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CHEMISTRY, M2 EQ-Bank 1 MC

A glucose solution is prepared by dissolving pure glucose \(\ce{C6H12O6}\) in distilled water. A 50 mL vial contains 12.0 milligrams of glucose.

What is the concentration of glucose in the solution?

  1. \(1.3 \times 10^{-3}\ \text{mol L}^{-1}\)
  2. \(2.2 \times 10^{-2}\ \text{mol L}^{-1}\)
  3. \(3.9 \times 10^{-3}\ \text{mol L}^{-1}\)
  4. \(6.7 \times 10^{-2}\ \text{mol L}^{-1}\)
Show Answers Only

\(A\)

Show Worked Solution

\(\ce{MM(C6H12O6)} = (6 \times 12.01) + (12 \times 1.008) + (6 \times 16.00) = 180.156\ \text{g mol}^{-1}\)

\(\ce{n(C6H12O6)} = \dfrac{12.0 \times 10^{-3}}{180.156}=6.66 \times 10^{-5}\ \text{mol}\)

\(c=\dfrac{\text{n}}{\text{V}}=\dfrac{6.66 \times 10^{-5}}{50 \times 10^{-3}}=1.3 \times 10^{-3}\ \text{mol L}^{-1}\)

\(\Rightarrow A\)

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}

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

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

Show Answers Only

a.   \(\text{Critical Path:}\ BEGI\)

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

c.  \(\text{Max time}\ =8\ \text{hours}\)

Show Worked Solution

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

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

\(\text{Critical Path:}\ BEGI\)
 

♦ Mean mark (a) 43%.

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

c.   \(\text{Since}\ C + F + H + I\ \text{is not a critical path:}\)

\(C + F + H + I = 18\ \text{or less (C.P. = 19 hours)}\)

\(3+F+H+7 = 18\ \text{or less}\)

\(\Rightarrow\ F+H = 8\ \text{or less}\)

\(\therefore\ \text{Max time from start of}\ F\ \text{to end of}\ H = 8\ \text{hours}\)

♦♦♦ Mean mark (c) 10%.

CHEMISTRY, M2 EQ-Bank 11

Iron forms a compound that contains iron (36.8%), sulfur (31.6%), and oxygen (31.6%). The compound boils at 120°C. For one mole of this compound, the density of its vapor at 150°C and 250 kPa is 42.0 g/L.

  1. Determine the empirical formula of the compound.   (2 mark)
  1. Calculate the molar mass of the compound.   (2 marks)
Show Answers Only

a.    \(\ce{Fe2S3O6}\)

b.    \(590.94\ \text{g mol}^{-1}\).

Show Worked Solution

a.   Divide each compound’s percentage by their molar masses:

\(\ce{Fe}: \dfrac{36.8%}{55.85} = 0.65 \ \Rightarrow \ \dfrac{0.659}{0.659}=1\)

\(\ce{S}: \dfrac{31.6%}{32.07} = 0.985 \ \Rightarrow \ \dfrac{0.985}{0.659}=\dfrac{3}{2}\)

\(\ce{O}: \dfrac{31.6%}{16.00} = 1.975 \ \Rightarrow \ \dfrac{1.975}{0.659}=3\)

  • Due to the fraction, each number must be doubled so there are only whole numbers.
  • Thus the empirical formula for the compound is \(\ce{Fe2S3O6}\).

b.    Using the Ideal Gas Law to find the volume of the vapour:

\(V=\dfrac{nRT}{P}=\dfrac{1 \times 8.314 \times (150+273)}{250}=14.07\ \text{L}\)

  • Molar mass of the vapour \(= 42 \times 14.07 = 590.94\ \text{g mol}^{-1}\).

Measurement, STD2 M6 2024 HSC 36

The diagram shows two vertical flagpoles, \(BE\) and \(CD\), set on sloping ground.
 

  1. What is the height of the flagpole \(BE\), correct to 1 decimal place?   (2 marks)
  2. What is the height of the flagpole \(CD\), correct to 1 decimal place?   (2 marks)
Show Answers Only

a.   \(BE=25.4\ \text{m}\)

b.   \(CD=19.7\ \text{m}\)

Show Worked Solution

a.   \(\text{Using the sine rule:}\)

\(\dfrac{BE}{\sin 27^{\circ}}\) \(=\dfrac{53.8}{\sin 106^{\circ}}\)  
\(\therefore BE\) \(=\dfrac{53.8 \times \sin 27^{\circ}}{\sin 106^{\circ}}\)  
  \(=25.408…\)  
  \(=25.4\ \text{m (1 d.p.)}\)  

 

b.   \(CD=EB-XB\)

\(\text{Consider}\ \Delta XBC:\)

\(\angle XBC=180-106=74^{\circ}, \ XC=ED=20\)

\(\tan 74^{\circ}\) \(=\dfrac{20}{XB}\)  
\(XB\) \(=\dfrac{20}{\tan 74^{\circ}}\)  
  \(=5.73\ \text{m}\)  

 
\(CD=25.4-5.73=19.7\ \text{m (1 d.p.)}\)

♦♦ Mean mark (b) 35%.

Statistics, STD2 S5 2024 HSC 35

A random variable is normally distributed with mean 0 and standard deviation 1. The table gives the probability that this random variable is less than \(z\).

\begin{array} {|c|c|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} z \rule[-1ex]{0pt}{0pt} & 0.6 & 0.7 & 0.8 & 0.9 & 1.0 & 1.1 & 1.2 & 1.3 & 1.4 \\
\hline
\rule{0pt}{2.5ex} \textit{Probability} \rule[-1ex]{0pt}{0pt} & 0.7257 & 0.7580 & 0.7881 & 0.8159 & 0.8413 & 0.8643 & 0.8849 & 0.9032 & 0.9192 \\
\hline
\end{array}

The probability values given in the table for different values of \(z\) are represented by the shaded area in the following diagram.
 

The scores in a university examination with a large number of candidates are normally distributed with mean 58 and standard deviation 15.

  1. By calculating a \(z\)-score, find the percentage of scores that are between 58 and 70.   (2 marks)
  2. Explain why the percentage of scores between 46 and 70 is twice your answer to part (a).   (1 mark)
  3. By using the values in the table above, find an approximate minimum score that a candidate would need to be placed in the top 10% of the candidates.   (2 marks)
Show Answers Only

a.   \(28.81%\)

b.   \(\text{Normal distribution is symmetrical about the mean (58).}\)

\(\text{Since 70 and 46 are both the same distance (12) from the mean, the}\)

\(\text{percentage of scores in this range will be twice the answer in part (a).}\)

c.   \(\text{Approx minimum score = 78%}\)

Show Worked Solution

a.   \(z\text{-score (58)}\ =\dfrac{x-\mu}{\sigma} = \dfrac{58-58}{15}=0\)

\(z\text{-score (70)}\ = \dfrac{70-58}{15}=0.8\)

\(\text{Using table:}\)

\(\text{% between 58–70}\ =0.7881-0.5=0.2881=28.81%\)
 

♦♦ Mean mark (a) 36%.

b.   \(\text{Normal distribution is symmetrical about the mean (58).}\)

\(\text{Since 70 and 46 are both the same distance (12) from the mean, the}\)

\(\text{percentage of scores in this range will be twice the answer in part (a).}\)
 

♦♦♦ Mean mark (b) 19%.

c.   \(z\text{-score 1.3 has a table value 0.9032}\)

\(1-0.9032=0.0968\ \Rightarrow\ \text{i.e. 9.68% of students score higher.}\)

\(\text{Find}\ x\ \text{for a}\ z\text{-score of 1.3:}\)

\(1.3\) \(=\dfrac{x-58}{15}\)  
\(x\) \(=1.3 \times 15 +58\)   
  \(=77.5\)  

 
\(\therefore\ \text{Approx minimum score = 78%}\)

♦♦ Mean mark (c) 30%.

Algebra, STD2 A4 2024 HSC 14 MC

Year 11 is making pizzas to raise money for a charity.

The cost \((C)\) and revenue \((R)\) in dollars, when \(x\) pizzas are sold are represented by the equations

\begin{aligned}
& C=2.5 x+6 \\
& R=8 x .
\end{aligned}

Enough pizzas are sold so that a profit is made.

By how much does the profit increase for each additional pizza sold?

  1. $2.50
  2. $5.50
  3. $6.00
  4. $8.00
Show Answers Only

\(B\)

Show Worked Solution

\(\text {Cost increases by } \$ 2.50 \text { for each extra pizza.}\)

\(\text {Revenue increases by } \$ 8 \text { for each extra pizza.}\)

\(\text {Profit (additional) }=8-2.5=\$ 5.50 \text { per pizza}\)

\(\Rightarrow B\)

♦ Mean mark 49%.

Financial Maths, STD2 F1 2024 HSC 13 MC

José takes out a loan of $9000. Simple interest is charged on the loan.

The loan and the interest charged will be repaid by making monthly repayments of $300 over 4 years.

What simple interest rate per annum, to the nearest percent, is charged on the loan?

  1. 15%
  2. 38%
  3. 40%
  4. 60% 
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Number of repayments}=4 \times 12=48\)

\(\text{Total repayments }=48 \times 300=\$ 14\,400\)

\(\text{Interest paid}=14\,400-9000=\$ 5400\)

  \(I\) \(=Prn\)
  \(5400\) \(=9000 \times r \times 4\)
  \(r\) \(=\dfrac{5400}{9000 \times4}=0.15\)

 
\(\Rightarrow A\)

♦ Mean mark 39%.

Statistics, STD2 S1 2024 HSC 28

Flowers were planted in two gardens (Garden A and Garden B).

On a particular day, 25 flowers were randomly selected from each garden and their heights measured in millimetres.

The data are represented in parallel box-plots.
 

