SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

ENGINEERING, AE 2022 HSC 27b

The top and front views of a transition piece in an aircraft air conditioning duct are shown.

Complete a half-pattern development of the transition piece, starting from the line `a`-1 given below.   (6 marks)
 

     

Show Answers Only

 

Notes on this drawing:

→ Ensure you find the true length of lines

→ Ensure fold lines are lighter than the outline

→ Since the half development cuts through a side, ensure you use a break line

→ Do not dimension

→ Only complete a half development and not a full development

Show Worked Solution

 

Notes on this drawing:

→ Ensure you find the true length of lines

→ Ensure fold lines are lighter than the outline

→ Since the half development cuts through a side, ensure you use a break line

→ Do not dimension

→ Only complete a half development and not a full development


♦ Mean mark 40%.

ENGINEERING, CS 2017 HSC 27b

An engineering team has been contracted to design a multi-function lifting device for a coastal container wharf.

The table shows some of the engineering design elements for this lifting device.
 

Explain how the lifting device can be tested and evaluated to determine if the criteria for the listed engineering elements are met.   (6 marks)

Show Answers Only

Show Worked Solution


♦ Mean mark 46%.

ENGINEERING, AE 2017 HSC 26c

In modern aircraft, the external skin is riveted to the frame using solution treated and quenched aluminium 4% copper alloy rivets. These rivets are used immediately to attach the external skin of the aircraft to the frame.

Describe the changes that occur to the structure and properties of these rivets after installation.   (3 marks)

Show Answers Only

Changes to structure and properties of the rivet

→ The microstructure of the rivet, after quenching, is a solid unstable solution of copper dissolved in aluminium.

→ On reaching room temperature, the heat energy created initiates the copper’s precipitation out of solid solution as finely distributed precipitates \(\ce{(CuAl2)} \), which strengthens the alloy considerably.

→ Work hardening is produced by cold working the metal when the rivet head is formed.

→ The rivet is significantly strengthened due to the installation, and its hardness will increase until precipitation hardening is finalised.

→ Corresponding decreases in the ductility of the metal will also be evident. 

Show Worked Solution

Changes to structure and properties of the rivet

→ The microstructure of the rivet, after quenching, is a solid unstable solution of copper dissolved in aluminium.

→ On reaching room temperature, the heat energy created initiates the copper’s precipitation out of solid solution as finely distributed precipitates \(\ce{(CuAl2)} \), which strengthens the alloy considerably.

→ Work hardening is produced by cold working the metal when the rivet head is formed.

→ The rivet is significantly strengthened due to the installation, and its hardness will increase until precipitation hardening is finalised.

→ Corresponding decreases in the ductility of the metal will also be evident. 


♦♦ Mean mark 27%.

ENGINEERING, TE 2017 HSC 23b

The digital TV receiver uses zener diodes.

Compare the operation of zener diodes with common diodes used in electrical circuits.   (3 marks)

Show Answers Only

Common diodes

→ Allows current to flow in a forward direction.

→ If exposed to a sufficiently large voltage, common diodes break down.

→ Produced by doping silicon.

→ When subjected to a reverse voltage, common diodes can leak current.

→ Used for rectification.
 

Zener diodes

→ Semiconductor device that allows current to flow in either a forward or reverse direction.

→ Produced by doping silicon using a different method to that of common diodes.

→ Used for voltage regulation.

→ When breakdown voltage is exceeded they provide a stable reference voltage over a large range of currents when connected in reverse in a circuit.

→ Reverse currents don’t damage zener diodes.

Show Worked Solution

Common diodes

→ Allows current to flow in a forward direction.

→ If exposed to a sufficiently large voltage, common diodes break down.

→ Produced by doping silicon.

→ When subjected to a reverse voltage, common diodes can leak current.

→ Used for rectification.
 

Zener diodes

→ Semiconductor device that allows current to flow in either a forward or reverse direction.

→ Produced by doping silicon using a different method to that of common diodes.

→ Used for voltage regulation.

→ When breakdown voltage is exceeded they provide a stable reference voltage over a large range of currents when connected in reverse in a circuit.

→ Reverse currents don’t damage zener diodes.


♦♦♦ Mean mark 29%.

ENGINEERING, AE 2017 HSC 22b

Explain how a carbon fibre bicycle frame is manufactured.   (3 marks)

Show Answers Only

→ A mould that is either a ‘bladder’ (deflated when frame cured) or a smooth steel mould is used.

→ Pre-impregnated carbon fibres are wrapped around the mould.

→ Fibre wrapping can utilise pre-manufactured sheets, hand laid over the mould or via roll-wrapping.

→ Alternatively, carbon fibre can be applied directly to the mould using filament-winding.

→ The frame is then put inside a pressure chamber (may be a heated autoclave) in a vacuum-proof bag.

→ A catalyst that may be UV light or heat sensitive, is then used to activate the resin.

Show Worked Solution

→ A mould that is either a ‘bladder’ (deflated when frame cured) or a smooth steel mould is used.

→ Pre-impregnated carbon fibres are wrapped around the mould.

→ Fibre wrapping can utilise pre-manufactured sheets, hand laid over the mould or via roll-wrapping.

→ Alternatively, carbon fibre can be applied directly to the mould using filament-winding.

→ The frame is then put inside a pressure chamber (may be a heated autoclave) in a vacuum-proof bag.

→ A catalyst that may be UV light or heat sensitive, is then used to activate the resin.


♦♦♦ Mean mark 23%.

ENGINEERING, PPT 2017 HSC 21b

Gears used in automotive engines can be manufactured using ferrous alloys.

  1. Sand casting and powder metallurgy are methods that can be used.
  2. Compare the properties of the gears manufactured by each method.   (4 marks)
  1. After manufacture the gears are case hardened.
  2. Describe how case hardening produces the required structure-property relationships for this application.   (3 marks)
Show Answers Only

i.   Sandcasting

→ Weaker dimensional stability than powder metallurgy.

→ If not poured correctly, grains can be columnar.

→ Fatigue can be initiated through poor surface finish.
 

Powder metallurgy

→ Greater dimensional stability.

→ A variety of gears that cannot be made using conventional methods can be produced using powder metallurgy.

→ A custom component can be initially produced as close as possible to the final shape of the product (near net shape forming).

→ Alloys can be produced with pores (self-lubricating).
 

ii.  Case hardening

→ Gears are heated to red heat in a nitrogen and/or carbon environment.

→ Nitrogen and carbon disperse into the steel’s surface at these high temperatures, increasing the carbon content to a level where it is hardened by quenching.

→ Martensite is formed when the steel is quenched.

→ Hence, the inside remains soft and tough whilst the outer case becomes wear resistant and hard.

