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CHEMISTRY, M7 2017 HSC 27

The boiling points and molar masses of three compounds are shown in the table.
 

Acetic acid, butan-1-ol and butyl acetate have very different molar masses but similar boiling points. Explain why in terms of the structure and bonding of the three compounds.   (5 marks)

Show Answers Only
  • Although the three listed compounds different molar masses, they have similar boiling points due to their different structures and resulting intermolecular forces.
  • Butyl acetate has the largest molar mass and therefore greatest dispersion forces but it is only slightly polar and has no hydrogen bonding.
  • Butan-1-ol has lower molar mass than butyl acetate and therefore smaller dispersion forces but it is polar and contains a hydrogen bound to an oxygen. Therefore, it exhibits hydrogen bonding resulting in strong intermolecular forces and a boiling point in the middle of the three compounds.
  • Acetic acid has the lowest molar mass and hence the weakest dispersion forces. It is however highly polar due to the presence of the carboxyl group \(\ce{(COOH)}\) and contains a hydrogen bound to an oxygen allowing the formation of hydrogen bonds between molecules.
  • The presence of a second oxygen in acetic acid increases the hydrogen bonding compared with butan-1-ol. 
  • These factors lead to acetic acid possessing the highest boiling point despite its molar mass being the lowest.
  • In summary, the totality of the intermolecular forces of all three molecules is similar and therefore similar boiling points.
Show Worked Solution
  • Although the three listed compounds different molar masses, they have similar boiling points due to their different structures and resulting intermolecular forces.
  • Butyl acetate has the largest molar mass and therefore greatest dispersion forces but it is only slightly polar and has no hydrogen bonding.
  • Butan-1-ol has lower molar mass than butyl acetate and therefore smaller dispersion forces but it is polar and contains a hydrogen bound to an oxygen. Therefore, it exhibits hydrogen bonding resulting in strong intermolecular forces and a boiling point in the middle of the three compounds.
  • Acetic acid has the lowest molar mass and hence the weakest dispersion forces. It is however highly polar due to the presence of the carboxyl group \(\ce{(COOH)}\) and contains a hydrogen bound to an oxygen allowing the formation of hydrogen bonds between molecules.
  • The presence of a second oxygen in acetic acid increases the hydrogen bonding compared with butan-1-ol. 
  • These factors lead to acetic acid possessing the highest boiling point despite its molar mass being the lowest.
  • In summary, the totality of the intermolecular forces of all three molecules is similar and therefore similar boiling points.

♦♦ Mean mark 41%.

CHEMISTRY, M7 2019 HSC 34

The following reaction scheme can be used to synthesise ethyl ethanoate.
 


 

Outline the reagents and conditions required for each step and how the product of each step could be identified.   (7 marks) 

Show Answers Only

Step 1:

  • To synthesise chloroethane (A) into ethanol (B), \(\ce{NaOH}\) is added and heated. \(\ce{KMnO4 / H+}\) is then added and heated.
  • The mixture is then treated with concentrated sulfuric acid and refluxed.
  • Ethanol (B) can be identified using infrared spectroscopy by looking for a broad absorption between 3230 cm ¯1 and 3550 cm ¯1, which indicates the presence of an \(\ce{O-H}\) bond. This absorption would not be present in chloroethane (A).
  • Alternative ways to identify ethanol include: mass spectrum analysis (single ion peak at m/z = 46), reactivity tests, and \( \ce{^1H NMR}\) spectrum analysis (3 signals vs 2 for chloroethane). 

Step 2:

  • Ethanol (B) can be converted into ethanoic acid (C) by combining it with a strong oxidant like sodium carbonate, which produces carbon dioxide bubbles, confirming the presence of a carboxylic acid.
  • Ethanol will not react as above and the compounds can be distinguished.
  • Alternative ways to identify ethanoic acid include: IR or \( \ce{^13C NMR}\) spectrum analysis, litmus indicators, mass spectrum analysis (ion peak at m/z = 60 vs m/z = 46) 

Step 3

  • Ethyl ethanoate (D) can be synthesised by heating a mixture of ethanol, ethanoic acid and concentrated sulfuric acid under reflux.
  • A \( \ce{^1H NMR}\) spectrum can be used to identify ethyl ethanoate as it will have 3 signals versus ethanol and ethanoic acid that will only have 2 each.
  • Alternative ways to identify ethyl ethanoate include: a distinct smell, no \(\ce{O-H}\) peaks in the IR spectrum or mass spectrum analysis (ion peak at m/z = 102).
Show Worked Solution

Step 1:

  • To synthesise chloroethane (A) into ethanol (B), \(\ce{NaOH}\) is added and heated. \(\ce{KMnO4 / H+}\) is then added and heated.
  • The mixture is then treated with concentrated sulfuric acid and refluxed.
  • Ethanol (B) can be identified using infrared spectroscopy by looking for a broad absorption between 3230 cm ¯1 and 3550 cm ¯1, which indicates the presence of an \(\ce{O-H}\) bond. This absorption would not be present in chloroethane (A).
  • Alternative ways to identify ethanol include: mass spectrum analysis (single ion peak at m/z = 46), reactivity tests, and \( \ce{^1H NMR}\) spectrum analysis (3 signals vs 2 for chloroethane). 

Step 2:

  • Ethanol (B) can be converted into ethanoic acid (C) by combining it with a strong oxidant like sodium carbonate, which produces carbon dioxide bubbles, confirming the presence of a carboxylic acid.
  • Ethanol will not react as above and the compounds can be distinguished.
  • Alternative ways to identify ethanoic acid include: IR or \( \ce{^13C NMR}\) spectrum analysis, litmus indicators, mass spectrum analysis (ion peak at m/z = 60 vs m/z = 46) 

Step 3

  • Ethyl ethanoate (D) can be synthesised by heating a mixture of ethanol, ethanoic acid and concentrated sulfuric acid under reflux.
  • A \( \ce{^1H NMR}\) spectrum can be used to identify ethyl ethanoate as it will have 3 signals versus ethanol and ethanoic acid that will only have 2 each.
  • Alternative ways to identify ethyl ethanoate include: a distinct smell, no \(\ce{O-H}\) peaks in the IR spectrum or mass spectrum analysis (ion peak at m/z = 102).

♦♦ Mean mark 38%.

CHEMISTRY, M5 2019 HSC 30

The following data apply to magnesium fluoride and magnesium chloride dissolving in water at 298 K.

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex} & \textit{Magnesium fluoride} \rule[-1ex]{0pt}{0pt}& \textit{Magnesium chloride}\\
\hline
\rule{0pt}{2.5ex}\Delta_{sol} H^{\ominus}\left(\text{kJ mol}^{-1}\right) \rule[-1ex]{0pt}{0pt}& -7.81 & -160 \\
\hline
\rule{0pt}{2.5ex}\Delta_{\text {sol }} S^{\ominus}\left(\text{J K}^{-1} mol^{-1}\right) \rule[-1ex]{0pt}{0pt}& -223 & -115 \\
\hline
\rule{0pt}{2.5ex}T \Delta_{\text {sol }} S^{\ominus}\left(\text{kJ mol}^{-1}\right) \rule[-1ex]{0pt}{0pt}& -66.4 & -34.2 \\
\hline
\rule{0pt}{2.5ex}\Delta_{\text {sol }} G^{\ominus}\left(\text{kJ mol}^{-1}\right) \rule[-1ex]{0pt}{0pt}& +58.6 & -125 \\
\hline
\end{array} 

Compare the effects of enthalpy and entropy on the solubility of these salts.   (3 marks)

Show Answers Only
  • Magnesium chloride dissolves in water spontaneously as it has a negative \(\triangle_{\text {sol }} G^{\ominus}\) ( – 125 kJ mol ¯1).
  • Magnesium flouride however does not dissolve in water spontaneously which is shown by its corresponding \(\triangle_{\text {sol }} G^{\ominus}\) of +58.6 kJ mol ¯1.
  • Both salts have a negative \(\triangle_{\text {sol }} S^{\ominus}\), resulting in a net positive \(-T \triangle_{\text {sol }} S^{\ominus}\) contribution to \(\triangle_{\text {sol }} G^{\ominus}\).
  • Both salts have a negative \(\triangle_{\text {sol }} H^{\ominus}\). It should be noted however that magnesium chloride’s negative value is significantly more negative at \(\triangle_{\text {sol }} H^{\ominus}\) ( – 160 kJ mol ¯1). This is greater than the \(-T \triangle_{\text {sol }} S^{\ominus}\) contribution (+34.2 kJ mol ¯1), resulting in a negative \(\triangle_{\mathrm{sol}} G^{\ominus}\).
  • This can be compared to magnesium fluoride that has a relatively small negative \(\triangle_{\text {sol }} H^{\ominus}\) ( – 7.81 kJ mol ¯1) which is smaller than the \(-T \triangle_{\text {sol }} S^{\ominus}\) contribution (+66.4 kJ mol ¯1), resulting in a positive \(\triangle_{\text {sol }} G^{\ominus}\).
Show Worked Solution
  • Magnesium chloride dissolves in water spontaneously as it has a negative \(\triangle_{\text {sol }} G^{\ominus}\) ( – 125 kJ mol ¯1).
  • Magnesium flouride however does not dissolve in water spontaneously which is shown by its corresponding \(\triangle_{\text {sol }} G^{\ominus}\) of +58.6 kJ mol ¯1.
  • Both salts have a negative \(\triangle_{\text {sol }} S^{\ominus}\), resulting in a net positive \(-T \triangle_{\text {sol }} S^{\ominus}\) contribution to \(\triangle_{\text {sol }} G^{\ominus}\).
  • Both salts have a negative \(\triangle_{\text {sol }} H^{\ominus}\). It should be noted however that magnesium chloride’s negative value is significantly more negative at \(\triangle_{\text {sol }} H^{\ominus}\) ( – 160 kJ mol ¯1). This is greater than the \(-T \triangle_{\text {sol }} S^{\ominus}\) contribution (+34.2 kJ mol ¯1), resulting in a negative \(\triangle_{\mathrm{sol}} G^{\ominus}\).
  • This can be compared to magnesium fluoride that has a relatively small negative \(\triangle_{\text {sol }} H^{\ominus}\) ( – 7.81 kJ mol ¯1) which is smaller than the \(-T \triangle_{\text {sol }} S^{\ominus}\) contribution (+66.4 kJ mol ¯1), resulting in a positive \(\triangle_{\text {sol }} G^{\ominus}\).

♦♦ Mean mark 34%.

CHEMISTRY, M8 2019 HSC 29

Stormwater from a mine site has been found to be contaminated with copper\(\text{(II)}\) and lead\(\text{(II)}\) ions. The required discharge limit is 1.0 mg L¯1 for each metal ion. Treatment of the stormwater with \(\ce{Ca(OH)2}\) solid to remove the metal ions is recommended.

  1. Explain the recommended treatment with reference to solubility. Include a relevant chemical equation.   (2 marks)

  2. Explain why atomic absorption spectroscopy can be used to determine the concentrations of \(\ce{Cu^2+}\) and \(\ce{Pb^2+}\) ions in a solution containing both species.   (2 marks)

  3. The data below were obtained after treatment of the stormwater.
     
         
     
    To what extent is the treatment effective in meeting the required discharge limit of 1.0 mg L¯1 for each metal ion? Support your conclusion with calibration curves and calculations.   (7 marks)
     
         

Show Answers Only

a.   Recommended Treatment:

  • Calcium hydroxide is a slightly soluble compound, while copper\(\text{(II)}\) hydroxide and lead\(\text{(II)}\) hydroxide are very insoluble in water.
  • When these compounds are added to water, the metal ions tend to precipitate out of solution.
  • For example, the addition of solid calcium hydroxide to water produces calcium ions \(\ce{Ca^2+}\) and hydroxide ions \(\ce{OH-}\), which can then react with lead\(\text{(II)}\) ions (\(\ce{Pb^2+})\) and copper\(\text{(II)}\) ions \(\ce{Cu^2+}\) to form precipitates of lead\(\text{(II}\) hydroxide and copper\(\text{(II)}\) hydroxide, respectively.
  • These reactions are represented by the equations:
  •    \(\ce{Pb^2+ + 2OH- -> Pb(OH)2, \ \ Cu^2+ + 2OH- -> Cu(OH)2}\) 

b.   Atomic absorption spectroscopy (AAS):

  • Can be used for determining the concentration of metal ions in a sample by measuring the absorbance of light at specific wavelengths that are characteristic of each metal.
  • AAS uses light wavelengths that correspond to atomic absorption by the element of interest, and since each element has unique wavelengths that are absorbed, the concentration of that element can be selectively measured in the presence of other species.
  • As a result, AAS can be used to independently measure the concentrations of different metal ions, such as lead\(\text{(II)}\) ions and copper\(\text{(II)}\) ions in a sample containing both types. 

c.   Concentrations of ions:

\begin{array} {|l|c|c|c|}
\hline  \text{Sample }& \ce{Cu^2+ \times 10^{-5} mol L^{-1}} & \ce{Pb^2+ \times 10^{-5} mol L^{-1}} \\
\hline \text{Water (pre-treatment)} & 5.95 & 4.75 \\
\hline \text{Water (post-treatment)} & 0.25 & 0.85 \\
\hline \end{array}

  • Concentrations of copper and lead have been significantly reduced.
  • Convert concentrations to compare with standard:

\begin{array} {ccc}
\ce{Cu^2+}: & 5.95 \times 10^{-5} \times 63.55 \times 1000 = 3.78\ \text{mg L}^{-1} \\
 & 0.25 \times 10^{-5} \times 63.55 \times 1000 = 0.16\ \text{mg L}^{-1} \\
& \\
\ce{Pb^2+}: & 4.75 \times 10^{-5} \times 207.2 \times 1000 = 9.84\ \text{mg L}^{-1} \\
& 0.85 \times 10^{-5} \times 207.2 \times 1000 = 1.76\ \text{mg L}^{-1} \end{array}

 

   

Conclusion:

  • The concentration of copper ions has been reduced to a level that is lower than the discharge limit (0.16 < 1.0) but the lead ion concentration has not (1.76 > 1.0).
  • The treatment has only been partially successful.
Show Worked Solution

a.   Recommended Treatment:

  • Calcium hydroxide is a slightly soluble compound, while copper\(\text{(II)}\) hydroxide and lead\(\text{(II)}\) hydroxide are very insoluble in water.
  • When these compounds are added to water, the metal ions tend to precipitate out of solution.
  • For example, the addition of solid calcium hydroxide to water produces calcium ions \(\ce{Ca^2+}\) and hydroxide ions \(\ce{OH-}\), which can then react with lead\(\text{(II)}\) ions (\(\ce{Pb^2+})\) and copper\(\text{(II)}\) ions \(\ce{Cu^2+}\) to form precipitates of lead\(\text{(II}\) hydroxide and copper\(\text{(II)}\) hydroxide, respectively.
  • These reactions are represented by the equations:
  •    \(\ce{Pb^2+ + 2OH- -> Pb(OH)2, \ \ Cu^2+ + 2OH- -> Cu(OH)2}\) 

