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CHEMISTRY, M1 EQ-Bank 18

Draw Lewis electron dot structures for the following ionic molecular compounds

  1. Sodium chloride  \((\ce{NaCl}\)   (1 mark)

  2. Aluminium fluoride \((\ce{GaF3}\)  (1 mark)

  3. Calcium nitride  \((\ce{Ca3N2}\)  (2 marks)

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i.    Sodium chloride  \((\ce{NaCl}\)

   

ii.  Aluminium Fluoride \((\ce{GaF3}\)

iii.  Calcium nitride   \((\ce{Ca3N2}\)

Show Worked Solution

i.    Sodium chloride  \((\ce{NaCl}\)

 

ii.  Aluminium Fluoride \((\ce{GaF3}\)

iii.  Calcium nitride   \((\ce{Ca3N2}\)

CHEMISTRY, M1 EQ-Bank 16

Draw Lewis electron dot structures for

  1. a nitrogen molecule \( (\ce{N2})\)   (1 mark)

  2. an oxygen molecule \( (\ce{O2}) \)  (1 mark)

Show Answers Only

i.    
       

ii.     
        

Show Worked Solution

i.    
     

ii.     
     

BIOLOGY, M2 EQ-Bank 9 MC

When proteins are broken down in chemical digestion, they become

  1. Lipids
  2. Glucose
  3. Glycerol
  4. Amino acids
Show Answers Only

\(D\)

Show Worked Solution

→ Proteins are broken down so as their beneficial components can be absorbed in the small intestine.

→ Amino acid chains (called polypeptide chains) make up all proteins, and are hence the smallest unit of a protein possible in chemical digestion.

\(\Rightarrow D\)

PHYSICS, M1 EQ-Bank 3

A soccer ball leaves the ground with a velocity of 10 ms\(^{-1}\) at 40° above the horizontal.

  1. Draw a vector diagram to represent the balls initial velocity.   (1 marks)

  2. Calculate, to 2 decimal place, the magnitude of the horizontal and vertical components of the vector.  (2 marks)

Show Answers Only

i. 
       

 

ii.   
       

\(v_v=6.43\ \text{ms}^{-1}\)

\(v_h=7.66\ \text{ms}^{-1}\)

Show Worked Solution

i. 
       

 

ii.   
       

\(v_v=10\sin40^{\circ}=6.43\ \text{ms}^{-1}\)

\(v_h=10\cos40^{\circ}=7.66\ \text{ms}^{-1}\)

PHYSICS, M1 EQ-Bank 1

A ball is thrown vertically upwards from ground level, it gains 50 metres vertically and then falls back to the ground.

  1. Determine the initial velocity, in metres per second, that the ball was thrown. Give your answer correct to one decimal place.   (2 marks)

  2. Calculate the total time of flight of the ball, in seconds, giving your answer correct to two decimal places.   (2 marks)

Show Answers Only

i.    \(u=31.3\ \text{ms}^{-1}\)

ii.   \(6.39\ \text{s}\).

Show Worked Solution

i.    \(\text{At 50 m}\ \Rightarrow\ v=0\ \text{ms}^{-1}\)

\(v^2\) \(=u^2+2as\)  
\(0^2\) \(=u^2+2 \times -9.8 \times 50\)  
\(u^2\) \(=980\)  
\(u\) \(=31.3\ \text{ms}^{-1}\)  

 

ii.   \(\text{Time to highest point}\ = \dfrac{1}{2}\ \text{time of flight} \)

\(v\) \(=u+at\)  
\(0\) \(=31.3-9.8t\)  
\(9.8t\) \(=31.3\)  
\(t\) \(=3.194\ \text{s}\)  

 

\(\text{Total time of flight}\ = 2\times 3.194 = 6.39\ \text{s (2 d.p.)}\)

CHEMISTRY, M3 EQ-Bank 1

Complete the table below by outlining one physical indication of each chemical change described below.   (3 marks)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{Chemical Reaction} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \textit{Physical Indication}\ \ \ \ \ \ \ \  \ \ \ \ \ \ \\
\hline
\rule{0pt}{2.5ex} \text{Precipitation} \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} \text{Combustion} \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} \text{Fermentation} \rule[-1ex]{0pt}{0pt} &  \\
\hline
\end{array}

Show Answers Only

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \textit{Chemical Reaction} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \ \ \ \ \ \textit{Physical Indication} \\
\hline
\rule{0pt}{2.5ex} \text{Precipitation} \rule[-1ex]{0pt}{0pt} & \text{Solid forms and falls out of solution} \\
\hline
\rule{0pt}{2.5ex} \text{Combustion} \rule[-1ex]{0pt}{0pt} & \text{Generation of heat and light} \\
\hline
\rule{0pt}{2.5ex} \text{Fermentation} \rule[-1ex]{0pt}{0pt} &  \text{Formation of gas (bubbles)} \\
\hline
\end{array}

