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ENGINEERING, AE 2019 HSC 15 MC

Titanium is used to manufacture aircraft undercarriages that support landing wheels.

Why is titanium used in preference to alloy steel for this purpose?

  1. It has lower density and lower strength.
  2. It has lower density and higher strength.
  3. It has higher density and lower strength.
  4. It has higher density and higher strength.
Show Answers Only

`B`

Show Worked Solution

→ Lower density and higher strength, means the material has a very high strength to weight ratio, or specific strength.

→ This is one of the most beneficial characteristics associated with titanium alloys.

`=>B`

ENGINEERING, AE 2019 HSC 12 MC

Which row of the table indicates the effects of retrofitting winglets to a plane?
 

Show Answers Only

`B`

Show Worked Solution

→ Winglets are designed to stop vortices and therefore decrease induced drag.

→ Winglets also increase lift by stopping air from moving from below the wing to above it around the wingtips.

`=>B`

ENGINEERING, CS 2019 HSC 9 MC

A sketch of a post-tensioned concrete beam is shown.
 

Which of the following shows the most likely loading condition the beam is designed to resist when viewed from the front?
 

Show Answers Only

`D`

Show Worked Solution

→ Concrete has high compressive strength, while the steel reinforcements have high tensile strength.

→ Hence, the most likely loading scenario will put the bottom of the beam in tension and the top in compression. 

`=>D`

ENGINEERING, TE 2019 HSC 2 MC

Which of the following identifies TWO properties of polymers that make them better suited than ceramics for the insulation of telecommunication cables?

  1. Ease of formability and higher density
  2. Ease of formability and higher flexibility
  3. Higher resistance to abrasion and higher density
  4. Higher resistance to abrasion and higher flexibility
Show Answers Only

`B`

Show Worked Solution

By Elimination:

→ Higher density is an undesirable property in cabling (not `A` or `C`)

→ Ceramics have higher resistance to abrasion (not `C` or `D`)

`=>B`

ENGINEERING, TE 2022 HSC 27a

Describe TWO different uses of polymers in telecommunications.   (4 marks)

Show Answers Only

Answers could include any two of the following:

Cable protection and insulation

→ Polymers offer protection for both the cable and anyone who may come in contact with it.

Circuit components

→ Polymers are widely used internally in many telecommunication devices.

Device casings and housings

→ Polymers are a common material used for casings for devices such as TV’s and mobile phones.

Show Worked Solution

Answers could include any two of the following:

Cable protection and insulation

→ Polymers offer protection for both the cable and anyone who may come in contact with it.

Circuit components

→ Polymers are widely used internally in many telecommunication devices.

Device casings and housings

→ Polymers are a common material used for casings for devices such as TV’s and mobile phones.

ENGINEERING, PPT 2022 HSC 26b

Different modes of transport emit different amounts of carbon dioxide `(\text{CO}_2)`. For example, petrol cars emit more `\text{CO}_2` than electric cars while in motion.

Describe engineering innovations that have led to lower carbon dioxide emissions from electric transportation.   (3 marks)

Show Answers Only

Answers should discuss one of the following:

Electric powered cars

→ Electric powered cars do not need heavy engine components to propel them as their energy comes from an electric source. There are also fewer moving parts resulting in a net decrease in weight.

→ Lighter engineering materials further contribute to the decreased weight, which improves fuel efficiency and lowers `\text{CO}_2` emissions.

→ Furthermore, combustion engines themselves produce large volumes of `\text{CO}_2`, hence their removal greatly decreases emissions.
 

Electric powered bikes

→ Place lower energy demands on a rider, which decreases the `\text{CO}_2` emissions from the cyclist.

→ Battery technology increases the bike’s range and decreases its emissions, enabling long distance travel without negative environmental impacts.

Show Worked Solution

Answers should discuss one of the following:

Electric powered cars

→ Do not need heavy engine components to propel them as their energy comes from an electric source. There are also fewer moving parts resulting in a net decrease in weight.

→ Lighter engineering materials further contribute to the decreased weight, which improves fuel efficiency and lowers `\text{CO}_2` emissions.

→ Furthermore, combustion engines themselves produce large volumes of `\text{CO}_2`, hence their removal greatly decreases emissions.
 

Electric powered bikes

→ Place lower energy demands on a rider, which decreases the `\text{CO}_2` emissions from the cyclist.

→ Battery technology increases the bike’s range and decreases its emissions, enabling long distance travel without negative environmental impacts.

ENGINEERING, AE 2022 HSC 25c

Explain how engineers have used their knowledge of both mechanics and hydraulics to contribute to aircraft design. Support your answer with specific examples.   (5 marks)

Show Answers Only

→ Modern aircraft encompass the consolidation of many advancements in mechanical principles and hydraulic systems.

→ Understanding of the mechanics of lift, thrust and drag through the use of wind tunnels and simulations has led to a better understanding of aerodynamics and flight.

→ This led to changes in aircraft design and shape to increase capacity, improve fuel efficiency and decrease drag in the form of wingtip vortices through the introduction of winglets. 

→ An increased understanding of hydraulics has led to improved braking power, allowing for heavier planes to be stopped safely.

→ Coupled with developments in computers, this understanding has also allowed for quick, computer controlled adjustments of control surfaces, with more power and precision than previously possible.

→ Mechanical developments to aircraft propulsion systems have also increased the  fuel efficiency and speed of modern day aircraft.

Show Worked Solution

→ Modern aircraft encompass the consolidation of many advancements in mechanical principles and hydraulic systems.

→ Understanding of the mechanics of lift, thrust and drag through the use of wind tunnels and simulations has led to a better understanding of aerodynamics and flight.

→ This led to changes in aircraft design and shape to increase capacity, improve fuel efficiency and decrease drag in the form of wingtip vortices through the introduction of winglets. 

→ An increased understanding of hydraulics has led to improved braking power, allowing for heavier planes to be stopped safely.

→ Coupled with developments in computers, this understanding has also allowed for quick, computer controlled adjustments of control surfaces, with more power and precision than previously possible.

→ Mechanical developments to aircraft propulsion systems have also increased the  fuel efficiency and speed of modern day aircraft.

ENGINEERING, AE 2022 HSC 24a

An aeronautical engineer is developing a prototype for a new propeller design.

Identify a suitable method of developing the prototype and outline the benefits of using this method.   (3 marks)

Show Answers Only

Possible methods include:

CAD software

→ Prototype can be created and tested using CAD software.

→ This software can quickly produce working models with little difficulty.

3D printing

→ Used alongside CAD, 3D printing allows you to hold a physical model and could allow for simple testing of the design.

CNC Milling

→ Can be used to create a physical prototype out of a variety of materials.

→ Significantly faster than traditional machining of pieces.

Scaled physical model

→ A simple smaller propeller can be used to test ideas and appearances.

Show Worked Solution

Possible methods include:

CAD software

→ Prototype can be created and tested using CAD software.

→ This software can quickly produce working models with little difficulty.

3D printing

→ Used alongside CAD, 3D printing allows you to hold a physical model and could allow for simple testing of the design.

CNC Milling

→ Can be used to create a physical prototype out of a variety of materials.

→ Significantly faster than traditional machining of pieces.

Scaled physical model

→ A simple smaller propeller can be used to test ideas and appearances.

PHYSICS, M8 2015 HSC 34d

In 1927, Davisson and Germer reported the results of an experiment in which they fired electrons at a crystal of nickel and observed how the electrons were scattered.

  1. State their conclusion, with reference to the results they obtained.   (2 marks)
  1. Explain the significance of this experiment to the Rutherford-Bohr model of the atom.   (3 marks)
Show Answers Only

i.    Experiment Results:

→ Davisson and Germer found that the electrons were not scattered in a random pattern, but formed an interference pattern after passage through the crystal.

→ This pattern is formed by constructive and destructive wave interference.

Conclusion:

→ Since interference is a phenomenon only observed with waves they concluded that electrons were also waves.
 

ii.   The Rutherford-Bohr model:

→ Postulated that electrons existed in fixed orbits. The model however, was unable to explain why only these orbits were stable.

→ The knowledge that electrons are waves provides a plausible explanation for this stability.

→ If electrons act as waves, as indicated by de Broglie, the electron can only exist in orbits where it experiences constructive interference.

Show Worked Solution

i.    Experiment Results:

→ Davisson and Germer found that the electrons were not scattered in a random pattern, but formed an interference pattern after passage through the crystal.

→ This pattern is formed by constructive and destructive wave interference.