♦ Mean mark (a) 46%.

b.   Atomic absorption spectroscopy (AAS):

  • Can be used for determining the concentration of metal ions in a sample by measuring the absorbance of light at specific wavelengths that are characteristic of each metal.
  • AAS uses light wavelengths that correspond to atomic absorption by the element of interest, and since each element has unique wavelengths that are absorbed, the concentration of that element can be selectively measured in the presence of other species.
  • As a result, AAS can be used to independently measure the concentrations of different metal ions, such as lead\(\text{(II)}\) ions and copper\(\text{(II)}\) ions in a sample containing both types. 

c.   Concentrations of ions:

\begin{array} {|l|c|c|c|}
\hline  \text{Sample }& \ce{Cu^2+ \times 10^{-5} mol L^{-1}} & \ce{Pb^2+ \times 10^{-5} mol L^{-1}} \\
\hline \text{Water (pre-treatment)} & 5.95 & 4.75 \\
\hline \text{Water (post-treatment)} & 0.25 & 0.85 \\
\hline \end{array}


♦♦ Mean mark (b) 32%.
  • Concentrations of copper and lead have been significantly reduced.
  • Convert concentrations to compare with standard:

\begin{array} {ccc}
\ce{Cu^2+}: & 5.95 \times 10^{-5} \times 63.55 \times 1000 = 3.78\ \text{mg L}^{-1} \\
 & 0.25 \times 10^{-5} \times 63.55 \times 1000 = 0.16\ \text{mg L}^{-1} \\
& \\
\ce{Pb^2+}: & 4.75 \times 10^{-5} \times 207.2 \times 1000 = 9.84\ \text{mg L}^{-1} \\
& 0.85 \times 10^{-5} \times 207.2 \times 1000 = 1.76\ \text{mg L}^{-1} \end{array}

   

Conclusion:

  • The concentration of copper ions has been reduced to a level that is lower than the discharge limit (0.16 < 1.0) but the lead ion concentration has not (1.76 > 1.0).
  • The treatment has only been partially successful.

♦ Mean mark (c) 53%.

CHEMISTRY, M6 2019 HSC 28

Assess the usefulness of the Brønsted-Lowry model in classifying acids and bases. Support your answer with at least TWO chemical equations.   (5 marks)

Show Answers Only
  • The Bronsted-Lowry model is a way of classifying acids and bases based on their ability to donate or accept protons.
  • This model is more comprehensive than the Arrhenius model, as it can explain the acid-base behaviour of more species, including those that do not contain \(\ce{OH-}\) ions, and non-aqueous acid-base reactions.
  • Consider the reaction \(\ce{NH3(g) + HCl(g) -> NH4Cl(s)}\) where a proton is transferred from hydrogen chloride to ammonia (according to the Bronsted-Lowry model). Ammonia is not an Arrhenius base as it doesn’t dissociate to produce \(\ce{OH-}\) ions and the reaction cannot be described using the Arrhenius model.
  • However, the Bronsted-Lowry model does have some limitation, such as its inability to explain the acidity of certain acidic oxides and their reactions with basic oxides.
  • e.g. \(\ce{CaO(s) + SO3(g) -> CaSO4(s)}\) is an acid-base reaction but since there is no proton transfer, it cannot be described using the Bonsted-Lowry model.
Show Worked Solution
  • The Bronsted-Lowry model is a way of classifying acids and bases based on their ability to donate or accept protons.
  • This model is more comprehensive than the Arrhenius model, as it can explain the acid-base behaviour of more species, including those that do not contain \(\ce{OH-}\) ions, and non-aqueous acid-base reactions.
  • Consider the reaction \(\ce{NH3(g) + HCl(g) -> NH4Cl(s)}\) where a proton is transferred from hydrogen chloride to ammonia (according to the Bronsted-Lowry model). Ammonia is not an Arrhenius base as it doesn’t dissociate to produce \(\ce{OH-}\) ions and the reaction cannot be described using the Arrhenius model.
  • However, the Bronsted-Lowry model does have some limitation, such as its inability to explain the acidity of certain acidic oxides and their reactions with basic oxides.
  • e.g. \(\ce{CaO(s) + SO3(g) -> CaSO4(s)}\) is an acid-base reaction but since there is no proton transfer, it cannot be described using the Bonsted-Lowry model.

♦♦♦ Mean mark 35%.
MARKER’S COMMENT: Two relevant equations required to achieve full marks.

CHEMISTRY, M6 2017 HSC 13 MC

25.0 mL of a 0.100 mol L ¯1 acid is to be titrated against a sodium hydroxide solution until final equivalence is reached.

Which of the following acids, if used in the titration, would require the greatest volume of sodium hydroxide?

  1. Acetic
  2. Citric
  3. Hydrochloric
  4. Sulfuric
Show Answers Only

`B`

Show Worked Solution
  • Citric acid is triprotic (i.e. ratio moles NaOH : acid = 3 : 1). It therefore requires the greates volume of NaOH.
  • Acetic acid and Hydrochloric acid are monoprotic (i.e. ratio moles NaOH : acid = 1 : 1)
  • Sulfuric acid is diprotic (i.e. ratio moles NaOH : acid = 2 : 1)

`=>B`


♦♦♦ Mean mark 29%.

PHYSICS, M6 2015 HSC 24

A part of a cathode ray oscilloscope was represented on a website as shown.
 

Electrons leave the cathode and are accelerated towards the anode.

  1. Explain why the representation of the path of the electron between the deflection plates is inaccurate.   (3 marks)
  1. Calculate the force on an electron due to the electric field between the cathode and the anode.   (2 marks)
  1. Calculate the velocity of an electron as it reaches the anode.   (2 marks)
Show Answers Only

a.    Inaccuracies include:

  • The electron is shown to travel towards the negative plate. This is inaccurate as the electron is a negatively charged particle so will be attracted towards the positive plate.
  • The electron is shown to only deflect once it has travelled halfway through the plates. This is inaccurate as the plate will deflect the electron immediately after it enters the uniform electric field. 
  • The electron is shown to follow a straight path. This is inaccurate as the electron has a constant acceleration towards the positive plate and a constant horizontal velocity, it will follow a parabolic path.

b.   `F=4xx10^(-14)  text{N}`

c.   `v=4.19 xx10^(7)  text{ms}^(-1)`

Show Worked Solution

a.    Inaccuracies include:

  • The electron is shown to travel towards the negative plate. This is inaccurate as the electron is a negatively charged particle so will be attracted towards the positive plate.
  • The electron is shown to only deflect once it has travelled halfway through the plates. This is inaccurate as the plate will deflect the electron immediately after it enters the uniform electric field. 
  • The electron is shown to follow a straight path. This is inaccurate as the electron has a constant acceleration towards the positive plate and a constant horizontal velocity, it will follow a parabolic path. 

b.    Using  `E=(F)/(q)`  and  `E=(V)/(d):`

`(F)/(q)` `=(V)/(d)`  
`F` `=(Vq)/(d)`  
  `=(5000 xx1.6 xx10^(-19))/(0.02)`  
  `=4xx10^(-14)  text{N}`  

 
c.    `a=(F)/(m)=(4xx10^(-14))/(9.1 xx10^(-31))=4.4 xx10^(16)  text{ms}^(-1)`
 

`v^(2)` `=u^(2)+2as`  
`:.v` `=sqrt(2xx4.4 xx10^(16)xx0.02)`  
  `=4.19 xx10^(7)  text{ms}^(-1)`  

♦♦♦ Mean mark (c) 20%.

PHYSICS, M5 2015 HSC 19 MC

An astronaut working outside a spacecraft in orbit around Earth is not attached to it.

Why does the astronaut NOT drift away from the spacecraft?

  1. The force of gravity acting on the astronaut and spacecraft is negligible.
  2. The spacecraft and the astronaut are in orbit around the Sun with the Earth.
  3. The forces due to gravity acting on both the astronaut and the spacecraft are the same.
  4. The accelerations of the astronaut and the spacecraft are inversely proportional to their respective masses.
Show Answers Only

`D`

Show Worked Solution

By Elimination:

  • Gravity is the force keeping the spacecraft and the astronaut in orbit, so it is not negligible (eliminate A).
  • Earths rotation around the sun is not relevant to the motion of the astronaut and the spacecraft relative to Earth (eliminate B).
  • Using  `F=(GMm)/(r^2)`, the force due to gravity acting on both is proportional to their respective masses. Since the spacecraft is significantly heavier, it will experience a greater force due to gravity (eliminate C).
  • The accelerations, `g=(F)/(m)` of the spacecraft and the astronaut are inversely proportional to their respective masses. As they both experience a force, `F`, due to gravity proportional to their masses, their accelerations are the same. Therefore, the astronaut will not drift from the spacecraft.

`=>D`


♦♦♦ Mean mark 15%.

PHYSICS, M5 2015 HSC 11 MC

Which of the following diagrams correctly represents the force(s) acting on a satellite in a stable circular orbit around Earth?
 

 

Show Answers Only

`D`

Show Worked Solution
  • In a stable, circular orbit the satellite requires no propulsion force to keep it in orbit. This is due to it undergoing uniform circular motion with gravity acting as its centripetal force.
  • A force of gravity acts on the satellite, with direction towards Earth’s centre of mass. 
  • No ‘reaction force’ acts on the satellite. This is because any ‘reaction’ to the gravitational force exerted on the satellite by the Earth is an equal and opposite force exerted on the Earth by the satellite (Newton’s Third Law).
  • i.e. The reaction force acts on the Earth, not the satellite.

`=>D`


♦♦♦ Mean mark 29%.

PHYSICS, M6 2016 HSC 30

The following makeshift device was made to provide lighting for a stranded astronaut on Mars.

The mass of Mars is `6.39 ×10^(23) \ text {kg}`.
 

The 2 kg mass falls, turning the DC generator, which supplies energy to the light bulb. The mass falls from a point that is 3 376 204 m from the centre of Mars.

  1. Calculate the maximum possible energy released by the light bulb as the mass falls through a distance of one metre.   (3 marks)
  1. Explain the difference in the behaviour of the falling mass when the switch is open.   (3 marks)
Show Answers Only

a.   7.48 J

b.   When switch is opened:

  • There is no induced current opposing the downwards motion of the mass (Lenz’s Law).
  • Hence, the mass will fall more quickly.
Show Worked Solution

a.   `DeltaE=U_(f)-U_(i)`

`=((-Gm_(1)m_(2))/(r_(f)))-((-Gm_(1)m_(2))/(r_(i)))`

`=(-6.67 xx10^(-11)xx6.39 xx10^(23)xx2)/(3\ 376\ 203)-((-6.67 xx10^(-11)xx6.39 xx10^(23)xx2))/(3\ 376\ 204)`

`=-7.48\ text{J}`
 

  • 7.48 J is lost by the falling mass.
  • The light bulb released 7.48 J of energy.

♦ Mean mark (a) 52%.

b.   When switch is opened:

  • There is no induced current opposing the downwards motion of the mass (Lenz’s Law).
  • Hence, the mass will fall more quickly.

♦♦♦ Mean mark (b) 27%.

PHYSICS, M6 2016 HSC 20 MC

In the motor shown, the rotor spins clockwise, as viewed from point `P`, when connected to a DC supply.
 

What happens when the motor is connected to an AC supply?

  1. There is no movement of the rotor.
  2. The rotor produces clockwise movement only.
  3. The rotor vibrates at the frequency of the AC supply.
  4. The rotor continuously turns half a rotation clockwise, then half a rotation anticlockwise.
Show Answers Only

`B`

Show Worked Solution
  • Initially, the electromagnet on the left has a magnetic south pole on its side closest to the coil, the electromagnet on the right has a north pole on its side closest to the coil and the current through the coil is anticlockwise as viewed from above.
  • Using the right hand palm rule, the rotor will rotate clockwise.
  • When the direction of AC current changes, the direction of current through the coil swaps, and the polarity of the electromagnets swaps.
  • This means that the direction of torque is maintained and the rotor continues to rotate clockwise.
  • The split-ring commutator functions as normal to maintain the direction of torque once the rotor has rotated through half a turn preventing the rotor from continuously turning half a rotation clockwise, then half a rotation anticlockwise.
  • The rotor produces clockwise movement only.

`=>B`


♦♦♦ Mean mark 17%.

PHYSICS, M6 2016 HSC 16 MC

The cone of a speaker is pushed so that the coil moves in the direction shown.
 

Which row of the table correctly identifies the behaviour of the speaker and the direction of the current through the conductor?

\begin{align*}
\begin{array}{l}
\rule{0pt}{1.5ex}\textit{} & \textit{} \\
\textit{}\rule[1ex]{0pt}{0pt}& \textit{} \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{1.5ex}\textit{The speaker} & \textit{The direction of the} \\
\textit{behaves like a ...}\rule[1ex]{0pt}{0pt}& \textit{current is from ...} \\
\hline
\rule{0pt}{2.5ex}\text{generator}\rule[-1ex]{0pt}{0pt}&\text{\(X\) to \(Y\)}\\
\hline
\rule{0pt}{2.5ex}\text{generator}\rule[-1ex]{0pt}{0pt}& \text{\(Y\) to \(X\)}\\
\hline
\rule{0pt}{2.5ex}\text{motor}\rule[-1ex]{0pt}{0pt}& \text{\(X\) to \(Y\)} \\
\hline
\rule{0pt}{2.5ex}\text{motor}\rule[-1ex]{0pt}{0pt}& \text{\(Y\) to \(X\)} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution
  • Motion of the coil causes the induction of a current causing the speaker to behave like a generator.
  • A magnetic north pole is induced on the right hand side of the speaker coil in order to oppose the motion of the coil away from the magnet by creating an attractive force between the coil and the magnet (Lenz’s Law).
  • Using the right hand grip rule the direction of induced current is from \(X\)  to \(Y\).

\(\Rightarrow A\)


♦♦♦ Mean mark 25%.

PHYSICS, M6 2017 HSC 28

Contrast the design of transformers and magnetic braking systems in terms of the effects that eddy currents have in these devices.   (6 marks)

Show Answers Only

Transformers:

  • A transformer involves primary and secondary coils wound around a laminated iron core.
  • When an AC current is applied to the primary coil, changes in magnetic flux induce a current in the secondary coil.
  • Eddy currents have undesirable effects in transformers as the iron core is a conductor.
  • So, there is induction of unwanted eddy currents which leads to energy losses in the form of heat.
  • Lamination of the iron core minimises these eddy currents and subsequent energy loss.