Show Worked Solution

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \textit{Chemical Reaction} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \ \ \ \ \ \textit{Physical Indication} \\
\hline
\rule{0pt}{2.5ex} \text{Precipitation} \rule[-1ex]{0pt}{0pt} & \text{Solid forms and falls out of solution} \\
\hline
\rule{0pt}{2.5ex} \text{Combustion} \rule[-1ex]{0pt}{0pt} & \text{Generation of heat and light} \\
\hline
\rule{0pt}{2.5ex} \text{Fermentation} \rule[-1ex]{0pt}{0pt} &  \text{Formation of gas (bubbles)} \\
\hline
\end{array}

CHEMISTRY, M1 EQ-Bank 9

Outline both a physical and chemical property of the elements contained in the highlighted periodic table groups below.   (4 marks)
 

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Group \(\text{A:}\) Alkali metals

→ Physical properties: Group 1 metals are soft and shiny.

→ Chemical properties: They are highly reactive and have low melting/boiling points.

Group \(\text{B:}\) Noble gases

→ Physical properties: Group 8 elements are colourless gases with no odour or taste.

→ Chemical properties: They are virtually unreactive and all have cryogenic boiling points.

Show Worked Solution

Group \(\text{A:}\) Alkali metals

→ Physical properties: Group 1 metals are soft and shiny.

→ Chemical properties: They are highly reactive and have low melting/boiling points.
 

Group \(\text{B:}\) Noble gases

→ Physical properties: Group 8 elements are colourless gases with no odour or taste.

→ Chemical properties: They are virtually unreactive and all have cryogenic boiling points.

CHEMISTRY, M1 EQ-Bank 5

Calculate the mass of \(\ce{H}\) in 15.2 grams of \(\ce{H2O}\). Give your answer in grams correct to two decimal places.   (2 marks)

Show Answers Only

\(1.70\ \text{grams}\)

Show Worked Solution

\(\ce{MM(H2O) = 2 \times 1.008 + 16.00 = 18.016\ \text{g mol}^{-1}}\)

\(\ce{\% of H = \dfrac{2.016}{18.016} \times 100 = 11.19\%}\)

\(\ce{m(H)\ \text{within 15.2 grams}\ H2O = 11.19\% \times 15.2 = 1.70\ \text{grams (to 2 d.p.)}} \)

BIOLOGY, M1 2020 VCE 3

Greenhouses have been used to generate higher crop yields than open-field agriculture. To encourage plant growth in greenhouses, the conditions required for photosynthesis are controlled. Commercial greenhouses, like the ones shown below, often use a lot of energy for heating, ventilation, lighting and water.

  1. Consider the reactions of photosynthesis. Why would it be important to maintain the temperature within narrow limits in a commercial greenhouse? Justify your answer.  (2 marks)

  2. Scientists are developing a new material to cover greenhouses, which can split incoming light and convert the rays from green wavelengths into red wavelengths.
  3. Explain how this new material increases crop yields.  (2 marks)

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a.    Temperature maintenance:

→ Photosynthesis is regulated by different enzymes which are intrinsic to numerous processes involved in photosynthesis.

→ If the temperature of a greenhouse is changed, then the enzymes may decrease in activity or denature, causing a reduction in the rate of photosynthesis or even ceasing it all together.

→ This can cause problems in the plants such as a reduction in growth or plant death.
 

b.    Crop yield increases:

→ Red wavelengths of light are used during photosynthesis, while green light is reflected off the plants.

→ This technology will allow an increase in crop yields by increasing the amount of red light which increases the rate of photosynthesis.

Show Worked Solution

a.    Temperature maintenance:

→ Photosynthesis is regulated by different enzymes which are intrinsic to numerous processes involved in photosynthesis.

→ If the temperature of a greenhouse is changed, then the enzymes may decrease in activity or denature, causing a reduction in the rate of photosynthesis or even ceasing it all together.

→ This can cause problems in the plants such as a reduction in growth or plant death.
 

b.    Crop yield increases:

→ Red wavelengths of light are used during photosynthesis, while green light is reflected off the plants.

→ This technology will allow an increase in crop yields by increasing the amount of red light which increases the rate of photosynthesis.

PHYSICS, M3 2017 VCE 14

A light ray from a laser passes from a glucose solution \((n=1.44)\) into the air \((n=1.00)\), as shown in Figure 12.
 