Compare the two datasets by examining the skewness of the distributions, and the measures of central tendency and spread.   (3 marks)

Show Answers Only

\(\text{Skewness comparison:}\)

\(\rightarrow\ \text{Garden A is negatively skewed while Garden B is positively skewed.}\)
  

\(\text{Measures of central tendency comparison:}\)

\(\rightarrow\ \text{IQR (A)}\ \approx 17\ \text{ is greater than IQR (B)}\ \approx 9\)

\(\rightarrow\ \text{This means the middle 50% of data points of B are more closely}\)

\(\text{clustered around the median than A.}\)
 

\(\text{Spread comparison:}\)

\(\rightarrow\ \text{Range (A)}\ \approx 39\ \text{is greater than range (B)}\ \approx 28\)

\(\rightarrow\ \text{This means the spread of data points of A is wider than the spread}\)

\(\text{of data points of B.}\)
 

\(\text{Overall, flowers from Garden A will tend to be higher than Garden B}\)

\(\text{flowers as well as exhibiting a larger range of heights.}\)

Show Worked Solution

\(\text{Skewness comparison:}\)

\(\rightarrow\ \text{Garden A is negatively skewed while Garden B is positively skewed.}\)
  

\(\text{Measures of central tendency comparison:}\)

\(\rightarrow\ \text{IQR (A)}\ \approx 17\ \text{ is greater than IQR (B)}\ \approx 9\)

\(\rightarrow\ \text{This means the middle 50% of data points of B are more closely}\)

\(\text{clustered around the median than A.}\)
 

\(\text{Spread comparison:}\)

\(\rightarrow\ \text{Range (A)}\ \approx 39\ \text{is greater than range (B)}\ \approx 28\)

\(\rightarrow\ \text{This means the spread of data points of A is wider than the spread}\)

\(\text{of data points of B.}\)
 

\(\text{Overall, flowers from Garden A will tend to be higher than Garden B}\)

\(\text{flowers as well as exhibiting a larger range of heights.}\)

♦ Mean mark 41%.

Algebra, STD2 A4 2024 HSC 26

A sheet of metal is folded to make a gutter, as shown. The cross-section of the gutter is a rectangle of width \(w\) cm and height \(h\) cm.
 

 

The area, \(A\) cm\(^{2}\), of the cross-section can be modelled by the quadratic formula

\(A=-0.5w^{2}+20w\)

A graph of this model is shown.
 

Find the width and height of the rectangle which will give the greatest possible area of the cross-section.   (3 marks)

Show Answers Only

\(w=20\ \text{cm}, h=10\ \text{cm}\)

Show Worked Solution

\(\text{Graph cuts}\ w\text{-axis at}\ \ w=0\ \ \text{and}\ \ w=40.\)

\(\text{By symmetry, maximum area occurs at}\ \ w=20.\)

\(A_{\text{max}}=-0.5 \times 20^2 + 20 \times 20 = 200\ \text{cm}^{2}\)

\(A_{\text{max}}\) \(=w \times h\)  
\(200\) \(=20 \times h\)  
\(h\) \(=10\ \text{cm}\)  

 
\(\therefore A_{\text{max}}\ \text{occurs when:}\ w=20\ \text{cm}, \ h=10\ \text{cm}\)

♦ Mean mark 43%.

Algebra, STD2 A1 2024 HSC 10 MC

Consider the formula  \(s=w t+\dfrac{p}{2}\).

Which of the following correctly shows \(p\) as the subject of the formula?

  1. \(p=2 w t-s\)
  2. \(p=2 w t-2 s\)
  3. \(p=2 s-w t\)
  4. \(p=2 s-2 w t\)
Show Answers Only

\(D\)

Show Worked Solution
  \(s\) \(=w t+\dfrac{p}{2}\)
  \(\dfrac{p}{2}\) \(=s-w t\)
  \(p\) \(=2(s-w t)\)
    \(=2s-2w t\)

 
\(\Rightarrow D\)

♦♦ Mean mark 36%.

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.

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

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

  1. where \(c\) and \(k\) are as found in part (a).
  2. 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)
Show Answers Only

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.   \(\text{Strategy 1}\)

\(\text{Billie’s carriage is 6 seconds behind Anna’s.}\)

\(\text{When}\ t=0,\ \text{Anna’s carriage is at the lowest point}\ = 21-18=3\)

\(\text{When}\ t=3,\ \text{by symmetry, both carriages are at the same height:}\)

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

\(\text{When}\ t=24,\ \text{Anna’s carriage is at the highest point}\ = 21+18=39\)

\(\text{When}\ t=27,\ \text{by symmetry, both carriages are at the same height:}\)

   \(h_2=21-18\cos(\dfrac{9\pi}{8}) = 37.629… = 37.63\ \text{m (2 d.p.)}\)
 

\(\text{Strategy 2}\)

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

\(\text{By inspection:}\)
 

\(\text{Heights are the same:}\)

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

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

♦♦ Mean mark (c) 38%.

Calculus, 2ADV C3 2024 HSC 17

In a particular electrical circuit, the voltage \(V\) (volts) across a capacitor is given by

\(V(t)=6.5\left(1-e^{-k t}\right)\),

where \(k\) is a positive constant and \(t\) is the number of seconds after the circuit is switched on.

  1. Draw a sketch of the graph of \(V(t)\), showing its behaviour as \(t\) increases.   (2 marks)
     

  1. When \(t=1\), the voltage across the capacitor is 2.6 volts.
  2. Find the value of \(k\), correct to 3 decimal places.   (2 marks)
  1. Find the rate at which the voltage is increasing when \(t=2\), correct to 3 decimal places.   (2 marks)

Show Answers Only

a.   
       

b.   \(k=0.511\)

c.  \(1.195\ \text{V/s}\)

Show Worked Solution

a.   
       
♦ Mean mark (a) 49%.

b.   \(V=6.5(1-e^{-kt})\)

\(\text{When}\ \ t=1, V=2.6:\)

\(2.6\) \(=6.5(1-e^{-k})\)  
\(1-e^{-k}\) \(=0.4\)  
\(e^{-k}\) \(=0.6\)  
\(-k\) \(=\ln(0.6)\)  
\(k\) \(=0.5108…\)  
  \(=0.511\ \text{(3 d.p.)}\)  

 
c.   \(V=6.5-6.5e^{-kt}\)

\(\dfrac{dV}{dt}=6.5ke^{-kt}\)

\(\text{Find}\ \dfrac{dV}{dt}\ \text{when}\ \ t=2:\)

\(\dfrac{dV}{dt}\) \(=6.5 \times 0.511 \times e^{-2 \times 0.511}\)  
  \(=1.1953…\)  
  \(=1.195\ \text{V/s (3 d.p.)}\)  

Probability, 2ADV S1 2024 HSC 18

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

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

  2. 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)

Show Answers Only

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

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

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

\(\text{Find}\ n\ \text{such that:}\)

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

 
\(\therefore\ \text{Least}\ n=10\)

♦ Mean mark (b) 39%.

CHEMISTRY, M2 EQ-Bank 10

A sample of nitrogen gas at a pressure of 400 kPa and a temperature of 60.0°C occupies a volume of 35.0 L.

  1. Calculate the volume of this nitrogen gas at 298 K and 120 kPa.   (3 marks)
  1. Calculate the number of moles of nitrogen in this sample.   (2 marks)
  1. Calculate the mass of this nitrogen sample.   (2 marks)
Show Answers Only

a.    \(104.4\ \text{L}\)

b.    \(5.06\ \text{mol}\)

c.    \(141.8\ \text{g}\)

Show Worked Solution

a.    Using the Combined Gas Law: \(\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\)

\(T_1 = 60^{\circ}\text{C}=333\ \text{K}\)

\(V_2=\dfrac{P_1V_1T_2}{T_1P_2}=\dfrac{400 \times 35 \times 298}{333 \times 120}=104.4\ \text{L}\)

 

b.    Using the Ideal Gas Law: \(PV=nRT\) 

\(n=\dfrac{PV}{RT}=\dfrac{400 \times 35}{8.314 \times 333}=5.06\ \text{mol}\)

 

c.    Nitrogen gas \(\ce{(N2)}\) has a molar mass of \(28.02\ \text{g mol}^{-1}\)

\(\ce{m(N2)}= n \times MM = 5.06 \times 28.02 = 141.8\ \text{g}\)

CHEMISTRY, M2 EQ-Bank 9

Aluminium carbonate reacts with nitric acid to produce aluminium nitrate, carbon dioxide, and water.

What volume of carbon dioxide will be produced if 15.0 g of aluminium carbonate is reacted and the gas is collected at 25°C and 100 kPa?   (5 marks)

Show Answers Only

\(4.76\ \text{L}\) 

Show Worked Solution
  • The chemical equation for the reaction:
  •    \(\ce{Al2(CO3)3 + 6HNO3 -> 2Al(NO3)3 + 3CO2 + 3H2O}\)
  • \(\ce{n(Al2(CO3)3)}=\dfrac{m}{MM}=\dfrac{15.0}{2(26.98+3(12.01)+9(16)}=0.0641\ \text{mol}\)
  • The ratio of \(\ce{Al2(CO3)3:CO2 = 1:3}\)
  • \(\ce{n(CO2)}= 3 \times 0.0641 = 0.192\ \text{mol}\)
  • Using Avogadro’s Law:
  •    \(V=n \times V_{\text{molar}} = 0.192 \times 24.79 = 4.76\ \text{L}\) 

Financial Maths, 2ADV M1 2024 HSC 30

Suppose the geometric series  \(x+x^2+x^3+\ \cdots\) has a limiting sum.

By considering the graph  \(y=-1-\dfrac{1}{x-1}\), or otherwise, find the range of possible values of \(S\).   (3 marks)

Show Answers Only

\(S \in \Big( -\dfrac{1}{2}, \infty\Big ) \)

Show Worked Solution

\(x + x^2+x^3\ \cdots \Rightarrow a=x, r=x\)

\(S=\dfrac{a}{1-r} = \dfrac{x}{1-x}=\dfrac{-(1-x)+1}{1-x} = -1+\dfrac{1}{1-x}=-1-\dfrac{1}{x-1}\)

♦♦ Mean mark 31%.