→ The outer casing may be tempered back for improved surface toughness or the gear can be used with the martensite casing.

Show Worked Solution

i.   Sandcasting

→ Weaker dimensional stability than powder metallurgy

→ If not poured correctly, grains can be columnar

→ Fatigue can be initiated through poor surface finish.
 

Powder metallurgy

→ Greater dimensional stability

→ A variety of gears that cannot be made using conventional methods can be produced using powder metallurgy

→ A custom component can be initially produced as close as possible to the final shape of the product (near net shape forming)

→ Alloys can be produced with pores (self-lubricating).

 


♦♦♦ Mean mark (i) 26%.

ii.  Case hardening

→ Gears are heated to red heat in a nitrogen and/or carbon environment.

→ Nitrogen and carbon disperse into the steel’s surface at these high temperatures, increasing the carbon content to a level where it is hardened by quenching.

→ Martensite is formed when the steel is quenched.

→ Hence, the inside remains soft and tough whilst the outer case becomes wear resistant and hard.

→ The outer casing may be tempered back for improved surface toughness or the gear can be used with the martensite casing.


♦♦♦ Mean mark (ii) 23%.

ENGINEERING, PPT 2019 HSC 24b

Normalised high-tensile steel has been chosen for the manufacture of a wing support beam.

  1. Draw and label the microstructures of a normalised high-tensile steel and an annealed high-tensile steel.   (2 marks)
     

  1. Explain how the microstructure produced by normalising high-tensile steel improves the steel's suitability for this application.   (2 marks)
Show Answers Only

i.    Microstructures

 

ii.    → The strength of the steel is markedly increased.

→ Normalising produces finer and more uniform grains.

Show Worked Solution

i.    Microstructures


♦♦♦ Mean mark (i) 26%.

ii.    → The strength of the steel is markedly increased.

→ Normalising produces finer and more uniform grains.


♦ Mean mark (ii) 41%.

BIOLOGY, M8 2019 HSC 33b

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

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 .

  1.  What are multiple alleles?   (2 marks)

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

   

  1. Analyse the data to assess the risk of developing Alzheimer's disease associated with the e2, e3 and e4 alleles.   (4 marks)

Show Answers Only

i.     → Alleles are the different variations of the same gene.

→ While most genes only have two alleles, dominant and recessive, some genes have 3 or more versions of itself. This phenomena is referred to as the gene having “multiple alleles”.
  

ii.  Analysis of data 

→ The table indicates that the alleles follow a hierarchy and have influence over the risk of Alzheimer’s, with certain combinations masking effects of others or amplifying them.

→ The e2 allele in both a homozygous genotype and coupled with e3 reduces the risk of Alzheimer’s by 40%. This suggests that the e2 allele is the one which reduces the risk of Alzheimer’s and can mask the effect of e3.

→ However, when coupled with e4, the risk of developing Alzheimer’s is 2.6 times more likely. This suggests that e4 is the more dominant allele.

→ The e4/e3 genotype also makes Alzheimer’s 3.2 times more likely in those individuals, and the e4/e4 genotypes makes it 14.9 times more likely. We can then make the conclusion that the e4 allele makes Alzheimer’s much more common in any genotype where it is present.

→ The e3 allele seems to be completely neutral and almost completely masked by both e2 and e4. In its homozygous genotype, it has no effect on the risk of developing Alzheimer’s, and when heterozygous with either e2 or e4, has little to no effect on the risk in comparison the e2 and e4’s homozygous genotypes.

→ It would then be accurate to conclude that the allele hierarchy is e3<e2<e4, with e3 being neutral and having no known effect, e2 reducing the risk of Alzheimer’s and e4 greatly increasing the risk of developing Alzheimer’s.

Show Worked Solution

i.     → Alleles are the different variations of the same gene.

→ While most genes only have two alleles, dominant and recessive, some genes have 3 or more versions of itself. This phenomena is referred to as the gene having “multiple alleles”.
 


♦♦♦ Mean mark (i) 24%.

ii.  Analysis of data 

→ The table indicates that the alleles follow a hierarchy and have influence over the risk of Alzheimer’s, with certain combinations masking effects of others or amplifying them.

→ The e2 allele in both a homozygous genotype and coupled with e3 reduces the risk of Alzheimer’s by 40%. This suggests that the e2 allele is the one which reduces the risk of Alzheimer’s and can mask the effect of e3.

→ However, when coupled with e4, the risk of developing Alzheimer’s is 2.6 times more likely. This suggests that e4 is the more dominant allele.

→ The e4/e3 genotype also makes Alzheimer’s 3.2 times more likely in those individuals, and the e4/e4 genotypes makes it 14.9 times more likely. We can then make the conclusion that the e4 allele makes Alzheimer’s much more common in any genotype where it is present.

→ The e3 allele seems to be completely neutral and almost completely masked by both e2 and e4. In its homozygous genotype, it has no effect on the risk of developing Alzheimer’s, and when heterozygous with either e2 or e4, has little to no effect on the risk in comparison the e2 and e4’s homozygous genotypes.

→ It would then be accurate to conclude that the allele hierarchy is e3<e2<e4, with e3 being neutral and having no known effect, e2 reducing the risk of Alzheimer’s and e4 greatly increasing the risk of developing Alzheimer’s.


♦♦ Mean mark (ii) 40%.

ENGINEERING, CS 2019 HSC 27bi

The bracket and lock pin assembly shown is used to attach the repeater transmitters to the tower.
 

Using the data given, determine the minimum lock pin diameter to use.   (3 marks)
 

Show Answers Only

`text{7 mm}`

Show Worked Solution

`E = 210\ text{GPa, FoS = 5 , Total Load = 3.5 kN}`

`text{Shear Strength of Pin}\ (sigma_s) = 240\ text{MPa}`

`text{Allowable}\ sigma_s = 240/5 = 48`

`sigma_s` `=P/(2A)\ \ text{(double shear)}`  
`2A` `=P/(sigma_s)`  
`A` `=3500/(2 xx 48)`  
  `=36.4583\ text{mm}^2`  

 

`A` `=(pi xx d^2)/4`  
`d` `=sqrt((4A)/pi)`  
  `=sqrt((4 xx 36.4583)/pi)`  
  `=6.813\ text{mm}`  

   
`text{∴ Pin diameter would be 7 mm (next available > 6.813 mm)}`


♦♦♦ Mean mark 30%.

ENGINEERING, TE 2019 HSC 26b

A simple circuit diagram of the AM radio receiver is shown.
 

