over $$30\,000$ is payable in tax?

10 cents
12 cents
20 cents
23 cents

Show Answers Only

$C$

Show Worked Solution

$\text{The gradient of line represents the tax payable per dollar.}$

$\text{Tax payable per dollar}:$

$= \dfrac{\text{rise}}{\text{run}}$

$= \dfrac{7000-1000}{60\ 000-30\ 000}$

$=\dfrac{1}{5}=0.20$

$\therefore\ \text{20 cents per dollar is payable in tax after }$30\, 000.$

$\Rightarrow C$

Algebra, STD2 EQ-Bank 02 MC

If $w=2y^3-1$, what is the value of $y$ when $w=13$?

$\dfrac{\sqrt[3]{14}}{2}$
$\sqrt[3]{6}$
$\sqrt[3]{7}$
$\sqrt[3]{14}$

Show Answers Only

$C$

Show Worked Solution

$\text{When}\ w=13:$

$w$	$=2y^3-1$
$13$	$=2y^3-1$
$2y^3$	$=14$
$y^3$	$=\dfrac{14}{2}=7$
$y$	$=\sqrt[3]{7}$

$\Rightarrow C$

Algebra, STD2 EQ-Bank 01

Jerico is the manager of a weekend market in which there are 220 stalls for rent. From past experience, Jerico knows that if he charges $d$ dollars to rent a stall. then the number of stalls, $s$, that will be rented is given by:

$s=220-4d$

How many stalls will be rented if Jerico charges $7.50 per stall . (1 mark)

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Complete the following table for the function $s=220-4d$. (1 mark)

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\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \quad d\quad \rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex} \quad 10\quad\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex} \quad 30\quad & \rule{0pt}{2.5ex} \quad 50\quad \\
\hline
\rule{0pt}{2.5ex} \quad s\quad \rule[-1ex]{0pt}{0pt} & \ & \ & \\
\hline
\end{array}

Using an appropriate vertical scale and labelled axes, graph the function $s=220-4d$ on the grid below. (2 marks)

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Does it make sense to use the formula $s=220-4d$ to calculate the number of stalls rented if Jerico charges $60 per stall? Explain your answer. (2 marks)

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a. $190\ \text{stalls will be rented}$

d. $\text{When}\ d=60, s=220-4\times 60=-20$

$\therefore\ \text{It does not make sense to charge }$60\ \text{ per stall}$

$\text{as you cannot have a negative number of stalls.}$

Show Worked Solution

a.	$s$	$=220-4d$
		$=220-4\times 7.50$
		$=190$

$190\ \text{stalls will be rented}$

d. $\text{When}\ d=60, s=220-4\times 60=-20$

$\therefore\ \text{It does not make sense to charge }$60\ \text{ per stall}$

$\text{as you cannot have a negative number of stalls.}$

Functions, 2ADV EQ-Bank 03

Find $a$ and $b$ such that $a$ and $b$ are real and $\dfrac{2\sqrt{3}+2}{\sqrt{6}-\sqrt{2}} = a\,\sqrt{2} + b\,\sqrt{6}$. (2 marks)

Show Answers Only

$a=2\ \ \text{and}\ \ b=1 $

Show Worked Solution

$\dfrac{2\sqrt{3}+2}{\sqrt{6}-\sqrt{2}} \times \dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}} $

$=\dfrac{(2\sqrt{3}+2)(\sqrt{6}+\sqrt{2})}{6-2}$

$=\dfrac{2\sqrt{18}+2\sqrt{6}+2\sqrt{6}+2\sqrt{2}}{4}$

$=\dfrac{6\sqrt{2}+4\sqrt{6}+2\sqrt{2}}{4}$

$=\dfrac{8\sqrt{2}+4\sqrt{6}}{4}$

$=2\sqrt{2}+\sqrt{6}$

$\therefore a=2\ \ \text{and}\ \ b=1 $

Functions, 2ADV EQ-Bank 01

Find $x$ and $y$ such that $x$ and $y$ are real and $\dfrac{\sqrt{2}+1}{\sqrt{6}-\sqrt{3}} = x\,\sqrt{3} + y\,\sqrt{6}$. (2 marks)