\(\text{If limiting sum}\ \Rightarrow \abs{r} \lt 1\ \Rightarrow \ \abs{x} \lt 1\)

\(\Rightarrow \text{Possible values of}\ S =\ \text{range of}\ \ y=-1-\dfrac{1}{x-1}\ \text{in domain}\ \abs{x} \lt 1\)

\(\text{Consider}\ \ y=-1-\dfrac{1}{x-1}:\)

\(\text{Asymptote at}\ \ x=1.\)

\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & 0 & 1 & 2 \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & -\dfrac{2}{3} &  -\dfrac{1}{2} & \ \ 0\ \  & \infty & -2 \\
\hline
\end{array}

\(\text{As}\ x \rightarrow \infty, \ y \rightarrow -1\)

\(\text{As}\ x \rightarrow -\infty, \ y \rightarrow -1\)

\(\abs{x} \lt 1\ \Rightarrow \  y \in \Big( -\dfrac{1}{2}, \infty\Big ) \)

\(\therefore\ \text{Possible values of}\ S \in \Big( -\dfrac{1}{2}, \infty\Big ) \)

Calculus, 2ADV C3 2024 HSC 29

Consider the curve  \(y=ax^2+bx+c\), where  \(a \neq 0\).

At a particular point, the tangent and normal to the curve are given by  \(t(x)=2x+3\)  and  \(n(x)=-\dfrac{1}{2}x-2\)  respectively.

The curve has a minimum turning point at  \(x=-4\).

Find the values of \(a, b\) and \(c\).   (4 marks)

Show Answers Only

\(a=\dfrac{1}{2}, b=4, c=5\)

Show Worked Solution
\(y\) \(=ax^{2}+bx+c\)  
\(y^{′}\) \(=2ax+b\)  
♦♦ Mean mark 35%.

\(\text{SP’s when}\ \ y^{′}=0\ \text{and}\ \ x=-4:\)

\(-8a+b=0\ …\ (1)\)

\(\text{Find point where}\ t(x), n(x)\ \text{and curve intersect:}\)

\(2x+3\) \(=-\dfrac{1}{2}x-2\)  
\(4x+6\) \(=-x-4\)  
\(5x\) \(=-10\)  
\(x\) \(=-2\)  

 
\(\Rightarrow\ \text{Intersection at}\ (-2,-1)\)
 

\(\text{At}\ \ x=-2,\ m_{\text{tang}} = -4a+b\)

\(-4a+b=2\ …\ (2) \)

   \(\text{Subtract}\ (2)-(1):\)

\(4a=2\ \ \Rightarrow\ \ a=\dfrac{1}{2}\)

   \(\text{Substitute}\ \ a=\dfrac{1}{2}\ \ \text{into (1):}\)

\(\Rightarrow\ \ b=4\)

\(\text{Since}\ (-2,-1)\ \text{lies on}\ y: \)

\(-1=\dfrac{1}{2}(-2)^{2}+4(-2)+c\)

\(\Rightarrow c=5\)

\(\therefore a=\dfrac{1}{2}, b=4, c=5\)

Calculus, 2ADV C4 2024 HSC 27

  1. Find the derivative of  \(x^{2}\tan\,x\)   (2 marks)
  2. Hence, find \(\displaystyle \int (x\,\tan\,x+1)^2\ dx\)   (3 marks)
Show Answers Only

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

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

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

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

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

♦♦ Mean mark (b) 34%.

CHEMISTRY, M2 EQ-Bank 5 MC

Which of the following best describes the properties of an ideal gas?

  1. The molecules do not attract each other and occupy a negligible volume.
  2. The molecules experience strong repulsive forces and expand to fill the container.
  3. The gas particles have significant mass and their collisions reduce total energy.
  4. The pressure of the gas depends on the chemical identity of the particles.
Show Answers Only

\(A\)

Show Worked Solution
  • An ideal gas is defined by the assumption that its molecules have no intermolecular forces and occupy negligible volume relative to the container.
  • The gas particles are in constant random motion, and collisions between them are perfectly elastic, meaning no kinetic energy is lost.

\(\Rightarrow A\)

CHEMISTRY, M2 EQ-Bank 1-2 MC

Use the equation below to answer the following two questions: 

\(\ce{2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l)}\)
 

Part 1

What volume of carbon dioxide would be produced by the combustion of 3.0 L of ethane gas \(\ce{C2H6(g)}\) in excess oxygen? (Assume the temperature and pressure remain the same)

  1. 6.0 L
  2. 3.0 L
  3. 1.5 L
  4. 0.5 L


Part 2

Whose law is applied when determining the volume relationships in the reaction above?

  1. Boyle's Law
  2. Gay-Lussac's Law
  3. Avogadro's Law
  4. Charles' Law
Show Answers Only

Part 1: \(A\)

Part 2: \(C\)

Show Worked Solution

Part 1

  • From the chemical equation, the mole ratio of ethane to carbon dioxide is \(1:2\).
  • According to Avogadro’s Law, at the same temperature and pressure, the volume of gases is directly proportional to the number of moles.
  • Thus, from the equation, 1 volume of ethane produces 2 volumes of carbon dioxide.
  • The volume of carbon dioxide produced \(=2 \times 3.0 = 6.0\ \text{L}\)

\(\Rightarrow A\)

 

Part 2

  • When determining the volume relationships between gases in a chemical reaction, Avogadro’s Law is used.
  • Avogadro’s Law states that equal volumes of gases, at the same temperature and pressure, contain the same number of moles of gas.
  • The volumes of gas react in simple whole-number ratios, as shown in the balanced equation.

\(\Rightarrow C\)

Financial Maths, STD2 F1 2024 HSC 8 MC

A bill for servicing a car is made up of:

    • $242 for parts, which includes 10% GST
    • $100 for labour, excluding GST.

The mechanic needs to add 10% GST onto the labour charge.

How much GST does the customer pay in total?

  1. $22.00
  2. $24.20
  3. $32.00
  4. $34.20
Show Answers Only

\(C\)

Show Worked Solution

\(\text {Let}\ \ X=\text{ parts cost ex-GST}\)

  \(X+X \times 0.1\) \(=242\)
  \(1.1X\) \(=242\)
  \(X\) \(=220\)

 
\(\text{Total GST}=0.1(220+100)=\$ 32.00\)

\(\Rightarrow C\)

♦ Mean mark 43%.

Measurement, STD2 M7 2024 HSC 38

A cake is constructed using a cylinder and a cone-shaped top. The cylinder has diameter 30 cm and height 6 cm.
 

The ratio of the volume of the cylinder to the volume of the cone-shaped top is \(5:1\).

The cake is to be cut into equal slices with volume 212 cm\(^{3}\).

How many equal slices can be cut from the cake?   (3 marks)

Show Answers Only

\(24\ \text{slices}\)

Show Worked Solution
\(V_{\text{cylinder}}\) \(=\pi r^{2}h\)  
  \(=\pi \times 15^{2} \times 6\)  
  \(=4241.15\ \text{cm}^{3}\)  

 
\(V_{\text{cylinder}} : V_{\text{cone}} = 5:1\)

\(\Rightarrow V_{\text{cone}} = \dfrac{1}{5} \times 4241.15=848.23\ \text{cm}^{3}\)

\(\text{Volume of cake}\ = 4241.15+848.23=5089.4\ \text{cm}^{3}\)

\(\therefore\ \text{Number of slices}\ \) \(=\dfrac{5089.4}{212}\)  
  \(=24\ \text{slices}\)  
♦♦ Mean mark 36%.

Measurement, STD2 M2 2024 HSC 37

Sakura will travel from Sydney (UTC +10) to Rio de Janeiro (UTC –3).

The flight from Sydney to Rio de Janeiro will take 20 hours.

The flight will arrive in Rio de Janeiro at 3 pm on Wednesday 20 July.

On what day and at what time will Sakura leave Sydney?   (2 marks)

Show Answers Only

\(\text{8 am (Wed)}\)

Show Worked Solution

\(\text{Arrive in Rio: 3 pm (Wed) }\)

\(\Rightarrow\ \text{Sydney is 13 hours ahead of Rio}\)

\(\Rightarrow\ \text{Arrival time (Sydney) = 3 pm plus 13 hours = 4 am (Thu)}\)

\(\text{Flight took 20 hours.}\)

\(\text{Departure time (Sydney) = 4 am (Thu) less 20 hours = 8 am (Wed)}\)

♦ Mean mark 50%.

Measurement, STD2 M1 2024 HSC 34

A container for soccer balls is made using two half spheres joined to each end of a cylindrical body.
 

Three soccer balls fit exactly inside the container. Each ball has a diameter of 23 cm.

The hemispherical ends of the container just touch the surface of the soccer balls.

What is the total surface area of the container? Give your answer in square metres correct to one decimal place.   (4 marks)

Show Answers Only

\(\text{0.5 m}^{2}\)

Show Worked Solution

\(\text{Ball diameter = 23 cm}\ \Rightarrow \ r=11.5\ \text{cm}\)

\(\text{S.A. (2 half spheres)}\) \(=2 \times \dfrac{1}{2} \times 4 \pi \times 11.5^2\)  
  \(=1662\ \text{cm}^{2}\)  
♦ Mean mark 41%.

\(\text{Cylinder width}\ =\ \text{ball circumference}= 2 \times \pi \times 11.5\)

\(\text{Cylinder height}\ = 2 \times 23 = 46\ \text{cm}\)

\(\text{S.A. (cylinder)}\ = 2\pi rh = 2 \times \pi \times 11.5 \times 46 = 3324\ \text{cm}^{2}\)
 

\(\text{Note:}\ 1\ \text{m}^{2} = 100\ \text{cm} \times 100\ \text{cm}\ = 10\,000\ \text{cm}^{2}\)

\(\text{S.A. (container)}\) \(=1662 + 3324\)  
  \(=4986\ \text{cm}^{2}\)  
  \(=\dfrac{4986}{10\,000}\ \text{m}^{2}\)  
  \(=0.5\ \text{m}^{2}\ \text{(1 d.p.)}\)  

Networks, STD2 N2 2024 HSC 4 MC

The map shows regions within a country.
 

A network diagram is to be drawn to represent this map. Vertices will be used to indicate each region and edges will be used to represent a border shared between two regions.

How many edges will there be in the network diagram?

