Band 3 – Page 2

Networks, STD2 N2 2024 HSC 16

A network of towns and the distances between them in kilometres is shown.

What is the shortest path from $T$ to $H$ ? (2 marks)

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A truck driver needs to travel from $Y$ to $G$ but knows that the road from $C$ to $G$ is closed.
What is the length of the shortest path the truck driver can take from $Y$ to $G$ after the road closure? (2 marks)

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Show Answers Only

a. $TYWH$

b. $\text{Length of shortest path}\ (YWHMG) = 89 \text{km}$

Show Worked Solution

a. $TYH=30+38=68, \quad TYWH=30+15+20=65$

$\therefore \text{ Shortest Path is}\ TYWH.$

b. $Y W C M G=15+25+25+25=90$

$YWHMG=15+20+29+25=89$

$\Rightarrow \ \text{All other paths are longer.}$

$\therefore\text{ Length of shortest path = 89 km}$

Networks, STD2 N3 2024 HSC 39

A project involving nine activities is shown in the network diagram.

The duration of each activity is not yet known.

The following table gives the earliest start time (EST) and latest start time (LST) for three of the activities. All times are in hours.

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Activity} \rule[-1ex]{0pt}{0pt} & EST & LST \\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \ 0\ \ \ \ \ \ & \ \ \ \ \ \ 2\ \ \ \ \ \ \\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & 0 & 1 \\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} & 12 & 12 \\
\hline
\end{array}

What is the critical path? (1 mark)
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The minimum time required for this project to be completed is 19 hours.
What is the duration of activity $I$? (1 mark)

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The duration of activity $C$ is 3 hours.
What is the maximum amount of time that could occur between the start of activity $F$ and the end of activity $H$? (1 mark)

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a. $\text{Critical Path:}\ BEGI$

b. $\text{Duration of}\ I =7\ \text{hours}$

c. $\text{Max tim}\ =9\ \text{hours}$

Show Worked Solution

a. $\text{Activity}\ A\ \text{and}\ C: \ LST \gt EST$

$\Rightarrow\ \text{Activity}\ A\ \text{and}\ C\ \text{not on critical path.}$

$\text{Critical Path:}\ BEGI$

b. $\text{Duration of}\ I = 19-12=7\ \text{hours}$

c. $\text{Activity}\ I\ \text{must start at 12 hours (on critical path).}$

$\text{Given duration activity}\ C = 3:$

$\text{Max time from start of}\ F\ \text{to end of}\ H = 12-3=9\ \text{hours}$

Trigonometry, 2ADV T3 2024 HSC 28

Anna is sitting in a carriage of a Ferris wheel which is revolving. The height, $A(t)$, in metres above the ground of the top of her carriage is given by

$A(t)=c-k\,\cos\Big( \dfrac{\pi t}{24}\Big) $,

where $t$ is the time in seconds after Anna's carriage first reaches the bottom of its revolution and $c$ and $k$ are constants.

The top of each carriage reaches a greatest height of 39 metres and a smallest height of 3 metres.

Find the value of $c$ and $k$. (2 marks)
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How many seconds does it take for one complete revolution of the Ferris wheel? (1 mark)
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Billie is in another carriage. The height, $B(t)$, in metres above the ground of the top of her carriage is given by

$B(t)=c-k\,\cos\Big( \dfrac{\pi}{24}(t-6)\Big) $,

where $c$ and $k$ are as found in part (a).
During each revolution, there are two occasions when Anna's and Billie's carriages are at the same heights. At what two heights does this occur? Give your answer correct to 2 decimal places. (4 marks)
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a. $c=\ \text{centre of motion}\ = \dfrac{39+3}{2} = 21$

$k=\ \text{amplitude}\ = \dfrac{39-3}{2}=18$

b. $T= 48\ \text{seconds}$

c. $h_1=4.37\ \text{m,}\ h_2=37.63\ \text{m}$

Show Worked Solution

a. $c=\ \text{centre of motion}\ = \dfrac{39+3}{2} = 21$

$k=\ \text{amplitude}\ = \dfrac{39-3}{2}=18$

b. $A(t)=21-18\,\cos\Big( \dfrac{\pi t}{24}\Big)\ \Rightarrow\ \ n=\dfrac{\pi}{24} $

$T=\dfrac{2\pi}{n} = 2\pi \times \dfrac{24}{\pi} = 48\ \text{seconds}$

c. $\text{Billie’s carriage is 6 seconds behind Anna’s.}$

$\text{Angle between the 2 carriages}\ = \dfrac{\pi \times 6}{24} = \dfrac{\pi}{4} $

$\text{By inspection:}$

$\text{Heights are the same:}$

$h_1=21-18\cos(\dfrac{\pi}{8}) = 4.370… = 4.37\ \text{m (2 d.p.)}$

$h_2=21-18\cos(\dfrac{7\pi}{8}) = 37.629… = 37.63\ \text{m (2 d.p.)}$

Probability, 2ADV S1 2024 HSC 18

In a game, the probability that a particular player scores a goal at each attempt is 0.15.

What is the probability that this player does NOT score a goal in the first two attempts? (1 mark)

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Determine the least number of attempts that this player must make so that the probability of scoring at least one goal is greater than 0.8. (2 marks)

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a. $0.7225$

b. $n=10$

Show Worked Solution

a. $P(G)=0.15, \ \ P(\bar{G})=0.85$

$P(\bar{G}\bar{G}) = 0.85^2=0.7225$

b. $\text{2 attempts:}\ P(\text{at least 1 goal})=1-P(\bar{G}\bar{G})=1-0.85^{2}$

$\text{3 attempts:}\ P(\text{at least 1 goal})=1-0.85^{3}$

$\text{n attempts:}\ P(\text{at least 1 goal})=1-0.85^{n}$

$\text{Find}\ n\ \text{such that:}$

$1-0.85^{n}$	$\gt 0.8$
$0.85^{n}$	$\lt 0.2$
$n \times\ln(0.85)$	$\lt \ln(0.2)$
$n$	$\gt \dfrac{\ln(0.2)}{\ln(0.85)}$
	$\gt 9.9…$

$\therefore\ \text{Least}\ n=10$

CHEMISTRY, M2 EQ-Bank 8 MC

represents which gas law?

Boyle’s Law
Charles’ Law
Avogadro’s Law
Gay-Lussac’s Law

Show Answers Only

$B$

Show Worked Solution

→ The equation represents Charles’ Law, which states that the volume of a gas is directly proportional to its temperature, provided the pressure and amount of gas remain constant.

$\Rightarrow B$

CHEMISTRY, M2 EQ-Bank 7 MC

According to Gay-Lussac's Law, the pressure of a gas is directly proportional to which quantity, assuming the volume and the number of moles are constant?

Volume
Temperature
Moles
Pressure

Show Answers Only

$B$

Show Worked Solution

→ Gay-Lussac’s Law states that the pressure of a gas is directly proportional to its temperature, provided the volume and number of moles remain constant.

→ This can be written mathematically like $\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$.