Show Answers Only

$x=1\ \ \text{and}\ \ y=\dfrac{2}{3} $

Show Worked Solution

$\dfrac{\sqrt{2}+1}{\sqrt{6}-\sqrt{3}} \times \dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{3}} $

$=\dfrac{(\sqrt{2}+1)(\sqrt{6}+\sqrt{3})}{6-3}$

$=\dfrac{\sqrt{12}+2\sqrt{6}+\sqrt{3}}{3}$

$=\dfrac{3\sqrt{3}+2\sqrt{6}}{3}$

$= \sqrt{3}+\dfrac{2}{3} \sqrt{6}$

$\therefore x=1\ \ \text{and}\ \ y=\dfrac{2}{3} $

CHEMISTRY, M6 2025 HSC 31

Hydrazine is a compound of hydrogen and nitrogen. The complete combustion of 1.0 L of gaseous hydrazine requires 3.0 L of oxygen, producing 2.0 L of nitrogen dioxide gas and 2.0 L of water vapour. All volumes are measured at 400°C.

Use the chemical equation for the combustion of hydrazine to show that the molecular formula for hydrazine is $\ce{N2H4}$. (2 marks)

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The relationship between the acid equilibrium constant $\left(K_a\right)$ and the corresponding conjugate base equilibrium constant $\left(K_b\right)$ is shown.
1. 1. $K_a \times K_b=K_w$
Use a relevant chemical equation to calculate the pH of a 0.20 mol L$^{-1}$ solution of $\ce{N2H5+}$ using the following data:

- the $K_b$ of hydrazine is $1.7 \times 10^{-6}$ at 25°C
- $\ce{N2H5+}$ is the conjugate acid of $\ce{N2H4}$. (4 marks)

Show Answers Only

a. Using Avogadro’s law:

At the same temperature and pressure, gas volumes are proportional to moles (i.e. the volume ratio will be equal to the mole ratio in a balanced equation).
$\ce{N2H4(g) + 3O2(g) -> 2NO2(g) + 2H2O(g)}$
Mole ratio $\text{Hydrazine} : \ce{O2} : \ce{NO2} : \ce{H2O} = 1:3:2:2\ \ \Rightarrow\ $ matches the volume ratios given.
Therefore $\ce{N2H4}$ is the correct molecular formula for hydrazine.

b. $\text{pH} = 4.46$

Show Worked Solution

a. Using Avogadro’s law:

At the same temperature and pressure, gas volumes are proportional to moles (i.e. the volume ratio will be equal to the mole ratio in a balanced equation).
$\ce{N2H4(g) + 3O2(g) -> 2NO2(g) + 2H2O(g)}$
Mole ratio $\text{Hydrazine} : \ce{O2} : \ce{NO2} : \ce{H2O} = 1:3:2:2\ \ \Rightarrow\ $ matches the volume ratios given.
Therefore $\ce{N2H4}$ is the correct molecular formula for hydrazine.

b. $K_a(\ce{N2H5+}) = \dfrac{K_w}{K_b(\ce{N2H4})} = \dfrac{1 \times 10^{-14}}{1.7 \times 10^{-6}} = 5.88235 \times 10^{-9}$

The ionisation of $\ce{N2H5+}$ is given the chemical equation below:
$\ce{N2H5+(aq) + H2O(l) \leftrightharpoons N2H4(aq) + H3O+(aq)}$
$K_a = \dfrac{\ce{[H3O+][N2H4]}}{\ce{[N2H5+]}}$
Using an Ice Table where all numbers are in mol L$^{-1}$.

\begin{array} {|c|c|c|c|}
\hline & \ce{[N2H5+]} & \ce{[N2H4]} & \ce{[H3O+]} \\
\hline \text{Initial} & 0.20 & 0 & 0 \\
\hline \text{Change} & -x & +x & +x \\
\hline \text{Equilibrium} & 0.20 -x & x & x \\
\hline \end{array}

Substituting into the $K_a$ expression:

$\dfrac{x^2}{0.20-x}$	$=5.88235 \times 10^{-9}$
$\dfrac{x^2}{0.20}$	$=5.88235 \times 10^{-9}$, as $x$ is really small
$x$	$=3.42997 \times 10^{-5}$

$\text{pH}$	$=-\log_{10}(\ce{[H3O+]})$
	$=-\log_{10}(3.42997 \times 10^{-5})$
	$=4.46$

CHEMISTRY, M8 2025 HSC 30

Phosgene is used in industry as a starting material to synthesise useful polymers. Phosgene $\ce{(Cl2CO)}$ is a gas at room temperature and is highly toxic.