  1. Calculate the critical angle (total internal reflection) from the glucose solution to the air.  (1 mark)

  2. The light ray strikes the surface at an angle of incidence to the normal of less than the critical angle calculated in part a.
  3. On Figure 12, sketch the ray or rays that should be observed.  (2 marks)

  4. The angle to the normal is increased to a value greater than the critical angle. An observer at point \(\text{X}\) in Figure 13 says she cannot see the laser.
     

  1. Explain why the observer says she cannot see the laser.  (2 marks)

Show Answers Only

a.    \(\theta_c=44^{\circ}\)

b.    
       
→ As the light strikes the surface at an angle less than that of the critical angle, both reflection and refraction will be observed.
 
c.    Observer cannot see the laser:

→ When the light strikes the surface at an angle greater than the critical angle, Total Internal Reflection will occur.

→ As a result of the total internal reflection, all the light will be reflected off the boundary back into the glucose solution and therefore will not travel to the observer.

Show Worked Solution

a.     \(\sin\theta_c\) \(=\dfrac{n_2}{n_1}\)
  \(\theta_c\) \(=\sin^{-1}\Big{(}\dfrac{1.00}{1.44}\Big{)}\)
    \(=44^{\circ}\)

 

b.    
       
→ As the light strikes the surface at an angle less than that of the critical angle, both reflection and refraction will be observed.
♦♦ Mean mark (b) 37%.

c.    Observer cannot see the laser:

→ When the light strikes the surface at an angle greater than the critical angle, Total Internal Reflection will occur.

→ As a result of the total internal reflection, all the light will be reflected off the boundary back into the glucose solution and therefore will not travel to the observer.

BIOLOGY, M4 2021 VCE 8

Ancient cave drawings contribute evidence of cognitive changes that are unique to modern humans, Homo sapiens. Biologists argue that these unique cognitive changes in H. sapiens allowed cultural evolution that would have been impossible in other hominin species.

  1. Describe how ancient cave drawings contribute evidence of cultural evolution in H. sapiens.   (2 marks)
Cave drawings depicting animals have been found on stalagmites in the Ardales Cave in south-west Europe. A stalagmite is a rock formation that grows upwards from a cave floor, as shown in the image below. Over a long period of time, water, containing minerals, dripped from the cave roof, solidified and formed a transparent layer on top of the drawings on the stalagmites.
 

  1. A team of scientists dated the mineral deposits in the solid, transparent layer using uranium-thorium dating. They found that some of the mineral deposits are 65 000 years old.

    What conclusion can be made about the age of the ancient drawings on the stalagmites? Justify your response.   (2 marks)

  1. The table below shows the arrival dates of some hominin species to the European continent as well as their extinction dates. The dates are the best estimates based on fossil evidence from multiple sites and genetic data.
     

\( \textbf{Hominin species} \)

\( \textbf{First appearance in Europe} \)

\( \textbf{(years before present)} \)

\( \textbf{Time of extinction} \)

\( \textbf{(years before present)} \)
\( \text{Homo sapiens} \) \(45\ 000\) \( \text{not extinct}\)
\( \text{Homo neanderthalensis} \) \(130\ 000\) \(30\ 000\)
\( \text{Homo heidelbergensis} \) \(800\ 000-600\ 000\) \(400\ 000-200\ 000\)
\( \text{Homo erectus} \) \(1\ 200\ 000-600\ 000\) \(143\ 000\)
 
  1. How does the discovery of the drawings in the Ardales Cave challenge the view that the art of cave drawing first arose in H. sapiens? In your response, refer to the evidence provided in the table and above.   (2 marks)
Show Answers Only

a.    Answers could include two of the following:

→ Cave drawings show information being passed through generations.

→ The paintings often depict changing hunting techniques or dreaming stories over time.

→ Increasing complexity of these cave drawings over time shows the development of complex thought and symbolic relationships in H. sapien tribes.
 

b.    → The drawings must be a minimum of 65,000 years old.

→ This is because they must be at least as old as the mineral layer on top of them.
 

c.    → H. sapiens are believed to have arrived in Europe 45,000 years ago.

→ The dating shows the drawings are a minimum of 65,000 years old.

→ Therefore the data suggests that either H. neanderthalensis drew the art, or H. sapiens arrived earlier than currently believed.

Show Worked Solution

a.    Answers could include two of the following:

→ Cave drawings show information being passed through generations.

→ The paintings often depict changing hunting techniques or dreaming stories over time.

→ Increasing complexity of these cave drawings over time shows the development of complex thought and symbolic relationships in H. sapien tribes.
 

b.    → The drawings must be a minimum of 65,000 years old.

→ This is because they must be at least as old as the mineral layer on top of them.

♦ Mean mark (b) 49%.

c.    → H. sapiens are believed to have arrived in Europe 45,000 years ago.

→ The dating shows the drawings are a minimum of 65,000 years old.