$\Rightarrow B$

Calculus, 2ADV C4 2024 HSC 27

Find the derivative of $x^{2}\tan\,x$ (2 marks)
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Hence, find $\displaystyle \int (x\,\tan\,x+1)^2\ dx$ (3 marks)
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a. $\dfrac{dy}{dx}=2x\,\tan\,x + x^2\,\sec^{2}x$

b. $x^{2}\tan^{2}x-\dfrac{x^{3}}{3}+x+C$

Show Worked Solution

a. $y=x^{2}\tan\,x$

$\text{By product rule:}$

$\dfrac{dy}{dx}=2x\,\tan\,x + x^2 \sec^{2}x$

b. $\displaystyle \int (x\,\tan\,x+1)^{2}\,dx$

\[=\int x^2\tan^{2}x + 2x\,\tan\,x +1\ dx\]

\[=\int x^{2}(\sec^{2}x-1)+2x\,\tan\,x +1\,dx\]

\[=\int 2x\,\tan\,x + x^{2}\sec^{2}x-x^{2}+1\,dx\]

\[=x^{2}\tan\,x-\dfrac{x^{3}}{3}+x+C\]

Calculus, 2ADV C3 2024 HSC 11

The graph of the function $g(x)$ is shown.

Using the graph, complete the table with the words positive, zero or negative as appropriate. (3 marks)

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\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{$x$-value} \rule[-1ex]{0pt}{0pt} & \textit{First derivative of $g(x)$ at $x$} \rule[-1ex]{0pt}{0pt} & \textit{Second derivative of $g(x)$ at $x$} \\
\hline
\rule{0pt}{2.5ex} \text{$x=-3$} \rule[-1ex]{0pt}{0pt} & \text{ } \rule[-1ex]{0pt}{0pt} & \text{ } \\
\hline
\rule{0pt}{2.5ex} \text{$x=1$} \rule[-1ex]{0pt}{0pt} & \text{ } \rule[-1ex]{0pt}{0pt} & \text{ } \\
\hline
\rule{0pt}{2.5ex} \text{$x=5$} \rule[-1ex]{0pt}{0pt} & \text{ } \rule[-1ex]{0pt}{0pt} & \text{ } \\
\hline
\end{array}

Show Answers Only

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{$x$-value} \rule[-1ex]{0pt}{0pt} & \textit{First derivative of $g(x)$ at $x$} \rule[-1ex]{0pt}{0pt} & \textit{Second derivative of $g(x)$ at $x$} \\
\hline
\rule{0pt}{2.5ex} \text{$x=-3$} \rule[-1ex]{0pt}{0pt} & \text{positive} \rule[-1ex]{0pt}{0pt} & \text{negative} \\
\hline
\rule{0pt}{2.5ex} \text{$x=1$} \rule[-1ex]{0pt}{0pt} & \text{zero} \rule[-1ex]{0pt}{0pt} & \text{zero} \\
\hline
\rule{0pt}{2.5ex} \text{$x=5$} \rule[-1ex]{0pt}{0pt} & \text{positive} \rule[-1ex]{0pt}{0pt} & \text{positive} \\
\hline
\end{array}

Show Worked Solution

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{$x$-value} \rule[-1ex]{0pt}{0pt} & \textit{First derivative of $g(x)$ at $x$} \rule[-1ex]{0pt}{0pt} & \textit{Second derivative of $g(x)$ at $x$} \\
\hline
\rule{0pt}{2.5ex} \text{$x=-3$} \rule[-1ex]{0pt}{0pt} & \text{positive} \rule[-1ex]{0pt}{0pt} & \text{negative} \\
\hline
\rule{0pt}{2.5ex} \text{$x=1$} \rule[-1ex]{0pt}{0pt} & \text{zero} \rule[-1ex]{0pt}{0pt} & \text{zero} \\
\hline
\rule{0pt}{2.5ex} \text{$x=5$} \rule[-1ex]{0pt}{0pt} & \text{positive} \rule[-1ex]{0pt}{0pt} & \text{positive} \\
\hline
\end{array}

Calculus, 2ADV C4 2024 HSC 5 MC

What is $ {\displaystyle \int(6 x+1)^3 d x} $ ?

$ \dfrac{1}{24}(6 x+1)^4+C $
$ \dfrac{1}{4}(6 x+1)^4+C $
$ \dfrac{2}{3}(6 x+1)^4+C $
$ \dfrac{3}{2}(6 x+1)^4+C $

Show Answers Only

$ A $

Show Worked Solution

\[ \int(6 x+1)^3 dx\]	$=\dfrac{1}{4} \cdot \dfrac{1}{6}(6 x+1)^4+C$
	$=\dfrac{1}{24}(6 x+1)^4+C$

$ \Rightarrow A $

CHEMISTRY, M2 EQ-Bank 6

A piece of zinc weighing 3.20 grams is placed into a beaker containing 300.0 mL of 0.7500 mol/L hydrochloric acid.

$\ce{Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)}$

Determine the limiting reagent (show all working). (2 marks)

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Calculate the volume of gas produced in this reaction at 25°C and 100 kPa. (2 marks)

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a. Zinc is the limiting reagent.

b. $1.21\ \text{L}$

Show Worked Solution

a. $\ce{n(Zn)} = \dfrac{m}{MM} = \dfrac{3.20}{65.38}=0.0489\ \text{mol}$

$\ce{n(HCl)} = c \times V = 0.75 \times 0.3 = 0.225$

→ Based on the mole ratio, $0.0489\ \text{mol}$ of Zinc would require $0.0978\ \text{mol}$ of hydrochloric acid.

→ As there is excess hydrochloric acid, Zinc is the limiting reagent.

b. The Mole ratio of $\ce{Zn:H2}$ is $1:1$

→ $0.0489\ \text{mol}$ of $\ce{H2(g)}$ is produced.

→ At $25^{\circ}\text{C}$ and $100\ \text{kPa}$, $1$ mole of gas $=24.79\ \text{L}$

→ $\ce{V(H2(g))} = 0.0489 \times 24.79 = 1.21\ \text{L}$

Statistics, 2ADV S3 2024 HSC 3 MC

Pia's marks in Year 10 assessments are shown. The scores for each subject were normally distributed.

\begin{array}{|l|c|c|c|}
\hline & \textit {Pia's mark} & \textit {Year 10 mean} & \textit {Year 10 standard} \\
&&&\textit {deviation}\\
\hline \text {English} & 78 & 66 & 6 \\
\hline \text {Mathematics} & 80 & 71 & 10 \\
\hline \text {Science} & 77 & 70 & 15 \\
\hline \text {History} & 85 & 72 & 9 \\
\hline
\end{array}

In which subject did Pia perform best in comparison with the rest of Year 10?

English
Mathematics
Science
History

Show Answers Only

$A$

Show Worked Solution

$\text {Consider the z-score of each option:}$

$z \text {-score (English)}=\dfrac{78-66}{6}=2$

$z \text {-score (Maths)}=\dfrac{80-71}{10}=0.9$

$z \text {-score (Science) }=\dfrac{77-70}{15}=0.46 \ldots$

$z \text {-score (History})=\dfrac{85-72}{9}=1.4 \ldots$

$\Rightarrow A$

Calculus, 2ADV C4 2024 HSC 14

The curves `y=(x-1)^2` and `y=5-x^2` intersect at two points, as shown in the diagram.