Conclusion:

→ Since interference is a phenomenon only observed with waves they concluded that electrons were also waves.
 

ii.   The Rutherford-Bohr model:

→ Postulated that electrons existed in fixed orbits. The model however, was unable to explain why only these orbits were stable.

→ The knowledge that electrons are waves provides a plausible explanation for this stability.

→ If electrons act as waves, as indicated by de Broglie and experimentally shown by Davisson and Germer, the electron can only exist in orbits where it experiences constructive interference.


Mean mark (ii) 54%.

PHYSICS, M8 2015 HSC 34aii

The mass of a polonium-218 nucleus is 218.00897 u, the mass of a lead-214 nucleus is 213.99981 u, and the mass of an alpha particle is 4.00260 u.

Calculate the energy released by this alpha decay.   (3 marks)

Show Answers Only

`9.8 xx10^(-13)\ text{J}`

Show Worked Solution

\(\text{Mass}_{i} = 218.00897\ u \)

\(\text{Mass}_{f} = 213.99981 + 4.00260 = 218.00241\ u \)

\(\text{Difference} = 218.00897-218.00241 = 0.00656\ u \)

\(\text{Difference (kg)} = 0.00656 \times 1.661 \times 10^{-27} = 1.08962 \times 10^{-29}\ \text{kg} \)
 

`E` `=mc^2`  
  `=1.08962xx10^(-29)xx(3 xx10^8)^2`  
  `=9.8 xx10^(-13)\ text{J}`  

PHYSICS, M8 2015 HSC 33d

The position of the Sun, star `W` and star `Z` are shown on the H-R diagram.
 

The curves `A` and `B` show intensity versus frequency for star `W` and the Sun, measured from the same distance.
 

  1. Identify which curve (`A` or `B`) represents star `W` and justify your choice.   (2 marks)
  1. Account for differences between stars `W` and `Z` that can be deduced from the H-R diagram.   (3 marks)
Show Answers Only

i.    → Curve A represents star `W`.

→ `W` is hotter so will emit more energy overall, and peak frequency will be higher.
 

ii.   Differences displayed in H-R graph:

→ Star `W` lies in the main sequence and is therefore fusing hydrogen to helium, most likely via CNO cycle as it is a large dense star.

→ Star `Z` is past main sequence and is therefore fusing larger elements.

→ Since star `Z` is cooler than `W` (refer to its H-R diagram location ) but has similar luminosity, we can deduce it is much larger than `W`.
 

Other possible answers could include:

→ Differences in fuel source, size and temperature of the stars.

Show Worked Solution

i.    → Curve A represents star `W`.

→ `W` is hotter so will emit more energy overall, and peak frequency will be higher.
 

ii.   Differences displayed in H-R graph:

→ Star `W` lies in the main sequence and is therefore fusing hydrogen to helium, most likely via CNO cycle as it is a large dense star.

→ Star `Z` is past main sequence and is therefore fusing larger elements.

→ Since star `Z` is cooler than `W` (refer to its H-R diagram location ) but has similar luminosity, we can deduce it is much larger than `W`.
 

Other possible answers could include:

→ Differences in fuel source, size and temperature of the stars.

PHYSICS, M8 2016 HSC 34bii

Calculate the initial energy level of an electron in a hydrogen atom if it emitted `4.089 × 10^{-19}` J on transition to the `n` = 2 level.   (3 marks)

Show Answers Only

`n_i = 4`

Show Worked Solution

`E/(hc) = 1/lambda`

`E/(hc)` `=R(1/((n_f)^2)-1/((n_i)^2))`  
  `=R(1/(2^2)-1/((n_i)^2))`  
`1/(n_i)^2` `=1/4-(E)/(Rhc)`  
  `=1/4-(4.089 xx 10^{-19})/(1.097 xx 10^7 xx 6.626 xx 10^(-34) xx 3 xx 10^8)`  
  `=0.06248`  

 
`n_i=sqrt(1/0.06248) = 4`

PHYSICS, M8 2016 HSC 34bi

Outline features of the hydrogen spectrum that Bohr's model could not explain.   (3 marks)

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→ Bohr’s model could not provide an adequate explanation of why some spectral lines were observed to be more intense than others.

→ This result suggested that certain transitions occurred much more frequently than others which Bohr’s model could not explain.

→ With more sensitive instruments, hyperfine lines were discovered to also exist in the spectrum.

→ This indicates there is some splitting of the energy level that Bohr’s model cannot explain.

Show Worked Solution

→ Bohr’s model could not provide an adequate explanation of why some spectral lines were observed to be more intense than others.

→ This result suggested that certain transitions occurred much more frequently than others which Bohr’s model could not explain.

→ With more sensitive instruments, hyperfine lines were discovered to also exist in the spectrum.

→ This indicates there is some splitting of the energy level that Bohr’s model cannot explain.

PHYSICS, M8 2016 HSC 34aii

Explain the stability of `\ _(2)^(4)text{He}` nuclei in terms of TWO forces.   (3 marks)

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Repulsive electrostatic force:

→ The nucleus of the helium atom given contains 2 protons and two neutrons.

→ Protons carry a positive charge so that there is an electrostatic repulsive force that occurs between the protons in the nucleus.
 

Attractive strong nuclear force:

→ A strong nuclear force of attraction occurs between proton-proton, neutron-neutron and protons-neutrons.

→ It is a short range force that is strongest when the nucleons are 0.5 to 1.0 the diameter of a proton apart.

→ Inside this range, it is much stronger than the repulsive electrostatic force and is responsible for keeping the nucleus together. 

Show Worked Solution

Repulsive electrostatic force:

→ The nucleus of the helium atom given contains 2 protons and two neutrons.

→ Protons carry a positive charge so that there is an electrostatic repulsive force that occurs between the protons in the nucleus.
 

Attractive strong nuclear force:

→ A strong nuclear force of attraction occurs between proton-proton, neutron-neutron and protons-neutrons.

→ It is a short range force that is strongest when the nucleons are 0.5 to 1.0 the diameter of a proton apart.

→ Inside this range, it is much stronger than the repulsive electrostatic force and is responsible for keeping the nucleus together. 

PHYSICS, M8 2017 HSC 34c

The diagrams show features of the hydrogen emission spectrum.
 

With reference to Bohr's postulates, explain how the line at 434.0 nm in the hydrogen emission spectrum is produced. Support your answer with calculations.   (4 marks)

Show Answers Only

→ Bohr’s postulates state that an electron orbits an atom’s nucleus at fixed energy levels.

→ If an electron jumps from a higher energy level to a lower energy level, it will emit a photon that contains an exact amount of energy.

→ The Rydberg equation can be used to calculate a photon’s wavelength.

→ The blue light is part of the visible Balmer series and therefore `n_f=2`.

→ Using `n_f=2`, `lambda` = 434.0 nm: 

`1/lambda` `=R(1/((n_f)^2)-1/((n_i)^2))`  
  `=R(1/(2^2)-1/((n_i)^2))`  
`1/(n_i)^2` `=1/4-1/(lambdaR)`  
`(n_i)^2` `=1/(1/4-1/(1.097 xx 10^7 xx 434 xx 10^9))`  
`n_i` `=5`  

 
→ The equation shows that a 434 nm wavelength for the emitted photon corresponds to a jump from the 5th energy level to the 2nd energy level.

Show Worked Solution

→ Bohr’s postulates state that an electron orbits an atom’s nucleus at fixed energy levels.

→ If an electron jumps from a higher energy level to a lower energy level, it will emit a photon that contains an exact amount of energy.

→ The Rydberg equation can be used to calculate a photon’s wavelength.

→ The blue light is part of the visible Balmer series and therefore `n_f=2`.

→ Using `n_f=2`, `lambda` = 434.0 nm: 

`1/lambda` `=R(1/((n_f)^2)-1/((n_i)^2))`  
  `=R(1/(2^2)-1/((n_i)^2))`  
`1/(n_i)^2` `=1/4-1/(lambdaR)`  
`(n_i)^2` `=1/(1/4-1/(1.097 xx 10^7 xx 434 xx 10^9))`  
`n_i` `=5`  

 
→ The equation shows that a 434 nm wavelength for the emitted photon corresponds to a jump from the 5th energy level to the 2nd energy level.


Mean mark 56%.

PHYSICS, M8 2017 HSC 34bii

The diagram shows some components of a nuclear reactor.
 

Explain how the labelled components work together to produce a controlled nuclear reaction.   (3 marks)

Show Answers Only

→ Fuel elements are composed of a fissile material that release energy and high-velocity neutrons when atoms are split.

→ These high speed neutrons are slowed down by the moderator material. 

→ In this way, the neutrons are able to be captured by more fissile atoms, which in turn split.