→ Therefore the data suggests that either H. neanderthalensis drew the art, or H. sapiens arrived earlier than currently believed.

PHYSICS, M3 2017 VCE 15 MC

Lee listens while a police car with a loud siren comes towards her, travels past her and then continues on away from her.

Compared with the sound she would hear from the siren if the police car were stationary, the sound has.

  1. a higher frequency as the car comes towards her and a lower frequency when the car moves away.
  2. a lower frequency as the car comes towards her and a higher frequency when the car moves away.
  3. a lower intensity as the car comes towards her and a greater intensity when the car moves away.
  4. the same frequency at all times.
Show Answers Only

\(A\)

Show Worked Solution

→ By the application of the doppler effect, the frequencies of sounds increase when moving towards an object and decrease when moving away from an object.

\(\Rightarrow A\)

PHYSICS, M1 2017 VCE 9*

A model car of mass 2.0 kg is propelled from rest by a rocket motor that applies a constant horizontal force of 4.0 N, as shown below. Assume that friction is negligible.
 

With the same rocket motor, the car accelerates from rest for 10 seconds.

Calculate the the final speed of the model car?   (2 marks)

Show Answers Only

\(20\ \text{ms}^{-1}\)

Show Worked Solution

\(a=\dfrac{F}{m} = \dfrac{4.0}{2.0} = 2\ \text{ms}^{-2} \)

\(v\) \(=u+at\)  
  \(=0+2 \times 10\)  
  \(=20\ \text{ms}^{-1}\)  

PHYSICS, M2 2017 VCE 8 MC

A model car of mass 2.0 kg is propelled from rest by a rocket motor that applies a constant horizontal force of 4.0 N, as shown below. Assume that friction is negligible.
 

Which one of the following best gives the magnitude of the impulse given to the car by the rocket motor in the first 5.0 seconds?

  1. \(4.0 \text{ N s}\)
  2. \(8.0 \text{ N s}\)
  3. \(20 \text{ N s}\)
  4. \(40 \text{ N s}\)
Show Answers Only

\(C\)

Show Worked Solution
\(I\) \(=Ft\)  
  \(=4 \times 5\)  
  \(=20 \text{ N s}\)  

 

\(\Rightarrow C\)

PHYSICS, M4 2017 VCE 3*

Two large charged plates with equal and opposite charges are placed close together, as shown in the diagram below.

A distance of 5.0 mm separates the plates. The electric field between the plates is equal to 1000 N C\(^{-1}\).
 

Calculate the voltage difference between the plates.   (2 marks)

Show Answers Only

Voltage difference is \(5\ \text{V}\).

Show Worked Solution
\(V\) \(=Ed\)  
  \(=1000 \times 5.0 \times 10^{-3}\)  
  \(=5\ \text{V}\)  

PHYSICS, M3 2018 VCE 11

Figure 13 shows two speakers, \(\text{A}\) and \(\text{ B}\), facing each other. The speakers are connected to the same signal generator/amplifier and the speakers are simultaneously producing the same 340 Hz sound.
 

Take the speed of sound to be 340 ms\( ^{-1}\).

  1. Calculate the wavelength of the sound.  (1 mark)

  2. A student stands in the centre, equidistant from speakers \(\text{A}\) and \(\text{B}\). He then moves towards speaker \(\text{B}\) and experiences a sequence of loud and quiet regions. He stops at the second region of quietness.
  3. How far has the student moved from the centre? Explain your reasoning.  (3 marks)

Show Answers Only

a.     \(\lambda=1\ \text{m}\)

b.    → At the centre of the speakers, there is an area of relative loudness

→ This is due to a path difference of 0 metres between the wavelengths of sounds coming from both speaker \(\text{A}\) and speaker \(\text{B}\). 

→ As the student moves towards speaker \(\text{B}\), the two sound waves will experience constructive (loud) and destructive interference (quiet).

→ The peak of speaker \(\text{A}\) will move into the trough of speaker \(\text{B}\) where he experiences the first region of quietness (node) which is 1 quarter wavelength (0.25m) from the centre.

→ The second region of quietness (node) will be half of a wavelength (0.5m) from the first region of quietness. Thus, the total distance from the centre to the second region of quietness is 0.75m.

Show Worked Solution

a.    \(\lambda=\dfrac{v}{f}=\dfrac{340}{340}=1\ \text{m}\)

 

b.    → At the centre of the speakers, there is an area of relative loudness

→ This is due to a path difference of 0 metres between the wavelengths of sounds coming from both speaker \(\text{A}\) and speaker \(\text{B}\). 

→ As the student moves towards speaker \(\text{B}\), the two sound waves will experience constructive and destructive interference.

→ The peak of speaker \(\text{A}\) will move into the trough of speaker \(\text{B}\) where he experiences the first region of quietness (node) which is 1 quarter wavelength (0.25m) from the centre.