Find the `x`-coordinates of the points of intersection of the two curves. (1 mark)

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Find the area enclosed by the two curves. (3 marks)

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a. `x=2\ \text{and}\ -1`

b. `9\ \text{units}^2`

Show Worked Solution

a. `y=(x-1)^2, \ y=5-x^2`

$\text{Intersection occurs when:}$

`(x-1)^2`	`=5-x^2`
`x^2-2x+1`	`=5-x^2`
`2x^2-2x-4`	`=0`
`2(x-2)(x+1)`	`=0`

`x=2\ \text{and}\ -1`

b.	`\text{Area}`	`= \int_{-1}^{2} (5-x^2)-(x-1)^2\ dx`
		`=\int_{-1}^{2} 5-x^2-x^2+2x-1\ dx`
		`=\int_{-1}^{2}4-2x^2+2x\ dx`
		`=[4x-\frac{2}{3}x^3+x^2]_{-1}^{2}`
		`=[(8-\frac{16}{3}+4)-(-4+\frac{2}{3}+1)]`
		`=\frac{20}{3}-(-\frac{7}{3})`
		`=9\ \text{u}^2`

Financial Maths, 2ADV M1 2024 HSC 12

Find the sum of the terms in the arithmetic series

$50 + 57 + 64 +\ ...\ +2024$ (3 marks)

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$293\,471$

Show Worked Solution

$\text{AP where}\ \ a=50, d=57-50=7$

$\text{Last term}$	$=a + (n-1)d$
$2024$	$=50+(n-1)d$
$n-1$	$=\dfrac{2024-50}{7}$
$n$	$=283$

$S_{283}$	$=\dfrac{n}{2}(a + l)$
	$=\dfrac{283}{2}(50+2024)$
	$=293\,471$

Probability, 2ADV S1 2024 HSC 2 MC

In a group of 60 students, 38 play basketball, 35 play hockey and 5 do not play either basketball or hockey.

How many students play both basketball and hockey?

Show Answers Only

$B$

Show Worked Solution

$\text{Method 1:}$

$\text{Method 2:}$

$n(B \cup H)$	$=n(B) + n(H)-n(B \cap H) $
$55$	$=38+35-n(B \cap H) $
$n(B \cap H) $	$=18$

$ \Rightarrow B $

Functions, 2ADV F1 2024 HSC 1 MC

Consider the function shown.

Which of the following could be the equation of this function?

$y=2 x+3$
$y=2 x-3$
$y=-2 x+3$
$y=-2 x-3$

Show Answers Only

$C$

Show Worked Solution

$\text {Gradient is negative (top left } \rightarrow \text { bottom right)}$

$y \text{-intercept = 3 (only positive option)}$

$\Rightarrow C$

Algebra, STD2 A1 2024 HSC 1 MC

If $x=-2.531$, what is the value of $x^2$ rounded to 2 decimal places?

$-6.41$
$-6.40$
$6.40$
$6.41$

Show Answers Only

$D$

Show Worked Solution

	$ x^2$	$=(-2.531)^2 $
		$=6.405\ldots $
		$=6.41$

$\Rightarrow D$

COMMENT: Brain dead calculator usage will cause issues here.

CHEMISTRY, M3 EQ-Bank 7 MC

Metal	Reaction when heated with oxygen	Reaction when heated with water
$\ce{Mg}$	$\text{burns readily if powered to form oxides}$	$\ce{\text{forms}\ OH\ \text{ions and hydrogen gas}}$
$\ce{Al}$		$\text{reacts with steam to form oxide ions and hydrogen gas}$
$\ce{Zn}$
$\ce{Fe}$

Using the table above, which of the following equations correctly represents the reaction of aluminium with water?

$\ce{Al(s) + H2O(l) -> Al(OH)3(aq) + H2(g)}$
$\ce{2Al(s) + 3H2O(g) -> Al2O3(s) + H2(g)}$
$\ce{Al(s) + 3H2O(l) -> Al(OH)3(aq) + 3H2(g)}$
$\ce{2Al(s) + 3H2O(g) -> Al2O3(s) + 3H2(g)}$

Show Answers Only

$D$

Show Worked Solution

→ From the table, aluminium reacts with gaseous water to form aluminium oxide and hydrogen gas.

$\Rightarrow D$

BIOLOGY, M2 EQ-Bank 17

The image below shows a dinosaur fossil found in South Africa believed to be 200 million years old.

What type of diet did this dinosaur likely consume? Explain your answer. (1 mark)
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Discuss two features that could be observed in the dinosaur’s digestive tract. (3 marks)
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a. Diet: herbivore

→ The presence of flat teeth in a dinosaur species strongly indicates a herbivorous diet, as these teeth are well-suited for grinding and processing plant material.

b. Digestive tract features:

→ Herbivorous dinosaurs likely possessed specialised digestive tracts adapted for processing plant material.

→ One key feature would be an enlarged caecum, a pouch-like structure connected to the large intestine, which housed symbiotic bacteria to break down cellulose through fermentation. This process would have allowed dinosaurs to extract more nutrients from tough plant matter.

→ Additionally, these dinosaurs may have had elongated intestines to increase the surface area for nutrient absorption and provide more time for the digestion of fibrous plant material.

Show Worked Solution

a. Diet: herbivore

→ The presence of flat teeth in a dinosaur species strongly indicates a herbivorous diet, as these teeth are well-suited for grinding and processing plant material.

b. Digestive tract features:

→ Herbivorous dinosaurs likely possessed specialised digestive tracts adapted for processing plant material.

→ Additionally, these dinosaurs may have had elongated intestines to increase the surface area for nutrient absorption and provide more time for the digestion of fibrous plant material.

BIOLOGY, M4 EQ-Bank 7

Examine the diagram provided, which depicts the biological relationships within an ecosystem.

Describe how food webs differ from food chains in representing the flow of energy within an ecosystem. (2 marks)

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Name a consumer from the second trophic level in the diagram. (1 mark)
Poaching has pushed impala populations towards extinction in certain African regions. Describe the potential ecological consequences for other species in the food web shown. (2 marks)

Show Answers Only

a. Food webs vs food chains:

→ Food webs and food chains both represent energy flow in ecosystems, but differ in complexity and scope.

→ While a food chain shows a single, linear path of energy transfer from one organism to another, a food web illustrates multiple interconnected food chains within an ecosystem.

→ Food webs thus provide a more comprehensive and realistic depiction of the complex feeding relationships and energy transfers that occur among various species.

b. Second trophic level (\Rightarrow\) first order consumer

Zebra or Impala

c. Ecological consequences:

→ The decline or extinction of impalas would impact lions and vultures as they are an important food source of both.

→ Zebras would benefit from increased available vegetation due to reduced competition from impalas, leading to potential population growth.

→ The potential increase in zebra population would provide a more abundant food source for lions and vultures eventually, mitigating but not replacing the loss of impalas to these predator.

→ This scenario illustrates the complex interdependencies within the ecosystem and the cascading effects of species loss on different trophic levels.

Show Worked Solution

a. Food webs vs food chains:

→ Food webs and food chains both represent energy flow in ecosystems, but differ in complexity and scope.

→ While a food chain shows a single, linear path of energy transfer from one organism to another, a food web illustrates multiple interconnected food chains within an ecosystem.

→ Food webs thus provide a more comprehensive and realistic depiction of the complex feeding relationships and energy transfers that occur among various species.

b. Second trophic level (\Rightarrow\) first order consumer

Zebra or Impala

c. Ecological consequences:

→ The decline or extinction of impalas would impact lions and vultures as they are an important food source of both.

→ Zebras would benefit from increased available vegetation due to reduced competition from impalas, leading to potential population growth.

→ The potential increase in zebra population would provide a more abundant food source for lions and vultures eventually, mitigating but not replacing the loss of impalas to these predator.

→ This scenario illustrates the complex interdependencies within the ecosystem and the cascading effects of species loss on different trophic levels.

BIOLOGY, M4 EQ-Bank 3

Consider a grassland ecosystem with a population of rabbits, foxes, and various grass species.