  1. Complete the table giving the function of each of the circuit components listed.   (5 marks)
     

  1. Complete the table by drawing the waveform at positions B and C.   (2 marks)
     

  1. This AM radio receiver produced a demodulated electrical signal with a small current. The speakers in the earphones converted this signal into sound.
  2. Explain why the speakers in the earphones required a high impedance (resistance) in order to produce sound of sufficient volume to be heard.   (2 marks)
Show Answers Only

“

Show Worked Solution

i.  


♦ Mean mark (i) 47%.

ii.  


♦♦♦ Mean mark (ii) 22%.

iii.   → An external power source is not used to power this radio.

→ The signal picked up by the aerial powers the speakers.

→ If the impedance (resistance) is high an audible signal can be generated by the speaker at a low current.

→ Because the earphone impedance is high, the sensitivity of the coil improves, resulting in a more accurately tuned signal.

→ The sound would be inaudible if the impedance is low, as the power to the earpiece would also be low.


♦♦♦ Mean mark (iii) 15%.

BIOLOGY, M5 2020 HSC 32b

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.
 

 

  1. Use the information provided in Diagram 1 to explain why the rabies virus cannot be classified as a cellular pathogen.   (3 marks)

  2. After infection the virus reproduces in muscle cells near the bite site and in the central nervous system. This requires the single-stranded rabies RNA to be transcribed, translated and replicated in the cytoplasm of host cells. These processes are shown in Diagram 2.
  3. Use the information provided in Diagrams 1 and 2 to explain the role of viral RNA polymerase in the reproduction of the virus.   (5 marks)

Show Answers Only

i.   Rabies virus cannot be a singular pathogen:

→ This pathogen contains only a single strand of RNA which only codes for five proteins.

→ Cellular pathogens such as bacteria contain a much larger genome in the form of DNA that allows the pathogen to perform complex processes without relying on a host.
 

ii.   The Role of Viral RNA 

→ The viral RNA polymerase (which is made from L and P proteins) is responsible for the production of the viral proteins and RNA, components of the rabies viruses.

→ RNA polymerase is responsible for transcription of viral RNA into mRNA, which is then used by the host’s ribosomes to produce respective viral proteins.

→ RNA polymerase is also responsible for the replication of the viral RNA. In this process, a complementary RNA strand is produced from the original RNA strand. This strand is then used as a template for RNA polymerase to rapidly produce more RNA, complementary to the template. The new RNA will therefore be identical to the original.

→ In this way, RNA polymerase is essential in producing viral proteins and new RNA strands which form new rabies virus particles.

Show Worked Solution

i.   Rabies virus cannot be a singular pathogen:

→ This pathogen contains only a single strand of RNA which only codes for five proteins.

→ Cellular pathogens such as bacteria contain a much larger genome in the form of DNA that allows the pathogen to perform complex processes without relying on a host.


♦♦ Mean mark (i) 33%.

ii.   The Role of Viral RNA 

→ The viral RNA polymerase (which is made from L and P proteins) is responsible for the production of the viral proteins and RNA which are components of the rabies viruses.

→ RNA polymerase is responsible for transcription of viral RNA into mRNA, which is then used by the host’s ribosomes to produce viral proteins.

→ RNA polymerase is also responsible for the replication of the viral RNA. In this process, a complementary RNA strand is produced from the original RNA strand. This strand is then used as a template for RNA polymerase to rapidly produce more RNA, complementary to the template. The new RNA will therefore be identical to the original.

→ In this way, RNA polymerase is essential in producing viral proteins and new RNA strands which form new rabies virus particles.


♦♦♦ Mean mark (ii) 29%.

ENGINEERING, CS 2019 HSC 22c

The diagram shows some dimensions and forces associated with a telecommunications tower.
 

By considering any necessary reaction, calculate the magnitude of the forces in members `M` and `N`. State the nature of each force. Ignore the weight of the tower.   (6 marks)

Show Answers Only

“

Show Worked Solution

Forces at Joint `A`

Horizontal forces `=0`

`:.` To calculate vertical force at `A ` → use moments.


♦♦ Mean mark 36%.
\({\circlearrowright}\)`+SigmaM_C` `=0`  
`0` `=-(12xx4)+(R_Axx12)-(10xx7)-(3xx18)`  
`12R_A` `=48+70+54`  
`R_A` `=172/12`  
  `=14.33\ text{kN}↑`  

    
Forces in Member `N` → method of joints at `A`

→ No horizontal forces

→ Member `AC` redundant and carrying no load

→ `F_(up) = F_(down)`

`:.` Member `AB` in compression (the force acting down on joint `A` from member `AB` is 14.33 kN)

`:.` Force in N = 14.33 kN (compression)

  
Using Method of Sections → take moments about Joint `H`

Find the perpendicular distance `d`

`BH^2` `=18^2+6^2`  
`BH` `=sqrt{360}`  
`sin\ 40.6º` `=d/(sqrt{360})`  
`d` `=sqrt{360}xx sin\ 40.6º`  
`d` `=12.348\ text{m}`  

 

\({\circlearrowright}\)`+SigmaM_H` `=0`
`0` `=+(12xx2)+(Mxx12.348)-(7xx4)`
`12.348M` `=-24+28`
`M` `=4/12.348`
`M` `=0.324\ text{kN (tension)}`

  

`:.\ ` `M` `=0.324\ text{kN (tension)}`
  `N` `=14.33\ text{kN (compression)}`

ENGINEERING, PPT 2019 HSC 19 MC

A webbed flange is shown.
 

Which image correctly represents section A–A?
 

Show Answers Only

`C`

Show Worked Solution

By Elimination:

→ Webs do not get sectioned (eliminate `B` and `D`)

→ The upright cylinder itself must be sectioned (eliminate `A`)

`=>C`


♦♦♦ Mean mark 26%.

ENGINEERING, CS 2019 HSC 5 MC

Using computer-aided drawing (CAD) software a new outline was produced 18.32 mm away from the original outline of an item as shown.
 

Which CAD command can produce this result most efficiently?

  1. Trim
  2. Offset
  3. Mirror
  4. Expand
Show Answers Only

`B`

Show Worked Solution

→ Although there is some disparity between certain CAD programs, this is generally achieved using the offset tool.

`=>B`


♦♦♦ Mean mark 20%.

ENGINEERING, CS 2022 HSC 26a

  1.  The diagram shows a tower crane being used in the construction of a building.
     

  1. Determine the number of 100 kg concrete blocks required to place the boom arm in equilibrium.   (2 marks)
  1. Under a different set of conditions, a wind force is applied, as shown in the diagram.
     