  1. \(8\)
  2. \(7\)
  3. \(6\)
  4. \(5\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Network diagram:}\)
 

\(\text{Network has 7 edges.}\)

\(\Rightarrow B\)

♦ Mean mark 49%.

Statistics, STD2 S1 2024 HSC 3 MC

Shoppers are invited to complete an online survey.

What type of sampling is this?

  1. Self-selecting
  2. Simple random
  3. Stratified
  4. Systematic
Show Answers Only

\(A\)

Show Worked Solution

\(\text{By elimination}\)

\(\text{Option A: Individuals voluntarily choose to participate.}\ \checkmark\)

\(\text{Option B: Simple random sampling requires every individual in the population}\)

\(\text{to have an equal chance of being selected.}\ \cross \)

\(\text{Options C and D: Both options refer to a systematic process for those included}\)

\(\text{in the survey.}\ \cross \)

\(\Rightarrow A\)

♦ Mean mark 42%.

Measurement, STD2 M6 2024 HSC 32

A regular pentagon \(ABCDE\) is drawn inside a circle with a radius 30 cm.

\(O\) is the centre of the circle.

What is the area of the shaded region of the circle. Answer correct to 2 significant figures.   (4 marks)

Show Answers Only

\(\text{690 cm}^{2}\)

Show Worked Solution

\(\text{Method 1}\)

\(\text{Area}\ \Delta ODC\) \(= \dfrac{1}{2} ab \sin C\)  
  \(=\dfrac{1}{2} \times 30 \times 30 \times  \sin 72^{\circ} \)  
  \(=427.98\ \text{cm}^{2}\)  

 

\(\text{Shaded Area}\) \(=\ \text{Area of circle}-5 \times\ \text{Area}\ \Delta ODC\)  
  \(=(\pi \times 30^2)-(5 \times 427.98)\)  
  \(=687.5\)  
  \(=690\ \text{cm}^{2}\ \text{(2 sig.fig)}\)  

  

♦ Mean mark 48%.
\(\text{Method 2}\)

\(\text{Consider}\ \Delta ODC:\)

\(\text{Area}=427.98\ \text{cm}^{2}\ \ \text{(see above)}\)

 \(\text{Area of sector}\ ODC = \dfrac{72}{360} \times \pi \times 30^{2} = 565.49\ \text{cm}^{2}\)

\(\text{Shaded area (total)}\) \(=(565.49-427.98) \times 5\)  
  \(=690\ \text{cm}^{2}\ \text{(2 sig.fig)}\)  

Probability, STD2 S2 2024 HSC 31

A coin is biased so that it is twice as likely to show a head than a tail.

  1. What is the probability of obtaining a head with one throw of this coin?   (1 mark)
  2. In two throws of this coin, what is the probability of obtaining at least one head?   (2 marks)
Show Answers Only

a.   \(P(H) = \dfrac{2}{3}\)

b.   \(P\text{(at least 1 head)}\ = \dfrac{8}{9}\)

Show Worked Solution

a.   \(P(H) = \dfrac{2}{3}\)

b.   \(P(T) = 1-P(H)=1-\dfrac{2}{3}=\dfrac{1}{3}\)

  \(P\text{(at least 1 head)}\) \(=1-P\text{(no heads)}\)
    \(=1-P(TT)\)
    \(=1-\dfrac{1}{3} \times \dfrac{1}{3}\)
    \(=\dfrac{8}{9}\)
♦ Mean mark (a) 51%.
♦♦ Mean mark (b) 30%.

Functions, 2ADV F2 2024 HSC 4 MC

The parabola  \(y=(x-3)^2-2\)  is reflected about the \(y\)-axis. This is then reflected about the \(x\)-axis.

What is the equation of the resulting parabola?

  1. \(y=(x+3)^2+2\)
  2. \(y=(x-3)^2+2\)
  3. \(y=-(x+3)^2+2\)
  4. \(y=-(x-3)^2+2\)
Show Answers Only

\( C \)

Show Worked Solution

\(y=(x-3)^2-2\)

\(\text{Reflect in the}\ y\text{-axis}\ (f(-x)):\)

\(y=(-x-3)^2-2\)

\(\text{Reflect in the}\ x\text{-axis}\ (-f(-x)):\)

\(y\) \(=-\left[(-x-3)^2-2\right]\)  
  \(=-(x+3)^2+2\)  

 
\( \Rightarrow C \)

♦ Mean mark 54%.

Algebra, STD2 A1 2024 HSC 1 MC

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

  1. \(-6.41\)
  2. \(-6.40\)
  3. \(6.40\)
  4. \(6.41\)
Show Answers Only

\(D\)

Show Worked Solution
  \( x^2\) \(=(-2.531)^2 \)
    \(=6.405\ldots \)
    \(=6.41\)

 
\(\Rightarrow D\)

♦♦ Mean mark 29%!
COMMENT: Brain dead calculator usage will cause issues here.

CHEMISTRY, M3 EQ-Bank 10

A student conducted an experiment to measure the voltage generated by using various combinations of metals in an electrolyte solution.
 

   

  1. Complete the table below.   (2 marks)

\begin{array} {|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Experiment} \rule[-1ex]{0pt}{0pt} & \textbf{Half cell A} & \textbf{Half cell B} & \textbf{Cell Potential (V)}\\
\hline
\rule{0pt}{2.5ex} \textbf{1} \rule[-1ex]{0pt}{0pt} & \ce{Zn(s) | Zn^{2+}(aq)} & \ce{Pb(s) | Pb^{2+}(aq)} &  \\
\hline
\rule{0pt}{2.5ex} \textbf{2} \rule[-1ex]{0pt}{0pt} & \ce{Cu(s) | Cu^{2+}(aq)} & \ce{Pb(s) | Pb^{2+}(aq)} &  \\
\hline
\end{array}

  1. Explain whether the Zinc electrode is the anode or the cathode.   (2 marks)
  1. Write the net ionic equation for Experiment 2.   (2 marks)
Show Answers Only

a.  Cell potential of Experiment 1:

\(\ce{Zn(s) -> Zn^{2+} + 2e^-} \quad V=0.76\ \text{V}\)

\(\ce{Pb^{2+} + 2e^- -> Pb(s)} \quad V=-0.13\ \text{V}\)

\(E^{\circ}_{\text{cell}}=0.76 + -0.13 =0.63\ \text{V}\)
 

Cell potential of Experiment 2:

\(\ce{Pb(s) -> Pb^{2+} + 2e^-} \quad V=0.13\ \text{V}\)

\(\ce{Cu^{2+} + 2e^- -> Cu(s)} \quad V=0.34\ \text{V}\)

\(E^{\circ}_{\text{cell}}=0.13 + 0.34 =0.47\ \text{V}\)
 

b.    Zinc electrode is the anode.

  • Zinc is a more reactive metal than lead and therefore will undergo oxidation while lead undergoes reduction.
  • Since oxidation occurs at the anode, the zinc electrode will be the anode in this cell.

 
c.    \(\ce{Pb(s) + Cu^{2+}(aq) -> Pb^{2+}(aq) + Cu(s)}\)

Show Worked Solution

a.  Cell potential of experiment 1:

\(\ce{Zn(s) -> Zn^{2+} + 2e^-} \quad V=0.76\ \text{V}\)

\(\ce{Pb^{2+} + 2e^- -> Pb(s)} \quad V=-0.13\ \text{V}\)

\(E^{\circ}_{\text{cell}}=0.76 + -0.13 =0.63\ \text{V}\)
 

Cell potential of experiment 2:

\(\ce{Pb(s) -> Pb^{2+} + 2e^-} \quad V=0.13\ \text{V}\)

\(\ce{Cu^{2+} + 2e^- -> Cu(s)} \quad V=0.34\ \text{V}\)

\(E^{\circ}_{\text{cell}}=0.13 + 0.34 =0.47\ \text{V}\)
 

b.    Zinc electrode is the anode.

  • Zinc is a more reactive metal than lead and therefore will undergo oxidation while lead undergoes reduction.
  • Since oxidation occurs at the anode, the zinc electrode will be the anode in this cell.

 
c.    \(\ce{Pb(s) + Cu^{2+}(aq) -> Pb^{2+}(aq) + Cu(s)}\)

CHEMISTRY, M3 EQ-Bank 9

A galvanic cell was created as seen below:
 

   

  1. State the direction that the electrons run through the conducting wire.   (1 mark)
  1. State which electrode is the anode and which electrode is the cathode.   (1 mark)
  1. Explain the direction of the anion flow from the salt bridge.   (2 marks)
Show Answers Only

a.    The electrons will run from the nickel electrode to the copper electrode.

  • The electrons run from the more active metal (oxidises) to the less active metal (reduces).

b.    Nickel electrode  \(\Rightarrow\)  is the anode (undergoes oxidation).

Copper electrode  \(\Rightarrow\)  cathode (undergoes reduction).
 

c.    The nitrate ions in the salt bridge will flow to the nickel solution. 

  • At the anode, oxidation occurs, producing positive ions (cations) that increase the positive charge.
  • To balance this, anions move to the nickel solution, neutralising the excess positive charge and maintaining electrical neutrality in the solution.
Show Worked Solution

a.    The electrons will run from the nickel electrode to the copper electrode.

  • The electrons run from the more active metal (oxidises) to the less active metal (reduces).

b.    Nickel electrode  \(\Rightarrow\)  is the anode (undergoes oxidation).

Copper electrode  \(\Rightarrow\)  cathode (undergoes reduction).
 

c.    The nitrate ions in the salt bridge will flow to the nickel solution. 

  • At the anode, oxidation occurs, producing positive ions (cations) that increase the positive charge.
  • To balance this, anions move to the nickel solution, neutralising the excess positive charge and maintaining electrical neutrality in the solution.

CHEMISTRY, M3 EQ-Bank 6 MC

Which of the following species decreases in oxidation number?