Describe one example of predation in this ecosystem. (1 mark)
Explain two ways in which competition might occur between the rabbits. (2 marks)
Evaluate how the removal of foxes might affect both the rabbit population and the grass species. (1 mark)

Show Answers Only

a. → Predation occurs when foxes hunt and eat rabbits.

b. Competition among rabbits could result from (choose two):

→ limited resources such as food or water

→ suitable burrow sites

→ mating partners

c. Removing foxes:

→ could lead to an increase in the rabbit population.

→ this growing population might then overgraze the grass species, potentially causing a decline in grass biodiversity and altering the ecosystem’s structure.

Show Worked Solution

a. → Predation occurs when foxes hunt and eat rabbits.

b. Competition among rabbits could result from (choose two):

→ limited resources such as food or water

→ suitable burrow sites

→ mating partners

c. Removing foxes:

→ could lead to an increase in the rabbit population.

→ this growing population might then overgraze the grass species, potentially causing a decline in grass biodiversity and altering the ecosystem’s structure.

BIOLOGY, M4 EQ-Bank 2

Is soil pH a biotic or abiotic factor? (1 mark)
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Describe how soil pH can affect plant growth. (1 mark)
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Explain one way in which plants might alter soil pH. (1 mark)
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a. Abiotic factor

b. Soil pH:

→ Affects plant growth by influencing nutrient availability and solubility.

→ Most plants thrive in slightly acidic to neutral soils (pH 6.0-7.0) where essential nutrients are most accessible.

c. Plants alter soil pH (choose 1):

→ through root exudates, which release organic acids into the soil.

→ through the decomposition of their leaf litter, which can increase soil acidity over time.

Show Worked Solution

a. Abiotic factor

b. Soil pH:

→ Affects plant growth by influencing nutrient availability and solubility.

→ Most plants thrive in slightly acidic to neutral soils (pH 6.0-7.0) where essential nutrients are most accessible.

c. Plants alter soil pH (choose 1):

→ through root exudates, which release organic acids into the soil.

→ through the decomposition of their leaf litter, which can increase soil acidity over time.

BIOLOGY, M4 EQ-Bank 2

Our actions as a human species are inadvertently altering the evolutionary trajectories of countless organisms.

Explain two distinct mechanisms by which human activities exert selection pressures on other species. For each mechanism, provide a specific example of a species affected by this pressure. (4 marks)

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Show Answers Only

Answers could include two of the following.

Mechanism: urbanisation

→ Human activities exert selection pressure on other species through urbanisation. As cities expand, isolated pockets of natural habitat are created, forcing species to adapt to smaller, disconnected areas.

→ For example, the San Diego deer mouse has shown rapid evolution in its size and fur colour to better camouflage against urban environments, with urban mice becoming larger and darker than their rural counterparts.

Mechanism: pollution from fossil fuel burning

→ Humans exert selection pressure on other species through pollution from fossil fuel burning.

→ The emission of sulfur-dioxide from coal-burning power plants has led to acid rain, which changes soil and water pH levels. This has resulted in strong selection pressure on aquatic organisms.

→ For instance, in some Scandinavian lakes, the European perch has evolved increased tolerance to acidic conditions in order to survive.

Mechanism: use of pesticides

→ Another mechanism is the widespread use of pesticides, which creates strong selection pressures for resistance. In agriculture, the overuse of pesticides has led to the evolution of resistance in many insect pest species.

→ A specific example is the green peach aphid, which has developed resistance to multiple classes of insecticides. This adaptation makes it increasingly difficult to control these crop pests without resorting to even more potent chemicals.

Show Worked Solution

Answers could include two of the following.

Mechanism: urbanisation

Mechanism: pollution from fossil fuel burning

→ Humans exert selection pressure on other species through pollution from fossil fuel burning.

→ The emission of sulfur-dioxide from coal-burning power plants has led to acid rain, which changes soil and water pH levels. This has resulted in strong selection pressure on aquatic organisms.

→ For instance, in some Scandinavian lakes, the European perch has evolved increased tolerance to acidic conditions in order to survive.

Mechanism: use of pesticides

CHEMISTRY, M3 EQ-Bank 7 MC

Which of the following observations indicates a chemical change has occurred?

Dissolving sugar in water
Change in temperature when two solutions are mixed
Melting of ice into water
Mixing sand with iron filings

Show Answers Only

$B$

Show Worked Solution

→ A chemical change is one where a new substance is created and cannot be reversed.

→ The changing temperature of a mixed solution is the result of bonds breaking and forming new products thus indicating a chemical change.

→ All other options can be reversed by physical processes and therefore represent physical changes only.

$\Rightarrow B$

BIOLOGY, M4 EQ-Bank 10

Ecosystems are dynamic, shaped not only by physical forces but also by the living organisms within them. Including a specific example, explain one biotic factor that has significantly impacted past ecosystems:

Over a relatively short timescale (within a few decades or centuries) (2 marks)
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Over an extended geological timescale (millions of years) (2 marks)
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a. Short timescale biotic factor: introduction of invasive species

→ The introduction of cane toads in Australia in 1935 has rapidly altered local ecosystems within decades.

→ Cane toads have caused declines in native predator populations that attempt to eat the toxic toads, leading to cascading effects throughout the food web.

b. Geological timescale biotic factor: evolution of land plants

→ The development of land plants around 470 Mya led to increased oxygen production, soil formation, and the creation of new habitats.

→ This gradual but profound change altered atmospheric composition and weather patterns. This reshaped terrestrial ecosystems and paved the way for the evolution of terrestrial animal life.

Show Worked Solution

a. Short timescale biotic factor: introduction of invasive species

→ The introduction of cane toads in Australia in 1935 has rapidly altered local ecosystems within decades.

→ Cane toads have caused declines in native predator populations that attempt to eat the toxic toads, leading to cascading effects throughout the food web.

b. Geological timescale biotic factor: evolution of land plants

→ The development of land plants around 470 Mya led to increased oxygen production, soil formation, and the creation of new habitats.

→ This gradual but profound change altered atmospheric composition and weather patterns. This reshaped terrestrial ecosystems and paved the way for the evolution of terrestrial animal life.

CHEMISTRY, M3 EQ-Bank 17

A student tested how soluble silver salts are by reacting a 0.1 mol L$^{-1}$ silver nitrate solution with 0.1 mol L$^{-1}$ solutions of calcium hydroxide, calcium chloride, and calcium sulfate. The results are shown below:

\begin{array} {|l|l|}
\hline \ \ \ \ \ \text{Compound} & \ \ \ \ \ \text{Observation} \\
\hline \text{calcium hydroxide} & \text{No reaction} \\
\hline \text{calcium chloride} & \text{White precipitate} \\
\hline \text{calcium sulfate} & \text{No reaction} \\
\hline \end{array}

Write a balanced chemical equation for the reaction with calcium chloride. (2 marks)

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Name the white precipitate. (1 mark)

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a. $\ce{CaCl2(aq) + 2AgNO3(aq) -> Ca(NO3)2(aq) + 2AgCl(s)}$

b. The white precipitate is $\ce{AgCl}$ → silver chloride.

Show Worked Solution

a. $\ce{CaCl2(aq) + 2AgNO3(aq) -> Ca(NO3)2(aq) + 2AgCl(s)}$

b. The white precipitate is $\ce{AgCl}$ → silver chloride.

BIOLOGY, M4 EQ-Bank 6

The graph below shows the concentration of $\ce{CO2}$ in the earth's atmosphere over the last 800 years.

How would scientists obtain these historical levels of $\ce{CO2}$? (1 mark)

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Explain the shape of the graph over the 800 years of data presented. (3 marks)

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a. Data would be obtained by gas analysis within ice cores.

b. Graph shape over the past 800 years:

→ Atmospheric levels of $\ce{CO2}$ were very steady between the period 1200-1800 at approximately 280 ppm.