→ The second region of quietness (node) will be half of a wavelength (0.5m) from the first region of quietness. Thus, the total distance from the centre to the second region of quietness is 0.75m.

♦♦♦ Mean mark (b) 20%.

PHYSICS, M3 2018 VCE 11 MC

Alex hears the siren from a stationary fire engine.

Compared with the sound Alex hears from the stationary fire engine, the sound Alex will hear as the fire engine approaches him will have increased

  1. speed.
  2. period.
  3. amplitude.
  4. frequency.
Show Answers Only

\(C\) and \(D\)

Show Worked Solution

→ \(D\) is correct by the doppler effect.

→ \(C\) is correct because as the fire engine approaches Alex, the sound will become louder which corresponds to an increase in wave amplitude.

PHYSICS, M2 2018 VCE 8-9 MC

A railway truck \(\text{X}\) of mass 10 tonnes, moving at 6.0 m s\(^{-1}\), collides with a stationary railway truck \(\text{Y}\) of mass 5.0 tonnes. After the collision the trucks are joined together and move off as one. The situation is shown below.
 

   

\(\text{Question 8}\)

The final speed of the joined railway trucks after the collision is closest to

  1. \( 2.0 \text{ m s}^{-1}\)
  2. \( 3.0 \text{ m s} ^{-1}\)
  3. \( 4.0 \text{ m s} ^{-1}\)
  4. \( 6.0 \text{ m s} ^{-1}\)
     

\(\text{Question 9}\)

The collision of the railway trucks is best described as one where

  1. kinetic energy is conserved but momentum is not conserved.
  2. kinetic energy is not conserved but momentum is conserved.
  3. neither kinetic energy nor momentum is conserved.
  4. both kinetic energy and momentum are conserved.
Show Answers Only

\(\text{Question 8:}\ C\)

\(\text{Question 9:}\ B\)

Show Worked Solution

\(\text{Question 8}\)

→ By the law of conservation of momentum:

\(m_Xu_X+m_Yu_Y\) \(=m_Xv_X+m_Yv_Y\)  
  \(=v(m_X + m_Y)\ \ \ (v_X=v_Y) \)  
\(10\ 000 \times 6 +0\) \(= v(10\ 000 + 5000)\)  
\(60\ 000\) \(=15\ 000v\)  
\(v\) \(=4\ \text{ms}^{-1}\)   

 
\( \Rightarrow C\)
 

\(\text{Question 9}\)

→ By the law of conservation of momentum, momentum in the collision is conserved.

→ Kinetic energy conservation:

       \(KE_{\text{init}}\) \(=\dfrac{1}{2}m_Xu_X^2+\dfrac{1}{2}m_Yu_Y^2\)
   

\(=\dfrac{1}{2} \times 10\ 000 \times 6^2 + \dfrac{1}{2} \times 5000 \times 0^2\)
    \(=180\ 000\ \text{J}\)

 

\(KE_{\text{final}}\) \(=\dfrac{1}{2}(m_X+m_Y)v^2\)  
  \(=\dfrac{1}{2} \times 15\ 000 \times 4^2\)   
  \(=120\ 000\ \text{J}\)  

 
→ The kinetic energy of the system decreases after the collision and so is not conserved.

\(\Rightarrow B\)

PHYSICS, M4 2018 VCE 3 MC

A straight wire carries a current of 10 A.

Which one of the following diagrams best shows the magnetic field associated with this current?
 

Show Answers Only

\(A\)

Show Worked Solution

→ Using the right hand grip rule, the thumb points up and the fingers curl in an anticlockwise direction as seen from above.

\(\Rightarrow A\)

PHYSICS, M3 2019 VCE 13

In an experimental set-up used to investigate standing waves, a 6.0 m length of string is fixed at both ends, as shown in Figure 12. The string is under constant tension, ensuring that the speed of the wave pulses created is a constant 40 ms\(^{-1}\).
 


In an initial experiment, a continuous transverse wave of frequency 7.5 Hz is generated along the string.

  1. Determine the wavelength of the transverse wave travelling along the string.  (1 mark)

  2. Will a standing wave form? Give a reason for your answer.  (2 marks)

Show Answers Only

a. \(\lambda=5.3\ \text{m}\)

b.    → A standing wave will not form.

→ A standing wave will form only if the wavelength of the wave is equal to an integer multiple \(\dfrac{\lambda}{2}\), as 5.3 m is not multiple of 3 m, a standing wave will not form.

Show Worked Solution

a.    \(\lambda = \dfrac{v}{f} = \dfrac{40}{7.5} = 5.3\ \text{m} \) 

 
b.    → A standing wave will not form.