  1. Determine the magnitude and nature of the internal reaction in member A.   (6 marks)
Show Answers Only

i.   `12`

ii.  `A=25\ text{kN in compression}`

Show Worked Solution

i.   Let the total weight of the concrete blocks `= x`

`50x` `=9.23xx65`  
`50x` `=600`  
`x` `=600/50`  
  `=12\ text{kN ↓}`  
  `=12\ 000\ text{N ↓}`  
`m` `=1200\ text{kg}`  

 
∴ 12 × 100 kg concrete blocks are needed for the counterweight.


♦ Mean mark (i) 46%.

ii.   Magnitude and nature of internal reaction

\( \circlearrowright+\Sigma M_R \) `=0`  
`0` `=-(6xx5.5)+(R_Lxx1)+(10xx6)-(10.4xx7)`  
`R_L` `=45.8\ text{kN}↑`  

  
Taking the horizontal section shown:


  

\( \circlearrowright + \Sigma M_P \) \(= 0\)  
`0` `=(Axx1)+(45.8xx1)-(10.4xx2)`  
`A` `=-25\ text{kN}`  

  
∴ `A=25\ text{kN in compression}`


♦ Mean mark (ii) 43%.

PHYSICS, M8 2015 HSC 34e

Assess the impact of THREE advances in knowledge about particles and forces on the understanding of the atomic nucleus.   (6 marks)

Show Answers Only

Three of many possible developments are included below.

Advance One:

→ The discovery of the neutron allowed scientists to understand the masses of the nuclei.

→ This discovery enabled scientists to better identify trends in both the periodic table.

Advance Two:

→ Knowledge of the strong nuclear force helps us to explain the interaction between protons and neutrons in the nucleus, and how this force can overcome the electrostatic  force of repulsion.

→ This discovery helps to explain why certain isotopes are unstable.

Advance Three:

→ Knowledge that protons and neutrons are made from different combinations of two types of quarks.

→ This helped to unify our understanding of subatomic particles, informing our base knowledge of quantum physics through the development of the Standard Model.

Show Worked Solution

Three of many possible developments are included below.

Advance One:

→ The discovery of the neutron allowed scientists to understand the masses of the nuclei.

→ This discovery enabled scientists to better identify trends in both the periodic table.

Advance Two:

→ Knowledge of the strong nuclear force helps us to explain the interaction between protons and neutrons in the nucleus, and how this force can overcome the electrostatic  force of repulsion.

→ This discovery helps to explain why certain isotopes are unstable.

Advance Three:

→ Knowledge that protons and neutrons are made from different combinations of two types of quarks.

→ This helped to unify our understanding of subatomic particles, informing our base knowledge of quantum physics through the development of the Standard Model.


♦♦♦ Mean mark 32%.

PHYSICS, M8 2016 HSC 33e

Describe how the distribution of stars on a Hertzsprung-Russell diagram relates to the processes that occur during their evolution.   (6 marks)

Show Answers Only

→ An H-R diagram distributes stars into different groupings that relate to the processes that occurred during their evolution.

→ Hydrogen fusion is the primary source of energy of stars on the main sequence.

→ Hydrogen fusion is replaced by helium fusion as the main source of energy in the star’s next evolutionary phase. This grouping of stars is also known as red giants.

→ After helium fusion, the next evolutionary stage for most stars involves gravitational collapse. At this stage, the surface of the star recedes and gravitational potential energy is converted to radiant energy. This grouping is known as white dwarfs.

→ A star’s transition between evolutionary phases occurs quickly relative to time spent in each group. 

→ This transition speed results in fewer stars being distributed in areas outside of these groups on the H-R diagram.
 

Other possible answers could include:

→ Reference to other groups such as protostars, supergiants

→ Sketch of H-R diagram

→ Reference to properties of stars related to their distribution within particular groups

→ Globular and open clusters.

Show Worked Solution

→ An H-R diagram distributes stars into different groupings that relate to the processes that occurred during their evolution.

→ Hydrogen fusion is the primary source of energy of stars on the main sequence.

→ Hydrogen fusion is replaced by helium fusion as the main source of energy in the star’s next evolutionary phase. This grouping of stars is also known as red giants.

→ After helium fusion, the next evolutionary stage for most stars involves gravitational collapse. At this stage, the surface of the star recedes and gravitational potential energy is converted to radiant energy. This grouping is known as white dwarfs.

→ A star’s transition between evolutionary phases occurs quickly relative to time spent in each group. 

→ This transition speed results in fewer stars being distributed in areas outside of these groups on the H-R diagram.
 

Other possible answers could include:

→ Reference to other groups such as protostars, supergiants

→ Sketch of H-R diagram

→ Reference to properties of stars related to their distribution within particular groups

→ Globular and open clusters.


♦♦ Mean mark 44%.

BIOLOGY, M8 2018 HSC 30

The graph shows the expected life span (the age to which people are expected to live in years) for people of different ages during the 20th century in one country.
 

There have been many biological developments that have contributed to our understanding of the identification, treatment and prevention of disease.

Evaluate the impact of these developments on the expected life span. In your answer, include reference to trends in the data provided.  (8 marks)

Show Answers Only

→ For all ages listed in the graph, life expectancy increased during the 20th century.

→ The lifespan from birth has increased more significantly than other ages ~ 48 to 74 years.

→ The smallest increase being for 60 year olds at ~ 5 years.

→ The ability to understand pathogens and the causes of infectious disease (Koch and Pasteur) has led to early identification and treatment of childhood illnesses such as rubella, polio and whooping cough.

→ Koch and Pasteur established germ theory, culture techniques and a set of postulates to follow in order to create the link between a particular pathogen and disease.

→ Vaccines to combat childhood illnesses were developed through a knowledge of germ theory.

→ The infant/childhood mortality rate has improved significantly, and hence life expectancy, due to the immunity provided by vaccines.

→ An understanding of inherited disorders has also improved lifespans with early diagnosis and prenatal genetic screening for genetic disorders and illnesses.

→ Antibiotic remedies were developed to combat bacterial diseases such as Staphylococcus aureus, due to an understanding of the difference between prokaryotic and eukaryotic cells.

→ With the use of antibiotics many diseases were then no longer life threatening, leading to improved mortality rates across all ages.

→ However, bacterial resistance has resulted with the overuse of antibiotics, so some diseases are now unresponsive to antibiotic treatment.

→ Epidemiology studies involving intricate planning and design, control groups and large scale analysis of data have lead to improvements in the treatment of non-infectious diseases such as cancer.

→ For example the discovery of links between smoking and lung cancer, sun exposure and melanoma, obesity and type II diabetes, has lead to widespread public health campaigns to inform people of the health risks and lowered the associated mortality rates.

→ Improved hygiene, food storage and preservation, and water filtration also occurred in the 20th century leading to fewer preventable diseases and hence increased life spans for all age groups.