→ In the period 1800-2000, $\ce{CO2}$ levels increased exponentially from 280 ppm to 370 ppm.

→ This increase coincided with industrialisation which saw the widespread use of fossil fuels like coal, oil, and natural gas for energy.

→ The use of these energy sources released large amounts of (\ce{CO2}\) into the atmosphere and is regarded as a major contributing factor to rise in atmospheric $\ce{CO2}$ levels over this period.

Show Worked Solution

a. Data would be obtained by gas analysis within ice cores.

b. Graph shape over the past 800 years:

→ Atmospheric levels of $\ce{CO2}$ were very steady between the period 1200-1800 at approximately 280 ppm.

→ In the period 1800-2000, $\ce{CO2}$ levels increased exponentially from 280 ppm to 370 ppm.

→ This increase coincided with industrialisation which saw the widespread use of fossil fuels like coal, oil, and natural gas for energy.

BIOLOGY, M4 EQ-Bank 5

Scientists analyse the ratio of $\ce{^{16}O}$ to $\ce{^{18}O}$ isotopes in various geological samples to reconstruct past climatic conditions.

Describe one method used to obtain oxygen isotope data from ancient samples. (1 mark)

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Explain how the relationship between $\ce{^{16}O}$ and $\ce{^{18}O}$ ratios provides evidence of past changes in ecosystems. (3 marks)

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a. Scientists analyse gas trapped within ice cores.

b. $\ce{^{16}O : ^{18}O}$ ratios

→ The ratio of $\ce{^{16}O : ^{18}O}$ in geological samples reflects past temperature and precipitation patterns, which are crucial factors in shaping ecosystems.

→ During warmer periods, there is a higher proportion of the heavier $\ce{^{18}O}$ isotope in precipitation.

→ During cooler periods, there is a relative increase in lighter $\ce{^{16}O}$.

→ These isotopic signatures allow scientists to reconstruct past climate conditions and infer associated ecosystem changes.

Show Worked Solution

a. Scientists analyse gas trapped within ice cores.

b. $\ce{^{16}O : ^{18}O}$ ratios

→ The ratio of $\ce{^{16}O : ^{18}O}$ in geological samples reflects past temperature and precipitation patterns, which are crucial factors in shaping ecosystems.

→ During warmer periods, there is a higher proportion of the heavier $\ce{^{18}O}$ isotope in precipitation.

→ During cooler periods, there is a relative increase in lighter $\ce{^{16}O}$.

→ These isotopic signatures allow scientists to reconstruct past climate conditions and infer associated ecosystem changes.

CHEMISTRY, M3 EQ-Bank 3 MC

Which of the following are the products of the complete combustion of propane, $\ce{C3H8}$?

Carbon monoxide, soot and water
Carbon dioxide, carbon monoxide, soot and water
Carbon monoxide and water
Carbon dioxide and water

Show Answers Only

$D$

Show Worked Solution

→ A reaction which under goes complete combustion will produce carbon dioxide and water.

→ Carbon monoxide is a product of incomplete combustion which occurs when there is a lack of oxygen available.

$\Rightarrow D$

CHEMISTRY, M3 EQ-Bank 2 MC

Which of the following equations involving sulfur compounds represents a synthesis reaction?

$\ce{2ZnS(s) + 3O2(g) -> 2ZnO(s) + 2SO2(g)}$
$\ce{SO2(g) + O2(g) -> 2SO3(g)}$
$\ce{SO2(g) -> S(s) + O2(g)}$
$\ce{H2SO4(aq) -> SO3(g) + H2O(l)}$

Show Answers Only

$B$

Show Worked Solution

→ A synthesis reaction is one where two or more reactants (sulfur dioxide and oxygen) combine to form a more complex product (sulfur trioxide).

$\Rightarrow B$

BIOLOGY, M3 EQ-Bank 2

Define convergent evolution within the context of Darwin and Wallace's Theory of Evolution by Natural Selection. (1 mark)

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Describe two key principles of natural selection that lead to convergent evolution. (2 marks)

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Provide two examples of convergent evolution, one involving Australian fauna and one non-Australian, explaining how each demonstrates the process of natural selection. (2 marks)

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a. Convergent evolution definition:

→ The process by which unrelated organisms develop similar traits or features as a result of adapting to similar environmental pressures or ecological niches.

b. Key principles that lead to convergent evolution:

→ Natural selection acts on existing variations within populations, favouring traits that enhance survival and reproduction in specific environments.

→ When unrelated organisms face similar environmental challenges, natural selection can lead to the evolution of similar adaptations, even in distantly related species.

→ This process occurs independently in each lineage, resulting in analogous structures or behaviours that serve similar functions but have different evolutionary origins.

c. Australian fauna (example):

→ The similarity between the marsupial Tasmanian tiger (thylacine) and the placental grey wolf is an example of convergent evolution.

→ Despite their distant relationship, both evolved similar body shapes, jaw structures, and striped patterns due to adapting to similar predatory lifestyles.

Non-Australian (example):

→ The similar body shapes of sharks and dolphins is another example of convergent evolution.

→ Though one is a fish and the other a mammal, both have evolved streamlined bodies, dorsal fins, and tail flukes as adaptations for efficient swimming in marine environments.

→ This demonstrates how natural selection can produce similar outcomes in response to comparable environmental pressures.

Show Worked Solution

a. Convergent evolution definition:

→ The process by which unrelated organisms develop similar traits or features as a result of adapting to similar environmental pressures or ecological niches.

b. Key principles that lead to convergent evolution:

→ Natural selection acts on existing variations within populations, favouring traits that enhance survival and reproduction in specific environments.

→ When unrelated organisms face similar environmental challenges, natural selection can lead to the evolution of similar adaptations, even in distantly related species.

→ This process occurs independently in each lineage, resulting in analogous structures or behaviours that serve similar functions but have different evolutionary origins.

c. Australian fauna (example):

→ The similarity between the marsupial Tasmanian tiger (thylacine) and the placental grey wolf is an example of convergent evolution.

→ Despite their distant relationship, both evolved similar body shapes, jaw structures, and striped patterns due to adapting to similar predatory lifestyles.

Non-Australian (example):

→ The similar body shapes of sharks and dolphins is another example of convergent evolution.

→ Though one is a fish and the other a mammal, both have evolved streamlined bodies, dorsal fins, and tail flukes as adaptations for efficient swimming in marine environments.

→ This demonstrates how natural selection can produce similar outcomes in response to comparable environmental pressures.

BIOLOGY, M3 EQ-Bank 2 MC

Which of the following is an example of a behavioural adaptation in Australian wildlife?

The waxy cuticle on eucalyptus leaves .
The salt glands in marine birds.
The burrowing of the desert frill-neck lizard during extreme heat.
The pouch of a kangaroo.

Show Answers Only

$C$

Show Worked Solution

→ The burrowing of the desert frill-neck lizard during extreme heat is a behavioural adaptation because it’s a learned action that helps the animal cope with environmental challenges, rather than a physical feature or internal process.

$\Rightarrow C$

CHEMISTRY, M3 EQ-Bank 5 MC

Which of the following best explains how an increase in temperature affects the rate of a chemical reaction?

It increases the energy of the products, making the reaction proceed faster.
It increases the frequency and energy of collisions between reactant molecules.
It decreases the activation energy needed for the reaction.
It decreases the concentration of the reactants, slowing the reaction.