Magnetic Braking Systems:

  • In contrast, eddy currents are beneficial in magnetic braking systems.
  • Magnetic breaking involves using eddy currents to produce a force that stops a moving vehicle by converting kinetic energy into heat energy.
  • In order to maximise induced eddy currents, and thus the breaking effect, magnetic breaks are designed with large sheets or discs of conductive material such as copper.
Show Worked Solution

Transformers:

  • A transformer involves primary and secondary coils wound around a laminated iron core.
  • When an AC current is applied to the primary coil, changes in magnetic flux induce a current in the secondary coil.
  • Eddy currents have undesirable effects in transformers as the iron core is a conductor.
  • So, there is induction of unwanted eddy currents which leads to energy losses in the form of heat.
  • Lamination of the iron core minimises these eddy currents and subsequent energy loss.

Magnetic Braking Systems:

  • In contrast, eddy currents are beneficial in magnetic braking systems.
  • Magnetic breaking involves using eddy currents to produce a force that stops a moving vehicle by converting kinetic energy into heat energy.
  • In order to maximise induced eddy currents, and thus the breaking effect, magnetic breaks are designed with large sheets or discs of conductive material such as copper.

♦ Mean mark 49%.

PHYSICS, M6 2017 HSC 27

The diagram shows an electric circuit in a magnetic field directed into the page. The graph shows how the flux through the conductive loop changes over a period of 12 seconds.
 

  1. Calculate the maximum magnetic field strength within the stationary loop during the 12-second interval.   (2 marks)
  1. Calculate the maximum voltage generated in the circuit by the changing flux. In your answer, indicate the polarity of the terminals `P` and `Q` when this occurs.   (3 marks)
Show Answers Only

a.   `2.1\ text{T}`

b.   `0.3  text{V}`

  • Terminal `P` will be positive and terminal `Q` will be negative (Lenz’s Law).
Show Worked Solution
a.       `Phi` `=BA`
  `B` `=(Phi)/(A)=(Phi)/(pir^(2))`
  `B_max` `=(0.6)/(pi xx(0.3)^(2))=2.1\ text{T}`

 


♦♦ Mean mark (a) 38%.

b.   Voltage (emf) = time rate of flux

  • The induced emf is at a maximum when the rate of change of flux is a maximum.
  • From the graph, this occurs at  t = 10 – 12 s (steepest gradient).
  •   `epsilon=-(Delta Phi)/(Delta t)=-((-0.6)/(2))=0.3\ text{V}`
  • Terminal `P` will be positive and terminal `Q` will be negative (Lenz’s Law).

♦♦ Mean mark (b) 27%.

PHYSICS, M6 2017 HSC 19 MC

Earth's magnetic field is shown in the following diagram.
 

Two students standing a few metres apart on the equator at points `X` and `Y`, where Earth's magnetic field is parallel to the ground, hold a loop of copper wire between them. Part of the loop is rotated like a skipping rope as shown, while the other part remains motionless on the ground.
 

At what point during the rotation of the wire does the maximum current flow in a direction from `P` to `Q` through the moving part of the wire?

  1. `A`
  2. `B`
  3. `C`
  4. `D`
Show Answers Only

`C`

Show Worked Solution
  • The magnetic field through the loop is going into the page.
  • Using the right hand palm rule, the maximum current flow from `P` to `Q` produces an upwards force on the wire.
  • By Lenz’s Law, this occurs when the wire is moving downwards.

`=>C`


♦♦ Mean mark 28%.

PHYSICS, M6 2017 HSC 17 MC

A magnet rests on an electronic balance. A rigid copper rod runs horizontally through the magnet, at right angles to the magnetic field. The rod is anchored so that it cannot move.
 

Which expression can be used to calculate the balance reading when the switch is closed?

  1. `0.200\ text{kg} +BIl`
  2. `0.200\ text{kg} +(BIl)/9.8`
  3. `0.200\ text{kg} -BIl`
  4. `0.200\ text{kg} -(BIl)/9.8`
Show Answers Only

`B`

Show Worked Solution
  • Using the right hand palm rule, the force on the conductor due to the magnet is up.
  • By Newton’s Third Law, an equal force is exerted downwards on the magnet, adding to the reading on the scale.
  • Since the scale shows mass, not force, `F=BIl`  is divided by 9.8 to convert to a mass reading.

`=>B`


♦♦♦ Mean mark 27%.

PHYSICS, M6 2018 HSC 30

The diagram shows a model of a system used to distribute energy from a power station through transmission lines and transformers to houses.
 

During the evening peak period there is an increase in the number of electrical appliances being turned on in houses.

Explain the effects of this increased demand on the components of the system, with reference to voltage, current and energy.   (6 marks)

Show Answers Only

Current and voltage:

  • An increase in the number of electrical appliances causes an increase in energy demand, and a greater power requirement.
  • As `P=VI`, and `V` is constant at approximately 240`V` at the secondary coil of the step down transformer (T2), a greater current is drawn.
  • Similarly, this increases the current in the transmission lines as well as the power station.
  • As the voltage of T2 is approximately fixed at 240`V` and the ratio of coils in each transformer stays constant, the output voltage from the power station, the transmission line voltage and the supply voltage to houses will also remain approximately constant.

Energy losses:

  • As the current increases, the heat produced by resistance in the wires increases leading to losses in energy according to `P_(loss)=I^2R`.
  • Energy loss is also caused by the formation of eddy currents in the transformer cores, and this increases as current increases.
  • This power loss will cause slight voltage drops along the transmission lines leading to slight decreases in voltage inputs and outputs at each of the transformers.
Show Worked Solution

Current and voltage:

  • An increase in the number of electrical appliances causes an increase in energy demand, and a greater power requirement.
  • As `P=VI`, and `V` is constant at approximately 240`V` at the secondary coil of the step down transformer (T2), a greater current is drawn.
  • Similarly, this increases the current in the transmission lines as well as the power station.
  • As the voltage of T2 is approximately fixed at 240`V` and the ratio of coils in each transformer stays constant, the output voltage from the power station, the transmission line voltage and the supply voltage to houses will also remain approximately constant.

Energy losses:

  • As the current increases, the heat produced by resistance in the wires increases leading to losses in energy according to `P_(loss)=I^2R`.
  • Energy loss is also caused by the formation of eddy currents in the transformer cores, and this increases as current increases.
  • This power loss will cause slight voltage drops along the transmission lines leading to slight decreases in voltage inputs and outputs at each of the transformers.

♦♦ Mean mark 42%.

Measurement, STD1 M3 2022 HSC 32

The diagram shows two right-angled triangles, `A B C` and `A B D`,

where `A C=35 \ text{cm}`, `B D=93 \ text{cm}`, `/_ A C B=41^@` and ` /_ A D B=\theta`.
 


 

Calculate the size of angle `\theta`, to the nearest minute.  (4 marks)

Show Answers Only

`19^@6`′

Show Worked Solution

`text{In}\ Delta ABC:`

`tan 41^@` `=(AB)/35`  
`AB` `=35xxtan 41^@`  
  `=30.425…`  

  
`text{In}\ Delta ABD:`

`sin theta` `=(AB)/(BD)`  
  `=(30.425…)/93`  
`:.theta` `=sin^(-1)((30.425…)/93)`  
  `=19.09…`  
  `=19^@6`′`\ \ text{(nearest minute)}`  

♦♦♦ Mean mark 17%.

Statistics, STD1 S1 2022 HSC 29

The ages of the 10 members in a tennis club are

`{:[24,25,27,33,34,34,35,39,47,59.]:}`

Could the age `59` be considered an outlier? Justify your answer with calculations.  (3 marks)

Show Answers Only

`Q_1=27, \ Q_3=39`

`IQR = Q_3-Q_1 = 39-27 = 12`
 

`text{Upper limit:}`

`Q_3 + 1.5 xx IQR` `=39 + 1.5 xx 12`  
  `=39 + 1.5 xx 12`  
  `=57`  

  
`:.\ text{S}text{ince  59 > 57,  59 is an outlier.}`

Show Worked Solution

`Q_1=27, \ Q_3=39`

`IQR = Q_3-Q_1 = 39-27 = 12`
 

`text{Upper limit:}`

`Q_3 + 1.5 xx IQR` `=39 + 1.5 xx 12`  
  `=39 + 1.5 xx 12`  
  `=57`  

  
`:.\ text{S}text{ince  59 > 57,  59 is an outlier.}`


♦♦♦ Mean mark 11%.

BIOLOGY, M5 2019 HSC 30

Experiments were conducted to obtain data on the traits 'seed shape' in plants and 'feather colour' in chickens. In each case, the original parents were pure breeding and produced the first generation (F1). The frequency data diagrams below relate to the second generation offspring (F2), produced when the F1 generations were bred together.
 

Explain the phenotypic ratios of the F2 generation in both the plant and chicken breeding experiments. Include Punnett squares and a key to support your answer.   (5 marks)

Show Answers Only
  • Graph A shows a 3:1 phenotypic ratio. This is typical of a dominant/recessive allele phenotypic ratio of two heterozygous parents.
  • The Punnet square below supports this argument, where R refers to the dominant seed shape (e.g. round) and r is the recessive allele, producing another seed shape (e.g. wrinkled). The offspring have a 3:1 ratio of dominant : recessive seed shape.

\begin{array} {|c|c|c|}\hline  & \text{R} & \text{r} \\ \hline \text{R} & \text{RR} & \text{Rr} \\ \hline \text{r} & \text{Rr} & \text{rr} \\ \hline \end{array}

Key: R = Round     r = wrinkled

  • Graph B shows a 1:2:1 phenotypic ratio. Because both parents are heterozygous, this ratio is typical of a co-dominant or incomplete dominant trait.
  • If B is an allele referring to black colour feathers and W is the allele for white colour feathers then both parents will be BW, which is either grey colour feathers (co-dominance) or both black and white feathers (incomplete dominance). A cross between these genotypes will produce a phenotypic ratio of the same seen in the graph.

\begin{array} {|c|c|c|}\hline  & \text{B} & \text{W} \\ \hline \text{B} & \text{BB} & \text{BW} \\ \hline \text{W} & \text{BW} & \text{WW} \\ \hline \end{array}

Key: B = Black Feathers     W= White Feathers

Show Worked Solution
  • Graph A shows a 3:1 phenotypic ratio. This is typical of a dominant/recessive allele phenotypic ratio of two heterozygous parents.
  • The Punnet square below supports this argument, where R refers to the dominant seed shape (e.g. round) and r is the recessive allele, producing another seed shape (e.g. wrinkled). The offspring have a 3:1 ratio of dominant : recessive seed shape.

\begin{array} {|c|c|c|}\hline  & \text{R} & \text{r} \\ \hline \text{R} & \text{RR} & \text{Rr} \\ \hline \text{r} & \text{Rr} & \text{rr} \\ \hline \end{array}

Key: R = Round     r = wrinkled

  • Graph B shows a 1:2:1 phenotypic ratio. Because both parents are heterozygous, this ratio is typical of a co-dominant or incomplete dominant trait.
  • f B is an allele referring to black colour feathers and W is the allele for white colour feathers then both parents will be BW, which is either grey colour feathers (co-dominance) or both black and white feathers (incomplete dominance). A cross between these genotypes will produce a phenotypic ratio of the same seen in the graph.

\begin{array} {|c|c|c|}\hline  & \text{B} & \text{W} \\ \hline \text{B} & \text{BB} & \text{BW} \\ \hline \text{W} & \text{BW} & \text{WW} \\ \hline \end{array}

Key: B = Black Feathers     W= White Feathers


♦♦ Mean mark 44%.

BIOLOGY, M8 2019 HSC 29

Describe ONE mechanism by which plants maintain internal water homeostasis.   (3 marks)

Show Answers Only
  • The widening and reduction of the size of stomata is one mechanism which controls the water movement out of the leaf; the wider the stomata, the more water will leave the plant via transpiration.
  • The width of the stomata is controlled by both environmental conditions and hormones within the plant.
  • When internal water is low, the stress hormone abscisic acid is released, causing the stomata to close.
Show Worked Solution
  • The widening and reduction of the size of stomata is one mechanism which controls the water movement out of the leaf; the wider the stomata, the more water will leave the plant via transpiration.
  • The width of the stomata is controlled by both environmental conditions and hormones within the plant.
  • When internal water is low, the stress hormone abscisic acid is released, causing the stomata to close.

♦♦♦ Mean mark 23%.

BIOLOGY, M5 2019 HSC 28

Huntington's disease is an autosomal dominant condition caused by a mutation of a gene on chromosome 4. It causes nerve cells to break down.

Stargardt disease is an autosomal recessive condition caused by a mutation of a different gene on chromosome 4 . It causes damage to the retina.

A patient is heterozygous for both Huntington's (Hh) and Stargardt disease (Rr). His father's extended family has numerous cases of both of these diseases. His mother does not have either disease and is homozygous for both genes.

  1. Complete the tables, showing the TWO alleles the patient inherited from each parent.   (2 marks)
     
    \begin{aligned}
    &\begin{array}{|l|}
    \hline \rule{0pt}{2.5ex}\quad \quad \textit{ Alleles from father }\quad \quad \rule[-1ex]{0pt}{0pt} \\
    \hline \rule{0pt}{2.5ex}\text{} \rule[-1ex]{0pt}{0pt}\\
    \hline
    \end{array}
    &\begin{array}{|c|}
    \hline \rule{0pt}{2.5ex}\quad \quad \textit{Alleles from mother } \quad \quad \rule[-1ex]{0pt}{0pt}\\
    \hline \rule{0pt}{2.5ex}\text{}\rule[-1ex]{0pt}{0pt}\\
    \hline
    \end{array}
    \end{aligned}
     
  2. The diagram shows the patient's homologous pair of chromosome 4 at various stages of meiosis.
  3. Add the relevant alleles to the diagram to model the production of possible gamete combinations. Include a key and an example of crossing over.   (4 marks)
     

Show Answers Only

a.     

\begin{array} {|c|c|}\hline \text{Alleles from father} & \text{Alleles from mother} \\
\hline \text{H, r} & \text{h, R} \\ \hline \end{array}

b.

Show Worked Solution

a.     

\begin{array} {|c|c|}\hline \text{Alleles from father} & \text{Alleles from mother} \\
\hline \text{H, r} & \text{h, R} \\ \hline \end{array}


♦ Mean mark (a) 47%.

b.


♦♦♦ Mean mark (b) 25%.

BIOLOGY, M6 2019 HSC 26

The map shows the percentage of adult indigenous populations able to digest lactose.
 