→ The control rods play an important role in the process by absorbing neutrons. This limits the number of atoms undergoing fission and hence they can be used to control the release of energy.

Show Worked Solution

→ Fuel elements are composed of a fissile material that release energy and high-velocity neutrons when atoms are split.

→ These high speed neutrons are slowed down by the moderator material. 

→ In this way, the neutrons are able to be captured by more fissile atoms, which in turn split.

→ The control rods play an important role in the process by absorbing neutrons. This limits the number of atoms undergoing fission and hence they can be used to control the release of energy.

PHYSICS, M8 2018 HSC 34bii

The following is a nuclear reaction that produces a neutron.

\({ }_2^4 \alpha+{ }_4^9 \text{Be} \rightarrow{ }_6^{12}\text{C} +{ }_0^1 \text{n}\)

The table shows the masses of the particles in the reaction.

\begin{array}{|c|c|}
\hline
\textit{ Particle } & \textit{Mass (u) } \\
\hline
\rule{0pt}{2.5ex}{ }_2^4 \alpha \rule[-1ex]{0pt}{0pt}& 4.0012 \\
\hline
\rule{0pt}{2.5ex}{ }_4^9 \text{Be} \rule[-1ex]{0pt}{0pt}& 9.0122 \\
\hline
{\rule{0pt}{2.5ex} }_6^{12} \text{C} \rule[-1ex]{0pt}{0pt}& 12.0000 \\
\hline
\rule{0pt}{2.5ex}{ }_0^1 \text{n} \rule[-1ex]{0pt}{0pt}& 1.0087 \\
\hline
\end{array}

Using the data from the table, calculate the energy released in this reaction. State your answer in joules.   (3 marks)

Show Answers Only

`7.0 xx10^(-13)\ text{J}`

Show Worked Solution

\(\text{Mass}_{i} = 4.0012 + 9.0122 = 13.0134\ u \)

\(\text{Mass}_{f} = 12.000 + 1.0087 = 13.0087\ u \)

\(\text{Difference} = 13.0087-13.0134 = –0.0047\ u \)

\(\text{Difference (kg)} = 0.0047 \times 1.661 \times 10^{-27} = 7.8067 \times 10^{-30}\ \text{kg} \)
 

`E` `=mc^2`  
  `=7.8067xx10^(-30)xx(3 xx10^8)^2`  
  `=7.0 xx10^(-13)\ text{J}`  

PHYSICS, M8 2018 HSC 34bi

The diagram shows apparatus used to investigate subatomic particles.
 

     

How did Chadwick use a law of physics to identify a property of `X`?   (2 marks)

Show Answers Only

→ Chadwick measured the momentum of both `X` and the ejected protons.

→ Using the law of conservation of momentum he was able to calculate the mass of `X` which was similar to the mass of a proton.

Show Worked Solution

→ Chadwick measured the momentum of both `X` and the ejected protons.

→ Using the law of conservation of momentum he was able to calculate the mass of `X` which was similar to the mass of a proton.

BIOLOGY, M5 2018 HSC 28

  1. Define the terms genotype and phenotype.  (2 marks)

  2. A population of mammals living in a very cold climate have long, thick fur which assists their survival. At times, a widespread infestation of mites causes their fur to become thin, and bald patches appear in their coats. In this population some individuals are resistant to the mites. This is due to the presence of a co-dominant allele `(text{M}^text(R))`. Individuals that are homozygous for this allele `(text{M}^text(R)text{M}^text(R))` are infertile.
  3. Explain how both genotype and phenotype influence the inheritance of genes and natural selection in this population.  (4 marks)

Show Answers Only

a.   Definitions:

→ Genotype refers to the genetic makeup or combination of genes for a specific trait in an organism.

→ Phenotype refers to the physical manifestation of the genotype or how it appears, develops or behaves in the organism.
  

b.   Genotype and Phenotype influences:

→ The selection pressure for the mammals is long, thick fur to enable them to survive the cold climate.

→ Therefore the allele for long, thick hair is carried by most of the population.

→ The phenotype of these mammals has changed because of the mite infestation.

→ Even though they still have genes for long fur their survival fitness has been compromised as they develop bald patches.

→ The survival of the mammals should be improved during infestations because they possess the mite resistant gene.

→ However, they become infertile when they carry the two alleles – homozygous `(text{M}^text{R}text{M}^text{R})`.

→ Even though they are genetically better suited to the conditions, having long fur and mite resistance, they cannot pass the genes on as they are sterile.

→ Heterozygotes will have some mite resistance as the allele for mite resistance is co-dominant, so their survival rates will be better than those without the allele.

→ The allele can be kept in the population as it can be passed by heterozygotes to future generations.

Show Worked Solution

a.   Definitions:

→ Genotype refers to the genetic makeup or combination of genes for a specific trait in an organism.

→ Phenotype refers to the physical manifestation of the genotype or how it appears, develops or behaves in the organism.
  

b.   Genotype and Phenotype influences:

→ The selection pressure for the mammals is long, thick fur to enable them to survive the cold climate.

→ Therefore the allele for long, thick hair is carried by most of the population.

→ The phenotype of these mammals has changed because of the mite infestation.

→ Even though they still have genes for long fur their survival fitness has been compromised as they develop bald patches.

→ The survival of the mammals should be improved during infestations because they possess the mite resistant gene.

→ However, they become infertile when they carry the two alleles – homozygous `(text{M}^text{R}text{M}^text{R})`.

→ Even though they are genetically better suited to the conditions, having long fur and mite resistance, they cannot pass the genes on as they are sterile.

→ Heterozygotes will have some mite resistance as the allele for mite resistance is co-dominant, so their survival rates will be better than those without the allele.

→ The allele can be kept in the population as it can be passed by heterozygotes to future generations.


♦ Mean mark (b) 44%.

BIOLOGY, M7 2018 HSC 24

A new flu vaccine is prepared each year to protect the population against the current strains of influenza virus. The effectiveness of flu vaccines varies from year to year and can be measured using the overall vaccination effectiveness (VE) index. A VE of 60% means that a vaccinated individual's chance of getting the flu is reduced by 60%.

The following data show the VE over a 10-year period.

\begin{array}{|c|c|}
\hline \rule{0pt}{2.5ex}\textit{Influenza season } & \rule[-1ex]{0pt}{0pt}\quad \textit{VE(%)} \quad \\
\hline \rule{0pt}{2.5ex}2006-2007 & \rule[-1ex]{0pt}{0pt}52 \\
\hline \rule{0pt}{2.5ex}2008-2009 & \rule[-1ex]{0pt}{0pt}41 \\
\hline \rule{0pt}{2.5ex}2010-2011 & \rule[-1ex]{0pt}{0pt}60 \\
\hline \rule{0pt}{2.5ex}2012-2013 & \rule[-1ex]{0pt}{0pt}49 \\
\hline \rule{0pt}{2.5ex}2014-2015 & \rule[-1ex]{0pt}{0pt}19 \\
\hline
\end{array}

  1. Draw an appropriate graph to represent the data on the following grid.   (3 marks)
     
     

     
  2. Provide a possible explanation for the vaccination effectiveness (VE) index in 2014-2015.   (3 marks)

Show Answers Only

a.

   

b.   → The vaccine was less effective between 2014-15 (at 19%).

→ When vaccines are administered the production of antibodies specific to a particular antigen is triggered.

→ If the pathogen mutates after the production of the vaccine the vaccine will not be effective.

→ Therefore, mutation of the virus may be responsible for the vaccine being less effective in 2014-15 and hence the lower VE %.

Show Worked Solution

a.

   

b.   → The vaccine was less effective between 2014-15 (at 19%).

→ When vaccines are administered the production of antibodies specific to a particular antigen is triggered.

→ If the pathogen mutates after the production of the vaccine the vaccine will not be effective.

→ Therefore, mutation of the virus may be responsible for the vaccine being less effective in 2014-15 and hence the lower VE %.


♦ Mean mark (b) 40%.

BIOLOGY, M7 2018 HSC 22b

Describe the contribution of Robert Koch to our understanding of disease.  (3 marks)

Show Answers Only

→ Robert Koch demonstrated a relationship between microbes and infectious disease.

→ Koch revealed that specific infectious diseases were caused by specific microbes.

→ He developed Koch’s postulates which allow for the identification of specific diseases.

Show Worked Solution

→ Robert Koch demonstrated a relationship between microbes and infectious disease.

→ Koch revealed that specific infectious diseases were caused by specific microbes.

→ He developed Koch’s postulates which allow for the identification of specific diseases.


Mean mark 58%.