Show Answers Only

$B$

Show Worked Solution

→ Increasing the temperature raises the kinetic energy of the reactant molecules, which increases both the frequency of collisions and the energy of these collisions.

→ More frequent and higher-energy collisions result in a higher likelihood of successful collisions that overcome the activation energy, thereby speeding up the reaction.

→ Therefore, the rate of reaction increases with temperature.

$\Rightarrow B$

BIOLOGY, M3 EQ-Bank 1

Describe the concept of structural adaptation in living organisms. In your answer, one example of a structural adaptation in a plant species and one example in an animal species. (3 marks)

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→ A structural adaptation refers to a physical feature of an organism that has evolved over time to enhance its survival in a specific environment.

→ Plant example: Many Australian eucalyptus species have a thick, waxy cuticle on their leaves which helps reduce water loss in arid conditions.

→ Animal example: The streamlined body shape of dolphins is a structural adaptation that allows for efficient movement through water, reducing drag and enabling high-speed swimming.

Show Worked Solution

→ A structural adaptation refers to a physical feature of an organism that has evolved over time to enhance its survival in a specific environment.

→ Plant example: Many Australian eucalyptus species have a thick, waxy cuticle on their leaves which helps reduce water loss in arid conditions.

→ Animal example: The streamlined body shape of dolphins is a structural adaptation that allows for efficient movement through water, reducing drag and enabling high-speed swimming.

CHEMISTRY, M3 EQ-Bank 11

Brian measured the reaction rate of 0.5 g of sodium metal in 2.0 mol/L nitric acid. The volume of hydrogen gas produced per minute was recorded both without a catalyst and with copper as a catalyst.

\begin{array} {|c|c|c|}
\hline \text{Time (minutes)} & \text{Volume of hydrogen gas (mL)} & \text{Volume of hydrogen gas (mL)} \\ & \text{no catalyst} & \text{catalyst} \\
\hline \text{0} & 0 & 0 \\
\hline \text{1} & 16 & 7 \\
\hline \text{2} & 30 & 13\\
\hline \text{3} & 43 & 19 \\
\hline \text{4} & 44 & 26 \\
\hline \text{5} & 44 & 32 \\
\hline \text{6} & 44 & 38 \\
\hline \text{7} & 44 & 44 \\
\hline \end{array}

On the grid below, plot a line graph for the data in the table. (3 marks)
Write a conclusion for the experiment. (2 marks)

--- 3 WORK AREA LINES (style=blank) ---

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b. Experiment conclusion:

→ The rate of reaction was significantly faster when copper was used as a catalyst compared to when no catalyst was present.

→ This is because the copper catalyst provided an alternative reaction pathway with a lower activation energy, allowing the reactant particles to collide more frequently and effectively.

Show Worked Solution

b. Experiment conclusion:

→ The rate of reaction was significantly faster when copper was used as a catalyst compared to when no catalyst was present.

→ This is because the copper catalyst provided an alternative reaction pathway with a lower activation energy, allowing the reactant particles to collide more frequently and effectively.

BIOLOGY, M3 EQ-Bank 2

Describe, providing an example, an abiotic factor that could act as a selection pressure in

a desert ecosystem. (2 marks)
--- 5 WORK AREA LINES (style=lined) ---
an aquatic ecosystem. (2 marks)
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a. Desert ecosystem:

→ An abiotic factor acting as a selection pressure is extremely low rainfall/moisture.

→ For example, the thorny devil lizard has evolved a unique skin structure that allows it to collect and channel water to its mouth from any part of its body, enabling it to survive in extremely arid conditions.

b. Aquatic ecosystem:

→ In an aquatic ecosystem, water temperature can act as a significant abiotic selection pressure.

→ The Great Barrier Reef, for example, experiences coral bleaching events when water temperatures rise. These events stress the coral and if prolonged, can potentially kill the coral.

Show Worked Solution

a. Desert ecosystem:

→ An abiotic factor acting as a selection pressure is extremely low rainfall/moisture.

b. Aquatic ecosystem:

→ In an aquatic ecosystem, water temperature can act as a significant abiotic selection pressure.

→ The Great Barrier Reef, for example, experiences coral bleaching events when water temperatures rise. These events stress the coral and if prolonged, can potentially kill the coral.

BIOLOGY, M3 EQ-Bank 3 MC

In a coral reef ecosystem, which of the following scenarios represents a biotic selection pressure rather than an abiotic one?

An increase in water temperature due to climate change.
An outbreak of coral-eating crown-of-thorns starfish.
A rise in sea level causing deeper water over the reef.
An increase in water acidity due to higher atmospheric $\ce{CO2}$.

Show Answers Only

$B$

Show Worked Solution

→ An outbreak of coral-eating crown-of-thorns starfish is a biotic selection pressure because it involves the direct interaction between living organisms (the starfish and the coral).

→ The other options are abiotic factors related to physical or chemical changes in the environment.

$\Rightarrow B$

Functions, EXT1 F1 EQ-Bank 5

Given $f(x)=x(x+2)(2-x)$

On the graph, sketch the graphs of $y=f(\abs{x})$ (2 marks)

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Show Worked Solution

Functions, EXT1 F1 EQ-Bank 6

The diagram shows the graph of $f(x)=x(x+2)(2-x)$

On the graph, sketch the graph of $y=\abs{f(x)}$ (2 marks)

--- 0 WORK AREA LINES (style=blank) ---

Show Answers Only

Show Worked Solution

CHEMISTRY, M3 EQ-Bank 2 MC

Select the correct characteristic of a catalyst from the options below:

A catalyst slows down the rate of reaction
A catalyst doubles during a chemical reaction
A catalyst is used up in a chemical reaction
A catalyst speeds up the rate of reaction

Show Answers Only

$D$

Show Worked Solution

→ A catalyst speeds up the rate of a reaction by providing an alternative pathway with lower activation energy.

→ It remains chemically unchanged at the end of the reaction and is not consumed during the process.

$\Rightarrow D$

CHEMISTRY, M3 EQ-Bank 10

A student conducted a series of investigations where 8.50 g of sodium carbonate was reacted with excess nitric acid $\ce{(HNO3)}$ at a temperature of 25°C and 100 kPa. The volume of carbon dioxide gas produced was measured at regular intervals during each investigation. In experiment A, sodium carbonate was provided as large crystals, and in experiment B, it was supplied in powdered form.

Both reactions produced 1.988 L of $\ce{CO2(g)}$ however experiment B finished reacting before experiment A finished reacting.

Explain why experiment B had a faster rate of reaction than experiment A. (1 mark)

--- 2 WORK AREA LINES (style=blank) ---

Using the volume of $\ce{CO2(g)}$ produced, calculate the maximum mass of carbon dioxide produced in the reaction. (3 marks)

--- 2 WORK AREA LINES (style=blank) ---

Explain why both experiments produced in the same volume of carbon dioxide. (1 mark)

--- 2 WORK AREA LINES (style=blank) ---

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a. Experiment B faster than experiment A:

→ In experiment B, the surface area of the sodium carbonate was greater than in experiment A due to it being in a powdered form.

→ Thus, there are a greater number of collisions able to occur in experiment B, leading to a greater number of successful collisions.

→ Overall this allows the rate of reaction of experiment B to be greater than that of experiment A.

b. $3.52\ \text{g}$

c. Reasons why same volume $\ce{CO2}$ produced:

→ Both experiments used the same amount of sodium carbonate reacting with excess hydrochloric acid.

→ The maximum amount of carbon dioxide produced is dependent on how much sodium carbonate reacted.