→ Improved quarantine requirements have helped prevent the spread of plant, animal and human diseases via international travel.

→In conclusion, developments in biology have lead to increased life expectancy across all age groups, with the biggest improvements for babies and children.

→ These benefits are not necessarily a worldwide phenomenon as poor living conditions and access to medical treatment is not available in many poor socioeconomic communities.

Show Worked Solution

→ For all ages listed in the graph, life expectancy increased during the 20th century.

→ The lifespan from birth has increased more significantly than other ages ~ 48 to 74 years.

→ The smallest increase being for 60 year olds at ~ 5 years.

→ The ability to understand pathogens and the causes of infectious disease (Koch and Pasteur) has led to early identification and treatment of childhood illnesses such as rubella, polio and whooping cough.

→ Koch and Pasteur established germ theory, culture techniques and a set of postulates to follow in order to create the link between a particular pathogen and disease.

→ Vaccines to combat childhood illnesses were developed through a knowledge of germ theory.

→ The infant/childhood mortality rate has improved significantly, and hence life expectancy, due to the immunity provided by vaccines.

→ An understanding of inherited disorders has also improved lifespans with early diagnosis and prenatal genetic screening for genetic disorders and illnesses.

→ Antibiotic remedies were developed to combat bacterial diseases such as Staphylococcus aureus, due to an understanding of the difference between prokaryotic and eukaryotic cells.

→ With the use of antibiotics many diseases were then no longer life threatening, leading to improved mortality rates across all ages.

→ However, bacterial resistance has resulted with the overuse of antibiotics, so some diseases are now unresponsive to antibiotic treatment.

→ Epidemiology studies involving intricate planning and design, control groups and large scale analysis of data have lead to improvements in the treatment of non-infectious diseases such as cancer.

→ For example the discovery of links between smoking and lung cancer, sun exposure and melanoma, obesity and type II diabetes, has lead to widespread public health campaigns to inform people of the health risks and lowered the associated mortality rates.

→ Improved hygiene, food storage and preservation, and water filtration also occurred in the 20th century leading to fewer preventable diseases and hence increased life spans for all age groups.

→ Improved quarantine requirements have helped prevent the spread of plant, animal and human diseases via international travel.

→In conclusion, developments in biology have lead to increased life expectancy across all age groups, with the biggest improvements for babies and children.

→ These benefits are not necessarily a worldwide phenomenon as poor living conditions and access to medical treatment is not available in many poor socioeconomic communities.


♦♦ Mean mark 35%.

BIOLOGY, M5 2018 HSC 29

The diagram models the process of meiosis.
 

  1. Describe the process that accounts for the changes shown in the model during interphase.   (2 marks)

  2. Explain the structure and behaviour of chromosomes in the first division of meiosis. Include detailed reference to the model.   (5 marks)

Show Answers Only

a.   Interphase:

→ In preparation for cell division, DNA is replicated during interphase with the cell.

→ During replication the bonds between the hydrogen bases are broken and the DNA strands (double helix) are unwound into separate strands.

→ Polymerase enzymes add complementary nucleotides to each strand until the strands are identical DNA copies.

→ Half of the original strand is contained in each copy.
 

b.   The first division of meiosis:

→ Homologous chromosomes are represented in the model as the same size.

→ Different shading is used to distinguish between paternal or maternal origins.

→ Individual chromosomes form homologous pairs during the first division of meiosis.

→ These pairs have the same genes but the alleles are not identical.

→ Pairing results in crossing over and an exchange in genetic material between non-sister chromatids takes place.

→ The shading in the diagram illustrates these new combinations.

→ Half the number of chromosomes are possessed by the two resultant daughter cells (diploid cells ⇒ haploid cells).

→ These are randomly assigned, one from each homologous pair.

→ Therefore, when they return to the diploid number, new combinations of chromosomes and greater genetic variation results.

Show Worked Solution

a.   Interphase:

→ In preparation for cell division, DNA is replicated during interphase with the cell.

→ During replication the bonds between the hydrogen bases are broken and the DNA strands (double helix) are unwound into separate strands.

→ Polymerase enzymes add complementary nucleotides to each strand until the strands are identical DNA copies.

→ Half of the original strand is contained in each copy.
 


♦♦ Mean mark (a) 23%.

b.   The first division of meiosis:

→ Homologous chromosomes are represented in the model as the same size.

→ Different shading is used to distinguish between paternal or maternal origins.

→ Individual chromosomes form homologous pairs during the first division of meiosis.

→ These pairs have the same genes but the alleles are not identical.

→ Pairing results in crossing over and an exchange in genetic material between non-sister chromatids takes place.

→ The shading in the diagram illustrates these new combinations.

→ Half the number of chromosomes are possessed by the two resultant daughter cells (diploid cells ⇒ haploid cells).

→ These are randomly assigned, one from each homologous pair.

→ Therefore, when they return to the diploid number, new combinations of chromosomes and greater genetic variation results.


♦ Mean mark (b) 45%.

BIOLOGY, M5 2017 HSC 24

  1. Three genes are arranged along a homologous pair of chromosomes as shown.
     

   

    1. What is the individual's genotype before crossing over occurs?   (1 mark)

    2. Label, on the diagram below, the alleles after crossing over has occurred.   (1 mark)
       

   

  1. Explain the effect of independent assortment of chromosomes on the genotype of the offspring.   (2 marks)

 

Show Answers Only

a.i   `text{Aa Bb Gg}`

a.ii


 

b.  Independent Assortment

→ A random alignment of homologous chromosomes takes place during meiosis.

→ The possible number of chromosome combinations is consequently increased.

→ Therefore, the genetic variation of offspring increases.

Show Worked Solution

a.i   `text{Aa Bb Gg}`

a.ii


♦♦♦ Mean mark (a)(i) 11% 
♦♦ Mean mark (a)(ii) 35%.

b.  Independent Assortment

→ A random alignment of homologous chromosomes takes place during meiosis.

→ The possible number of chromosome combinations is consequently increased.

→ Therefore, the genetic variation of offspring increases.


♦ Mean mark (b) 48%.

BIOLOGY, M8 2017 HSC 9 MC

An experiment was planned to investigate the effect of the enzyme, amylase, on starch.

The following combination of test tubes was considered.
 
     

Two drops of iodine will be added to each test tube.

Which combination of test tubes would ensure that the experiment is valid?
 

Show Answers Only

`A`

Show Worked Solution

→ `W` and `Y` only differ by the addition of the amylase, so the experiment is valid.

`=>A`


♦♦♦ Mean mark 25%.

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.
 

  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, M8 2016 HSC 19-20 MC

Refer to the following information to answer Questions 19 and 20.
 