BIOLOGY, M8 2018 HSC 21

  1. Identify TWO responses of a named endotherm to a decrease in body temperature.  (2 marks)

  2. Outline the role of the nervous system of an endotherm in maintaining homeostasis when its body temperature changes.  (2 marks)

Show Answers Only

a.   Vasoconstriction and shivering.

b.   Body temperature maintenance in endotherms:

→ This is achieved via negative feedback loops.

→ Temperature receptors in the brain and skin detect changes in body temperature.

→ The central nervous system receives a signal via the receptors.

→ Muscles or glands (effectors) then receive nerve signals sent from the central nervous system.

Show Worked Solution

a.   Vasoconstriction and shivering.
 

b.   Body temperature maintenance in endotherms:

→ This is achieved via negative feedback loops.

→ Temperature receptors in the brain and skin detect changes in body temperature.

→ The central nervous system receives a signal via the receptors.

→ Muscles or glands (effectors) then receive nerve signals sent from the central nervous system.

BIOLOGY, M7 2018 HSC 17 MC

The diagram shows the life cycle of the malaria parasite, Plasmodium sp. Five stages in this life cycle are numbered on the diagram.
 


 

To prevent malaria, the following four strategies have been used:

    • taking anti-malarial drugs
    • spraying swamps with insecticides
    • using mosquito nets over beds
    • administering a malaria vaccine.

Which row in the table shows the stage in the life cycle in which each of these strategies would be most effective?
 

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ & \\
 \rule[-2ex]{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|}
\hline
\rule{0pt}{2.5ex}\textit{Taking}\rule[-1ex]{0pt}{0pt}& \textit{Spraying swamps} & \textit{Using mosquito}& \textit{Adminstering a}\\
\textit{anti-malarial drugs}\rule[-1.5ex]{0pt}{0pt}& \textit{with insecticides} & \textit{nets over beds}& \textit{malaria vaccine}\\
\hline
\rule{0pt}{2.5ex}5\rule[-1ex]{0pt}{0pt}&1& 3& 2\\
\hline
\rule{0pt}{2.5ex}2\rule[-1ex]{0pt}{0pt}&5& 1& 3\\
\hline
\rule{0pt}{2.5ex}1\rule[-1ex]{0pt}{0pt}& 4& 2& 5 \\
\hline
\rule{0pt}{2.5ex}2\rule[-1ex]{0pt}{0pt}& 3 & 5&1\\
\hline
\end{array}
\end{align*}

Show Answers Only

\(D\)

Show Worked Solution

→ Vaccines are most effective when administered before the parasite enters the cell (1).

→ Mosquito nets are only useful if utilised prior to the injection of the parasite into a human (5).

→ Insecticides help prevent the vectors from breeding (3).

→ Anti-malarial drugs may be useful once the disease is present (2).

\(\Rightarrow D\)

BIOLOGY, M7 2017 HSC 30a

What is the role of ONE type of T lymphocyte in the immune response?  (2 marks)

Show Answers Only

“

Show Worked Solution

Answers could include one of the following:

→ Helper T cells secrete cytokines to regulate interaction between B and T cells.

→ Killer T cells kill infected cells.

→ Regulatory T cells make B and T cells inactive after the pathogen is destroyed.

→ Memory T cells reactivate quickly in the body once the same pathogen is detected providing acquired immunity.

BIOLOGY, M5 2017 HSC 28

A pedigree chart of an inherited characteristic is shown.
 

 

Subsequent genetic analysis showed `text{I-2}` does not have the recessive allele.

Explain the inheritance of this characteristic.   (3 marks)

Show Answers Only

“

Show Worked Solution

→ `text{I-1}` must be homozygous for the condition given that both male offspring `text{(II-3}` and `text{II-5)}` are also affected.

→ The condition must be recessive as `text{III-1}` is affected and both parents `text{(II-1}` and `text{II-2)}` are unaffected.

→ As `text{I-2}` does not have the recessive allele the condition is sex-linked, as autosomal recessive would require both parents to be carriers.

→ `text{II-2}` must be an unaffected carrier (possess the recessive gene) of the sex-linked allele.

BIOLOGY, M8 2017 HSC 26

A controlled experiment was performed to investigate the effect of substrate concentration on the rate of an enzyme-catalysed reaction. Data were collected and are presented in the graph.
 

  1. What is the independent variable in this experiment?   (1 mark)

  2. Explain the trends shown in the graph.   (3 marks)

Show Answers Only

a.  Substrate concentration.

b.  Trends in the graph

→ Initially the rate of reaction increases sharply with the substrate concentration increase as there is more enzyme available to bind and promote the reaction.

→ As the enzyme available is depleted and it is no longer available to catalyse the reaction, the rate of reaction plateaus.

Show Worked Solution

a.  Substrate concentration.

b.  Trends in the graph

→ Initially the rate of reaction increases sharply with the substrate concentration increase as there is more enzyme available to bind and promote the reaction.

→ As the enzyme available is depleted and it is no longer available to catalyse the reaction, the rate of reaction plateaus.

BIOLOGY, M7 2017 HSC 22a

For each type of disease in the following table, name a specific disease and its cause.   (2 marks)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Type of disease} \rule[-1ex]{0pt}{0pt} & \quad \quad\text{Name of disease} \quad\quad &\quad \quad\text{Cause of disease}\quad\quad\\
\hline
\rule{0pt}{2.5ex} \text{Infectious disease} \rule[-1ex]{0pt}{0pt} & \text{} &\text{}\\
\rule{0pt}{2.5ex} \text{} \rule[-1ex]{0pt}{0pt} & \text{} \text{}\\
\rule{0pt}{2.5ex} \text{} \rule[-1ex]{0pt}{0pt} & \text{} \text{}\\
\hline
\rule{0pt}{2.5ex} \text{Non-infectious} \rule[-1ex]{0pt}{0pt} & \text{} \text{}\\
\text{disease} \rule[-1ex]{0pt}{0pt} & \text{} \text{}\\
\rule{0pt}{2.5ex} \text{} \rule[-1ex]{0pt}{0pt} & \text{} \text{}\\
\hline
\end{array} 

Show Answers Only

Answers could include one example of the two given:

Show Worked Solution

Answers could include one example of the two given:

BIOLOGY, M5 2017 HSC 15 MC

A plant breeder crosses two plants of the same species. They are both pure-breeds for flower colour. The colour of the flowers of all the offspring is the same, but different to that of the parents.

What type of inheritance does this result show?

  1. Sex-linked
  2. Recessive
  3. Dominance
  4. Co-dominance
Show Answers Only

`D`

Show Worked Solution

→ Co-dominance occurs when both alleles are fully expressed in the genotype, and a colour different from either parent is produced.

`=>D`

CHEMISTRY, M6 2018 HSC 17 MC

Increasing amounts of carbon dioxide were dissolved in two beakers, one containing water and one a mixture of water and a buffer. The pH in each beaker was measured and the results graphed.

Which graph best represents the results?
 

 

Show Answers Only

`A`

Show Worked Solution

By Elimination:

→ A buffer will resist changes in pH caused by the addition of an acidic or basic solution.

→ If the water and buffer solution is decreasing in pH, the water solution will be declining in pH quicker and its graph will lie below (eliminate B and D).

→ If the water and buffer solution is increasing in pH, the water solution should be increasing at a greater rate (eliminate C).

`=>A`

BIOLOGY, M7 2016 HSC 28

Bacteriophages are viruses that use bacteria as host cells. One type of bacteriophage infects only one species of bacterial cell. The infected cells are destroyed as the bacteriophage particles burst out of them, completing their life cycle.

Bacteriophages can be used to fight bacterial infections. As bacteria evolve, bacteriophages also evolve.

  1. Explain TWO advantages of using bacteriophage treatment compared to antibiotic treatment for bacterial infections.   (4 marks)

  2. Describe a possible disadvantage of using bacteriophage treatment.   (2 marks)

Show Answers Only

a.   Include two of the following advantages:

→ Bacterial pathogens have become resistant to antibiotics and so antibiotics can be less effective.

→ Bacteriophages can mutate, therefore phenotypes can evolve that stay pathogenic to mutated bacteria.

→ There is the risk of killing useful bacteria when using antibiotics as well as pathogens.

→ Bacteriophages specifically target the pathogen.
  

b.   Include one of the following disadvantages:

→ Residual bacteriophages may have unknown side effects.

→ As they can mutate there is the possibility that bacteriophages could damage and infect a non-target bacteria or human cells, leading to disease.

Show Worked Solution

a.   Include two of the following advantages:

→ Bacterial pathogens have become resistant to antibiotics and so antibiotics can be less effective.