\(\ce{Fe^{2+}(aq) + MnO4^-(aq) + 8H^+(aq) -> 5Fe^{3+} + Mn^{2+}(aq) + 4H2O(l)}\)

  1. \(\ce{Fe^{2+}(aq)}\)
  3. \(\ce{H2O(l)}\)
  4. \(\ce{H^+(aq)}\)
Show Answers Only

\(B\)

Show Worked Solution
  • The oxidation number of  \(\ce{Fe^{2+}(aq)}\)  increases from +2 to +3.
  • The oxidation number of \(\ce{Mn}\) in \(\ce{MnO4^-}\) 
  •    \(\ce{Mn} +4(-2)=-1\ \ \Rightarrow\ \ \ce{Mn}=7\)
  • \(\ce{Mn}\) decreases from an oxidation number of 7 to 2 during the reaction.

\(\Rightarrow B\)

CHEMISTRY, M3 EQ-Bank 4 MC

Which set of oxidation and reduction reactions would result in a spontaneous process?

\begin{align*}
\begin{array}{l}
\ & \\
\textbf{A.}\\
\textbf{B.}\\
\textbf{C.}\\
\textbf{D.}\\
\end{array}
\begin{array}{|c|c|}
\hline
\textit{Oxidation} & \textit{Reduction} \\
\hline
\ce{Mg(s) -> Mg^{2+}(aq) + 2e^-} & \ce{Cu^{2+}(aq) +2e^- -> Cu(s)} \\
\hline
\ce{Ag(s) -> Ag^+(aq) + e^-} & \ce{Fe^{3+}(aq) +e^- -> fe^{2+}} \\
\hline
\ce{Cu(s) -> Cu^{2+}(aq) + 2e^-} & \ce{Zn^{2+}(aq) +2e^- -> Zn(s)} \\
\hline
\ce{Pb^{2+}(aq) +2e^- -> Pb(s)} & \ce{Cu(s) -> Cu^{2+}(aq) + 2e^-} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution
  • For a redox reaction to be spontaneous, the cell potential must be a positive number.
  • Option A:  \(E^{\circ}_{\text{cell}} = 0.34-(-2.36) = 2.70\ \text{V}\)  → Spontaneous
  • Options B and C: both have negative cell potentials and will therefore be non-spontaneous.
  • Option D: has the reduction and oxidation equations around the wrong way, going against the natural direction for a spontaneous reaction.

\(\Rightarrow A\)

BIOLOGY, M4 EQ-Bank 8

A 3000 hectare koala sanctuary was created in 1980 and the koala population over the next 35 years was monitored and the data graphed below.
 

Identify and explain the ecological significance of the parts of the graph labelled A, B and C.   (5 marks)

Show Answers Only

Curve at point A:

  • The curve at point A shows exponential population growth.
  • The koala population is rapidly increasing due to abundant resources and minimal limiting factors, allowing for maximum reproductive success. 

Curve at point B:

  • At point B the population growth begins to slow.
  • As the koala population grows, competition for resources like food and habitat increases.
  • While the population still grows, it does so with a decreasing population growth rate. 

Curve at point C:

  • Point C shows the population reaching the sanctuary’s carrying capacity at around 1300 koalas.
  • The koala population has levelled off with births and deaths in balance.
  • At this stage, the environment is supporting the maximum sustainable number of individuals.
Show Worked Solution

Curve at point A:

  • The curve at point A shows exponential population growth.
  • The koala population is rapidly increasing due to abundant resources and minimal limiting factors, allowing for maximum reproductive success. 

Curve at point B:

  • At point B the population growth begins to slow.
  • As the koala population grows, competition for resources like food and habitat increases.
  • While the population still grows, it does so with a decreasing population growth rate. 

Curve at point C:

  • Point C shows the population reaching the sanctuary’s carrying capacity at around 1300 koalas.
  • The koala population has levelled off with births and deaths in balance.
  • At this stage, the environment is supporting the maximum sustainable number of individuals.

BIOLOGY, M2 EQ-Bank 17

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

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

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.
  • 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.

CHEMISTRY, M3 EQ-Bank 8

A galvanic cell has been set up as illustrated in the diagram below.

  1. The standard potential for this reaction is 0.78 V. Use half equations to determine the unknown electrode.   (2 marks)
  1. The unknown solution is light green in colour. Explain what will happen to the colour of the unknown solution as the reaction proceeds.   (2 marks)
  1. After some time, a solid deposit formed on the copper electrode was removed and dried. The mass of the deposit was 0.150 g. Determine the final concentration of the copper nitrate solution.   (3 marks)
Show Answers Only

a.    \(\ce{Fe(s) -> Fe^{2+} + 2e^-}\)

b.    As the reaction progresses:

  • The solution will darken, taking on a more intense green colour.
  • This change occurs because nickel undergoes oxidation to form \(\ce{Fe^{2+}}\) ions, which are released into the solution, thereby increasing its colour intensity.

c.    \(0.137\ \text{mol L}^{-1}\)

Show Worked Solution
a.     \(E^{\circ}_{\text{cell}}\) \(=E^{\circ}_{\text{cathode}}-E^{\circ}_{\text{anode}}\)
  \(0.78\) \(=0.34-E^{\circ}_{\text{anode}}\)
  \(E^{\circ}_{\text{anode}}\) \(=0.34-0.78\)
  \(E^{\circ}_{\text{anode}}\) \(=-0.44\)
     
  • \(\ce{Fe^{2+} + 2e^- -> Fe(s)} \qquad -0.44\ \text{V}\)
  • \(\ce{Fe^{2+}}\) is undergoing oxidation. The correct half equation is: \(\ce{Fe(s) -> Fe^{2+} + 2e^-}\)
     

b.    As the reaction progresses:

  • The solution will darken, taking on a more intense green colour.
  • This change occurs because nickel undergoes oxidation to form \(\ce{Fe^{2+}}\) ions, which are released into the solution, thereby increasing its colour intensity.
     

c.    Moles of solid copper formed on electrode: \(\dfrac{m}{MM}=\dfrac{0.175}{63.55}=2.36 \times 10^{-3}\ \text{mol}\)

Moles of copper taken out of solution: \(2.36 \times 10^{-3}\ \text{mol}\)

Moles of copper remaining in solution: \((0.15 \times 0.18)-2.36 \times 10^{-3}= 0.0246\ \text{mol}\)

Final concentration: \(c=\dfrac{n}{V}=\dfrac{0.0246}{0.18}=0.137\ \text{mol L}^{-1}\)

BIOLOGY, M4 EQ-Bank 2 MC

Which of the following relationships is an example of parasitism?

  1. A cleaner wrasse removing parasites from a larger fish.
  2. A lichen formed by a fungus and an alga growing together on a rock.
  3. An orchid growing on the branch of a rainforest tree without harming it.
  4. A cuckoo laying its eggs in another bird's nest for the host to raise.
Show Answers Only

\(D\)

Show Worked Solution
  • Option A is an example of mutualism (both benefit).
  • Option B describes mutualism in the form of a lichen. The fungus provides structure and moisture, while the alga produces food through photosynthesis.
  • Option C is an example of commensalism. The orchid benefits from the support and access to sunlight, while the tree is unaffected.
  • Option D is an example of parasitism. The cuckoo benefits by having its offspring raised without expending energy, while the host bird is harmed by wasting resources on raising another species’ young.

\(\Rightarrow D\)

BIOLOGY, M4 EQ-Bank 1 MC

Commensalism is a form of symbiotic relationship. Which of the following best describes a commensal relationship between two organisms?

  1. Both organisms benefit from the interaction.
  2. One organism benefits while the other is unaffected.
  3. One organism benefits at the expense of the other.
  4. Neither organism is affected by the interaction.
Show Answers Only

\(B\)

Show Worked Solution
  • In a commensal relationship, one organism benefits from the interaction while the other is neither helped nor harmed.

\(\Rightarrow B\)

BIOLOGY, M4 EQ-Bank 4

Describe the differences between mutualistic and parasitic symbiotic relationships in ecosystems. In your answer, provide one specific example of each type of relationship from nature.   (3 marks)

Show Answers Only
  • Mutualistic and parasitic relationships are both types of symbiosis involving a relationship between two different species.
  • In a mutualistic relationship, both organisms benefit from the interaction.
  • Example: the relationship between clownfish and sea anemones, where the clownfish receive protection from predators among the anemone’s tentacles, while the anemone benefits from the fish’s waste products.
  • In contrast, in a parasitic relationship, one organism (the parasite) benefits at the expense of the other (the host). 
  • Example: tapeworms living in a dog’s intestine, where the tapeworm gains nutrients from the dog’s digested food, causing potential health issues for the dog without providing any benefits. 
Show Worked Solution
  • Mutualistic and parasitic relationships are both types of symbiosis involving a relationship between two different species.
  • In a mutualistic relationship, both organisms benefit from the interaction.
  • Example: the relationship between clownfish and sea anemones, where the clownfish receive protection from predators among the anemone’s tentacles, while the anemone benefits from the fish’s waste products.
  • In contrast, in a parasitic relationship, one organism (the parasite) benefits at the expense of the other (the host). 
  • Example: tapeworms living in a dog’s intestine, where the tapeworm gains nutrients from the dog’s digested food, causing potential health issues for the dog without providing any benefits. 

BIOLOGY, M4 EQ-Bank 2

  1. Is soil pH a biotic or abiotic factor?   (1 mark)
  2. Describe how soil pH can affect plant growth.   (1 mark)
  3. Outline one way in which plants might alter soil pH.   (1 mark)
Show Answers Only

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 3 MC

A scientist is using a population growth model to predict the future of a species in a changing environment. Which of the following factors would be LEAST likely to be included in this model?

  1. Birth rates and death rates of the species.
  2. Carrying capacity of the habitat.
  3. Genetic diversity within the population.
  4. Average daily temperature fluctuations.
Show Answers Only

\(D\)

Show Worked Solution
  • Population growth models typically focus on factors directly related to population dynamics such as birth rates, death rates, and carrying capacity.
  • While genetic diversity might be included in more complex models, average daily temperature fluctuations are too fine-grained for most population models and would be more relevant in climate models.

\(\Rightarrow D\)

BIOLOGY, M4 EQ-Bank 8

Indigenous land management practices are increasingly recognised for their effectiveness in ecosystem restoration.