Justify a suitable precaution when using phosgene. (2 marks)

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Phosgene is synthesised by the reaction of carbon monoxide $\ce{(CO)}$ and chlorine $\ce{(Cl2)}$ in the gas phase.
1. 1. $\ce{Cl2(g) + CO(g) \rightleftharpoons Cl2CO(g)}$
Explain why an excess of carbon monoxide and a catalyst are used in the industrial synthesis of phosgene. (3 marks)

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a. Precaution when using phosgene inside a fume hood:

As phosgene is a highly toxic gas at room temperature. working in a fume hood prevents inhalation by safely removing the gas from the breathing zone and venting it outside the laboratory.
This measure significantly reduces the risk of poisoning and exposure.

b. Role of excess carbon monoxide:

An excess of carbon monoxide is used to increase the rate of production of phosgene.
According to Le Chatelier’s principle, increasing the concentration of $\ce{CO}$ shifts the equilibrium to the right, favouring the formation of $\ce{Cl2CO}$ and increasing the overall yield of phosgene.

Role of catalyst:

A catalyst is used to increase the rate of reaction by providing an alternative reaction pathway with a lower activation energy.
This allows phosgene to be produced more rapidly and efficiently without affecting the equilibrium position.
In industrial settings, this increases production speed and reduces energy costs.

Show Worked Solution

a. Precaution when using phosgene inside a fume hood:

As phosgene is a highly toxic gas at room temperature. working in a fume hood prevents inhalation by safely removing the gas from the breathing zone and venting it outside the laboratory.
This measure significantly reduces the risk of poisoning and exposure.

b. Role of excess carbon monoxide:

An excess of carbon monoxide is used to increase the rate of production of phosgene.
According to Le Chatelier’s principle, increasing the concentration of $\ce{CO}$ shifts the equilibrium to the right, favouring the formation of $\ce{Cl2CO}$ and increasing the overall yield of phosgene.

Role of catalyst:

A catalyst is used to increase the rate of reaction by providing an alternative reaction pathway with a lower activation energy.
This allows phosgene to be produced more rapidly and efficiently without affecting the equilibrium position.
In industrial settings, this increases production speed and reduces energy costs.

CHEMISTRY, M7 2025 HSC 28

Kevlar and polystyrene are two common polymers.

A section of their structures is shown.

Kevlar is produced through a reaction of two different monomers, one of which is shown. Draw the missing monomer in the box provided. (1 mark)

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Kevlar chains are hard to pull apart, whereas polystyrene chains are not.
With reference to intermolecular forces, explain the difference in the physical properties of the two polymers. (3 marks)

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b. The physical differences between the two polymers are:

Kevlar chains are very hard to pull apart because the polymer contains many amide groups that can form strong hydrogen bonds between neighbouring chains. These strong forces hold the chains tightly together, making Kevlar rigid and very strong.
The close packing of the chains also gives Kevlar a high melting point, because a large amount of energy is required to break the hydrogen bonds.
Polystyrene, on the other hand, does not contain groups that can form hydrogen bonds. Its polymer chains are mostly non-polar, so the only forces between the chains are weak dispersion forces.
These weaker attractions mean the chains can slide past each other, making polystyrene much softer, brittle, and it also has a lower melting point than Kevlar. Because the forces between chains are weak, polystyrene is much easier to pull apart compared to Kevlar.