The ability to digest lactose is due to the presence of an enzyme (lactase) which can metabolise the sugar (lactose) present in milk. The gene responsible for producing lactase is usually permanently switched off at some time between the ages of 2 and 5 years. However, some people remain able to digest lactose throughout their lives.

With reference to evolution and DNA, provide possible reasons for the distribution shown in the map.   (5 marks)

Show Answers Only
  • The ability to digest lactose is most likely due to a mutation where the gene is not shut off after age 5, and the body continues to produce lactase.
  • The differentiation in the graph above can also be viewed as the frequency of this mutation in certain areas.
  • In areas such as Europe, where over 90% of the adult indigenous population can digest lactose, this mutation is highly prevalent, suggesting it may have played a role in survival advantage.
  • In these areas, milk may make up a high portion of the diet and therefore the ability to digest lactose would prove advantageous and the resulting selective pressure would see it passed on (natural selection).
  • The opposite is true when looking at countries such as Australia where < 20% of indigenous adults can digest lactose.
  • We can deduce that milk is not essential to their traditional diet as the mutation is negligible and not passed on (i.e. it doesn’t constitute a survival advantage).
Show Worked Solution
  • The ability to digest lactose is most likely due to a mutation where the gene is not shut off after age 5, and the body continues to produce lactase.
  • The differentiation in the graph above can also be viewed as the frequency of this mutation in certain areas.
  • In areas such as Europe, where over 90% of the adult indigenous population can digest lactose, this mutation is highly prevalent, suggesting it may have played a role in survival advantage.
  • In these areas, milk may make up a high portion of the diet and therefore the ability to digest lactose would prove advantageous and the resulting selective pressure would see it passed on (natural selection).
  • The opposite is true when looking at countries such as Australia where < 20% of indigenous adults can digest lactose.
  • We can deduce that milk is not essential to their traditional diet as the mutation is negligible and not passed on (i.e. it doesn’t constitute a survival advantage).

♦♦ Mean mark 33%.

Algebra, STD1 A3 2022 HSC 10 MC

The diagram shows a container, closed at the base. It is to be filled with water at a constant rate.
 

 

Which graph best shows the depth of water in the container as time varies?
 


  

Show Answers Only

`A`

Show Worked Solution

The container will fill at a constant rate for the cylindrical section of the container → a straight line (linear graph).

The container will fill more slowly at a decreasing rate for the conical section of the container → a curved line (non-linear graph).

Therefore, `A` and `B` are the only options.

It cannot be graph `B` as it shows the depth increasing at an increasing rate after the straight line section.

`=>A`


♦♦♦ Mean mark 24%.

BIOLOGY, M7 2019 HSC 33d

Alzheimer's disease causes destruction of brain tissue, dementia and eventually death.

The diagram shows the effect of Alzheimer's disease on the brain.

Amyloid beta protein is produced in the human brain throughout life. In people with Alzheimer's disease, it accumulates in excessive amounts.

The gene with the greatest known effect on the risk of developing late-onset Alzheimer's disease is called APOE. It is found on chromosome 19.

The APOE gene has multiple alleles, including e2, e3 and e4 .

The table shows the risk of developing Alzheimer's disease for various APOE genotypes compared to average risk in the population.
 

  

A large epidemiological study was conducted. It used historical data to investigate the association between Herpes simplex virus (HSV) infection and dementia. Dementia is caused by a variety of brain illnesses. Alzheimer's disease is the most common cause of dementia.

The study used the records of 8362 patients with HSV infection and 25086 randomly selected sex- and age-matched control patients without HSV infection. Some of the patients with HSV had been treated with antiviral medication.

The graph below shows some results of the study.
 

Diseases are classified as infectious or non-infectious.

Evaluate whether Alzheimer's disease should be classified as an infectious disease or a non-infectious disease. In your answer, include reference to the information and data provided above.   (8 marks)

Show Answers Only

Infectious vs non-infection disease classification

  • Infectious diseases are a result of pathogens, biological agents of disease, which transmit disease between hosts. A pathogen is a cause of a certain disease if it meets the criteria in Koch’s postulates.
  • The study above shows the association between HSV and Alzheimer’s.
  • The study is conducted over a long period and includes a large and controlled sample size, so the assumption can be made that the findings are valid.
  • HSV is an infectious disease as it is caused by a pathogen, the virus Herpes simplex.
  • The findings show that treating HSV with antiviral medication also reduces the risk of developing Alzheimer’s.
  • This may mean that Alzheimer’s is also a virus and therefore an infectious disease. 
  • Non-infectious diseases are not contagious and do not spread from person to person. They are a result of environmental factors or genetic conditions.
  • Alzheimer’s is the result of a build-up of the amyloid beta protein, which is produced in the brain.
  • The synthesis of this protein is regulated by the APOE gene. This gene also has various alleles, each of which in different combinations can increase or decrease a individual’s risk of developing Alzheimer’s.
  • This indicates that Alzheimer’s is a non-infectious disease, as it is not transmitted by a pathogen. Rather, it results from a natural build up of a specific protein, which may be accelerated or reduced based on genotype. 

Conclusion

  • From the information provided it is not possible to accurately classify Alzheimer’s as either an infectious or non-infectious disease.
  • There is evidence to support that the risk of developing Alzheimer’s can be linked to both antiviral and virus traits as well as genotype.
Show Worked Solution

Infectious vs non-infection disease classification

  • Infectious diseases are a result of pathogens, biological agents of disease, which transmit disease between hosts. A pathogen is a cause of a certain disease if it meets the criteria in Koch’s postulates.
  • The study above shows the association between HSV and Alzheimer’s.
  • The study is conducted over a long period and includes a large and controlled sample size, so the assumption can be made that the findings are valid.
  • HSV is an infectious disease as it is caused by a pathogen, the virus Herpes simplex.
  • The findings show that treating HSV with antiviral medication also reduces the risk of developing Alzheimer’s.
  • This may mean that Alzheimer’s is also a virus and therefore an infectious disease. 
  • Non-infectious diseases are not contagious and do not spread from person to person. They are a result of environmental factors or genetic conditions.
  • Alzheimer’s is the result of a build-up of the amyloid beta protein, which is produced in the brain.
  • The synthesis of this protein is regulated by the APOE gene. This gene also has various alleles, each of which in different combinations can increase or decrease a individual’s risk of developing Alzheimer’s.
  • This indicates that Alzheimer’s is a non-infectious disease, as it is not transmitted by a pathogen. Rather, it results from a natural build up of a specific protein, which may be accelerated or reduced based on genotype. 

Conclusion

  • From the information provided it is not possible to accurately classify Alzheimer’s as either an infectious or non-infectious disease.
  • There is evidence to support that the risk of developing Alzheimer’s can be linked to both antiviral and virus traits as well as genotype.

♦♦ Mean mark 46%.

BIOLOGY, M6 2019 HSC 25

A human karyotype that shows evidence of chromosomal mutation is shown.

  1. Identify the evidence of chromosomal mutation in the karyotype.  (1 mark)

  2. Explain how cell division and fertilisation could lead to the production of this karyotype.  (4 marks)

Show Answers Only

a.   There is only one copy of the sex chromosome (Monosomy, Turner Syndrome).

b.    Process of producing karyotype:

  • During meiosis to produce a gamete, homologous chromosomes (during meiosis I) or sister chromatids (during meiosis II) may fail to separate.
  • This can occur with sex chromosomes, producing one gamete with 2 copies of the sex chromosome, and one without.
  • When the gamete without any sex cells undergoes fertilisation with a normal gamete with only one sex cell, the zygote will have only one sex cell, like the karyotype above.
Show Worked Solution

a.   There is only one copy of the sex chromosome (Monosomy, Turner Syndrome).

b.    Process of producing karyotype:

  • During meiosis to produce a gamete, homologous chromosomes (during meiosis I) or sister chromatids (during meiosis II) may fail to separate.
  • This can occur with sex chromosomes, producing one gamete with 2 copies of the sex chromosome, and one without.
  • When the gamete without any sex cells undergoes fertilisation with a normal gamete with only one sex cell, the zygote will have only one sex cell, like the karyotype above.

♦ Mean mark (a) 47%
♦♦♦ Mean mark (b) 24%.

Vectors, EXT2 V1 2022 HSC 9 MC

Let \(A\) and \(B\) be two distinct points in three-dimensional space. Let \(M\) be the midpoint of \(A B\).

Let \(S_1\) be the set of all points \(P\) such that  \(\overrightarrow{AP} \cdot \overrightarrow{BP}=0\). 

Let \(S_2\) be the set of all points \(N\) such that \(\Big|\overrightarrow{AN}\Big|=\Big| \overrightarrow{MN} \Big| \).

The intersection of \(S_1\) and \(S_2\) is the circle \(S\).

What is the radius of the circle \(S\) ?

  1. \(\dfrac{\Big| \overrightarrow{AB} \Big|}{2} \)
  2. \(\dfrac{\Big| \overrightarrow{AB} \Big|}{4} \)
  3. \(\dfrac{\sqrt3 \Big| \overrightarrow{AB} \Big|}{2} \)
  4. \(\dfrac{\sqrt3 \Big| \overrightarrow{AB} \Big|}{4} \)
Show Answers Only

\(D\)

Show Worked Solution

`text{Diagram below is a 2-D sliced image of the 3-D geometry:}`

\(\overrightarrow{AP} \cdot \overrightarrow{BP}=0\). 

\(r=\dfrac{\Big| \overrightarrow{AB} \Big|}{2} \)

\(S_2\ \text{includes}\ N\ \text{where}\ \Big| \overrightarrow{AN} \Big|=\Big| \overrightarrow{MN} \Big| = \dfrac{r}{2} \)  

\(\text{Let}\ r_s=\ \text{radius of}\ S\)

\(\text{Point}\ P\ \text{is intersection of}\ S_1\ \text{and}\ S_2 \)

\(\text{By Pythagoras (see diagram):}\)

\(r_s^2\) \(=r^2-(\dfrac{r}{2})^2 \)  
  \(=\dfrac{3r^2}{4}\)  
\(r_s\) \(=\dfrac{\sqrt3}{2} \times r\)  
  \(=\dfrac{\sqrt3}{2} \cdot \dfrac{\Big| \overrightarrow{AB} \Big|}{2} \)  
  \(=\dfrac{\sqrt3 \Big| \overrightarrow{AB} \Big|}{4} \)  

 
\(=>D\)


♦♦♦ Mean mark 20%.

Complex Numbers, EXT2 N2 2022 HSC 16d

Find all the complex numbers `z_1, z_2, z_3` that satisfy the following three conditions simultaneously.  (3 marks)

`{[|z_(1)|=|z_(2)|=|z_(3)|],[z_(1)+z_(2)+z_(3)=1],[z_(1)z_(2)z_(3)=1]:}`

Show Answers Only

`z_1,z_2,z_3=1,i,-i \ \ text{(in any order)}`

Show Worked Solution

`z_1z_2z_3=1\ \ =>\ \ abs(z_1) abs(z_2) abs(z_3)=1`

`text{Given}\ \ abs(z_1) = abs(z_2) = abs(z_3)`

`=>abs (z_1)^3=1 \ \ => \ \ abs(z_1)=1`

`:.abs(z_1) = abs(z_2) = abs(z_3)=1`
 


♦♦♦ Mean mark 12%.

`text{Consider}\ \ z_1=e^(itheta):`

`1/z_1=e^(-itheta)=bar(e^(itheta))=bar(z)_1`

`text{Similarly,}`

`1/z_2=bar(z)_2, \ 1/z_3=bar(z)_3`

`=>z_1z_2z_3=1`

  
`text{Consider}\ \ z_1z_2:`

`z_1z_2=1/z_3=barz_3`

`text{Similarly,}`

`z_2z_3=1/z_1=barz_1, \ z_1z_3=1/z_2=barz_2`

`z_1z_2+z_1z_3+z_2z_3` `=barz_3+barz_2+barz_1`  
  `=bar(z_1+z_2+z_3)`  
  `=1`  

 
`z_1, z_2,z_3\ \ text{are zeros of polynomial:}`

`z^3-z^2+z-1` `=0`  
`(z-1)(z^2+1)` `=0`  

 
`z_1,z_2,z_3=1,i,-i \ \ text{(in any order)}`

Proof, EXT2 P1 2022 HSC 16c

It is given that for positive numbers `x_(1),x_(2),x_(3),dots,x_(n)` with arithmetic mean `A`,
 

               `(x_(1)xxx_(2)xxx_(3)xx cdots xxx_(n))/(A^(n)) <= 1`    (Do NOT prove this.)

Suppose a rectangular prism has dimensions  `a,b,c`  and surface area `S`.

  1. Show that  `abc <= ((S)/(6))^((3)/(2))`.  (2 marks)

  2. Using part (i), show that when the rectangular prism with surface area `S` is a cube, it has maximum volume.  (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
Show Worked Solution

i.    `text{Show}\ \ abc <= ((S)/(6))^((3)/(2))`

`S=2(ab+bc+ac)`

`A=(ab+bc+ac)/3`

`text{Using given relationship, where}\ x_1=ab, x_2=bc, …`


♦♦♦ Mean mark (i) 26%.
`(ab xx bc xx ca)/((ab+bc+ac)/3)^3` `<=1`  
`(ab xx bc xx ca)` `<=((ab+bc+ac)/3)^3`  
`(abc)^2` `<=((2(ab+bc+ac))/6)^3`  
`abc` `<=(S/6)^(3/2)`  

 

ii.   `text{If prism is a cube}\ \ a=b=c`

`=> V=a^3, \ \ S=6a^2`

`(S/6)^(3/2)=((6a^2)/6)^(3/2)=a^3`
 

`text{In the case of a cube:}`

`V=(S/6)^(3/2)`


♦♦♦ Mean mark (ii) 26%.

`text{Also,}\ \ V<=(S/6)^(3/2)\ \ \ text{(using part (i))}`

`:.\ text{For rectangular prisms with a given surface area, a cube}`

`text{has the maximum volume.}`

Complex Numbers, EXT2 N2 2022 HSC 16a

A square in the Argand plane has vertices

        `5+5i,quad5-5i,quad-5-5i`  and  `-5+5i`.

The complex numbers `z_A=5+i, z_B` and `z_C` lie on the square and form the vertices of an equilateral triangle, as shown in the diagram.
 
 
                 
 
Find the exact value of the complex number `z_B`.  (4 marks)

Show Answers Only

`(5-16/sqrt3)+5i`

Show Worked Solution

`z_A=5+i, \ \ z_B=b+5i, \ \ z_C=c-5i`

`z_B-z_A=(b-5)+4i`

`z_C-z_A=(c-5)-6i`
 


♦♦♦ Mean mark 21%.