Describe one specific case study where traditional knowledge has been applied to heal a damaged ecosystem in Australia. In your answer, explain the concept of 'Country' or 'Place' in this context and outline two specific restoration strategies used, highlighting their cultural significance.   (5 marks)

Show Answers Only

Case study 1: Pilliga forest in NSW (one example of many)

  • In this context, ‘Country’ encompasses not just the physical landscape of the forest but the spiritual and cultural connections of the Aboriginal people.
  • Strategy 1: One key restoration strategy employed is the reintroduction of cultural burning practices. This traditional fire management technique helps reduce fuel loads, promote biodiversity, and encourages the growth of culturally significant plants.
  • Strategy 2: The restoration of natural water flows and wetlands, guided by traditional knowledge of the landscape was another restoration strategy used.
  • This approach not only improves water quality and habitat for native species but also revitalises culturally important water sites.
  • These strategies are culturally significant as they represent a continuation of ancestral practices and reinforce the Aboriginal people’s role as custodians of the land. 

Case study 2: Restoration of the Gulgalda in Tasmania

  • Gulgalda is a critically endangered plant species sacred to the Tasmanian Aboriginal people.
  • In this context, ‘Country’ refers to not just the physical landscape but the interconnected relationships between land, plants and people, including spiritual and cultural elements.
  • Strategy 1: The use of cultural burning, a controlled fire management technique that stimulates Gulgalda germination and reduces competition from other plants.
  • Strategy 2: The project incorporated traditional harvesting practices, where plant material is collected sustainably to propagate new individuals while maintaining the spiritual connection to Country.
  • These strategies hold cultural significance as they preserve ancestral traditions and reaffirm the Aboriginal people’s stewardship over the land.
Show Worked Solution

Case study 1: Pilliga forest in NSW (one example of many)

  • In this context, ‘Country’ encompasses not just the physical landscape of the forest but the spiritual and cultural connections of the Aboriginal people.
  • Strategy 1: One key restoration strategy employed is the reintroduction of cultural burning practices. This traditional fire management technique helps reduce fuel loads, promote biodiversity, and encourages the growth of culturally significant plants.
  • Strategy 2: The restoration of natural water flows and wetlands, guided by traditional knowledge of the landscape was another restoration strategy used.
  • This approach not only improves water quality and habitat for native species but also revitalises culturally important water sites.
  • These strategies are culturally significant as they represent a continuation of ancestral practices and reinforce the Aboriginal people’s role as custodians of the land. 

Case study 2: Restoration of the Gulgalda in Tasmania

  • Gulgalda is a critically endangered plant species sacred to the Tasmanian Aboriginal people.
  • In this context, ‘Country’ refers to not just the physical landscape but the interconnected relationships between land, plants and people, including spiritual and cultural elements.
  • Strategy 1: The use of cultural burning, a controlled fire management technique that stimulates Gulgalda germination and reduces competition from other plants.
  • Strategy 2: The project incorporated traditional harvesting practices, where plant material is collected sustainably to propagate new individuals while maintaining the spiritual connection to Country.
  • These strategies hold cultural significance as they preserve ancestral traditions and reaffirm the Aboriginal people’s stewardship over the land.

BIOLOGY, M4 EQ-Bank 7

Agricultural intensification has led to widespread land degradation in many parts of the world. 

Identify two major forms of land degradation resulting from agricultural practices. In your answer, describe a specific restoration technique used to address each form of degradation.   (4 marks)

Show Answers Only

Soil erosion:

  • Soil erosion is a major form of land degradation that is caused by intensive cropping and overgrazing.
  • It can be addressed through crop rotation and destocking. These strategies help to stabilise soil structure, increase organic matter, and reduce erosion. 

Soil salinisation:

  • Soil salinisation is a form of land degradation whereby excessive salts accumulate in the soil, reducing its fertility and hindering plant growth.
  • It is often caused by poor irrigation practices and can be mitigated through the use of salt-tolerant crops and improved drainage systems.
Show Worked Solution

Soil erosion:

  • Soil erosion is a major form of land degradation that is caused by intensive cropping and overgrazing.
  • It can be addressed through crop rotation and destocking. These strategies help to stabilise soil structure, increase organic matter, and reduce erosion. 

Soil salinisation:

  • Soil salinisation is a form of land degradation whereby excessive salts accumulate in the soil, reducing its fertility and hindering plant growth.
  • It is often caused by poor irrigation practices and can be mitigated through the use of salt-tolerant crops and improved drainage systems.

BIOLOGY, M4 EQ-Bank 6

The concept of 'Country' holds deep significance in Aboriginal culture.

Explain the meaning of 'Country' from an Aboriginal perspective and describe a restoration practice that aligns with Aboriginal understanding of Country and could be used in post-mining landscapes.   (3 marks)

Show Answers Only
  • In Aboriginal culture, ‘Country’ refers to a holistic concept that encompasses not just the physical landscape, but also spiritual and cultural connections.
  • It’s a living entity that Aboriginal people have a responsibility to care for and maintain.

Restoration process (could include one of the following):

  • A restoration practice that aligns with this understanding is the use of cultural burning, a traditional land management technique. This controlled, low-intensity burning, guided by Indigenous knowledge, can be applied in post-mining landscapes to promote traditional biodiversity and reduce wildfire risk. 
  • The promotion and funding of Aboriginal custodians to teach the skills of bush regeneration to younger generations.
Show Worked Solution
  • In Aboriginal culture, ‘Country’ refers to a holistic concept that encompasses not just the physical landscape, but also spiritual and cultural connections.
  • It’s a living entity that Aboriginal people have a responsibility to care for and maintain.

Restoration process (could include one of the following):

  • A restoration practice that aligns with this understanding is the use of cultural burning, a traditional land management technique. This controlled, low-intensity burning, guided by Indigenous knowledge, can be applied in post-mining landscapes to promote traditional biodiversity and reduce wildfire risk. 
  • The promotion and funding of Aboriginal custodians to teach the skills of bush regeneration to younger generations.

BIOLOGY, M4 EQ-Bank 5

Mining activities often leave significant impacts on ecosystems. Environmental scientists and ecologists work to develop and implement restoration practices to heal these damaged landscapes.

Describe two specific restoration practices commonly used in post-mining landscapes. In your answer, identify one challenge faced in the restoration process and how it might be overcome.   (4 marks)

Show Answers Only

Restoration practices (choose 2)

  • Practice 1: Topsoil replacement involves carefully storing the original topsoil during mining operations and then reapplying it during restoration, which helps to preserve the soil’s seed bank and beneficial microorganisms.
  • Practice 2: Native species revegetation focuses on planting local plant species to re-establish the area’s natural ecosystem.
  • Practice 2: Hydrological restoration aims to re-establish natural water flows and drainage patterns disrupted by mining activities. This can involve reshaping the landscape to mimic natural contours and/or creating wetlands. 

Restoration challenge/response:

  • A significant challenge in the restoration process is soil compaction, which can occur due to heavy machinery used during mining and restoration activities.
  • This challenge can be addressed through techniques such as deep ripping, where the soil is mechanically loosened to improve its structure, or by using native plant species with deep root systems that can penetrate and gradually improve compacted soils over time.
Show Worked Solution

Restoration practices (choose 2)

  • Practice 1: Topsoil replacement involves carefully storing the original topsoil during mining operations and then reapplying it during restoration, which helps to preserve the soil’s seed bank and beneficial microorganisms.
  • Practice 2: Native species revegetation focuses on planting local plant species to re-establish the area’s natural ecosystem.
  • Practice 2: Hydrological restoration aims to re-establish natural water flows and drainage patterns disrupted by mining activities. This can involve reshaping the landscape to mimic natural contours and/or creating wetlands. 

Restoration challenge/response:

  • A significant challenge in the restoration process is soil compaction, which can occur due to heavy machinery used during mining and restoration activities.
  • This challenge can be addressed through techniques such as deep ripping, where the soil is mechanically loosened to improve its structure, or by using native plant species with deep root systems that can penetrate and gradually improve compacted soils over time.

BIOLOGY, M4 EQ-Bank 4

While models provide valuable insights into potential biodiversity changes, they are not without their shortcomings.

Identify and explain two limitations of models used to predict future impacts on biodiversity. In your answer suggest a way scientists might address or mitigate these limitations.   (4 marks)

Show Answers Only

Answers could include two of the following:

Limitation: inadequate data

  • Model accuracy requires adequate data which can be expensive and difficult to obtain.
  • To address this, scientists can use developing technology to capture more data and use machine learning techniques to better interpret that data. 

Limitation: inaccurate model inputs (projections)

  • Models often require projections of human behaviour, such as greenhouse gas emissions, which can be highly unpredictable.
  • This uncertainty can significantly affect the accuracy of long-term biodiversity predictions.
  • One way to mitigate this limitation is to use scenario-based modelling, where multiple possible futures are simulated based on different human activity projections. 

Limitation: complexity of the system being modelled

  • Ecological interactions are complex systems with unpredictable feedback loops. These systems are extremely difficulty to model.
  • Scientists need to be aware of and incorporate other research into their model. For example, (unexpected) massive methane releases as permafrost melts in the Arctic should be included in any climate model as the science becomes known.
Show Worked Solution

Answers could include two of the following:

Limitation: inadequate data

  • Model accuracy requires adequate data which can be expensive and difficult to obtain.
  • To address this, scientists can use developing technology to capture more data and use machine learning techniques to better interpret that data. 

Limitation: inaccurate model inputs (projections)

  • Models often require projections of human behaviour, such as greenhouse gas emissions, which can be highly unpredictable.
  • This uncertainty can significantly affect the accuracy of long-term biodiversity predictions.
  • One way to mitigate this limitation is to use scenario-based modelling, where multiple possible futures are simulated based on different human activity projections. 

Limitation: complexity of the system being modelled

  • Ecological interactions are complex systems with unpredictable feedback loops. These systems are extremely difficulty to model.
  • Scientists need to be aware of and incorporate other research into their model. For example, (unexpected) massive methane releases as permafrost melts in the Arctic should be included in any climate model as the science becomes known.