→ Bacteriophages can mutate, therefore phenotypes can evolve that stay pathogenic to mutated bacteria.

→ There is the risk of killing useful bacteria when using antibiotics as well as pathogens.

→ Bacteriophages specifically target the pathogen.
  


Mean mark (a) 59%.

b.   Include one of the following disadvantages:

→ Residual bacteriophages may have unknown side effects.

→ As they can mutate there is the possibility that bacteriophages could damage and infect a non-target bacteria or human cells, leading to disease.


Mean mark (b) 56%.

BIOLOGY, M5 2016 HSC 26

Students conducted preliminary experiments across different species to analyse their DNA base composition.

The table shows the experimental data collected.

\begin{array}{|c|c|c|}
\hline \rule{0pt}{2.5ex}\textit{Species} \rule[-1ex]{0pt}{0pt}& \textit{% Adenine} & \textit{% Guanine} \\
\hline \rule{0pt}{2.5ex}\text{A} \rule[-1ex]{0pt}{0pt}& 38 & 12 \\
\hline \rule{0pt}{2.5ex}\text{B} \rule[-1ex]{0pt}{0pt}& 26 & 22 \\
\hline \rule{0pt}{2.5ex}\text{C} \rule[-1ex]{0pt}{0pt}& 8 & 40 \\
\hline \rule{0pt}{2.5ex}\text{D} \rule[-1ex]{0pt}{0pt}& 20 & 32 \\
\hline \rule{0pt}{2.5ex}\text{E} \rule[-1ex]{0pt}{0pt}& 33 & 18 \\
\hline
\end{array}

 

  1. On the grid below, plot the % Adenine vs % Guanine of the species analysed AND draw a suitable line of best fit.   (3 marks)
     
     

     
  2. Identify the relationship shown by the data.   (1 mark)

  3. Explain the relationship shown by the data.   (3 marks)

Show Answers Only

a.  

b.  As % Adenine increases, the % Guanine decreases.
  

c.    Explanation of data relationship:

→ Adenine and Thymine are complementary base pairs. Likewise with Guanine and Cytosine.

→ It would therefore be expected that as the percentage of adenine increases, the percentage of thymine would also increase.

→ Adenine + Thymine + Cytosine + Guanine = 100%

→ It follows that as adenine increases the amount of guanine should decrease.

Show Worked Solution

a.  

b.  As % Adenine increases, the % Guanine decreases.
  

c.    Explanation of data relationship:

→ Adenine and Thymine are complementary base pairs. Likewise with Guanine and Cytosine.

→ It would therefore be expected that as the percentage of adenine increases, the percentage of thymine would also increase.

→ Adenine + Thymine + Cytosine + Guanine = 100%

→ It follows that as adenine increases the amount of guanine should decrease.


♦♦♦ Mean mark (c) 28%.

BIOLOGY, M6 2016 HSC 18 MC

How does the production of a new transgenic species have the potential to alter the path of evolution?

  1. The creation of new genes increases biodiversity.
  2. The removal of genes from a species decreases biodiversity.
  3. The transfer of genes within a species increases biodiversity.
  4. The transfer of genes between two species increases biodiversity.
Show Answers Only

`D`

Show Worked Solution

→ A new transgenic species arises through the transfer of genes between species.

`=>D`

BIOLOGY, M7 2016 HSC 15 MC

The table shows four features that can be used to distinguish between types of pathogens.

Which row of the table is correct?
 

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Feature}\rule[-1ex]{0pt}{0pt}& \quad \textit{Prion} \quad & \ \textit{Protozoan} \ \\
\hline
\rule{0pt}{2.5ex}\text{Contains DNA}\rule[-1ex]{0pt}{0pt}&\checkmark& X \\
\hline
\rule{0pt}{2.5ex}\text{Cellular}\rule[-1ex]{0pt}{0pt}& X & \checkmark\\
\hline
\rule{0pt}{2.5ex}\text{Can reproduce}\rule[-1ex]{0pt}{0pt}& \checkmark & X \\
\hline
\rule{0pt}{2.5ex}\text{Composed of protein}\rule[-1ex]{0pt}{0pt}& X & \checkmark \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(B\)

Show Worked Solution

By elimination:

→ Both contain DNA, can reproduce and are composed of protein therefore eliminate \(A\), \(C\) and \(D\)

→ Prion are non-cellular and protozoan are cellular, therefore \(B\).

\(\Rightarrow B\)


Mean mark 59%.

BIOLOGY, M8 2016 HSC 7 MC

The diagram shows a simplified model of a mammalian nephron and three processes labelled 1, 2 and 3.
 

Which row of the table correctly identifies each of the processes?
 

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\ \ \ \quad \quad 1 \rule[-1ex]{0pt}{0pt}& \ \ \quad \quad 2 & \ \ \quad \quad 3 \\
\hline
\rule{0pt}{2.5ex}\text{Reabsorption }\rule[-1ex]{0pt}{0pt}&\text{Filtration } &\text{Secretion }\\
\hline
\rule{0pt}{2.5ex}\text{Filtration }\rule[-1ex]{0pt}{0pt}& \text{Reabsorption } &\text{Secretion }\\
\hline
\rule{0pt}{2.5ex}\text{Secretion }\rule[-1ex]{0pt}{0pt}& \text{Reabsorption } &\text{Filtration }\\
\hline
\rule{0pt}{2.5ex}\text{Filtration }\rule[-1ex]{0pt}{0pt}& \text{Secretion } &\text{Reabsorption}\\
\hline
\end{array}
\end{align*}

Show Answers Only

\(B\)

Show Worked Solution

→ The only correct sequence is filtration, reabsorption and secretion.

\(\Rightarrow B\)

CHEMISTRY, M6 2018 HSC 6 MC

Sodium hydrogen carbonate is often used to clean up large spills of acids and alkalis.

Why is it a suitable chemical for this application?

  1. It is diprotic and is readily neutralised.
  2. It is amphiprotic, stable and easily handled.
  3. It is diprotic and easily cleaned up when neutralised.
  4. It is amphiprotic and only small quantities are required.
Show Answers Only

`B`

Show Worked Solution

→ Sodium hydrogen carbonate can donate and accept protons (amphiprotic).

→ This allows it to neutralise both acids and bases.

→ Sodium hydrogen carbonate usage also benefits from its stability.

`=>B`

CHEMISTRY, M7 2018 HSC 5 MC

Cellulose extracted from biomass is able to be used as a raw material in the manufacture of polymers because it

  1. is a condensation polymer.
  2. is a strong flexible molecule.
  3. produces carbon dioxide when burnt.
  4. contains a basic carbon-chain structure.
Show Answers Only

`D`

Show Worked Solution

→ The cellulose molecule can be used in polymer manufacture as it contains a basic carbon chain structure.

`=>D`

CHEMISTRY, M6 2018 HSC 29

The concentration of hydrochloric acid in a solution was determined by an acid base titration using a standard solution of sodium carbonate.

  1. Explain why sodium carbonate is a suitable compound for preparation of a standard solution.  (2 marks)

  2. A 25.00 mL sample of 0.1050 mol L¯1 sodium carbonate solution was added to a conical flask and three drops of methyl orange indicator added. The mixture was titrated with the hydrochloric acid and the following readings were recorded.
     

     
    Using the data from the table, calculate the concentration of the hydrochloric acid.  (3 marks)

  3. Explain the effect on the calculated concentration of hydrochloric acid if phenolphthalein is used as the indicator instead of methyl orange.  (2 marks)

Show Answers Only

a.    → \(\ce{Na2CO3}\) is a stable compound.

→ \(\ce{Na2CO3}\) is a pure solid that will not readily absorb water from the atmosphere.

→ An accurate weight of \(\ce{Na2CO3}\) can therefore be obtained in the experiment’s measurements.
 

b.    0.2425 mol L–¹
 

c.    → This is a strong acid / weak base titration.

→ Its equivalence point will occur at a pH less than seven and phenolphthalein changes colour in the pH range 10 – 8.3.

→ Phenolphthalein indicator would therefore signal the end point before equivalence (i.e. with a lower volume of acid). 

→ The calculated concentration of \(\ce{HCl}\) would be higher than the correct concentration.

Show Worked Solution

a.    → \(\ce{Na2CO3}\) is a stable compound.

→ \(\ce{Na2CO3}\) is a pure solid that will not readily absorb water from the atmosphere.

→ An accurate weight of \(\ce{Na2CO3}\) can therefore be obtained in the experiment’s measurements.
 