`text{Internal angles of equilateral triangle} = pi/3:`

`=>\ (z_B-z_A)\ text{is an anti-clockwise rotation of}\ (z_C-z_A)\ text{by}\ pi/3`
 

`e^(i pi/3)(z_B-z_A)=z_C-z_A`

`(1/2+i sqrt3/2)((b-5)+4i)=(c-5)-6i`

`(b-5)/2+2i+((b-5)sqrt3)/2 i-2sqrt3` `=(c-5)-6i`  
`(b-5-4sqrt3)/2 + i((4+(b-5)sqrt3)/2)` `=(c-5)-6i`  

 
`text{Equating imaginary parts:}`

`(4+(b-5)sqrt3)/2` `=-6`  
`(b-5)sqrt3` `=-16`  
`b-5` `=-16/sqrt3`  
`b` `=5-16/sqrt3`  

 
`:.z_B=(5-16/sqrt3)+5i`

ENGINEERING, PPT 2020 HSC 25b

Members of steel trusses are often connected using metal rivets as shown.
 

Describe the hot working process that the rivet has undergone to create the resulting structure. Support your answer by completing and labelling the diagram of the sectioned rivet.   (3 marks)

Show Answers Only
  • For hot working to be successful, the rivet has to be heated above the recrystallisation temperature of the material.
  • A riveting gun is then used to hot forge the head of the rivet into place.
  • This causes grain flow to form around the head of the rivet, as shown in the image above.

Show Worked Solution
  • For hot working to be successful, the rivet has to be heated above the recrystallisation temperature of the material.
  • A riveting gun is then used to hot forge the head of the rivet into place.
  • This causes grain flow to form around the head of the rivet, as shown in the image above.


♦♦♦ Mean mark 26%.

BIOLOGY, M6 2019 HSC 19-20 MC

Use the following diagram to answer Questions 19-20.

The diagram shows how CRISPR/Cas9 can be used as a new tool for genetic engineering. This technology has dramatically improved scientists' ability to successfully modify genomes.
 

Question 19

What type of structure must Cas9 be?

  1. Enzyme
  2. mRNA
  3. Ribosome
  4. tRNA

 
Question 20

Scientists have been able to use biotechnology to 'cut and paste' DNA for decades.

Why would the new CRISPR/Cas9 technology have improved the scientists' success in cutting DNA of specific genes?

  1. Cas9 is able to combine with specific DNA.
  2. Cas 9 has an active site that cuts target DNA.
  3. gRNA has the same nucleotides as the target DNA.
  4. gRNA has nucleotides complementary to the target DNA.
Show Answers Only

Question 19: `A`

Question 20: `D`

Show Worked Solution

Question 19

  • Cas9 performs a ‘cutting mechanism’, a property of DNA helicase, a known enzyme.
  • Cas9 also contains a recognition site which it uses to perform its tasks, another property of enzymes.

`=>A`
  

Question 20

  • Cas9 has gRNA with complementary bases to the target gene, as displayed in the diagram above.
  • The improvement is due to this feature, as scientists can now use multiple gRNA sequences to cut various amounts of genes and various recognition sites.

`=>D`


♦♦♦ Mean mark (Q20) 23%.

BIOLOGY, M5 2019 HSC 17 MC

The pedigree shows the inheritance of a genetic disorder.
 

Which row of the table correctly identifies the two possible types of inheritance for this disorder?
 

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \textit{Autosomal} \quad &\quad \textit{Autosomal} \quad & \quad \textit{Sex-linked} \quad & \quad \textit{Sex-linked}\quad \\
\textit{dominant} &\textit{recessive} \rule[-1ex]{0pt}{0pt}& \textit{dominant} & \textit{recessive}\\
\hline
\rule{0pt}{2.5ex}\checkmark \rule[-1ex]{0pt}{0pt}& & \checkmark & \\
\hline
\rule{0pt}{2.5ex}\checkmark \rule[-1ex]{0pt}{0pt}& & & \checkmark \\
\hline
\rule{0pt}{2.5ex}& \checkmark \rule[-1ex]{0pt}{0pt}& \checkmark & \\
\hline
\rule{0pt}{2.5ex}& \checkmark \rule[-1ex]{0pt}{0pt}& & \checkmark \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution

By Elimination:

  • If the trait is autosomal recessive, then it would be impossible to not be affected if both parents are affected (aa x aa).
  • This is not the case for the first individual in generation 3 (eliminate C and D).
  • If the trait is sex-linked recessive, then anytime an affected female has male children, they must be affected, as the recessive allele from the mother must couple with the Y chromosome.
  • This is not the case for the first individual in generation 3 (eliminate B).

\(\Rightarrow A\)


♦♦♦ Mean mark 24%.

CHEMISTRY, M8 2019 HSC 19 MC

Compound `text{X}` shows three signals in its \(\ce{^{13}C NMR}\) spectrum.

Treatment of `text{X}` with hot acidified potassium permanganate produces a compound `text{Y}`. Compound `text{Y}` turns blue litmus red.

Compound `text{X}` produces compound `text{Z}` upon reaction with hot concentrated sulfuric acid.

Which of the following correctly identifies compounds `text{X}`, `text{Y}` and `text{Z}`?
 

Show Answers Only

`D`

Show Worked Solution
  • The given information suggests that compound `text{Y}` is a carboxylic acid because it is produced through the oxidation of compound `text{X}`, a primary alcohol, with an oxidising agent and turns blue litmus red.
  • The treatment of compound `text{X}` with hot concentrated sulfuric acid results in a dehydration reaction.
  • In summary, the information provided suggests that compound `text{Y}` is a carboxylic acid and compound `text{X}` is a primary alcohol, and that the treatment of compound `text{X}` with hot concentrated sulfuric acid results in a dehydration reaction.

`=>D`


♦♦♦ Mean mark 33%.

CHEMISTRY, M5 2019 HSC 18 MC

Consider the following equilibrium.

\(\ce{HF(aq) + CF3COO-(aq) \rightleftharpoons F-(aq) + CF3COOH(aq) \ \ \ \ \ \ $K_{eq}$ = 3.80 \times 10^{-4}} \)

Which row of the table correctly identifies the strongest acid and the strongest base in this system?
 

  \(\text{Strongest acid}\) \(\text{Strongest base}\)
A.   \(\ce{CF3OOH(aq)}\) \(\ce{F-(aq)}\)
B. \(\ce{CF3OOH(aq)}\) \(\ce{CF3OO-(aq)}\)
C. \(\ce{HF(aq)}\) \(\ce{F-(aq)}\)
D. \(\ce{HF(aq)}\) \(\ce{CF3OO-(aq)}\)

Show Answers Only

`A`

Show Worked Solution
  • The small size of `K_(eq)` means that there is a higher concentration of reactants than products at equilibrium.
  • This shifts the equilibrium towards the reactants, which means that the reverse reaction is more likely to occur.
  • Because \(\ce{F-}\) is more likely to accept a proton than \(\ce{CF3COO-}\), it is a stronger base.
  • On the other hand, \(\ce{CF3COO-}\) is more likely to donate a proton than \(\ce{HF}\), making it the stronger acid.

`=>A`


♦♦♦ Mean mark 23%.

BIOLOGY, M6 2022 HSC 32

Researchers have identified a gene that determines the inflammatory response of lung cells to infection with a virus. An allele of this gene is associated with increased inflammation and increased chance of death from the virus.

The table shows the percentage presence of the allele in people with different ancestries.
 

Explain how mutation, natural selection, genetic drift and gene flow could have led to these differences in the gene pools of populations with differing ancestry.   (7 marks)

Show Answers Only

Mutation

  • A mutation is a mechanism of change which permanently alters DNA, changing genotype and phenotype of the ‘host’.
  • In this case, a mutation has occurred which has altered the gene responsible for the inflammatory response in the lungs in response to a virus, making death more common for people who contract the virus as the gene is now faulty.
  • Other mechanisms of variation such as natural selection, gene flow and genetic drift, are responsible for the allele frequency distribution.

Natural selection

  • Refers to the process where selection pressures such as predators, climate, or in this case the lung virus, favour certain individuals, which then means that their alleles can be carried into further generations.
  • It is highly likely that the virus was never present in South Asia, hence the high frequency in South Asian people (60.3%); the allele was negligible.
  • The virus most likely was present in Europe, Africa and East Asia (<15%) and was responsible for almost eliminating the allele in these populations.

Gene flow

  • Gene Flow is the movement of alleles into new populations, usually via migration.
  • Interbreeding between the migrated individual in the new population with the allele, in conjunction with processes such as natural selection can cause the allele to be prominent in the new population.
  • The allele could have originated in South Asia, and gene flow may be responsible for the other frequencies, especially in Europe.

Genetic drift

  • This occurs when populations experience drastic changes in allele frequencies due to random chance events.
  • These events, such as a natural disaster or re-establishing a new population, can eliminate or change the allele frequency and do not cater towards any genotype. 
  • It is possible genetic drift is responsible for the comparatively high allele frequency in South Asia, or the relatively low frequency in Africa and East Asia.
Show Worked Solution

Mutation

  • A mutation is a mechanism of change which permanently alters DNA, changing genotype and phenotype of the ‘host’.
  • In this case, a mutation has occurred which has altered the gene responsible for the inflammatory response in the lungs in response to a virus, making death more common for people who contract the virus as the gene is now faulty.
  • Other mechanisms of variation such as natural selection, gene flow and genetic drift, are responsible for the allele frequency distribution.

Natural selection

  • Refers to the process where selection pressures such as predators, climate, or in this case the lung virus, favour certain individuals, which then means that their alleles can be carried into further generations.
  • It is highly likely that the virus was never present in South Asia, hence the high frequency in South Asian people (60.3%); the allele was negligible.
  • The virus most likely was present in Europe, Africa and East Asia (<15%) and was responsible for almost eliminating the allele in these populations.

Gene flow

  • Gene Flow is the movement of alleles into new populations, usually via migration.
  • Interbreeding between the migrated individual in the new population with the allele, in conjunction with processes such as natural selection can cause the allele to be prominent in the new population.
  • The allele could have originated in South Asia, and gene flow may be responsible for the other frequencies, especially in Europe.

Genetic drift

  • This occurs when populations experience drastic changes in allele frequencies due to random chance events.
  • These events, such as a natural disaster or re-establishing a new population, can eliminate or change the allele frequency and do not cater towards any genotype. 
  • It is possible genetic drift is responsible for the comparatively high allele frequency in South Asia, or the relatively low frequency in Africa and East Asia.

♦♦♦ Mean mark 32%.

BIOLOGY, M5 2022 HSC 28c

Outline the ways in which the DNA of prokaryotes and eukaryotes differ.   (3 marks)

Show Answers Only
  • DNA of Prokaryotes is found in loops whereas the DNA of Eukaryotes is linear.
  • DNA of Prokaryotes is found in the nucleoid region whereas the DNA of Eukaryotes is found as multiple chromosomes in the nucleus (membrane).
  • Prokaryotes have extra “non-nucleoid” DNA in form of plasmids, non-coding fragments found within the cytoplasm, whereas Eukaryotes have non-nuclear DNA within mitochondria (animals) or chloroplasts (plants).
  • Prokaryotes contain nucleoid associated proteins which make DNA form looped structures, while Eukaryotic DNA contains histones, proteins which DNA can wrap around to condense into chromatin and chromosomes.
  • Eukaryotes have Introns, sections of DNA which are non-coding for polypeptides, where Prokaryotic DNA does not contain any non-coding regions.
Show Worked Solution
  • DNA of Prokaryotes is found in loops whereas the DNA of Eukaryotes is linear.
  • DNA of Prokaryotes is found in the nucleoid region whereas the DNA of Eukaryotes is found as multiple chromosomes in the nucleus (membrane).
  • Prokaryotes have extra “non-nucleoid” DNA in form of plasmids, non-coding fragments found within the cytoplasm, whereas Eukaryotes have non-nuclear DNA within mitochondria (animals) or chloroplasts (plants).
  • Prokaryotes contain nucleoid associated proteins which make DNA form looped structures, while Eukaryotic DNA contains histones, proteins which DNA can wrap around to condense into chromatin and chromosomes.
  • Eukaryotes have Introns, sections of DNA which are non-coding for polypeptides, where Prokaryotic DNA does not contain any non-coding regions.

♦♦♦ Mean mark 24%.

BIOLOGY, M8 2022 HSC 20 MC

Renal dialysis involves passing blood from a patient past a dialysate solution in order to remove waste such as urea from the blood.

Which diagram correctly shows possible concentrations of urea and the direction of flow of both solutions in a dialysis machine?
 

Show Answers Only

`D`

Show Worked Solution

By Elimination:

  • The blood and dialysate must flow in opposite directions in a dialysis machine to improve diffusion efficiency (eliminate A and C).
  • Initially, the blood will have a high concentration of urea and the dialysate will have a low concentration.

`=>D`


♦♦♦ Mean mark 28%.

ENGINEERING, CS 2020 HSC 9 MC

A cured cylindrical concrete specimen is tested under a compressive load through its vertical axis, until it fails.
 

Which of the following is the most likely shape of the front view of the failed specimen?
 

Show Answers Only

`A`

Show Worked Solution
  • There are 3 common fracture patterns in concrete under compressive testing.
  • Of the options given, `A` is the only one displaying an example of these patterns.

`=>A`


♦♦♦ Mean mark 14%.

ENGINEERING, PPT 2020 HSC 6 MC

Which electronic component can be used to increase or decrease the speed of a train powered by an electric motor?

  1. Diode
  2. Capacitor
  3. Transistor
  4. Potentiometer
Show Answers Only

`D`

Show Worked Solution
  • Potentiometers can act as variable resistors.
  • These can increase or decrease current and therefore adjust the speed of an electric motor.

`=>D`


♦♦♦ Mean mark 20%.

Vectors, EXT2 V1 2022 HSC 14a

  1. The two non-parallel vectors `\vec{u}` and `\vec{v}` satisfy  `lambda vec(u)+mu vec(v)= vec(0)`  for some real numbers `\lambda` and `\mu`.
  2. Show that  `\lambda=\mu=0`.  (2 marks)

  3. The two non-parallel vectors `\vec{u}` and `\vec{v}` satisfy  `\lambda_1 \vec{u}+\mu_1 \vec{v}=\lambda_2 \vec{u}+\mu_2 \vec{v}`  for some real numbers `\lambda_1, \lambda_2, \mu_1` and `\mu_2`.
  4. Using part (i), or otherwise, show that  `\lambda_1=\lambda_2`  and  `\mu_1=\mu_2`. (1  mark)

The diagram below shows the tetrahedron with vertices `A, B, C` and `S`.

The point `K` is defined by  `vec(SK)=(1)/(4) vec(SB)+(1)/(3) vec(SC)`, as shown in the diagram.