→ A standing wave will only form if the length of the string is equal to an integer multiple of \(\dfrac{\lambda}{2}\).

→ Since 6.0 m is not an integer multiple of \(\dfrac{5.3}{2}\) m, a standing wave will not form.

♦♦ Mean mark (b) 38%.

BIOLOGY, M4 2021 VCE 22 MC

The graph below shows when the five major mass extinction events and several other mass extinction events occurred. It also shows the percentage of species that were lost in each event.
 

According to the graph, to be classified as a major mass extinction event, the percentage of species that are lost has to be at least

  1. 10%
  2. 50%
  3. 74%
  4. 97%
Show Answers Only

\(C\)

Show Worked Solution

→ The lowest % of species lost in a mass extinction event is 74% within this data set.

→ It follows that for an extinction event to be classified as major, at least 74% of species must have been lost.

\(\Rightarrow C\)

BIOLOGY, M1 2021 VCE 6 MC

In a series of experiments, the rate of photosynthesis in plant cells was measured in environments with different concentrations of carbon dioxide. All other variables were kept constant.

Which one of the following graphs reflects the trend that would be shown by the results of these experiments?
 


 

 

Show Answers Only

\(C\)

Show Worked Solution

→ As carbon dioxide is a reactant in photosynthesis, increasing the concentration will lead to a higher rate of photosynthesis initially.

→ This rate of increase will plateau and eventually stop as all the active sites of involved enzymes are being used.

→ If this relationship was graphed, the shape seen would most accurately resemble that in graph \(C\).

\(\Rightarrow C\)

CHEMISTRY, M4 EQ-Bank 3

A 3.1g sample of \(\ce{CaCO3_{(s)}}\) decomposes into \(\ce{CaO_{(s)}}\) and \(\ce{CO2_{(g)}}\). Entropy values  for these chemicals are given below and the molar enthalpy for the reaction is 360 kJ/mol.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex}\text{Substance}\rule[-1ex]{0pt}{0pt} & \text{Standard Entropy}\ (\Delta S) \\
\hline
\rule{0pt}{2.5ex}\ce{CaCO3}\rule[-1ex]{0pt}{0pt} & \text{92.88 J/K} \\
\hline
\rule{0pt}{2.5ex}\ce{CaO(s)}\rule[-1ex]{0pt}{0pt} & \text{39.75 J/K} \\
\hline
\rule{0pt}{2.5ex}\ce{CO2(g)}\rule[-1ex]{0pt}{0pt} & \text{213.6 J/K} \\
\hline
\end{array}

  1. Write a chemical equation for the above reaction.  (1 mark)

  2. Calculate the entropy change for this reaction.  (2 marks)

  3. Determine whether the reaction is spontaneous or non-spontaneous at room temperature.  (2 marks)

Show Answers Only
  1. \(\ce{CaCO3(s) \rightarrow CaO(s) + CO2(g)}\)
  2. \(\ce{4.97 J K^{-1}}\)
  3. \(\text{Since}\ \Delta G > 0,\ \text{the reaction is not spontaneous.}\)
Show Worked Solution

i.    \(\ce{CaCO3(s) \rightarrow CaO(s) + CO2(g)}\)

ii.     \(\Delta S\) \(=\Sigma S_{\text{products}}-\Sigma S_{\text{reactants}}\) 
    \(= 213.6 + 39.75-92.88\)
    \(= 160.47\ \text{J mol}^{-1}\ \text{K}^{-1}\)

 
\(\ce{n(CaCO3)}= \dfrac{\text{m}}{\text{MM}} = \dfrac{3.1}{100.09} = 0.03097\ \text{mol} \)

\(\text{Entropy change}\ = 160.47 \times 0.03097 = 4.97\ \text{J K}^{-1}\)
 

iii.   \(\text{Room Temperature = 298.15 K}\)

\(\Delta G\) \(=\Delta H-T \Delta S\)  
  \(=360-(298.15 \times 0.16047) \)  
  \(= 312.179\ \text{kJ}\)  

 
\(\text{Since}\ \Delta G > 0,\ \text{the reaction is not spontaneous.}\)

CHEMISTRY, M4 EQ-Bank 2

In the following reactions, predict whether entropy will increase or decrease, giving reasons. (3 marks)

  1. \(\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)}\)   (1 mark)
  2. \(\ce{CaCO3(s) \rightleftharpoons CaO(g) + CO2(g)}\)   (1 mark)
  3. \(\ce{2HCl(aq) + Zn(s) \rightleftharpoons ZnCl2(aq) + H2(g)}\)   (1 mark)
Show Answers Only

i.     → Decrease

→ 4 moles of gas becomes 2 moles of gas on the product side, causing the reaction to become more ordered and decreasing in entropy.

ii.    → Increase

→ A solid decomposes into 2 moles of gas. This reaction becomes more disordered, and the phase change involves an increase in entropy.

iii.   → Increase

→ An aqueous and solid reactant becomes an aqueous and gaseous product. The phase changes overall involve an increase in entropy.