Question 19

What can be inferred from the scientists' discovery?

  1. Cancer cells carry unique antigens.
  2. Self-antigens are not present on cancer cells.
  3. The melanoma patient has a dysfunctional immune system.
  4. The body cannot mount an immune response against cancer cells.

 
Question 20

The effect of the melanoma vaccine is to stimulate

  1. T cells which produce antibodies.
  2. cytotoxic T cells which activate B cells.
  3. cell division to produce more lymphocytes.
  4. production of B cells which destroy melanoma cells.
Show Answers Only

Q19.  `A`

Q20.  `C`

Show Worked Solution

Q19. 

→ The vaccine can create an immune response against the cancer cells if self-antigens are not present in the cells.

`=>A`
 

Q20. 

→ Lymphocyte development would be promoted by the vaccine.

`=>C`


Mean mark (Q19) 55%
♦♦♦ Mean mark (Q20) 16%

BIOLOGY, M5 2016 HSC 13-14 MC

Refer to the following information to answer Questions 13 and 14.

The diagram shows some chromosomes during some stages of meiosis.
 

Question 13

When does the segregation of homologous chromosomes occur?

  1. Before stage (1)
  2. Between stages (1) and (2)
  3. Between stages (2) and (3)
  4. Between stages (1) and (2) and again between stages (2) and (3)

 
Question 14

The chromosomes shown carry

  1. different genes and different alleles.
  2. different genes and the same alleles.
  3. the same genes and different alleles.
  4. the same genes and the same alleles.
Show Answers Only

Q13.  `B`

Q14.  `C`

Show Worked Solution

Q13. 

→ Stage 1 shows homologous chromosomes, therefore segregation occurs between stages 1 and 2 (meiosis).

`=>B`
 


♦♦ Mean mark (Q13) 24%.

Q14. 

→ The chromosomes carry the same genes as they are homologous, however, an exchange of alleles has taken place.

`=>C`


Mean mark (Q14) 51%.

CHEMISTRY, M5 2018 HSC 30

Over the last 50 years, scientists have recorded increases in the following:

  • the amount of fossil fuels burnt
  • atmospheric carbon dioxide levels
  • average global air temperature and ocean temperature
  • the volume of carbon dioxide dissolved in the oceans.

Analyse the factors that affect the equilibrium between carbon dioxide in the air and carbon dioxide in the oceans. In your answer, make reference to the scientists' observations and include relevant equations.  (7 marks)

Show Answers Only

Fossil fuel combustion:

→ Combustion of fossil fuels releases \(\ce{CO2}\) and heat energy, both of which are released into the atmosphere.

→ Increased burning of fossil fuels will contribute to further rises in atmospheric \(\ce{CO2}\), as described in the equation for the combustion of octane

\(\ce{2C8H18 + 25O2 -> 16CO2 + 18H2O\qquad \triangle H –ve}\)
 

Carbon dioxide and other climate interactions:

→ \(\ce{CO2}\) combines with water according to the following equilibrium in an exothermic reaction.

\(\ce{CO2(g) + H2O(l) \rightleftharpoons H2CO3(aq) \rightleftharpoons H+(aq) + HCO^3-(aq)\qquad   \triangle H –ve}\)

→ This is an equilibrium and by Le Chatelier’s principle when a system is changed, the system will adjust to oppose the change.

→ Factors that affect equilibrium in this system are temperature, pressure and concentration of reactants and products.

→ The increase of \(\ce{CO2}\) in the air due to the combustion of fossil fuels described above, increases the pressure due to \(\ce{CO2}\) in the system.  By Le Chatelier’s principle, the system will oppose this by absorbing more \(\ce{CO2}\) into the oceans.

→ Scientists have been measuring the level of \(\ce{CO2}\) in oceans due to this effect and confirmed the increase in \(\ce{CO2}\).

→ However, this equilibrium is exothermic and as it causes temperature rises, by Le Chatelier’s principle, the reverse reaction may be subsequently favoured. This would have the effect of decreasing the amount of \(\ce{CO2}\) dissolving in the oceans.

→ In summary, if global temperatures continue to rise and \(\ce{CO2}\) in the atmosphere becomes stable or reduces, the system may adjust so that oceans may release \(\ce{CO2}\) rather than absorbing it.

Show Worked Solution

Fossil fuel combustion:

→ Combustion of fossil fuels releases \(\ce{CO2}\) and heat energy, both of which are released into the atmosphere.

→ Increased burning of fossil fuels will contribute to further rises in atmospheric \(\ce{CO2}\), as described in the equation for the combustion of octane

\(\ce{2C8H18 + 25O2 -> 16CO2 + 18H2O\qquad \triangle H –ve}\)
 

Carbon dioxide and other climate interactions:

→ \(\ce{CO2}\) combines with water according to the following equilibrium in an exothermic reaction.

\(\ce{CO2(g) + H2O(l) \rightleftharpoons H2CO3(aq) \rightleftharpoons H+(aq) + HCO^3-(aq)\qquad   \triangle H –ve}\)

→ This is an equilibrium and by Le Chatelier’s principle when a system is changed, the system will adjust to oppose the change.

→ Factors that affect equilibrium in this system are temperature, pressure and concentration of reactants and products.

→ The increase of \(\ce{CO2}\) in the air due to the combustion of fossil fuels described above, increases the pressure due to \(\ce{CO2}\) in the system.  By Le Chatelier’s principle, the system will oppose this by absorbing more \(\ce{CO2}\) into the oceans.

→ Scientists have been measuring the level of \(\ce{CO2}\) in oceans due to this effect and confirmed the increase in \(\ce{CO2}\).

→ However, this equilibrium is exothermic and as it causes temperature rises, by Le Chatelier’s principle, the reverse reaction may be subsequently favoured. This would have the effect of decreasing the amount of \(\ce{CO2}\) dissolving in the oceans.

→ In summary, if global temperatures continue to rise and \(\ce{CO2}\) in the atmosphere becomes stable or reduces, the system may adjust so that oceans may release \(\ce{CO2}\) rather than absorbing it.


♦♦ Mean mark 41%.

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%.

BIOLOGY, M8 2018 HSC 12 MC

The graph shows four possible relationships between ambient temperature and body temperature.
 

Which line on the graph represents the relationship between ambient temperature and body temperature for an endotherm in a terrestrial environment?

  1. `W`
  2. `X`
  3. `Y`
  4. `Z`
Show Answers Only

`B`

Show Worked Solution

→ Endotherms are animals which perform complex processes to maintain a constant internal body temperature over a variety of external temperatures.

`=>B`


♦♦♦ Mean mark 25%.