→ As the initial mass of sodium carbonate is the same in both reaction and both reactions went until completion, the volume of carbon dioxide will be the same despite the different rates of reaction between experiment A and B.

Show Worked Solution

a. Experiment B faster than experiment A:

→ In experiment B, the surface area of the sodium carbonate was greater than in experiment A due to it being in a powdered form.

→ Thus, there are a greater number of collisions able to occur in experiment B, leading to a greater number of successful collisions.

→ Overall this allows the rate of reaction of experiment B to be greater than that of experiment A.

b. The maximum volume of carbon dioxide produced is 1.988 L.

→ At SLC (standard laboratory conditions), the 1 mole of gas takes up 24.79 L.

→ $\ce{n(CO2(g))}= \dfrac{1.988}{24.79} = 0.08\ \text{mol}$

→ $\ce{m(CO2(g))} = n \times MM = 0.08 \times 44.01 =3.52\ \text{g}$

c. Reasons why same volume $\ce{CO2}$ produced:

→ Both experiments used the same amount of sodium carbonate reacting with excess hydrochloric acid.

→ The maximum amount of carbon dioxide produced is dependent on how much sodium carbonate reacted.

Functions, EXT1 F1 EQ-Bank 4

Given $f(x)=3 x-1$, determine the domain and range for $y=\sqrt{f(x)}$. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Sketch the graph $y=\sqrt{f(x)}$, clearly identifying any axis intercepts. (1 mark)

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i.	$f(x)$	$=3x-1$
	$y$	$=\sqrt{3x-1}$

$\text{Domain:}\ \ 3 x-1 \ \geqslant 0 \ \Rightarrow \ x \geqslant \dfrac{1}{3}$

$\text {Range:}\ \ y \geqslant 0$

ii.

Show Worked Solution

i.	$f(x)$	$=3x-1$
	$y$	$=\sqrt{3x-1}$

$\text{Domain:}\ \ 3 x-1 \ \geqslant 0 \ \Rightarrow \ x \geqslant \dfrac{1}{3}$

$\text {Range:}\ \ y \geqslant 0$

ii.

Trigonometry, EXT1 T1 EQ-Bank 5

State the domain and range for $f(x)=4 \sin ^{-1}(2 x-3)$. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Sketch the function $f(x)=4 \sin ^{-1}(2 x-3)$. (1 mark)

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a.	$\text{Domain:}\ \ -1 \leqslant$	$2 x-3 \leqslant 1 $
	$ 2 \leqslant$	$2 x \leqslant 4$
	$ 1 \leqslant$	$x \leqslant 2$

	$\text{Range:}\ \ -\dfrac{\pi}{2} \leqslant$	$y \leqslant \dfrac{\pi}{2}$
	$ -2 \pi \leqslant $	$4 y \leqslant 2 \pi$

Show Worked Solution

a.	$\text{Domain:}\ \ -1 \leqslant$	$2 x-3 \leqslant 1 $
	$ 2 \leqslant$	$2 x \leqslant 4$
	$ 1 \leqslant$	$x \leqslant 2$

	$\text{Range:}\ \ -\dfrac{\pi}{2} \leqslant$	$y \leqslant \dfrac{\pi}{2}$
	$ -2 \pi \leqslant $	$4 y \leqslant 2 \pi$

BIOLOGY, M2 EQ-Bank 3 MC

Which of the following plant tissues is responsible for transporting water and minerals from the roots to the leaves?

Phloem
Xylem
Epidermis
Palisade mesophyll

Show Answers Only

$B$

Show Worked Solution

→ Xylem is the vascular tissue responsible for transporting water and minerals from the roots to the leaves in plants.

$\Rightarrow B$

CHEMISTRY, M2 EQ-Bank 8v3

Consider the compounds ethanol ($\ce{C2H6O}$), formaldehyde ($\ce{CH2O}$), and acetic acid ($\ce{C2H4O2}$).
Identify which TWO of these compounds have the same empirical formula and justify your choice. (2 marks)

--- 4 WORK AREA LINES (style=blank) ---
The empirical formula of a compound is $\ce{C4H5N2}$ and its molar mass is determined to be 243.3 g mol$^{-1}$.
Calculate the molecular formula of this compound. (3 marks)

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a. Formaldehyde ($\ce{CH2O}$) and acetic acid ($\ce{C2H4O2}$) have the same empirical formula of $\ce{CH2O}$.

b. The molecular formula of the compound is $\ce{C12H15N6}$.

Show Worked Solution

a. Determine the empirical formula of each compound:

Ethanol ($\ce{C2H6O}$):
\[\text{Empirical formula} = \ce{C2H6O}\]

Formaldehyde ($\ce{CH2O}$):
\[\text{Empirical formula} = \ce{CH2O}\]

Acetic acid ($\ce{C2H4O2}$):
\[\text{Empirical formula} = \ce{CH2O}\]

Formaldehyde and acetic acid have the same empirical formula of $\ce{CH2O}$.

b. Calculate the molar mass of the empirical formula $\ce{C4H5N2}$:

\[\text{Molar mass of} \ \ce{C4H5N2} = 4 \times 12.01 + 5 \times 1.01 + 2 \times 14.01 = 81.1 \ \text{g/mol}\]

Determine the ratio of the molar mass of the compound to the molar mass of the empirical formula:

\[\text{Ratio} = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{243.3 \ \text{g/mol}}{81.1 \ \text{g/mol}} = 3\]

Multiply the subscripts in the empirical formula by 3 to get the molecular formula:

\[\text{Molecular formula} = \ce{(C4H5N2)} \times 3 = \ce{C12H15N6}\]

Thus, the molecular formula of the compound is $\ce{C12H15N6}$.

CHEMISTRY, M2 EQ-Bank 8v2

Consider the compounds glyceraldehyde ($\ce{C3H6O3}$), glycolic acid ($\ce{C2H4O3}$), and ribose ($\ce{C5H10O5}$).
Identify which TWO of these compounds have the same empirical formula and justify your choice. (2 marks)

--- 4 WORK AREA LINES (style=blank) ---
The empirical formula of a compound is $\ce{C3H5O}$ and its molar mass is determined to be 114.14 g mol$^{-1}$.
Calculate the molecular formula of this compound. (3 marks)

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a. glyceraldehyde ($\ce{C3H6O3}$) and ribose ($\ce{C5H10O5}$) have the same empirical formula of $\ce{CH2O}$.

b. The molecular formula of the compound is $\ce{C6H10O2}$.

Show Worked Solution

a. Determine the empirical formula of each compound:

Glyceraldehyde ($\ce{C3H6O3}$):
\[\text{Empirical formula} = \ce{CH2O}\]

Glycolic acid ($\ce{C2H4O3}$):
\[\text{Empirical formula} = \ce{C2H4O3}\]

Ribose ($\ce{C5H10O5}$):
\[\text{Empirical formula} = \ce{CH2O}\]

Acetone and ribose have the same empirical formula of $\ce{CH2O}$.

b. Calculate the molar mass of the empirical formula $\ce{C3H5O}$:

\[\text{Molar mass of} \ \ce{C3H5O} = 3 \times 12.01 + 5 \times 1.008 + 1 \times 16.00 = 57.07 \ \text{g/mol}\]

Determine the ratio of the molar mass of the compound to the molar mass of the empirical formula:

\[\text{Ratio} = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{114.14 \ \text{g/mol}}{57.07 \ \text{g/mol}} = 2\]

Multiply the subscripts in the empirical formula by 2 to get the molecular formula:

\[\text{Molecular formula} = \ce{(C3H5O)} \times 2 = \ce{C6H10O2}\]

Thus, the molecular formula of the compound is $\ce{C6H10O2}$.