Show Worked Solution

i.     → Decrease

→ 4 moles of gas becomes 2 moles of gas on the product side, causing the reaction to become more ordered and decreasing in entropy.
 

ii.    → Increase

→ A solid decomposes into 2 moles of gas. This reaction becomes more disordered, and the phase change involves an increase in entropy.
 

iii.   → Increase

→ An aqueous and solid reactant becomes an aqueous and gaseous product. The phase changes overall involve an increase in entropy.

CHEMISTRY, M3 EQ-Bank 1

Find the oxidation state of each element in the compound \(\ce{CuSO4}\).   (2 marks)

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\(\ce{Cu +2, O -2, S +6}\)

Show Worked Solution

\(\ce{Sulphate ion: SO4^{2-}}\)

\(\ce{Cu: +2\ \ \text{(compound is neutral)}}\)

\(\ce{O: -2}\)

\(\text{Let}\ \ x =\ \text{oxidation number of S}\)

\(2+x+(4 \times -2)\) \(=0\)  
\(x\) \(= +6\)  

CHEMISTRY, M3 2013 VCE 14-15 MC

\(\ce{Cu(s) + 4HNO3(aq)\rightarrow Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)}\)
 

\(\text{Question 14}\)

Which one of the following will not increase the rate of the above reaction?

  1. decreasing the size of the solid copper particles
  2. increasing the temperature of \(\ce{HNO3}\) by 20 °C
  3. increasing the concentration of \(\ce{HNO3}\)
  4. allowing \(\ce{NO2}\) gas to escape

 
\(\text{Question 15}\)

In the above reaction, the number of successful collisions per second is a small fraction of the total number of collisions.

The major reason for this is that

  1. the nitric acid is ionised in solution.
  2. some reactant particles have too much kinetic energy.
  3. the kinetic energy of the particles is reduced when they collide with the container’s walls.
  4. not all reactant particles have the minimum kinetic energy required to initiate the reaction.
Show Answers Only

\(\text{Question 14:}\ D\)

\(\text{Question 15:}\ D\)

Show Worked Solution

\(\text{Question 14}\)

→ Options \(A, B\) and \(C\) will all increase the rate of the given chemical reaction.

\(\Rightarrow D\)
 

\(\text{Question 15}\)

→ Successful collisions occur only if the particles involved have at least the minimum kinetic energy required and the correct orientation.

\(\Rightarrow D\)

CHEMISTRY, M2 2012 16*

A helium balloon is inflated to a volume of 5.65 L and a pressure of 10.2 atm at a temperature of 25°C.

Calculate the amount of helium, in moles, in the balloon.   (2 marks)

Show Answers Only

\(2.36\ \text{mol}\)

Show Worked Solution
\(\ce{n(He)}\) \(= \dfrac{pV}{RT} \)  
  \(= \dfrac{10.2 \times 101.3 \times 5.65}{8.31 \times 298} \)  
  \(= 2.36\ \text{mol} \)  

PHYSICS, M4 2019 VCE 6a

A home owner on a large property creates a backyard entertainment area. The entertainment area has a low-voltage lighting system. To operate correctly, the lighting system requires a voltage of 12 \(\text{V}\). The lighting system has a resistance of 12 \(\Omega\).

Calculate the power drawn by the lighting system.  (3 mark)

Show Answers Only

\(12\ \text{W}\)

Show Worked Solution

→ Using  \(P=IV\) and  \(I= \dfrac{I}{R} :\)

\(P\) \(=\dfrac{V^2}{R}\)  
  \(=\dfrac{12^2}{12}\)  
  \(=12\ \text{W}\)  

→ The power drawn from the lighting system is 12 \(\text{W}\).

PHYSICS, M4 2019 VCE 2

Figure 2 shows two equal positive stationary point charges placed near each other.

 

 
 

 

Sketch on Figure 2 the shape and direction of the electric field lines. Use at least eight field lines.  (2 marks) 

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Show Worked Solution

→ Field lines must not touch.

→ Fields will repel one another.

CHEMISTRY, M2 2014 VCE 9*

An aerosol can with a volume of 300.0 mL contains 2.80 g of propane gas as a propellant. The warning label says the aerosol may explode at temperatures above 60.0 °C.