The point `L` is the point of intersection of the straight lines `S K` and `B C`.
 
                       
 

  1. Using part (ii), or otherwise, determine the position of `L` by showing that  `vec(BL)=(4)/(7) vec(BC)`.  (2 marks)

  2. The point `P` is defined by  `vec(AP)=-6 vec(AB)-8 vec(AC)`.
  3. Does `P` lie on the line `A L`? Justify your answer.  (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solution)}`
  2. `text{Proof (See Worked Solution)}`
  3. `text{Proof (See Worked Solution)}`
  4. `P\ text{lies on}\ vec(AL)\ \ text{(See Worked Solution)}`
Show Worked Solution
i.    `lambda vecu+mu vecv` `=0`
  `lambda vecu` `=-mu vecv`

 
`lambda=0\ \ text{or}\ \ vecu=-(mu/lambda)vecv=k vecv\ \ (kinRR)`

`text{S}text{ince}\ \ vecu and vecv\ \ text{are not parallel}`

`=> lambda=mu=0`
 

ii.  `\lambda_1 \vec{u}+\mu_1 \vec{v}=\lambda_2 \vec{u}+\mu_2 \vec{v}`

`\lambda_1 \vec{u}-\lambda_2 \vec{u}+\mu_1 \vec{v}-\mu_2 \vec{v}` `=vec0`  
`(\lambda_1-\lambda_2)\vec{u}+(\mu_1-\mu_2)\vec{v}` `=vec0`  

 

`text{Using part (i):}`

`(\lambda_1-\lambda_2)=0 and (\mu_1-\mu_2)=0`

`:.\lambda_1=\lambda_2  and  \mu_1=\mu_2\ …\ text{as required}`


Mean mark (ii) 56%.
iii.   `vec(BL)` `=lambda vec(BC)`
    `=lambda(vec(BS)+vec(SC))\ \ \ …\ (1)`

 

`vec(BL)` `=vec(BS)+mu vec(SK)`  
  `=-vec(SB)+mu(1/4vec(SB)+1/3vec(SC))`  
  `=-vec(SB)+mu/4vec(SB)+mu/3vec(SC)`  
  `=mu/3vec(SC)+(mu/4-1)vec(SB)\ \ \ …\ (2)`  


♦♦♦ Mean mark (iii) 25%.

`text{Using}\ \ (1) = (2):`

`lambda vec(BS)+ lambda vec(SC)=mu/3vec(SC)+(mu/4-1)vec(SB)`

`mu/3=lambda, \ \ 1-mu/4=lambda`

`mu/3` `=1-mu/4`  
`(4mu+3mu)/12` `=1`  
`(7mu)/12` `=1`  
`mu` `=12/7`  

 
`lambda=(12/7)/3=4/7`

`:.vec(BL)=(4)/(7) vec(BC)\ \ text{… as required}`
 

iv.  `vec(AP)=-6 vec(AB)-8 vec(AC)`

`text{If}\ P\ text{lies on}\ AL, ∃k\ text{such that}\ \ vec(AP)=k vec(AL)`


♦♦♦ Mean mark (iv) 23%.
`-6 vec(AB)-8 vec(AC)` `=k vec(AL)`  
`-6 vec(AB)-8 (vec(AB)+vec(BC))` `=k(vec(AB)+vec(BL))`  
`-14 vec(AB)-8vec(BC)` `=k(vec(AB)+4/7vec(BC))\ \ text{(see part (iii))}`  
`-14 vec(AB)-8vec(BC)` `=kvec(AB)+(4k)/7vec(BC)`  

 
`k=-14, \ \ (4k)/7=-8\ \ =>\ \ k=-14`

`:.\ P\ text{lies on}\ \ AL.`

ENGINEERING, CS 2021 HSC 25b

A truss is fixed to a wall at `A` and `B` as shown. Ignore the mass of the truss.
 


 

  1. Determine the horizontal reaction at `A`.   (2 marks)
  1. Determine, using method of joints, the internal reaction in member `AC`. Indicate the nature of the force in the member.   (3 marks)
  1. Determine, using method of sections, the internal reaction in member `CE`. Indicate the nature of the force.   (3 marks)
Show Answers Only

i.    `3.75\ text{kN ←}`

ii.    `AC=4802\ text{N}`

iii.   `CE = 3\ text{kN (tension)}`

Show Worked Solution

i.   Horizontal reaction at `A`

\(\circlearrowright \Sigma \text{M}_\text{B}\) \(= 0\)  
`0` `= -(A_H xx 4\ text{m}) + (1500\ text{N} xx  10\ text{m})`  
`A_H` `= 3750\ text{N}= 3.75\ text{kN ←}`  

 

ii.   Method of joints at A
 
     

→Σ`F_H` `=0`  
`0` `= -3750+AC xx sin(51.34°)`  
`AC` `=3750/sin(51.34°)=4802\ text{N}`  

♦♦ Mean mark (ii) 41%.

iii.  Method of Sections

\(\circlearrowright \Sigma \text{M}_\text{B}\) \(=0\)  
`0` `= (1500 xx 10)-(CE xx 5)`  
`CE` `=(15\ 000)/5=3000\ text{N}=3\ text{kN (tension)}`  

♦♦♦ Mean mark (iii) 30%.

BIOLOGY, M7 2020 HSC 32c

Rabies is a disease that can affect all mammals and is caused by the rabies virus. It is transmitted by the bite of an infected animal. Without treatment it almost always results in death.
 
        

  1.  

The rabies virus is a single-stranded RNA virus. It contains and codes for only five proteins. The diagrams show the structure and reproduction of the virus.
 

 

Post exposure prophylaxis (PEP) is given to patients who have been bitten by a rabid animal.

PEP includes an injection of human rabies antibodies (HRIG) as well as injections of a rabies vaccine at 0, 3, 7 and 14 days after exposure to the virus.

The following graphs show a generalised response to rabies infection without and with PEP.
  

Explain how PEP prevents rabies developing after infection with the virus. Support your answer with reference to the information and data provided above.   (8 marks)

Show Answers Only

Once the rabies virus has entered the wound:

  • It will use the patient’s cells to replicate and the viral concentration will increase (as seen in the first five days).
  • Without PEP the virus will continue to replicate, migrate to the CNS (in the first graph this occurs by day 7), and eventually cause rabies and death.
  • Initially, the infected individual will not have the antibodies required to inactivate the virus.
  • A HRIG injection provides the required antibodies to inactivate the virus, through inhibiting replication or enhancing phagocytosis. 
  •  The PEP graph shows that these antibodies will only last up to 21 days, but are essential in inactivating the initial virus, as seen by a reduction in viral concentration after 6-8 days. 

The rabies vaccine works by:

  • containing an inactivated or weakened version of the rabies virus, which stimulates an immune response by the individual.
  • Initially, macrophages will display an MHC-antigen complex on its surface which helper T lymphocytes will bind to.
  • This then stimulates specific plasma B cells which can produce complementary antibodies, and memory B cells, which stay dormant and can rapidly differentiate into plasma B cells when exposed to the same virus.
  • The PEP graph shows the rapid production of antibodies on day 7, which coincides with rapid decrease in the virus concentration over the next few days. 
  • The antibodies then remain in the bloodstream and slowly decline over months, which allows quick diffusion if the virus is encountered within that timeframe.
Show Worked Solution

Once the rabies virus has entered the wound:

  • It will use the patient’s cells to replicate and the viral concentration will increase (as seen in the first five days).
  • Without PEP the virus will continue to replicate, migrate to the CNS (in the first graph this occurs by day 7), and eventually cause rabies and death.
  • Initially, the infected individual will not have the antibodies required to inactivate the virus.
  • A HRIG injection provides the required antibodies to inactivate the virus, through inhibiting replication or enhancing phagocytosis. 
  •  The PEP graph shows that these antibodies will only last up to 21 days, but are essential in inactivating the initial virus, as seen by a reduction in viral concentration after 6-8 days. 

The rabies vaccine works by:

  • containing an inactivated or weakened version of the rabies virus, which stimulates an immune response by the individual.
  • Initially, macrophages will display an MHC-antigen complex on its surface which helper T lymphocytes will bind to.
  • This then stimulates specific plasma B cells which can produce complementary antibodies, and memory B cells, which stay dormant and can rapidly differentiate into plasma B cells when exposed to the same virus.
  • The PEP graph shows the rapid production of antibodies on day 7, which coincides with rapid decrease in the virus concentration over the next few days. 
  • The antibodies then remain in the bloodstream and slowly decline over months, which allows quick diffusion if the virus is encountered within that timeframe.

♦♦ Mean mark 41%.

BIOLOGY, M8 2020 HSC 31

  1. The levels of glucose, insulin and glucagon were measured in the plasma of 24 healthy adults at intervals over a 5-hour period. After 1 hour at rest the patients ate a large carbohydrate meal. The results are shown.

  2. Use the data provided to explain how blood glucose is controlled in the body.   (6 marks)

  3. Outline how in humans, maintenance of temperature is different to the way that glucose is controlled.   (3 marks)

Show Answers Only

a.   Glucose Levels

  • Plasma levels within the first 60 minutes represent the resting glucose levels of the individuals.
  • After the 1 hour mark, the rise in glucose is a result of the absorption of glucose into the bloodstream from the gut after eating the carb rich meal.

Insulin Levels

  • Insulin is a hormone secreted by beta cells in the pancreas. It causes excess glucose in the bloodstream to be taken by the liver and stored as glycogen.
  • This can be seen when the insulin levels directly correlate to the amount of glucose in the body; glucose absorbed in the meal stimulates release of insulin.
  • The reduction in glucose by the insulin then causes insulin to also fall.

Glucagon Levels

  • Glucagon is a hormone secreted by alpha cells in the pancreas. It can almost be seen as the opposite of insulin, and forms the negative feedback loop responsible for controlling blood glucose. 
  • It causes glycogen stores in the liver to decompose into glucose and be absorbed by the bloodstream when blood glucose levels drop too low.
  • When glucose levels rise between the 1-2 hour mark, glucagon levels drop significantly, as the build up of glucose from the meal has meant that glycogen stores are not needed.
  • Gradually, as glucose levels drop, glucagon levels will increase as the glucose from the meal is depleted.

b.  Differences in temperature vs glucose maintenance

  • Temperature changes are detected by the hypothalamus and sensory neurons, but changes in blood glucose levels (BGL) are detected by the pancreas.
  • Responses are carried out by the nervous system when temperature changes are detected, but carried out by hormones when changes in BGL are detected.
Show Worked Solution

a.   Glucose Levels

  • Plasma levels within the first 60 minutes represent the resting glucose levels of the individuals.
  • After the 1 hour mark, the rise in glucose is a result of the absorption of glucose into the bloodstream from the gut after eating the carb rich meal.

Insulin Levels

  • Insulin is a hormone secreted by beta cells in the pancreas. It causes excess glucose in the bloodstream to be taken by the liver and stored as glycogen.
  • This can be seen when the insulin levels directly correlate to the amount of glucose in the body; glucose absorbed in the meal stimulates release of insulin.
  • The reduction in glucose by the insulin then causes insulin to also fall.

Glucagon Levels

  • Glucagon is a hormone secreted by alpha cells in the pancreas. It can almost be seen as the opposite of insulin, and forms the negative feedback loop responsible for controlling blood glucose. 
  • It causes glycogen stores in the liver to decompose into glucose and be absorbed by the bloodstream when blood glucose levels drop too low.
  • When glucose levels rise between the 1-2 hour mark, glucagon levels drop significantly, as the build up of glucose from the meal has meant that glycogen stores are not needed.
  • Gradually, as glucose levels drop, glucagon levels will increase as the glucose from the meal is depleted.

♦ Mean mark (a) 48%.

b.  Differences in temperature vs glucose maintenance

  • Temperature changes are detected by the hypothalamus and sensory neurons, but changes in blood glucose levels (BGL) are detected by the pancreas.
  • Responses are carried out by the nervous system when temperature changes are detected, but carried out by hormones when changes in BGL are detected.

♦♦ Mean mark (b) 36%.

BIOLOGY, M5 2020 HSC 28

  1. A student drew a diagram to model part of the process of meiosis.
     

  1. Explain the misunderstanding of meiosis shown in this model.   (3 marks)

  2. Explain the effect of meiosis on genetic variation.   (3 marks)

Show Answers Only

a.
 
       
  

  • In the diagram above, the paired homologous chromosomes are incorrectly drawn.
  • Prior to crossing over, each chromosome duplicates forming the sister chromatids with one being maternal and one being paternal.
  • In the model above, they are shown as a mix of both, when in actuality the sister chromatids should be identical.
     

b.   The Effect of Meiosis on Genetic Variation

Independent Assortment

  • when chromosomes are lined up along the cells equator in independent order and orientation to all other chromosomes.

Random Segregation  

  • different combinations of maternal and paternal chromosomes end up in resulting gametes, increasing variability amongst them.

Crossing Over

  • the process of exchanging genetic material between chromatids of homologous chromosomes during Meiosis I, leading to unique combinations of alleles on each chromatid.
Show Worked Solution

a.
 
       

  

  • In the diagram above, the paired homologous chromosomes are incorrectly drawn.
  • Prior to crossing over, each chromosome duplicates forming the sister chromatids with one being maternal and one being paternal.
  • In the model above, they are shown as a mix of both, when in actuality the sister chromatids should be identical.

♦♦♦ Mean mark (a) 25%.

b.   The Effect of Meiosis on Genetic Variation

Independent Assortment

  • when chromosomes are lined up along the cells equator in independent order and orientation to all other chromosomes.

Random Segregation  

  • different combinations of maternal and paternal chromosomes end up in resulting gametes, increasing variability amongst them.

Crossing Over

  • the process of exchanging genetic material between chromatids of homologous chromosomes during Meiosis I, leading to unique combinations of alleles on each chromatid.

♦ Mean mark (b) 43%.

CHEMISTRY, M6 2020 HSC 33

Excess solid calcium hydroxide is added to a beaker containing 0.100 L of 2.00 mol L¯1 hydrochloric acid and the mixture is allowed to come to equilibrium.