CHEMISTRY, M3 EQ-Bank 20

During a laboratory investigation, a student mixed two solutions and observed a sudden colour change, an increase in temperature, and the formation of bubbles.

  1. Explain why these observations indicate a chemical change.   (3 marks)
  1. Describe two types of chemical reactions that could cause at least two of these observations each.   (2 marks)
Show Answers Only

a.   Colour Change:

  • This suggests that new chemical compounds are forming that have different properties from the original reactants.

Temperature Increase (Exothermic Reaction):

  • The release of heat indicates that the reaction is exothermic, where energy is released as bonds are formed in the products.

Gas Production (Bubbles):

  • The formation of bubbles without boiling is a sign that a gas is being produced as a result of the reaction.
     

b.   Acid-Base Reaction:

  • When an acid reacts with a base, it can lead to a colour change if an indicator is present (e.g., phenolphthalein changes from colourless to pink). An exothermic reaction may also occur, causing a temperature increase.

Decomposition Reaction:

  • Certain decomposition reactions, such as the breakdown of hydrogen peroxide, produce gas (oxygen) and heat. The bubbling and increase in temperature can be observed in this type of reaction.

 

Show Worked Solution

a.   Colour Change:

  • This suggests that new chemical compounds are forming that have different properties from the original reactants.

Temperature Increase (Exothermic Reaction):

  • The release of heat indicates that the reaction is exothermic, where energy is released as bonds are formed in the products.

Gas Production (Bubbles):

  • The formation of bubbles without boiling is a sign that a gas is being produced as a result of the reaction.
     

b.   Acid-Base Reaction:

  • When an acid reacts with a base, it can lead to a colour change if an indicator is present (e.g., phenolphthalein changes from colourless to pink). An exothermic reaction may also occur, causing a temperature increase.

Decomposition Reaction:

  • Certain decomposition reactions, such as the breakdown of hydrogen peroxide, produce gas (oxygen) and heat. The bubbling and increase in temperature can be observed in this type of reaction.

CHEMISTRY, M3 EQ-Bank 8 MC

A student conducts an experiment by mixing an unknown metal powder with a solution of hydrochloric acid. The following observations are made: rapid bubbling, a slight rise in temperature, and a distinctive metallic odour.

Based on these observations, which of the following best identifies the indicators of a chemical change and explains what might have occurred?

  1. Formation of a precipitate, indicating the metal is insoluble in acid.
  2. Production of a gas and increase in temperature, suggesting the metal reacts to release hydrogen gas.
  3. Emission of light and production of an odor, implying the metal is highly reactive.
  4. Absorption of heat and production of a gas, indicating an endothermic reaction producing carbon dioxide.
Show Answers Only

\(B\)

Show Worked Solution
  • The rapid bubbling indicates the release of a gas, likely hydrogen, which is common when active metals react with acids. The slight rise in temperature suggests an exothermic reaction.
  • The presence of a metallic odour might be due to vaporization of trace compounds.
  • Therefore, the correct indicators are gas production and temperature change, which signify a chemical reaction.

\(\Rightarrow B\)

CHEMISTRY, M3 EQ-Bank 6 MC

Which of the following correctly identifies the gas or gases released when hydrochloric acid reacts with magnesium, potassium hydroxide, calcium carbonate, and ammonium carbonate, respectively?

\begin{align*}
\begin{array}{l}
\ & \\
\ & \\
\textbf{A.}\\
\textbf{}\\
\textbf{B.}\\
\textbf{C.}\\
\textbf{D.}\\
\end{array}
\begin{array}{|l|l|l|l|}
\hline
\textit{Magnesium} & \textit{Potassium} & \textit{Calcium} & \textit{Ammonium} \\
& \textit{hydroxide} & \textit{carbonate} & \textit{carbonate} \\
\hline
\text{No gas} & \text{Hydrogen} & \text{Carbon dioxide} & \text{Carbon dioxide} \\
& & & \text{and ammonia} \\
\hline
\text{Hydrogen} & \text{No gas} & \text{Carbon dioxide} & \text{Ammonia} \\
\hline
\text{Carbon dioxide} & \text{Hydrogen} & \text{No Gas} & \text{Carbon dioxide} \\
\hline
\text{Hydrogen} & \text{No gas} & \text{Carbon Dioxide} & \text{Carbon dioxide} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(D\)

Show Worked Solution
  • \(\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)}\)  (Hydrogen gas produced)
  • \(\ce{KOH(aq) + HCl(aq) -> KCl(aq) + H2O(l)}\)  (No gas produced)
  • \(\ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)}\)  (Carbon dioxide produced)
  • \(\ce{(NH4)2CO3(s) + 2HCl(aq) -> 2NH4Cl(aq) + CO2(g) + H2O(l)}\)  (Carbon dioxide produced)

\(\Rightarrow D\)

CHEMISTRY, M3 EQ-Bank 18

A student heats sodium metal, copper carbonate and propane gas \(\ce{(C3H8)}\) individually with a Bunsen burner. All of the substances react but only two of the substances react with the oxygen in the air.

Write a balanced chemical equation for each of the reactions that occurred.   (3 marks)

Show Answers Only

Combustion of sodium metal:  \(\ce{2Na(s) + O2(g) -> 2NaO(s)}\)

Combustion of propane gas:  \(\ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)}\)

Decomposition of copper carbonate:  \(\ce{CuCO3(s) -> CuO(s) + CO2(g)}\)

Show Worked Solution

Combustion of sodium metal:  \(\ce{2Na(s) + O2(g) -> 2NaO(s)}\)

Combustion of propane gas:  \(\ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)}\)

Decomposition of copper carbonate:  \(\ce{CuCO3(s) -> CuO(s) + CO2(g)}\)

  • The last reaction is endothermic and so requires the heat of the Bunsen burner to proceed where as the first two require the heat of the Bunsen burner to overcome their activation energies.

BIOLOGY, M4 EQ-Bank 8

Explain how the unique abiotic factors of the Australian continent have influenced the evolution of sclerophyll plants. In your answer, provide two specific adaptations.   (4 marks)

Show Answers Only
  • The evolution of sclerophyll plants in Australia is closely tied to the continent’s abiotic factors, particularly its nutrient-poor soils, arid climate, and frequent bushfires.
  • These plants have developed tough, leathery leaves with a thick cuticle to reduce water loss in the dry Australian environment, a key adaptation to water scarcity.
  • Sclerophyll species like eucalypts and banksias have also evolved strategies to cope with nutrient-poor soils, such as cluster roots that efficiently extract phosphorus.
  • Many sclerophyll plants have developed fire-resistant adaptations, including lignotubers for post-fire regeneration and fruits that release seeds after fire.
  • These adaptations demonstrate how the challenging abiotic conditions in Australia have acted as strong selective pressures, shaping the evolution of a unique flora highly specialised for survival in this harsh environment.
Show Worked Solution
  • The evolution of sclerophyll plants in Australia is closely tied to the continent’s abiotic factors, particularly its nutrient-poor soils, arid climate, and frequent bushfires.
  • These plants have developed tough, leathery leaves with a thick cuticle to reduce water loss in the dry Australian environment, a key adaptation to water scarcity.
  • Sclerophyll species like eucalypts and banksias have also evolved strategies to cope with nutrient-poor soils, such as cluster roots that efficiently extract phosphorus.
  • Many sclerophyll plants have developed fire-resistant adaptations, including lignotubers for post-fire regeneration and fruits that release seeds after fire.
  • These adaptations demonstrate how the challenging abiotic conditions in Australia have acted as strong selective pressures, shaping the evolution of a unique flora highly specialised for survival in this harsh environment.

BIOLOGY, M4 EQ-Bank 7

Analyse the evidence for the evolution of marsupials in Australia. In your answer, describe one piece of fossil evidence that supports the evolution of marsupials in Australia and provide one limitation of using fossil evidence to reconstruct evolutionary histories.   (4 marks)

Show Answers Only
  • Fossil evidence strongly supports the evolution of marsupials in Australia.
  • Riversleigh fossil deposits in Queensland provide important fossil evidence from 33 Mya. These deposits contain marsupial fossils of the ancestors of modern kangaroos, koalas, and wombats.
  • The diversity and age of these fossils demonstrate the long evolutionary history of marsupials on the Australian continent and provide insights into their adaptive radiation into various ecological niches. 
  • A limitation of using fossil evidence to reconstruct evolutionary histories is the incompleteness of the fossil record.
  • Many species may not fossilise due to their soft bodies or the absence of suitable preservation conditions, leading to gaps in our understanding of evolutionary transitions.
Show Worked Solution
  • Fossil evidence strongly supports the evolution of marsupials in Australia.
  • Riversleigh fossil deposits in Queensland provide important fossil evidence from 33 Mya. These deposits contain marsupial fossils of the ancestors of modern kangaroos, koalas, and wombats.
  • The diversity and age of these fossils demonstrate the long evolutionary history of marsupials on the Australian continent and provide insights into their adaptive radiation into various ecological niches. 
  • A limitation of using fossil evidence to reconstruct evolutionary histories is the incompleteness of the fossil record.
  • Many species may not fossilise due to their soft bodies or the absence of suitable preservation conditions, leading to gaps in our understanding of evolutionary transitions.

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.

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

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

Show Answers Only

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.

BIOLOGY, M4 EQ-Bank 2

Aboriginal rock paintings provide valuable insights into Australia's past ecosystems and the changes they've undergone over time.