♦ Mean mark (a) 43%.

b.   \(\ce{Na2CO3(aq) + 2HCl(aq) -> 2NaCl(aq) + H2O(l) + CO2(g)}\)

\[\ce{Average titre = \frac{21.65 + 21.70 + 21.60}{3} = 21.65 mL}\]

\[\ce{n(Na2CO3) = c \times V = 0.1050 \times 0.0250 = 0.002625 mol}\]

\(\ce{n(HCl) = 2 \times n(Na2CO3) = 0.005250 mol}\)

\[\ce{[HCl] = \frac{n}{V} = \frac{0.005250}{0.02165} = 0.2425 mol L^{-1}}\]  

c.    → This is a strong acid / weak base titration.

→ Its equivalence point will occur at a pH less than seven and phenolphthalein changes colour in the pH range 10 – 8.3.

→ Phenolphthalein indicator would therefore signal the end point before equivalence (i.e. with a lower volume of acid). 

→ The calculated concentration of \(\ce{HCl}\) would be higher than the correct concentration.


♦♦♦ Mean mark (c) 29%.

CHEMISTRY, M8 2018 HSC 28

A camp stove using butane as a fuel was used to heat a pot of water inside a small tent. Poisonous carbon monoxide `(text{CO})` gas can be released from these stoves.
 

An investigation was carried out to determine the carbon monoxide concentration in the tent when the clearance height of the pot above the flame was altered. The results are shown in the table.

\begin{array} {|l|c|c|c|c|}
\hline \text{Clearance height (mm)} & 35 & 40 & 45 & 50 \\
\hline \text{CO concentration (ppm)} & 120 & 87 & 50 & 18\\
\hline \end{array}

  1. Construct a graph of the data on the grid.   (3 marks)

  1. Air containing a `(text{CO})` concentration above 30 ppm is considered unsafe to breathe.  
  2. What is the minimum clearance height at which the pot should be placed?   (1 mark)

  3. Increasing the clearance height decreases the efficiency of the stove according to the following table. 

\begin{array} {|l|c|c|c|c|}
\hline \text{Clearance height (mm)} & 35 & 40 & 45 & 50 \\
\hline \text{Efficiency (%)} & 90 & 70 & 50 & 30\\
\hline \end{array}

  1. A bushwalker only has 15.0 g of butane with which to heat 1.0 L of water with a starting temperature of 20°C.
  2. Calculate the highest temperature of the water that could safely be achieved in the tent. (Molar heat of combustion of butane: `DeltaH_(c)` = 2877 kJ mol ¯1)  (4 marks)

Show Answers Only

a.   
     

b.  48 mm

c.   91.1°C

Show Worked Solution

a.   
     

b.  From the graph, the minimum height above the flame is 48 mm.

c.    → The lowest safe distance = 48 mm (part (b)).

→ Fuel efficiency is 40% at 48 mm 

→ `DeltaH_(c) = 0.4 xx 2877 = 1151\ text{kJ mol}^{-1}`

\[\ce{n(C4H10) = \frac{m}{MM} = \frac{15}{4 \times 12.01 + 10 \times 1.008} = \frac{15}{58.12} = 0.258 mol}\]

\(\ce{Energy (C4H10) = 0.258 \times 1151 = 297 kJ}\)

`q` `=mC DeltaT`  
`297\ 000` `=1.0 xx 4.18 xx 10^3 xx Delta T`  
`Delta T` `=(297\ 000)/(4.18 xx 10^3)`  
  `=71.1°text{C}`  

 
`:.` Highest temperature of water = 20.0 + 71.1 = 91.1°C


Mean mark (c) 54%.

CHEMISTRY, M5 2018 HSC 25

The graph shows the number of molecules of \(\ce{N2}\) and \(\ce{H2}\) that possess a certain kinetic energy at two different temperatures.
 

With reference to the graph, explain why changing the temperature and adding a catalyst would change the rate of production of ammonia.  (4 marks)

Show Answers Only

→ Gas particles need to attain activation energy \( \text{E}_\text{A} \) in order to react when they collide.

→ \( \text{E}_\text{A} \) is a level of kinetic energy that is sufficient for reactions to occur.

→ Some \(\ce{N2}\) and \(\ce{H2}\) particles have a kinetic energy above \( \text{E}_\text{A} \) when the temperature is lower at \(\text{T}_1\) (see graph above).

→ However, the number of \(\ce{N2}\) and \(\ce{H2}\) particles with enough energy to react and produce ammonia is significantly greater at the higher temperature \(\text{T}_2\), where the graph has shifted to the right.

→ A catalyst lowers the \( \text{E}_\text{A} \). On the graph, this would shift the dotted \( \text{E}_\text{A} \) line to the left.

→ In the presence of a catalyst, both graphs show there will be a greater number of particles that can react in a given time and therefore the rate of ammonia production will increase in both cases.

Show Worked Solution

→ Gas particles need to attain activation energy \( \text{E}_\text{A} \) in order to react when they collide.

→ \( \text{E}_\text{A} \) is a level of kinetic energy that is sufficient for reactions to occur.

→ Some \(\ce{N2}\) and \(\ce{H2}\) particles have a kinetic energy above \( \text{E}_\text{A} \) when the temperature is lower at \(\text{T}_1\) (see graph above).

→ However, the number of \(\ce{N2}\) and \(\ce{H2}\) particles with enough energy to react and produce ammonia is significantly greater at the higher temperature \(\text{T}_2\), where the graph has shifted to the right.

→ A catalyst lowers the \( \text{E}_\text{A} \). On the graph, this would shift the dotted \( \text{E}_\text{A} \) line to the left.

→ In the presence of a catalyst, both graphs show there will be a greater number of particles that can react in a given time and therefore the rate of ammonia production will increase in both cases.


Mean mark 53%.

CHEMISTRY, M7 2018 HSC 24

The following apparatus was set up to test the reaction rate of fermentation of glucose at different temperatures.
 

 
 

  1. Write a balanced equation for the fermentation of glucose.  (1 mark)

  2. After 24 hours, 5.5 mL of gas was collected at 25°C and 100 kPa.
  3. Calculate the mass of glucose that would have been reacted.  (3 marks)

Show Answers Only

a.    \(\ce{C6H12O6(aq) -> 2C2H5OH(l) + 2CO2(g)}\)
 

b.    \(\ce{Vol(CO2) = 5.5\ ÷\ 1000 = 0.0055 L}\)

\[\ce{n(CO2) = \frac{V(CO2)}{24.79} = 2.2186 \times 10^{-4} mol}\]

\[\ce{n(C6H12O6) = \frac{1}{2} \times n(CO2) = 1.1093 \times 10^{-4} mol}\]

\(\ce{MM(C6H12O6) = (6 \times 12.01 + 12 \times 1.008 + 6 \times 16.00) = 180.156}\)

\(\ce{m(C6H12O6) = 1.1093 \times 10^{-4} \times 180.156 = 0.020 g}\)

Show Worked Solution
a.    \(\ce{C6H12O6(aq) -> 2C2H5OH(l) + 2CO2(g)}\)
 

Mean mark (a) 57%.
 

b.    \(\ce{Vol(CO2) = 5.5\ ÷\ 1000 = 0.0055 L}\)

\[\ce{n(CO2) = \frac{V(CO2)}{24.79} = 2.2186 \times 10^{-4} mol}\]

\[\ce{n(C6H12O6) = \frac{1}{2} \times n(CO2) = 1.1093 \times 10^{-4} mol}\]

\(\ce{MM(C6H12O6) = (6 \times 12.01 + 12 \times 1.008 + 6 \times 16.00) = 180.156}\)

\(\ce{m(C6H12O6) = 1.1093 \times 10^{-4} \times 180.156 = 0.020 g}\)

CHEMISTRY, M8 2018 HSC 22

A bottle of solution is missing its label. It is either `text{Pb(NO}_(3))_(2), text{Ba(NO}_(3))_(2)` or `text{Fe(NO}_(3))_(2)`

Using only `text{HCl, NaOH}` and `text{H}_(2) text{SO}_(4)` solutions, outline a sequence of steps that could be followed to confirm the identity of the solution in the bottle. Include observed results and ionic equations in your answer.    (4 marks)

Show Answers Only

“

Show Worked Solution

Prepare 1 test tube of the unknown solution.

Step 1: Add \(\ce{HCl}\) to the solution.

→ If a white precipitate forms, \(\ce{Pb(NO3)2}\) is present.

→ \(\ce{Pb^2+(aq) + 2Cl-(aq) -> PbCl2(s)}\)
 

Step 2: Add \(\ce{H2SO4}\) to solution if no precipitate in Step 1.