Show Worked Solution

b. The physical differences between the two polymers are:

Kevlar chains are very hard to pull apart because the polymer contains many amide groups that can form strong hydrogen bonds between neighbouring chains. These strong forces hold the chains tightly together, making Kevlar rigid and very strong.
The close packing of the chains also gives Kevlar a high melting point, because a large amount of energy is required to break the hydrogen bonds.
Polystyrene, on the other hand, does not contain groups that can form hydrogen bonds. Its polymer chains are mostly non-polar, so the only forces between the chains are weak dispersion forces.
These weaker attractions mean the chains can slide past each other, making polystyrene much softer, brittle, and it also has a lower melting point than Kevlar. Because the forces between chains are weak, polystyrene is much easier to pull apart compared to Kevlar.

CHEMISTRY, M7 2025 HSC 27

Mixtures of hydrocarbons can be obtained from crude oil by the process of fractional distillation. Examples include petrol, diesel and natural gas.

Outline an environmental implication for a use of a named hydrocarbon mixture that is obtained from crude oil. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Ethene is a simple hydrocarbon obtained from crude oil.
Using structural formulae, write the chemical equation for the conversion of ethene to ethanol, including any other necessary reagents. (3 marks)

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When ethanol is reacted with ethanoic acid, ethyl ethanoate is formed, as shown by the equation.
1. 1. $\text{ethanol} \ + \ \text{ethanoic acid } \ \rightleftharpoons \ \text{ethyl ethanoate} \ +\ \text{water}$
The graph below shows the concentration of ethanol from the start of the reaction, $t_0$, up to a time $t_1$.
At time $t_1$, an additional amount of ethanol is added to the system.
Sketch on the graph the changes that occur in the concentration of ethanol between time $t_1$, and when the system reaches a new equilibrium before time $t_2$. (3 marks)
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a. Environmental implication:

The combustion of petrol, a hydrocarbon mixture obtained from crude oil, leads to the release of large amounts of carbon dioxide.
Increased $\ce{CO2}$ levels intensify the enhanced greenhouse effect, contributing to global warming, climate change, and associated environmental impacts such as rising sea levels and extreme weather patterns.
A typical component of petrol, octane, burns according to:
$\ce{2C8H18(l) + 25O2(g) -> 16CO2(g) + 18H2O(g)}$

Show Worked Solution

a. Environmental implication:

The combustion of petrol, a hydrocarbon mixture obtained from crude oil, leads to the release of large amounts of carbon dioxide.
Increased $\ce{CO2}$ levels intensify the enhanced greenhouse effect, contributing to global warming, climate change, and associated environmental impacts such as rising sea levels and extreme weather patterns.
A typical component of petrol, octane, burns according to:
$\ce{2C8H18(l) + 25O2(g) -> 16CO2(g) + 18H2O(g)}$

There will be a sudden increase at $t_1$ and then the concentration of ethanol will decrease smoothly until a new equlibrium concentration (greater than the original equilibrium concentration) is reached.

\(96\)	\(=k \times 24\)
\(k\)	\(=\dfrac{96}{24}\)
\(k\)	\(=4\ \text{(as required)}\)

\(96\)	\(=k \times 24\)
\(k\)	\(=\dfrac{96}{24}\)
\(k\)	\(=4\ \text{(as required)}\)

\(650\)	\(=4\times t\)
\(t\)	\( =\dfrac{650}{4}\)
	\(=162.5\ \text{minutes}\)

\(87.50\)	\(=k \times 250\)
\(k\)	\(=\dfrac{87.50}{250}\)
\(k\)	\(=0.35\ \text{(as required)}\)

\(87.50\)	\(=k \times 250\)
\(k\)	\(=\dfrac{87.50}{250}\)
\(k\)	\(=0.35\ \text{(as required)}\)

\(140\)	\(=0.35\times L\)
\(L\)	\( =\dfrac{140}{0.35}\)
	\(=400\ \text{m}\)

\(2100\)	\(=k \times 500\)
\(k\)	\(=\dfrac{2100}{500}\)
\(k\)	\(=4.2\)

\(27\)	\(=k \times 18\)
\(k\)	\(=\dfrac{27}{18}\)
\(k\)	\(=\dfrac{3}{2}\)

\(\dfrac{mv_B^2}{r}\)	\(=T+mg\)
\(v_B^2\)	\(=rg \ \ (T=0)\)
\(v_B\)	\(=\sqrt{r g}\)