Explain why Aboriginal rock paintings are considered a valid source of ecological information and how it complements other forms of paleontological data in understanding ecosystem changes.   (4 marks)

Show Answers Only
  • Aboriginal rock paintings provide direct visual records of flora and fauna from thousands of years ago.
  • These paintings were created by people who closely observed their environment.
  • A notable example is the Nawarla Gabarnmang rock art site in Arnhem Land, which features a painting of a thylacine (Tasmanian tiger) dated to be at least 17,000 years old.
  • This painting provides evidence that thylacines once inhabited mainland Australia, long after they had disappeared from the fossil record in this region.
  • In this way, Aboriginal rock art sites help to fill gaps in the paleontological record and provides insights into the distribution of this species before its mainland extinction.
  • By comparing such rock art with current ecosystems, scientists can infer changes in biodiversity, contributing to our understanding of long-term ecological changes in Australia.
Show Worked Solution
  • Aboriginal rock paintings provide direct visual records of flora and fauna from thousands of years ago.
  • These paintings were created by people who closely observed their environment.
  • A notable example is the Nawarla Gabarnmang rock art site in Arnhem Land, which features a painting of a thylacine (Tasmanian tiger) dated to be at least 17,000 years old.
  • This painting provides evidence that thylacines once inhabited mainland Australia, long after they had disappeared from the fossil record in this region.
  • In this way, Aboriginal rock art sites help to fill gaps in the paleontological record and provides insights into the distribution of this species before its mainland extinction.
  • By comparing such rock art with current ecosystems, scientists can infer changes in biodiversity, contributing to our understanding of long-term ecological changes in Australia.

BIOLOGY, M4 EQ-Bank 1

Describe two examples of paleontological evidence from Australia that provide insights into past changes in ecosystems.   (4 marks)

Show Answers Only

Example 1: Megafauna fossils in the Naracoorte Caves, South Australia

  • The Naracoorte Caves contain fossil record of Australian megafauna, including giant kangaroos and marsupial lions, dating back to the Pleistocene epoch.
  • These fossils provide evidence of a dramatic shift in Australia’s ecosystem, from one that supported large herbivores and their predators to the current environment dominated by smaller species.
  • The disappearance of these megafauna species from the fossil record around 46,000 years ago coincides with human arrival in Australia, suggesting a potential link between human activity and ecosystem change.  

Example 2: Plant microfossils from the Atherton Tablelands, Qld

  • Sediment cores from crater lakes in the Atherton Tablelands contain plant microfossils such as pollen grains that date back over 200,000 years.
  • Analysis of these microfossils reveals changes in vegetation types over time, indicating shifts between rainforest and dry forest dominance in response to climate fluctuations.
  • This evidence provides a detailed record of how Australian plant communities have responded to past climate changes, including glacial and interglacial periods. 
Show Worked Solution

Example 1: Megafauna fossils in the Naracoorte Caves, South Australia

  • The Naracoorte Caves contain fossil record of Australian megafauna, including giant kangaroos and marsupial lions, dating back to the Pleistocene epoch.
  • These fossils provide evidence of a dramatic shift in Australia’s ecosystem, from one that supported large herbivores and their predators to the current environment dominated by smaller species.
  • The disappearance of these megafauna species from the fossil record around 46,000 years ago coincides with human arrival in Australia, suggesting a potential link between human activity and ecosystem change.  

Example 2: Plant microfossils from the Atherton Tablelands, Qld

  • Sediment cores from crater lakes in the Atherton Tablelands contain plant microfossils such as pollen grains that date back over 200,000 years.
  • Analysis of these microfossils reveals changes in vegetation types over time, indicating shifts between rainforest and dry forest dominance in response to climate fluctuations.
  • This evidence provides a detailed record of how Australian plant communities have responded to past climate changes, including glacial and interglacial periods. 

BIOLOGY, M3 EQ-Bank 9

Researchers have observed significant changes in cane toad populations since their introduction to Australia.

Describe one modern-day adaptation discovered in cane toads how it relates to the process of natural selection.   (3 marks)

Show Answers Only
  • Recent studies have revealed that cane toads at the invasion front in Australia have evolved to disperse more rapidly than their ancestors.
  • These modern toads have developed longer legs, allowing them to hop faster,  cover greater distances and colonise new areas more quickly than initially predicted.
  • These changes support natural selection by demonstrating an inherited trait that is passed to offspring and becomes more common in the population.
  • However, researchers have also found that faster “front line” cane toads are more likely to be eaten, breed less and have more spine arthritis than their ancestors.
  • The natural selection principle of traits that enhance survival and reproductive success being passed on is seemingly contradicted by these findings.
Show Worked Solution
  • Recent studies have revealed that cane toads at the invasion front in Australia have evolved to disperse more rapidly than their ancestors.
  • These modern toads have developed longer legs, allowing them to hop faster,  cover greater distances and colonise new areas more quickly than initially predicted.
  • These changes support natural selection by demonstrating an inherited trait that is passed to offspring and becomes more common in the population.
  • However, researchers have also found that faster “front line” cane toads are more likely to be eaten, breed less and have more spine arthritis than their ancestors.
  • The natural selection principle of traits that enhance survival and reproductive success being passed on is seemingly contradicted by these findings.

BIOLOGY, M3 EQ-Bank 8

Absolute dating techniques provide crucial information about the age of fossils and rocks, but different methods are suitable for different time periods.

Describe two absolute dating methods used by scientists, each appropriate for a different time scale.   (4 marks)

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Carbon-14 dating:

  • Carbon-14 dating is an absolute dating method used for relatively recent organic materials. It’s based on the decay of radioactive carbon-14 to nitrogen-14.
  • Carbon-14 has a half-life of 5730 years and measuring the ratio of carbon-14 to stable carbon-12, scientists can determine when the organism died.
  • Carbon-14 dating is most effective for materials up to about 50,000 years old, making it useful for dating recent fossils. 

Potassium-40 dating:

  • Potassium-40 dating is an absolute dating technique that can be used for much older materials.
  • It’s based on the decay of radioactive potassium-40 to argon-40. This method is effective for dating rocks and minerals containing potassium, typically those of igneous or metamorphic origin.
  • Potassium-40 has a half-life of about 1.3 billion years, making this method suitable for dating ancient fossils and rock layers from 50,000 to billions of years old.
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Carbon-14 dating:

  • Carbon-14 dating is an absolute dating method used for relatively recent organic materials. It’s based on the decay of radioactive carbon-14 to nitrogen-14.
  • Carbon-14 has a half-life of 5730 years and measuring the ratio of carbon-14 to stable carbon-12, scientists can determine when the organism died.
  • Carbon-14 dating is most effective for materials up to about 50,000 years old, making it useful for dating recent fossils. 

Potassium-40 dating:

  • Potassium-40 dating is an absolute dating technique that can be used for much older materials.
  • It’s based on the decay of radioactive potassium-40 to argon-40. This method is effective for dating rocks and minerals containing potassium, typically those of igneous or metamorphic origin.
  • Potassium-40 has a half-life of about 1.3 billion years, making this method suitable for dating ancient fossils and rock layers from 50,000 to billions of years old.

BIOLOGY, M3 EQ-Bank 6

Paleontologists use various methods to determine the age of fossils, providing crucial information about Earth's history and the evolution of life.

Describe two techniques that can be used to date fossils. In your answer, discuss one advantage and one limitation of each method.   (4 marks)

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Stratigraphy:

  • Stratigraphy is a relative dating technique based on the principle that sedimentary rocks are deposited in layers, with older layers at the bottom and younger layers on top.
  • An advantage of stratigraphy is that it can provide relative ages for fossils without requiring specialised equipment.
  • However, a limitation is that it only provides relative dates, not absolute ages, and can be disrupted by geological processes like folding or faulting. 

Radiocarbon dating:

  • Radiocarbon dating is an absolute dating technique used to determine the age of organic materials up to about 50,000 years old by measuring the decay of radioactive carbon-14 in fossils.
  • An advantage of this method is that it can provide precise absolute dates for relatively recent fossils.
  • However, a limitation is its restricted time range; it cannot be used for fossils older than about 50,000 years because most of the carbon-14 will have decayed.
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Stratigraphy:

  • Stratigraphy is a relative dating technique based on the principle that sedimentary rocks are deposited in layers, with older layers at the bottom and younger layers on top.
  • An advantage of stratigraphy is that it can provide relative ages for fossils without requiring specialised equipment.
  • However, a limitation is that it only provides relative dates, not absolute ages, and can be disrupted by geological processes like folding or faulting. 

Radiocarbon dating:

  • Radiocarbon dating is an absolute dating technique used to determine the age of organic materials up to about 50,000 years old by measuring the decay of radioactive carbon-14 in fossils.
  • An advantage of this method is that it can provide precise absolute dates for relatively recent fossils.
  • However, a limitation is its restricted time range; it cannot be used for fossils older than about 50,000 years because most of the carbon-14 will have decayed.

BIOLOGY, M3 EQ-Bank 3

Ginkgo biloba, often called a 'living fossil', is the only surviving species of the division Ginkgophyta. While native to China today, fossils of ginkgo-like plants have been found on every continent except Antarctica.

These fossils date back to the Permian period, over 270 million years ago. Ginkgo fossils have been discovered in locations as diverse as North America, Europe, and Australia.

Explain how this widespread fossil distribution of Ginkgo, compared to its limited native range today, supports the theory of evolution by natural selection.   (4 marks)

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  • The presence of Ginkgo fossils on multiple continents can be explained by the existence of the supercontinent Pangaea, which began breaking up about 200 million years ago.
  • Ginkgo-like plants were already present before this breakup, allowing their fossils to be distributed across what would become separate continents.
  • As continents drifted apart and climates changed over millions of years, Ginkgo species faced varying selection pressures in different regions.
  • In many areas, these pressures led to the extinction of local Ginkgo populations, demonstrating natural selection in action.
  • The survival of Ginkgo biloba shows the process of evolutionary adaptation. Its survival is likely due to traits that were advantageous within its specific environment, illustrating how environmental changes can drive both extinction and adaptation, key concepts in Darwin and Wallace’s theory.
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  • The presence of Ginkgo fossils on multiple continents can be explained by the existence of the supercontinent Pangaea, which began breaking up about 200 million years ago.
  • Ginkgo-like plants were already present before this breakup, allowing their fossils to be distributed across what would become separate continents.
  • As continents drifted apart and climates changed over millions of years, Ginkgo species faced varying selection pressures in different regions.
  • In many areas, these pressures led to the extinction of local Ginkgo populations, demonstrating natural selection in action.
  • The survival of Ginkgo biloba shows the process of evolutionary adaptation. Its survival is likely due to traits that were advantageous within its specific environment, illustrating how environmental changes can drive both extinction and adaptation, key concepts in Darwin and Wallace’s theory.