CHEMISTRY, M6 2016 HSC 29

A solution of hydrochloric acid was standardised by titration against a sodium carbonate solution using the following procedure.

  • All glassware was rinsed correctly to remove possible contaminants.
  • Hydrochloric acid was placed in the burette.
  • 25.0 mL of sodium carbonate solution was pipetted into the conical flask.

The titration was performed and the hydrochloric acid was found to be 0.200 mol L¯1.

  1. Identify the substance used to rinse the conical flask and justify your answer.   (2 marks)

  2. Seashells contain a mixture of carbonate compounds. The standardised hydrochloric acid was used to determine the percentage by mass of carbonate in a seashell using the following procedure.

• A 0.145 g sample of the seashell was placed in a conical flask.

• 50.0 mL of the standardised hydrochloric acid was added to the conical flask.

• At the completion of the reaction, the mixture in the conical flask was titrated with 0.250 mol L¯1 sodium hydroxide.

  1. The volume of sodium hydroxide used in the titration was 29.5 mL.
  2. Calculate the percentage by mass of carbonate in the sample of the seashell.   (4 marks)

Show Answers Only

a.   → Distilled water.

→ This should be used to rinse the conical flask as this will not change the number of moles of \(\ce{Na2CO3}\) placed in it.

b.   54.3%

Show Worked Solution

a.   → Distilled water.

→ This should be used to rinse the conical flask as this will not change the number of moles of \(\ce{Na2CO3}\) placed in it.
 


♦ Mean mark (a) 42%.

b.   \(\ce{HCl(aq) + NaOH(aq) -> NaCl (aq) + H2O(l)}\)

\(\ce{n(NaOH) = c \times V = 0.250 \times 0.0295 = 7.375 \times 10^{-3} mol}\)

\(\ce{n(HCl)_{after} = n(NaOH) = 7.375 \times 10^{-3} mol}\)

\(\ce{n(HCl)_{orig} = c \times V = 0.200 \times 0.0500 = 0.010 mol}\)

\(\ce{(HCl)_{used} = 0.0100-7.375 \times 10^{-3} = 2.625 \times 10^{-3} mol}\)
 

\(\ce{2HCl(aq) + CO3^2-(aq) -> CO2(g) + 2Cl-(aq)}\)

\[\ce{n(CO3)^2- = \frac{1}{2} \times 2.625 \times 10^{-3} = 1.3125 \times 10^{-3} mol}\]

\(\ce{m(CO3^2-) = 1.3125 \times 10^{-3} \times (12.01 + 3 \times 16.00) = 0.07876 g}\)

\[\therefore \ce{\text{%}(CO3^2-) = \frac{0.07876}{0.145} \times 100 = 54.3\text{%}}\]


♦♦ Mean mark (b) 40%.

CHEMISTRY, M7 2016 HSC 22

This apparatus was set up to produce methyl butanoate.
 

  1. Identify a safety issue in this experiment.   (1 mark)

  2. Using structural formulae, write the equation for the production of methyl butanoate.   (2 marks)

  3. Justify the use of apparatus `X` in this experiment.   (2 marks)

Show Answers Only

a.   Flame could ignite one of reagents which is flammable.

b.   
     

c.    → Esterification is a relatively slow reaction.

→ Heating the reaction makes it go faster. However, the low boiling points of the reactants make them volatile as they readily convert into gas.

→ The cooling condenser `X` prevents the gas reactants from escaping the experiment by condensing them back into the reaction mixture. This process allows the reaction to proceed at higher temperatures.

Show Worked Solution

a.   Flame could ignite one of reagents which is flammable.
 

b.   
     


 


♦ Mean mark (b) 52%.

c.    → Esterification is a relatively slow reaction.

→ Heating the reaction makes it go faster. However, the low boiling points of the reactants make them volatile as they readily convert into gas.

→ The cooling condenser `X` prevents the gas reactants from escaping the experiment by condensing them back into the reaction mixture. This process allows the reaction to proceed at higher temperatures.


♦♦♦ Mean mark (c) 14%.

CHEMISTRY, M7 2016 HSC 11 MC

What is the IUPAC name of the following compound?
 

  1. 1-bromo-1-chloro-2,2,2-trifluoroethane
  2. 1-chloro-1-bromo-2,2,2-trifluoroethane
  3. 2-chloro-2-bromo-1,1,1-trifluoroethane
  4. 2-bromo-2-chloro-1,1,1-trifluoroethane
Show Answers Only

`D`

Show Worked Solution

By Elimination:

→ Halogen substituents must be in alphabetical order (eliminate B and C)

→ Numbers are allocated using the first point of difference rule (eliminate A)

`=>D`


♦♦♦ Mean mark 28%.

ENGINEERING, TE 2022 HSC 13 MC

Four students (A, B, C, D) completed a table regarding the orbit of a GPS satellite.

Which student is correct?
 

Show Answers Only

`C`

Show Worked Solution

By Elimination:

→ GPS satellites need to cover most of earths surface, not just the equator, therefore not `A` or `D`.

→ GPS satellites are in medium earth orbit and therefore are not geostationary (do not match earths rotation), so not `A` or `B`.

`=>C`


♦♦♦ Mean mark 24%.

ENGINEERING, CS 2022 HSC 9 MC

Which of the following is an advantage of modelling in 3D computer aided drawing (CAD)?

  1. It allows for rapid prototyping.
  2. The design process is automated.
  3. The models can be intricate and complicated.
  4. It allows for finer tolerances than instrument drawing.
Show Answers Only

`A`

Show Worked Solution

→ Rapid prototyping allows designers to quickly create new concepts on a theme and show visual representations to interested parties. This is a key advantage of CAD drawing.

`=>A`


♦♦♦ Mean mark 30%.

BIOLOGY, M8 2015 HSC 31

'Renal dialysis and kidney transplants are very different treatments for the same medical condition. Each treatment was developed from a new application of biological knowledge.'

Justify these statements.  (8 marks)

Show Answers Only

Consider each element of the statement separately:

Medical Conditions

→ Kidney failure is a recognised medical condition that can be treated using both kidney transplants or renal dialysis.
 

Different Treatments

→ Renal dialysis is the cleansing of wastes from the blood externally using a dialysis machine.

→ During a kidney transplant the diseased organ is removed and replaced with a healthy organ provided by a donor.
 

New Application of biological information: Renal dialysis

→ The movement of substances from regions of high concentration to regions of low concentration is known as diffusion.

→ Diffusion can occur across a semi-permeable membrane.

→The dialysis machine allows blood to flow through tubing which is permeable to urea.

→ The solution around the tubing is continually replaced to maintain a steep concentration gradient so the urea can be removed out of the blood through diffusion.
 