→ If a white precipitate forms, \(\ce{Ba(NO3)2}\) is present.

→ \(\ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)}\)
 

Step 3: Add \(\ce{NaOH}\) to solution if no precipitate in Step 1 and 2.

→ If a green precipitate forms, \(\ce{Fe(NO3)2}\) is present.

→ \(\ce{Fe^2+(aq) + 2OH-(aq) -> Fe(OH)2(s)}\)


Mean mark 56%.

CHEMISTRY, M7 2018 HSC 21

Ethylene can be readily transformed into many useful products.

  1. What is the name of the industrial process by which ethylene is obtained from long chain alkanes?   (1 mark)

  2. Ethylene can be converted into vinyl chloride.
  3. Draw structural formulae for vinyl chloride and its polymer, polyvinyl chloride.   (2 marks)

Show Answers Only

a.   Cracking

b.   Vinyl chloride:


      

Polyvinyl chloride:
 
     

Show Worked Solution

a.   Cracking

b.   Vinyl chloride:


      

Polyvinyl chloride:
 
     

CHEMISTRY, M7 2018 HSC 3 MC

An esterification reaction is to be performed.

Which of the following substances, when added, would increase the yield of the product?

  1. Water
  2. Boiling chips
  3. More alkanol
  4. Dilute sulfuric acid
Show Answers Only

`C`

Show Worked Solution

→ The addition of more alkanol increases one of the reactants in a reversible reaction.

→ By Le Chatelier’s principle, this would shift the equilibrium to the right hand side (increase yield).

`=>C`

BIOLOGY, M8 2018 HSC 9 MC

Sunken stomata can be found in the leaves of some Australian plants. A section of such a leaf is shown.
 

How do sunken stomata assist the plant to conserve water in a dry environment?

  1. They trap moist air, reducing humidity.
  2. They prevent entry of gases into the leaf.
  3. They accumulate moist air, reducing transpiration.
  4. They increase the surface area available for transpiration.
Show Answers Only

`C`

Show Worked Solution

→ Sunken stomata reduce air flow around it, which leads to the accumulation of moist air and thus a reduction in transpiration.

`=>C`

BIOLOGY, M8 2018 HSC 7 MC

Students dissected a mammalian kidney and drew the following diagram.
 

Which row in the table shows the correct labels for \(X, Y\) and \(Z\)?
 

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\quad \quad \quad \quad X\rule[-1ex]{0pt}{0pt}& \quad \quad \quad  \quad Y & \ \quad \quad \quad Z\\
\hline
\rule{0pt}{2.5ex}\text{Cortex}\rule[-1ex]{0pt}{0pt}&\text{Medulla} &\text{Ureter}\\
\hline
\rule{0pt}{2.5ex}\text{Renal pyramid}\rule[-1ex]{0pt}{0pt}& \text{Bowman's capsule}&\text{Renal artery} \quad \quad \\
\hline
\rule{0pt}{2.5ex}\text{Medulla}\rule[-1ex]{0pt}{0pt}& \text{Cortex} &\text{Ureter}\\
\hline
\rule{0pt}{2.5ex}\text{Bowman's capsule}\rule[-1ex]{0pt}{0pt}& \text{Cortex} &\text{Renal artery}\\
\hline
\end{array}
\end{align*}

Show Answers Only

\(C\)

Show Worked Solution

By Elimination:

→ The outer layer \(Y\) is the cortex, which contains glomeruli and Bowman’s capsules. (Eliminate A and B).

→ \(Z\)is the ureter, which takes urine to the bladder. The renal artery is the feature above it, which takes unfiltered blood to the kidney. (Eliminate D).

\(\Rightarrow C\)

ENGINEERING, CS 2022 HSC 23d

A solar panel is to be attached to the roof of a residential garage. A beam of mass 40 kg is shown. The solar panel generates a 25 kg point load as shown.   
 

Complete the shear force and bending moment diagrams for the beam, using an appropriate scale.   (4 marks)
 

Show Answers Only

Notes:

→ Ensure each point is labelled

→ Label each axis with units so they do not have to be repeated at every point

→ Ensure consistent scale and correct distances between forces

Show Worked Solution

Notes:

→ Ensure each point is labelled

→ Label each axis with units so they do not have to be repeated at every point

→ Ensure consistent scale and correct distances between forces

CHEMISTRY, M8 2016 HSC 19 MC

Excess barium nitrate solution is added to 200 mL of 0.200 mol L¯1 sodium sulfate.

What is the mass of the solid formed?

  1. 4.65 g
  2. 8.69 g
  3. 9.33 g
  4. 31.5 g
Show Answers Only

`C`

Show Worked Solution

\(\ce{n(Na_2SO_4) = c\ × V = 0.2\ ×\ 0.2 = 0.04 mol}\)

\(\ce{n(BaSO_4) = 0.04 mol (1 : 1 mole ratio)}\)

\(\ce{n(Ba_2SO_4) = n × MM = 0.04\ × (137.3 + 32.07 + 16\ × 4) =9.3348 g = 9.33 (3 sig fig)}\)

`=>C`

CHEMISTRY, M6 2016 HSC 18 MC

40 mL of 0.10 mol L¯1 \(\ce{NaOH}\) is mixed with 60 mL of 0.10 mol L¯1 \(\ce{HCl}\).

What is the pH of the resulting solution?

  1. 7.0
  2. 1.7
  3. 1.4
  4. 1.2
Show Answers Only

`B`

Show Worked Solution

\(\ce{n(HCl)_{excess} = 0.006-0.004=0.002 mol}\)

\[\ce{[HCl] = \frac{0.002}{0.040+0.060} = 0.0200 mol L^{-1}}\]

\(\ce{[HCl] = [H+]}\)

\(\ce{pH = -log(0.0200) = 1.7}\)

`=> B`

ENGINEERING, TE 2022 HSC 22c

An engineer is selecting a suitable material for the exterior housing of a telecommunications device such as the smart watch. Justify material properties that should be considered.   (3 marks)

Show Answers Only

Properties to be considered:

→ Hardness: To resist abrasions/scratches encountered in daily life

→ Toughness: To resist damage from sudden impacts such as falling over

→ Water resistance: Protect internal electronics from exposure to water

→ Hypoallergenic: Avoid damage to user’s skin

→ Corrosion resistance and chemical stability: Increase lifespan of smartwatch and prevent injury to user

Show Worked Solution

Properties to be considered:

→ Hardness: To resist abrasions/scratches encountered in daily life

→ Toughness: To resist damage from sudden impacts such as falling over

→ Water resistance: Protect internal electronics from exposure to water

→ Hypoallergenic: Avoid damage to user’s skin

→ Corrosion resistance and chemical stability: Increase lifespan of smartwatch and prevent injury to user

ENGINEERING, TE 2022 HSC 22a

A smart watch is shown.
 

Many smart watches have in-built functions that provide useful health and safety features for the wearer, such as heart rate monitoring, location tracking and fall detection.

Discuss ONE legal or ethical implication that the telecommunications engineer would need to consider in the design, function or manufacture of such a device.   (3 marks)

Show Answers Only

Answers could include one the following.

Functionality:

→ If the product does not work as advertised/expected, this creates a legal/ethical concern as people’s lives could be at risk.

→ For example if the location services fail in an emergency situation the user may not be found.

Privacy:

→ If the user’s privacy is not protected or their data is shared/leaked, there could be legal and/or ethical consequences for the engineer.

Ease of use:

→ The device should be created with the user in mind. It should be simple to use, accessible, as cheap as possible and come with instructions for use if necessary.

Show Worked Solution

Answers could include one the following.

Functionality:

→ If the product does not work as advertised/expected, this creates a legal/ethical concern as people’s lives could be at risk.

→ For example if the location services fail in an emergency situation the user may not be found.

Privacy:

→ If the user’s privacy is not protected or their data is shared/leaked, there could be legal and/or ethical consequences for the engineer.

Ease of use:

→ The device should be created with the user in mind. It should be simple to use, accessible, as cheap as possible and come with instructions for use if necessary.

Mean mark 58%.

ENGINEERING, PPT 2022 HSC 21c

To change a tyre on a bus, a system of pulleys, as shown, is used to lift the wheel and tyre to re-fit them back on the bus. The wheel and tyre have a combined mass of 95 kg.
 

What effort is required to lift the wheel and tyre if the efficiency of the pulley system is 80%?   (3 marks)

Show Answers Only

`296.875\ text{N}`

Show Worked Solution
`W` `= 95 xx 10 = 950\ text{N}`  
`VR` `= 4\text{ (number of ropes to load)}`  
`eta` `= (MA)/(VR)`  
`MA` `= VRxxeta`  
  `= 4xx0.8`  
  `= 3.2`  
`MA` `= L/E`  
`E` `= L/(MA)`  
  `= 950/3.2`  
  `=296.875\ text{N}`  
Mean mark 56%.