  1. Show that the amount (in mol) of calcium hydroxide that reacts with the hydrochloric acid is 0.100 mol.   (2 marks)

  2. It is valid in this instance to make the simplifying assumption that the amount of calcium ions present at equilibrium is equal to the amount generated in the reaction in part (a).
  3. Calculate the pH of the resulting solution.   (4 marks)

Show Answers Only

a.   \(\ce{Ca(OH)2 (s) + 2 HCl (aq) -> CaCl2 (aq) + 2 H2O (l)}\)

`text{n(HCl)} = text{c} xx text{V} = 2.00 xx 0.100  = 0.200\ text{mol}`

\[\ce{n(Ca(OH)2) = \frac{n(HCL)}{2} = \frac{0.200}{2}= 0.100 mol}\]  

b.   `text{pH} = 11.35`

Show Worked Solution

a.   \(\ce{Ca(OH)2 (s) + 2 HCl (aq) -> CaCl2 (aq) + 2 H2O (l)}\)

`text{n(HCl)} = text{c} xx text{V} = 2.00 xx 0.100  = 0.200\ text{mol}`

\[\ce{n(Ca(OH)2) = \frac{n(HCL)}{2} = \frac{0.200}{2}= 0.100 mol}\]  

b.   \(\ce{Ca(OH)2(s) \rightleftharpoons Ca^2+ (aq) + 2 OH– (aq)}\)

`[text{Ca}^(2+)] = text{n} / text{V} = 0.100 / 0.100 = 1.00\ text{mol L}^-1`
 

\(\ce{K_{sp}}\) \( \ce{= [Ca^2+][OH– ]^2}\)  
`5.02 xx 10^(-6)` `= 1.00 xx [text{OH}^– ]^2`  
`[text{OH}^– ]` `=sqrt{5.02 xx 10^(-6)}=2.24 xx 10^(−3)\ text{mol L}^(-1)`  
`text{pOH }` `= −log_10(2.24 xx 10^(-3))= 2.650`  

 
`:.\ text{pH} = 14-2.650 = 11.35`


♦♦♦ Mean mark (b) 20%.

PHYSICS, M6 2018 HSC 18 MC

An experiment is set up as shown.
 

When the switch is closed, the reading on the spring balance changes immediately, then returns to the initial reading.

Which row of the table correctly shows the direction of the current through the straight conductor \(XY\) and the direction in which the pointer on the spring balance initially moves?

\begin{align*}
\begin{array}{l}
\textit{}& \textit{} \\
\textit{}& \textit{} \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\textit{Direction of current through}& \textit{Direction in which the} \\
\textit{the straight conductor}& \textit{pointer initially moves} \\
\hline
\rule{0pt}{2.5ex}\text{From \(X\) to \(Y\)}\rule[-1ex]{0pt}{0pt}&\text{Down}\\
\hline
\rule{0pt}{2.5ex}\text{From \(X\) to \(Y\)}\rule[-1ex]{0pt}{0pt}& \text{Up}\\
\hline
\rule{0pt}{2.5ex}\text{From \(Y\) to \(X\)}\rule[-1ex]{0pt}{0pt}& \text{Down} \\
\hline
\rule{0pt}{2.5ex}\text{From \(Y\) to \(X\)}\rule[-1ex]{0pt}{0pt}& \text{Up} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(B\)

Show Worked Solution
  • Using the right hand grip rule, when the switch is closed the current through the lower solenoid produces a south pole at its top.
  • So, a south pole is induced at the bottom of the solenoid connected to the spring balance (Lenz’s Law).
  • Using the right hand grip rule a second time. The current through the straight conductor goes from \(X\) to \(Y\).
  • As the two solenoids repel each other, the force on the higher solenoid counteracts its weight and the pointer moves up.

\(\Rightarrow B\)


♦♦♦ Mean mark 34%.

PHYSICS, M7 2018 HSC 16 MC

When a train is at rest in a tunnel, the train is slightly longer than the tunnel.
 

In a thought experiment, the train is travelling from left to right fast enough relative to the tunnel that its length contracts and it fits inside the tunnel.

An observer on the ground sets up two cameras, at `X` and `Y`, to take photos at exactly the same time. The photos show that both ends of the train are inside the tunnel.
 

A passenger travelling on the train at its centre can see both ends of the tunnel and is later shown the photos.

From the point of view of the passenger, what is observed and what can be deduced about the photos?

  1. The tunnel's length contracts so the train does not fit, and photo 2 is taken before photo 1.
  2. The tunnel's length contracts so the train does not fit, and photos 1 and 2 are taken at the same time.
  3. The tunnel appears to expand due to length contraction of the train, allowing it to fit in the tunnel, and photo 1 is taken before photo 2 .
  4. The tunnel appears to expand due to length contraction of the train, allowing it to fit in the tunnel, and photos 1 and 2 are taken at the same time.
Show Answers Only

`A`

Show Worked Solution
  • From the frame of reference of an observer on the train, the tunnel’s length will contract (all objects outside the train will appear length contracted).
  • Due to the relativity of simultaneity, the photo that the observer on the train sees as taken first is the photo taken at the end of the tunnel that the train was moving towards at that moment.

`=>A`


♦♦♦ Mean mark 36%.

ENGINEERING, PPT 2021 HSC 24d

A drive mechanism used to control the satellite dish of a mobile TV transmission van is shown.

Key dimensions of components are given.
 


 

The dimensions of each of the five holes in the gear are  \( \phi 10 \vee \phi 16 \times 90°\) on one side.

A partially sectioned drawing of the gear and shaft is provided below.

Complete the sectioned assembly of this drive mechanism to AS 1100 from the direction indicated by the arrow. Include break lines where appropriate.   (6 marks)
 

Show Answers Only

Show Worked Solution

Critical aspects of this drawing:

  • Do not section wedges
  • Do not section counterbores or holes
  • Break line must be included
  • Section lines of adjacent areas should go in different directions
  • Section lines should be at 45° and 3mm apart
  • Answer may be drawn half sectioned as it is symmetrical

♦♦ Mean mark 41%.

CHEMISTRY, M5 2020 HSC 19 MC

Nitrogen dioxide reacts to form dinitrogen tetroxide in a sealed flask according to the following equation.

`2 text{NO}_(2)(g) ⇌ text{N}_(2) text{O}_(4)(g) qquadqquad Delta H =-57.2 \ text{kJ  mol}^(-1)`

Which graph best represents the rates of both the forward and reverse reactions when an equilibrium system containing these gases is cooled at time `t` ?
 

 

Show Answers Only

`D`

Show Worked Solution
  • When the temperature decreases, both the forward and reverse reaction rates will decrease (eliminate A).
  • Exothermic reactions have a lower activation energy threshold (i.e. their reaction rates are less affected by cooling).
  • Therefore, the exothermic reaction rate will decrease to a smaller extent (eliminate B and C).

`=> D`


♦♦♦ Mean mark 22%.

CHEMISTRY, M6 2020 HSC 18 MC

An aqueous solution of sodium hydrogen carbonate has a pH greater than 7 .

Which statement best explains this observation?

  1. `text{H}_(2) text{O}(l)` is a stronger acid than `text{HCO}_(3)^(\ -)(aq)`.
  2. `text{HCO}_(3)^(\ -)(aq)` is a weaker acid than `text{H}_(2) text{CO}_(3)(aq)`.
  3. `text{Na}^(+)(aq)` reacts with water to produce the strong base `text{NaOH}(aq)`.
  4. The conjugate acid of `text{HCO}_(3)^(\ -)(aq)` is a stronger acid than `text{H}_(2)text{O}(l)`.
Show Answers Only

`A`

Show Worked Solution
  • \(\ce{HCO3-}\) can act as an acid or a base (amphiprotic).
  • \(\ce{H2O}\) is a stronger acid than \(\ce{HCO3-}\), so \(\ce{H2O}\) donates a proton to \(\ce{HCO3-}\).
  • As a result, this produces \(\ce{OH-}\) ions in solution, causing the pH to be greater than 7.

`=> A`


♦♦♦ Mean mark 23%.

CHEMISTRY, M8 2022 HSC 33

Analyse how a student could design a chemical synthesis process to be undertaken in the school laboratory. In your response, use a specific process relating to the synthesis of an organic compound, including a chemical equation, and refer to:

  • selection of reagent(s)
  • reaction conditions
  • any potential hazards and any safety precautions to minimise the risk
  • yield and purity of the product(s).   (8 marks)
Show Answers Only

Selecting reagents

  • The student could synthesise ethyl ethanoate through esterification between acetic acid and ethanol.
  • Both readily available in the school laboratory and are relatively safe.

Acetic acid + ethanol  ⇌  Ethyl ethanoate + water

  • Concentrated sulfuric acid should be used as the acid catalyst as it is a strong acid that is also readily available. 

Reaction conditions

  • Increasing the temperature of the system increases the reaction rate because it increases the average kinetic energy of the reactant molecules, and thus increases the likelihood of successful collisions, producing more product.
  • Additionally, the addition of reactants would increase the likelihood of successful collisions, thus increasing the reaction rate.
  • The reaction should also be undertaken under reflux allowing vaporised molecules to condense and return back to the reaction vessel, increasing the amount of reactants, and thus increasing the rate of reaction. 
  • Concentrated sulfuric acid should also be utilised as a catalyst in order to speed up the reaction and lower the activation energy. 

Potential hazards and safety precautions

  • The acetic acid and sulfuric acid used is corrosive and may cause skin and eye burns, therefore, appropriate lab coat and safety glasses should be utilised.
  • The organic reactants are highly flammable and may cause fires. The reaction mixture should be heated on a hot plate or heating mantle instead of a bunsen burner.
  • Refluxing may cause pressure build-up, therefore, ensure the reflux condenser is open.
  • Superheating and bumping may occur in apparatus. Boiling chips should be utilised to provide nucleation sites allowing liquids to boil smoothly. 

Yield and purity

  • Concentrated sulfuric acid, used as a catalyst, also acts as a dehydrating agent that removes water from the system and improves yield.
  • When the reaction reaches equilibrium, the ester can be separated from the mixture by adding excess sodium carbonate solution in order to neutralise the acid.
  • Transfer to a separation funnel to separate the organic layer (containing the ester) from the aqueous layer.
  • Then use fractional distillation to separate the ester from the organic layer.
Show Worked Solution

Selecting reagents

  • The student could synthesise ethyl ethanoate through esterification between acetic acid and ethanol.
  • Both readily available in the school laboratory and are relatively safe.

Acetic acid + ethanol  ⇌  Ethyl ethanoate + water

  • Concentrated sulfuric acid should be used as the acid catalyst as it is a strong acid that is also readily available. 

Reaction conditions

  • Increasing the temperature of the system increases the reaction rate because it increases the average kinetic energy of the reactant molecules, and thus increases the likelihood of successful collisions, producing more product.
  • Additionally, the addition of reactants would increase the likelihood of successful collisions, thus increasing the reaction rate.
  • The reaction should also be undertaken under reflux allowing vaporised molecules to condense and return back to the reaction vessel, increasing the amount of reactants, and thus increasing the rate of reaction. 
  • Concentrated sulfuric acid should also be utilised as a catalyst in order to speed up the reaction and lower the activation energy. 

Potential hazards and safety precautions

  • The acetic acid and sulfuric acid used is corrosive and may cause skin and eye burns, therefore, appropriate lab coat and safety glasses should be utilised.
  • The organic reactants are highly flammable and may cause fires. The reaction mixture should be heated on a hot plate or heating mantle instead of a bunsen burner.
  • Refluxing may cause pressure build-up, therefore, ensure the reflux condenser is open.
  • Superheating and bumping may occur in apparatus. Boiling chips should be utilised to provide nucleation sites allowing liquids to boil smoothly. 

Yield and purity

  • Concentrated sulfuric acid, used as a catalyst, also acts as a dehydrating agent that removes water from the system and improves yield.
  • When the reaction reaches equilibrium, the ester can be separated from the mixture by adding excess sodium carbonate solution in order to neutralise the acid.
  • Transfer to a separation funnel to separate the organic layer (containing the ester) from the aqueous layer.
  • Then use fractional distillation to separate the ester from the organic layer.

♦♦ Mean mark 52%.

CHEMISTRY, M6 2022 HSC 32

The concentration of citric acid, a triprotic acid, in a carbonated soft drink was to be determined.

Step 1: A solution of \( \ce{NaOH(aq)} \) was standardised by titrating it against 25.00 mL aliquots of a solution of the monoprotic acid potassium hydrogen phthalate \( \ce{(KHP)} \). The \( \ce{(KHP)} \) solution was produced by dissolving 4.989 g in enough water to make 100.0 mL of solution. The molar mass of \( \ce{(KHP)} \) is 204.22 g mol ¯1.

The results of the standardisation titration are given in the table.
 

Step 2: A 75.00 mL bottle of the drink was opened and the contents quantitatively transferred to a beaker. The soft drink was gently heated to remove \( \ce{CO2}\).

Step 3: The cooled drink was quantitatively transferred to a 250.0 mL volumetric flask and distilled water was added up to the mark.

Step 4: 25.00 mL samples of the solution were titrated with the \( \ce{NaOH(aq)}\) solution. The average volume of \( \ce{NaOH(aq)} \) used was 13.10 mL.

  1. Calculate the concentration of the triprotic citric acid in the soft drink.  (6 marks)
  1. Explain how your answer to part (a) would be different if the carbon dioxide was not removed from the soft drink.  (2 marks)
Show Answers Only

a.  `0.1298  text{mol L}^(–1)`

b.  \( \ce{CO2} \) can dissolve in water to produce \( \ce{H2CO3}\):

\( \ce{CO2(g) + H2O(l) \rightleftharpoons H2CO3(aq)} \)

  • This would enable \( \ce{NaOH} \) to react with \( \ce{H2CO3:} \)
  •    \( \ce{2NaOH(aq) + H2CO3(aq) -> Na2CO3(aq) + 2H2O(l)} \)
  • Therefore, if \( \ce{CO2} \) was not removed, more \( \ce{NaOH} \) would be required to reach the endpoint.
  • This would result in a higher citric acid concentration calculation.
Show Worked Solution

a.   \( \ce{KHP(aq) + NaOH(aq) -> NaKP(aq) + H2O(l)} \)

`text{n(HX)}= 4.989 / 204.22= 0.02443\ text{mol}`

`[text(HX)]= \text{n}/\text{V}= 0.02443 / 0.1000= 0.2443  text{mol L}^(–1)`

`text{n(HX) titrated} =  text{c} xx text{V}= 0.2443 xx 0.02500= 0.0006107\ text{mol}`

 `=>\ text{n(NaOH)}= 0.0006107  text{mol}`
 

Eliminate the first trial because it is an outlier.

`text{V}_(text(avg))text{(NaOH)}= 1 / 3 xx (27.40 + 27.20 + 27.60)= 27.40\ text{mL}`

`text{[NaOH]}= text{n} / text{V}= [6.107 xx 10^−3] / 0.02740 = 0.2229  text{mol L}^(–1)`

\( \ce{H3X(aq) + 3NaOH(aq) -> Na3X(aq) + 3H2O(l)} \)

`text{n(NaOH) titrated}= text{c} xx text{V}=0.2229 xx 0.01310= 2.920 xx 10^(−3)\ text{mol}`

`text{n(H}_3 text{X)}= 1/3 xx 2.920 xx 10^(−3)= 9.733 xx 10^(−4)\ text{mol}`

`text{[H}_3 text{X] diluted} = text{n} / text{V} = (9.733 xx 10^(−4) )/ 0.025= 0.03893 \ text{mol L}^(–1)`

`text{[H}_3 text{X] original}= 250.0 / 75.00 xx 0.03893= 0.1298  text{mol L}^(–1)`

Therefore, the concentration of citric acid in the soft drink is 0.1298 mol L¯1.