New Application of biological information: Kidney transplant

→ Increased knowledge of the immune system allowed for an improved understanding of organ rejection following organ transplantation.

→ B and T cells on donated organs are recognised by the immune system as foreign and it then attacks the transplanted organ.

→ Once an infection has been removed, suppressor T cells stop the immune cells and switch off the immune response.

→ This knowledge lead to the development of immunosuppressants or anti-rejection drugs designed to fight the immune symptoms’ response to reject donated organs.

→After transplantation, anti-rejection drugs are used by recipients for their lifetime.

Show Worked Solution

Medical Conditions

→ Kidney failure is a recognised medical condition that can be treated using both kidney transplants or renal dialysis.
 

Different Treatments

→ Renal dialysis is the cleansing of wastes from the blood externally using a dialysis machine.

→ During a kidney transplant the diseased organ is removed and replaced with a healthy organ provided by a donor.
 

New Application of biological information: Renal dialysis

→ The movement of substances from regions of high concentration to regions of low concentration is known as diffusion.

→ Diffusion can occur across a semi-permeable membrane.

→The dialysis machine allows blood to flow through tubing which is permeable to urea.

→ The solution around the tubing is continually replaced to maintain a steep concentration gradient so the urea can be removed out of the blood through diffusion.
 

New Application of biological information: Kidney transplant

→ Increased knowledge of the immune system allowed for an improved understanding of organ rejection following organ transplantation.

→ B and T cells on donated organs are recognised by the immune system as foreign and it then attacks the transplanted organ.

→ Once an infection has been removed, suppressor T cells stop the immune cells and switch off the immune response.

→ This knowledge lead to the development of immunosuppressants or anti-rejection drugs designed to fight the immune symptoms’ response to reject donated organs.

→After transplantation, anti-rejection drugs are used by recipients for their lifetime.


♦♦ Mean mark 41%.

CHEMISTRY, M6 2015 HSC 24

  1. Explain why the salt, sodium acetate, forms a basic solution when dissolved in water. Include an equation in your answer.   (2 marks)

  2. A solution is prepared by using equal volumes and concentrations of acetic acid and sodium acetate.
  3. Explain how the pH of this solution would be affected by the addition of a small amount of sodium hydroxide solution. Include an equation in your answer.   (3 marks)

Show Answers Only

a.   \(\ce{CH3COO-(aq) + H2O(l) \rightleftharpoons CH3COOH(aq) + OH-(aq)}\)

→ Sodium acetate is a basic salt.

→ Acetate is a strong base that accepts a proton, producing hydroxide.

→ The presence of \(\ce{OH-}\) ions produced by the hydrolysis of \(\ce{CH3COO-}\) increases the pH, producing a basic solution.
 

b.  \(\ce{CH3COO-(aq) + H3O+(l) \rightleftharpoons CH3COOH(aq) + H2O(l)}\)

→ The \(\ce{OH-}\) ions introduced into the solution will react with the \(\ce{H3O+}\) ions, reducing their concentration in the equilibrium mixture.

→ By Le Chatelier’s principle, this will subsequently move the reaction to the left to increase the \(\ce{H3O+}\) ions, thus minimising any change in pH.

Show Worked Solution

a.   \(\ce{CH3COO-(aq) + H2O(l) \rightleftharpoons CH3COOH(aq) + OH-(aq)}\)

→ Sodium acetate is a basic salt.

→ Acetate is a strong base that accepts a proton, producing hydroxide.

→ The presence of \(\ce{OH-}\) ions produced by the hydrolysis of \(\ce{CH3COO-}\) increases the pH, producing a basic solution.
 


♦♦ Mean mark (a) 37%.

b.  \(\ce{CH3COO-(aq) + H3O+(l) \rightleftharpoons CH3COOH(aq) + H2O(l)}\)

→ The \(\ce{OH-}\) ions introduced into the solution will react with the \(\ce{H3O+}\) ions, reducing their concentration in the equilibrium mixture.

→ By Le Chatelier’s principle, this will subsequently move the reaction to the left to increase the \(\ce{H3O+}\) ions, thus minimising any change in pH.


♦♦♦ Mean mark (b) 25%.

BIOLOGY, M7 2015 HSC 17 MC

The diagram shows an interaction between cells of the immune system.
 


 

What specific process is shown in the diagram?

  1. B cell encountering an antigen
  2. Activation of a macrophage by a helper T cell
  3. Stimulation of a B cell to become a plasma cell
  4. Cytotoxic T cell destroying a virus-infected cell
Show Answers Only

`C`

Show Worked Solution

Consider the activation of the macrophage:

→ The T cell is releasing cytokine to enable this process.

`=>C`


♦♦♦ Mean mark 8%!

BIOLOGY, M8 2015 HSC 16 MC

Why might epidemiology be considered more essential for the study of non-infectious diseases than for the study of infectious diseases?

  1. The causes of infectious diseases have already been determined.
  2. Only non-infectious diseases are affected by patterns of behaviour.
  3. Epidemiology cannot be used to find the causes of infectious diseases.
  4. Koch's postulates are not useful in finding the causes of non-infectious diseases.
Show Answers Only

`D`

Show Worked Solution

→ Options A – C can be shown to be false.

→ Koch’s postulates are only useful when identifying infectious diseases.

`=>D`


♦♦♦ Mean mark 23%.

BIOLOGY, M5 2015 HSC 15 MC

In a certain plant species, individual plants have either yellow, red or orange flowers.

Two plants, each with a different flower colour, were crossed in a breeding experiment like those carried out by Mendel. The F2 results were: 6 red, 11 orange and 5 yellow flowered plants.

What were the genotypes of the original parent plants?

  1. RY and RY
  2. RR and rr
  3. RR and YY
  4. Rr and RY
Show Answers Only

`=>C`

Show Worked Solution

→ The original parents were both homozygous because the F2 parents were both heterozygous.

`=>C`


♦♦♦ Mean mark 19%.

BIOLOGY, M7 2015 HSC 9 MC

A pathogen and a red blood cell are drawn to the same scale, with some features indicated.
 

What type of pathogen is this?

  1. A virus
  2. A prion
  3. A fungus
  4. A bacterium
Show Answers Only

→ Pathogen structure and size are indicative of a fungus.

`C`

Show Worked Solution

→ Pathogen structure and size are indicative of a fungus.

`=>C`


♦♦♦ Mean mark 27%.

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.
 


 

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.   There are a number of inaccuracies:

→ 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.   There are a number of inaccuracies:

→ 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)`  
`F` `=(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:

→ When the 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:

→ When the 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?
 

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`.

`=>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.

→ 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


♦♦ 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%.