CHEMISTRY, M4 EQ-Bank 7

Bond energies can be used to estimate the enthalpy change of a reaction. The equation for the combustion of methane $\ce{CH4}$ is shown below:

$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$

Given the following bond energies:

$\ce{C-H}$ bond: 412 kJ mol$^{-1}$
$\ce{O=O}$ bond: 498 kJ mol$^{-1}$
$\ce{C=O}$ bond: 805 kJ mol$^{-1}$
$\ce{O-H}$ bond: 463 kJ mol$^{-1}$

Calculate the total bond energy of the products. (2 marks)

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$3462\ \text{kJ mol}^{-1}$

Show Worked Solution

→ Total bond energy of 1 mole of $\ce{CO2} = 2 \times 805 = 1610\ \text{kJ mol}^{-1}$

→ Total bond energy of 2 mole of $\ce{H2O} = 4 \times 463 = 1852\ \text{kJ mol}^{-1}$

→ Thus the total bond energy of the products is $3462\ \text{kJ mol}^{-1}$

BIOLOGY, M2 EQ-Bank 10

Explain how physical and chemical digestion work together in mammals to improve the efficiency of nutrient absorption. (3 marks)

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→ Physical and chemical digestion work together synergistically to improve the efficiency of nutrient absorption in mammals.

→ For example, in the stomach, muscular contractions churn food, breaking it into smaller pieces and increasing its surface area.

→ This increased surface area allows the chemical digestive processes to work more effectively on the food particles.

→ The combined action continues in the small intestine, where physical segmentation movements mix the chyme with digestive enzymes, further breaking down nutrients chemically.

→ This teamwork between physical and chemical processes ensures that nutrients are broken down into their simplest forms, maximising their absorption through the intestinal wall.

Show Worked Solution

→ Physical and chemical digestion work together synergistically to improve the efficiency of nutrient absorption in mammals.

→ For example, in the stomach, muscular contractions churn food, breaking it into smaller pieces and increasing its surface area.

→ This increased surface area allows the chemical digestive processes to work more effectively on the food particles.

→ The combined action continues in the small intestine, where physical segmentation movements mix the chyme with digestive enzymes, further breaking down nutrients chemically.

→ This teamwork between physical and chemical processes ensures that nutrients are broken down into their simplest forms, maximising their absorption through the intestinal wall.

CHEMISTRY, M4 EQ-Bank 3

The combustion of methane $\ce{(CH4)}$ is represented by the following equation:

$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}\qquad \Delta H = -890\ \text{kJ mol}^{-1}$

Calculate the energy change when 3 moles of methane are combusted. (2 marks)

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$-2670\ \text{kJ mol}^{-1}$

Show Worked Solution

→ The enthalpy change for 1 mole of methane is given as $-890\ \text{kJ}$.

→ For 3 moles of methane, the energy change is:

$\Delta H = 3 \times -890 = -2670\ \text{kJ}$

BIOLOGY, M2 EQ-Bank 3

Write a word equation that describes the process of photosynthesis. (1 mark)
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Outline the distribution and uses of photosynthetic products in plants. (3 marks)
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b. Glucose:

→ The primary product of photosynthesis is glucose, which can be transported via the phloem to other parts of the plant, such as roots, fruits, or growing tissues.

→ Alternatively, it can be converted to starch for short-term storage in leaves or transported to roots or other storage organs for long-term storage.

Oxygen:

→ Oxygen is produced as a byproduct and released into the atmosphere through the stomata, although some is used by the plant for its own cellular respiration processes.

Show Worked Solution

b. Glucose:

→ The primary product of photosynthesis is glucose, which can be transported via the phloem to other parts of the plant, such as roots, fruits, or growing tissues.

→ Alternatively, it can be converted to starch for short-term storage in leaves or transported to roots or other storage organs for long-term storage.

Oxygen:

→ Oxygen is produced as a byproduct and released into the atmosphere through the stomata, although some is used by the plant for its own cellular respiration processes.

Functions, EXT1 F2 EQ-Bank 3 MC

When $2x^3-x^2+p x-6$ is divided by $x+2$ the remainder is $-4$. What is the value of $p$ ?

$-11$
$-7$
$-5$
$-2$

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$A$

Show Worked Solution

$\text{Since}\ (2x^3-x^2+p x-6)\ ÷\ (x+2)=-4 \ \ \Rightarrow\ \ P(-2)=-4$

$2\times (-2)^3-(-2)^2-2p-6$	$=-4$
$-16-4-2p-6$	$=-4$
$2p$	$=-22$
$p$	$=-11$

$\Rightarrow A$

BIOLOGY, M2 EQ-Bank 8

Epidermal tissues in plants can be compared to the epithelium in animals.

Describe one structural similarity between these tissue types. (1 mark)

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Explain one functional similarity they share. (1 mark)

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Discuss how both tissues contribute to their respective organism's interaction with the environment. (2 marks)

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a. Structural similarity could include one of the following:

→ Both plant epidermal tissue and animal epithelium form a single layer of tightly packed cells that cover the outer surfaces of the organism.

→ Both types of tissue are typically attached to a basement membrane that separates them from underlying tissues.

b. Functional similarity could include one of the following:

→ Both tissue types serve as a protective barrier, shielding the underlying tissues from physical damage.

→ Both tissue types can provide protection from pathogens and excessive water loss.

c. Interaction with the environment:

→ In plants, the epidermis regulates gas exchange through stomata and may produce structures like trichomes for protection or water conservation.

→ Similarly, in animals, the epithelium can be specialised for absorption, allowing the organism to respond to various environmental stimuli.

Show Worked Solution

a. Structural similarity could include one of the following:

→ Both plant epidermal tissue and animal epithelium form a single layer of tightly packed cells that cover the outer surfaces of the organism.

→ Both types of tissue are typically attached to a basement membrane that separates them from underlying tissues.

b. Functional similarity could include one of the following:

→ Both tissue types serve as a protective barrier, shielding the underlying tissues from physical damage.

→ Both tissue types can provide protection from pathogens and excessive water loss.

c. Interaction with the environment:

→ In plants, the epidermis regulates gas exchange through stomata and may produce structures like trichomes for protection or water conservation.

→ Similarly, in animals, the epithelium can be specialised for absorption, allowing the organism to respond to various environmental stimuli.

BIOLOGY, M2 EQ-Bank 5

Define cell differentiation. (1 mark)

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Explain one factor that influences cell differentiation. (1 mark)

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Discuss why cell differentiation is crucial for the functioning of complex organisms. (2 marks)

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a. Cell differentiation:

→ The process by which a less specialised cell becomes a more specialised cell type with a specific structure and function.

b. Factors that influence cell differentiation (include one of the following):

→ The timing of gene activation during development, which can determine what type of cell a stem cell becomes

→ The presence of specific growth factors or signalling molecules in the cell’s environment, which can activate or repress certain genes.

→ A cell’s position within an embryo or tissue, which exposes it to different chemical signals from neighbouring cells.

c. Reasons cell differentiation is crucial for the functioning of complex organisms:

→ It allows for the development of specialised tissues and organs, each performing specific functions more efficiently than generalised cells could.

→ This specialisation enables the division of labor within the organism, leading to more complex and sophisticated biological processes.

→ Cell differentiation also allows multicellular organisms to develop intricate body plans and respond more effectively to their environment, ultimately enhancing their survival and reproductive success.