♦ Mean mark (a) 48%.

b.   \( \ce{CO2} \) can dissolve in water to produce \( \ce{H2CO3}\):

\( \ce{CO2(g) + H2O(l) \rightleftharpoons H2CO3(aq)} \)

This would enable \( \ce{NaOH} \) to react with \( \ce{H2CO3:} \)

\( \ce{2NaOH(aq) + H2CO3(aq) -> Na2CO3(aq) + 2H2O(l)} \)

  • Therefore, if \( \ce{CO2} \) was not removed, more \( \ce{NaOH} \) would be required to reach the endpoint.
  • This would result in a higher citric acid concentration calculation.

♦♦ Mean mark (b) 28%.

CHEMISTRY, M7 2022 HSC 18 MC

A low molecular weight biopolymer is being investigated for its suitability for medical use. In one trial a molecular weight of `2900 pm 100\ text{g}\ text{mol}^(-1)` proved to be optimum.

A section of this biopolymer is shown.
 

Which will produce the suitable biopolymer?
 

Show Answers Only

`D`

Show Worked Solution

This polymer is a condensation polymer, meaning that it is formed through the reaction between monomers that consist of a carboxylic acid and/or an alcohol functional group, with the elimination of water.

Thus, if `text{n}` monomers react to form this polyester, `text{(n – 1)}` molecules of water would be eliminated.

Calculating their molar masses:

`text{n}\ xx\ text{monomers}` `= text{polymer} + text{(n – 1)} xx text{H}_2 text{O}`
`text{n} xx (90.078)` `= 2900 + text{(n – 1)} xx (18.016)`
`text{n}(90.078-18.016)` `=2900-18.016`
`:.\ text{n}` `=2881.984/72.062=39.99…`

 
`=> D`


♦♦ Mean mark 23%.

CHEMISTRY, M5 2022 HSC 17 MC

A 2.0 g sample of silver carbonate (MM = 275.81 g mol ¯1) was added to 100.0 mL of water in a beaker. The solubility of silver carbonate at this temperature is `1.2 × 10^(-4)` mol L ¯1. It was then diluted by adding another 100.0 mL of water.

What is the ratio of the concentration of silver ions in solution before and after dilution?

  1. `1: 1`
  2. `1: 2`
  3. `2: 1`
  4. `4: 1`
Show Answers Only

`A`

Show Worked Solution

The maximum moles of `text{Ag}_2 text{CO}_3` that can be dissolved in 100.0 mL is:

`text{n(Ag}_2 text{CO}_3 text{)}_max` `= 1.2 × 10^(−4) xx 0.1000`
  `= 1.2 xx 10^(−5)  text{mol}`

 
The number of moles of silver carbonate added to the water is:

`text{n(Ag}_2 text{CO}_3)` `= text{m} / text{MM}`
  `= 2.0 / 275.81`
  `= 7.2514 xx 10 ^(-3)  text{mol}`
  •  Thus, the moles of `text{Ag}_2 text{CO}_3`  added is more than the maximum moles of `text{Ag}_2 text{CO}_3` able to be dissolved. i.e. the solution would be saturated before and after dilution.
  • As a result, the solution would have the same `text{Ag}_2 text{CO}_3` concentration before and after, thus, the ratio is 1:1.

`=> A`


♦♦♦ Mean mark 15%.

Financial Maths, STD1 F2 2022 HSC 9 MC

In ten years, the future value of an investment will be $150 000. The interest rate is 4% per annum, compounded half-yearly.

Which equation will give the present value `(P V)` of the investment?
 

  1. `PV=(150\ 000)/((1+0.04)^(10))`
  2. `PV=(150\ 000)/((1+0.04)^(20))`
  3. `PV=(150\ 000)/((1+0.02)^(10))`
  4. `PV=(150\ 000)/((1+0.02)^(20))`
Show Answers Only

`D`

Show Worked Solution

`text{Compounding periods} = 10 xx 2 = 20`

`text{Compounding rate} = (4text{%})/2 = 2text{%} = 0.02`

`PV=(150\ 000)/((1+0.02)^(20))`

`=>D`


♦♦♦ Mean mark 16%.

CHEMISTRY, M6 2021 HSC 35

A manufacturer requires that its product contains at least 85% v/v ethanol.

The concentration of ethanol in water can be determined by a back titration. Ethanol is first oxidised to ethanoic acid using an excess of acidified potassium dichromate solution.

\(\ce{3C2H5OH($aq$) + 2Cr2O7^2-($aq$) + 16H^+($aq$) ->3CH3COOH($aq$) + 4Cr^3+($aq$) + 11H2O($l$)}\)   

The remaining dichromate ions are reacted with excess iodide ions to produce iodine \(\ce{(I2)}\)

\(\ce{Cr2O7^2-($aq$) + 14H^+($aq$) + 61^-($aq$) -> 2Cr^3+($aq$) + 7H2O($l$) + 3I2($aq$)}\)

The iodine produced is then titrated with sodium thiosulfate \(\ce{(Na2S2O3)}\).

\(\ce{I2($aq$) + 2S2O3^2-($aq$) -> 2I^-($aq$) + S4O6^2-($aq$)}\)   

A 25.0 mL sample of the manufacturer's product was diluted with distilled water to 1.00 L. A 25.0 mL aliquot of the diluted solution was added to 20.0 mL of 0.500 mol L¯1 acidified potassium dichromate solution in a conical flask. Potassium iodide (5.0 g) was added and the solution titrated with 0.900 mol L¯1 sodium thiosulfate. This was repeated three times.

The following results were obtained.
 


 

The density of ethanol is 0.789 g mL¯1.

Does the sample meet the manufacturer's requirements? Support your answer with calculations.   (7 marks)

Show Answers Only
\( \text{V}_\text{avg}\  (\ce{Na2S2O3}) \) `= (28.7 + 28.4 + 28.6) / 3`
  `= 28.5666…  text{mL}`
  `= 0.0285666…  text{L}`
   
  •  The first titration is an outlier and so is excluded from the average.

`text{n(Na}_2 text{S}_2 text{O}_3text{)}=\ text{c} xx text{V}= 0.900 xx 0.0285666…= 0.02571  text{mol}`

\( \ce{n(S2O3^2-) = n(Na2S2O3) = 0.02571\ \text{mol}} \)
 

`text{I}_2  text{and S}_2 text{O}_3  ^(2-)\ text{are in a}\ 1 : 2\ text{ratio:}`

`text{n(I}_2text{)} = 1/2 xx  text{n(S}_2 text{O}_3^( 2-) text{)}= 1/2 xx 0.02571= 0.012855\ text{mol}`
 

`text{Excess Cr}_2 text{O}_7^(\ \ 2-)  text{and I}_2  text{are in}\ 1:3\ text{ratio:}`

`text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) excess}= 1/3 xx  text{n(I}_2 text{)}= 1/3 xx 0.012855= 0.004285\ text{mol}`

`text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) initial}= text{c} xx text{V}= 0.500 xx 20/1000= 0.01\ text{mol}`

`text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) reacted with ethanol}`

`= text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) initial}-text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) excess}`

`= 0.01-0.004285`

`= 0.005715\ text{mol}`
 

`text{n(C}_2 text{H}_5 text{OH)}= 3/2 xx text{n(Cr}_2 text{O}_7^(\ \ 2-) text{)}= 3/2 xx 0.005715= 0.0085725\ text{mol}`

`text{m(C}_2 text{H}_5 text{OH)}= text{n} xx text{MM}=0.0085725 xx (2 xx 12.01 + 6 xx 1.008 + 16.00)= 0.3949\ text{g}`
 

  • Thus, 0.3949 g of ethanol is in a diluted 25 mL solution.

Find the mass of ethanol in the original solution:

`text{m(C}_2 text{H}_5 text{OH) original}= 0.3949… xx 1000/25= 15.796…\ text{g}`

`text{D}= text{m} / text{V}\ \ =>\ \ text{V}= text{m} / text{D}`

`text{V(C}_2 text{H}_5 text{OH)}= 15.796 / 0.789= 20.021\ text{mL}`

`text{% (C}_2text{H}_5text{OH)}= text{V(ethanol)} / text{V(sample)}= 20.021 / 25.0= 80.08… %\ text{v}//text{v}`
 

  • Therefore, the product doesn’t meet the manufacturer’s requirement as the concentration is less than 85%.
Show Worked Solution
\( \text{V}_\text{avg}\  (\ce{Na2S2O3}) \) `= (28.7 + 28.4 + 28.6) / 3`
  `= 28.5666…  text{mL}`
  `= 0.0285666…  text{L}`
   
  •  The first titration is an outlier and so is excluded from the average.

`text{n(Na}_2 text{S}_2 text{O}_3text{)}=\ text{c} xx text{V}= 0.900 xx 0.0285666…= 0.02571  text{mol}`

\( \ce{n(S2O3^2-) = n(Na2S2O3) = 0.02571\ \text{mol}} \)
 

`text{I}_2  text{and S}_2 text{O}_3  ^(2-)\ text{are in a}\ 1 : 2\ text{ratio:}`

`text{n(I}_2text{)} = 1/2 xx  text{n(S}_2 text{O}_3^( 2-) text{)}= 1/2 xx 0.02571= 0.012855\ text{mol}`
 

`text{Excess Cr}_2 text{O}_7^(\ \ 2-)  text{and I}_2  text{are in}\ 1:3\ text{ratio:}`

`text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) excess}= 1/3 xx  text{n(I}_2 text{)}= 1/3 xx 0.012855= 0.004285\ text{mol}`

`text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) initial}= text{c} xx text{V}= 0.500 xx 20/1000= 0.01\ text{mol}`

`text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) reacted with ethanol}`

`= text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) initial}-text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) excess}`

`= 0.01-0.004285`

`= 0.005715\ text{mol}`
 

`text{n(C}_2 text{H}_5 text{OH)}= 3/2 xx text{n(Cr}_2 text{O}_7^(\ \ 2-) text{)}= 3/2 xx 0.005715= 0.0085725\ text{mol}`

`text{m(C}_2 text{H}_5 text{OH)}= text{n} xx text{MM}=0.0085725 xx (2 xx 12.01 + 6 xx 1.008 + 16.00)= 0.3949\ text{g}`
 

  • Thus, 0.3949 g of ethanol is in a diluted 25 mL solution.

Find the mass of ethanol in the original solution:

`text{m(C}_2 text{H}_5 text{OH) original}= 0.3949… xx 1000/25= 15.796…\ text{g}`

`text{D}= text{m} / text{V}\ \ =>\ \ text{V}= text{m} / text{D}`

`text{V(C}_2 text{H}_5 text{OH)}= 15.796 / 0.789= 20.021\ text{mL}`

`text{% (C}_2text{H}_5text{OH)}= text{V(ethanol)} / text{V(sample)}= 20.021 / 25.0= 80.08… %\ text{v}//text{v}`
 

  • Therefore, the product doesn’t meet the manufacturer’s requirement as the concentration is less than 85%.

♦♦♦ Mean mark 39%.

CHEMISTRY, M6 2021 HSC 32

The molar enthalpies of neutralisation of three reactions are given.

Reaction 1:

\(\ce{HCl($aq$) + KOH($aq$) -> KCl($aq$) + H2O($l$)}\)                \(\ce{Δ$H$}\) \(\pu{=-57.6 kJ mol-1}\)

Reaction 2:

\(\ce{HNO3($aq$) + KOH($aq$) -> KNO3($aq$) + H2O($l$)}\)       \(\ce{Δ$H$}\) \(\pu{=-57.6 kJ mol-1}\)

Reaction 3:

\(\ce{HCN($aq$) + KOH($aq$) -> KCN($aq$) + H2O($l$)}\)            \(\ce{Δ$H$}\) \(\pu{=-12.0 kJ mol-1}\)

 
Explain why the first two reactions have the same enthalpy value but the third reaction has a different value.   (4 marks)

Show Answers Only
  • Reaction 1 and reaction 2 are both neutralisation reactions between strong acids and strong bases. These reactions completely ionise in solution when added to water.
  • Both reactions have the same net ionic equation:
  •   `text{H}^+ (aq) + text{OH}^-  (aq) →  text{H}_2 text{O} (l)`
  • Therefore, the enthalpy values obtained are the same for both reactions.
  • In reaction 3, `text{HCN}` is a weak acid that only partially ionises in an equilibrium reaction with water.
  •   `text{HCN} (aq) + text{H}_2 text{O} (l)  ⇋  text{CN}^-  (aq) + text{H}_3 text{O}^+ (aq).`
  • As the reaction continues, `text{HCN}` will further ionise as the equilibrium shifts to the right.
  • The bond-breaking is an endothermic process and thus will consume energy to break the bonds. As a result, the overall reaction is less exothermic than reaction 1 and reaction 2.
Show Worked Solution
  • Reaction 1 and reaction 2 are both neutralisation reactions between strong acids and strong bases. These reactions completely ionise in solution when added to water.
  • Both reactions have the same net ionic equation:
  •   `text{H}^+ (aq) + text{OH}^-  (aq) →  text{H}_2 text{O} (l)`
  • Therefore, the enthalpy values obtained are the same for both reactions.
  • In reaction 3, `text{HCN}` is a weak acid that only partially ionises in an equilibrium reaction with water.
  •   `text{HCN} (aq) + text{H}_2 text{O} (l)  ⇋  text{CN}^-  (aq) + text{H}_3 text{O}^+ (aq).`
  • As the reaction continues, `text{HCN}` will further ionise as the equilibrium shifts to the right.
  • The bond-breaking is an endothermic process and thus will consume energy to break the bonds. As a result, the overall reaction is less exothermic than reaction 1 and reaction 2.

♦ Mean mark 44%.

ENGINEERING, CS 2021 HSC 19 MC

The diagram shows a simply supported beam in equilibrium. It is loaded with a single force (`text{F}`) as shown.
 

Which of the following angles is closest to the angle of the reaction force to the horizontal at the fixed bearing?

  1. 7°
  2. 14°
  3. 31°
  4. 90°
Show Answers Only

`B`

Show Worked Solution
  • The fixed joint takes half of the vertical force and all of the horizontal force, therefore the angle will be half that of the original force `text{F}`.

`=>B`

♦♦♦ Mean mark 18%.