CHEMISTRY, M5 2016 HSC 28

A mixture of carbon monoxide, chlorine and phosgene \(\ce{(COCl_2)}\) gases was placed in a closed container. The concentrations of the gases were monitored over time.
 

  1. At what time does the system first reach equilibrium? Justify your answer.   (2 marks)

  2. At four minutes, the temperature of the container was increased.
  3. Explain, with reference to the graph, whether the decomposition of \(\ce{COCl_2}\) into \(\ce{CO}\) and \(\ce{Cl_2}\) is exothermic or endothermic.   (3 marks)

Show Answers Only

a.    At two minutes (where the concentrations stop changing)

b.    \(\ce{CO(g) + Cl2(g) \rightleftharpoons COCl2(g)}\)

→ From the graph, when the temperature is increased, the \(\ce{COCl2}\) concentration decreases while \(\ce{CO}\) and \(\ce{Cl2}\) concentrations increases.

→ The reaction is countering the increase in temperature by shifting to the left and absorbing heat (Le Chatelier’s principle).

→ Therefore the decomposition of \(\ce{COCl2(g)}\) is endothermic.

Show Worked Solution

a.    At two minutes (where the concentrations stop changing)

b.    \(\ce{CO(g) + Cl2(g) \rightleftharpoons COCl2(g)}\)

→ From the graph, when the temperature is increased, the \(\ce{COCl2}\) concentration decreases while \(\ce{CO}\) and \(\ce{Cl2}\) concentrations increases.

→ The reaction is countering the increase in temperature by shifting to the left and absorbing heat (Le Chatelier’s principle).

→ Therefore the decomposition of \(\ce{COCl2(g)}\) is endothermic.

CHEMISTRY, M8 2016 HSC 25

An unattended car is stationary with its engine running in a closed workshop. The workshop is 5.0 m × 5.0 m × 4.0 m and its volume is `1.0 × 10^(5)` L. The engine of the car is producing carbon monoxide in an incomplete combustion according to the following chemical equation:

\[\ce{C8H18(l) + \frac{17}{2}O2(g) -> 8CO(g) + 9H2O(l)} \]

Exposure to carbon monoxide at levels greater than 0.100 g L¯1 of air can be dangerous to human health.

6.0 kg of octane was combusted by the car in this workshop.

Using the equation provided, determine if the level of carbon monoxide produced in the workshop would be dangerous to human health. Support your answer with relevant calculations.   (4 marks)

Show Answers Only

\[\ce{n(C8H18) = \frac{m}{MM} = \frac{6000}{8 \times 12.01 + 18 \times 1.008} = \frac{6000}{114.224} = 52.53 mol}\]

\(\ce{n(CO) = 8 \times n(C8H18) = 420.24 mol}\)

\(\ce{m(CO) = n \times MM = 420.24 \times (12.01 + 16.00) = 11 771 g}\)

\[\ce{[CO] = \frac{11\ 771}{1 \times 10^{5}} = 0.118 g L^{-1}}\]

→ The carbon monoxide level is dangerous to human health as 0.118 g L¯1 exceeds the safe limit of 0.100 g L¯1 .

Show Worked Solution

\[\ce{n(C8H18) = \frac{m}{MM} = \frac{6000}{8 \times 12.01 + 18 \times 1.008} = \frac{6000}{114.224} = 52.53 mol}\]

\(\ce{n(CO) = 8 \times n(C8H18) = 420.24 mol}\)

\(\ce{m(CO) = n \times MM = 420.24 \times (12.01 + 16.00) = 11 771 g}\)

\[\ce{[CO] = \frac{11\ 771}{1 \times 10^{5}} = 0.118 g L^{-1}}\]

→ The carbon monoxide level is dangerous to human health as 0.118 g L¯1 exceeds the safe limit of 0.100 g L¯1 .

CHEMISTRY, M7 2016 HSC 23

A spirit burner containing ethanol was used to heat water in a conical flask for three minutes to measure the molar heat of combustion of ethanol.

The results from the investigation are shown.
 

  1. On the grid, draw a line graph to represent the data contained in the table.   (3 marks)
     

  1. The following values were also recorded during the investigation:
  2.       Initial mass of spirit burner = 236.14 g
  3.       Final mass of spirit burner = 235.56 g
  4.       Calculated experimental molar heat of combustion of ethanol = – 827 kJ mol ¯1.
  5. Using information from the previous page and the above values, determine the mass of water that was in the conical flask.   (3 marks)

Show Answers Only

a.   
     

b.  0.199 kg

Show Worked Solution

a.   


 

b.   Using the graph:

\(\triangle T = 31.0-18.5=12.5°\text{C} \)

\( \text{m(Ethanol burnt)} = 236.14-235.56=0.58\ \text{g} \)

\[ \ce{n(Ethanol) = \frac{m}{MM} = \frac{0.58}{2 \times 12.01 + 6 \times 1.008 + 16.00} = 0.0126 mol} \]

\(\ce{Energy released ($q$) = 0.0126 \times 827 = 10.420 kJ}\)

\[q = -mC\triangle T \ \ =>\ \ m= -\frac{q}{C\triangle T} \]

`:.m\text{(water)}` `=(10\ 420)/(4.18 xx 10^3 xx 12.5)` `=0.00019927\ text{g}`
  `=0.199\ text{kg}`  

Mean mark (b) 53%.

CHEMISTRY, M7 2016 HSC 15 MC

The table lists some properties of the straight-chained carbon compounds `W, X, Y` and `Z`.
 

Which row of the following table best identifies the compounds `W, X, Y` and `Z`?
 

\begin{align*} 
\begin{array}{c|c}
\rule{0pt}{2.5ex}\text{          }\rule[-1ex]{0pt}{0pt} \\
\rule{0pt}{2.5ex}\text{A.}\ \ \ \ \ \rule[-1ex]{0pt}{0pt} \\
\rule{0pt}{2.5ex}\text{B.}\ \ \ \ \ \rule[-1ex]{0pt}{0pt} \\
\rule{0pt}{2.5ex}\text{C.}\ \ \ \ \ \rule[-1ex]{0pt}{0pt} \\
\rule{0pt}{2.5ex}\text{D.}\ \ \ \ \ \rule[-1ex]{0pt}{0pt} \\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{W}\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex}\textit{X}\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex}\textit{Y}\rule[-1ex]{0pt}{0pt}& \rule{0pt}{2.5ex}\textit{Z}\rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex}\ \ \ \ce{C3H6}\ \ \ \rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex}\ \ \ \ce{C3H8}\ \ \ \rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex}\ce{CH3OH}\rule[-1ex]{0pt}{0pt}& \rule{0pt}{2.5ex}\ce{C4H9OH}\rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex}\ce{C3H8}\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex}\ce{C3H6}\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex}\ce{CH3OH}\rule[-1ex]{0pt}{0pt}& \rule{0pt}{2.5ex}\ce{C4H9OH}\rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex}\ce{C3H6}\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex}\ce{C3H8}\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex}\ce{C4H9OH}\rule[-1ex]{0pt}{0pt}& \rule{0pt}{2.5ex}\ce{CH3OH}\rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex}\ce{C3H8}\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex}\ce{C3H6}\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex}\ce{C4H9OH}\rule[-1ex]{0pt}{0pt}& \rule{0pt}{2.5ex}\ce{CH3OH}\rule[-1ex]{0pt}{0pt}\\
\hline
\end{array}
\end{align*}

Show Answers Only

`A`

Show Worked Solution

→ Compound W is unsaturated (C–C multiple bonds exist).

→ Compound X is saturated (all C–C bonds are single bonds).

→ Shorter alcohols, such as \(\ce{CH3OH}\) are more soluble.

`=>A`

CHEMISTRY, M5 2016 HSC 14 MC

Consider the following endothermic reaction taking place in a closed vessel.

\(\ce{N_2O_4($g$) \rightleftharpoons 2NO_2($g$)}\)

Which of the following actions would cause more \(\ce{N_2O_4}\) to be produced?

  1. Adding a catalyst
  2. Decreasing the volume
  3. Decreasing the pressure
  4. Increasing the temperature
Show Answers Only

`B`

Show Worked Solution

→ By decreasing the volume, the equilibrium will shift to the left so that less gas molecules are present (Le Chatelier’s principle).

`=>B`