What is the pressure in the can at a temperature of 60.0 °C?   (2 marks)

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\(5.87 \times 10^{2}\ \text{kPa}\)

Show Worked Solution

\(\ce{MM(C3H8) = 3 \times 12.0 + 8 \times 1.0 = 44.0\ \text{g mol}^{-1}} \)

\(\ce{n(C3H8) = \dfrac{2.80}{44.0} = 0.0636\ \text{mol}} \)

\(\ce{n(C3H8)}\) \( = \dfrac{pV}{RT} \)  
\(p(\ce{C3H8}) \) \(= \dfrac{nRT}{V} \)  
  \(=0.0636 \times 8.31 \times \dfrac{(60.0 + 273)}{300.0 \times 10^{-3}} \)  
  \(=5.87 \times 10^2\ \text{kPa} \)  

CHEMISTRY M2 2015 VCE 2*

When concentrated sulfuric acid is added to dry sucrose, \(\ce{C12H22O11}\), a black residue of pure carbon is produced.

An equation for the reaction is

\(\ce{2C12H22O11(s)+ 2H2SO4(aq) + O2(g)\rightarrow  22C(s) + 2CO2(g) + 24H2O(g) + 2SO2(g)}\)

\(\text{Molar mass}\ \ce{(C12H22O11) = 342.0 \text{ g mol}^{–1}}\)

Calculate the mass of carbon residue that could be produced by the reaction of 50.0 g of sucrose with excess concentrated sulfuric acid.   (2 marks) 

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\(\text{19.3 grams}\)

Show Worked Solution

\(\ce{n(C12H22O11)_{\text{react}} = \dfrac{50.0}{342.0} = 0.146\ \text{mol}} \)

\(\ce{n(C)_{\text{produced}} = 11 \times n(C12H22O11) = 11 \times 0.146 = 1.61\ \text{mol}} \)

\(\ce{m(C)_{\text{produced}} = 1.61 \times 12.0 = 19.3\ \text{g}} \)

PHYSICS, M4 2019 VCE 2*

The electric field between two parallel plates that are 1.0 × 10\(^{-2}\) m apart is 2.0 × 10\(^{-4}\) N C\(^{-1}\).

Calculate the voltage between the plates?   (2 marks)

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\(2.0 \times 10^{-6}\ \text{V} \)

Show Worked Solution
\(E\) \(=\dfrac{V}{d}\)  
\(V\) \(=E \times d\)  
  \(=2.0 \times 10^{-4} \times 1.0 \times 10^{-2}\)  
  \(=2.0 \times 10^{-6}\ \text{V} \)  

PHYSICS, M3 2020 VCE 13

A 0.8 m long guitar string is set vibrating at a frequency of 250 Hz. The standing wave envelope created in the guitar string is shown in Figure 12.
 

  1. Calculate the speed of the wave in the guitar string.  (2 marks)

  2. The frequency of the vibration in the guitar string is tripled to 750 Hz.
  3. On the guitar string below, draw the shape of the standing wave envelope now created. (2 marks)

 

 

 

Show Answers Only

a.   \(v=400\ \text{ms}^{-1}\)

b. 
       

Show Worked Solution
a.    \(v\) \(=f\lambda\)  
  \(=250 \times 1.6 \)  
  \(=400\ \text{ms}^{-1}\)  

  
b.
         

→ Tripling the frequency means decreasing the wavelength by a factor of 3, hence three half wavelengths will now fit on the string.

PHYSICS, M4 2020 VCE 1

Two bar magnets are placed close to each other, as shown in Figure 1.

Sketch the shape and the direction of at least four magnetic field lines between the two poles within the dashed border shown in Figure 1.  (2 marks)
 

Show Answers Only
 

Show Worked Solution

PHYSICS, M2 2020 VCE 9-10 MC

Two blocks of mass 5 kg and 10 kg are placed in contact on a frictionless horizontal surface, as shown in the diagram below. A constant horizontal force, \(F\), is applied to the 5 kg block.
 
 


 

\(\text{Question 9}\)

Which one of the following statements is correct?

  1. The net force on each block is the same.
  2. The acceleration experienced by the 5 kg block is twice the acceleration experienced by the 10 kg block.
  3. The magnitude of the net force on the 5 kg block is half the magnitude of the net force on the 10 kg block.
  4. The magnitude of the net force on the 5 kg block is twice the magnitude of the net force on the 10 kg block.

 
\(\text{Question 10}\)

If the force \(F\) has a magnitude of 250 N, what is the work done by the force in moving the blocks in a straight line for a distance of 20 m?

  1. \(5 \text{ kJ}\)
  2. \(25 \text{ kJ}\)
  3. \(50 \text{ kJ}\)
  4. \(500 \text{ kJ}\)
Show Answers Only

\(\text{Question 9:}\ C\)

\(\text{Question 10:}\ A\)

Show Worked Solution

\(\text{Question 9}\)

Using Newton’s second Law:  \(F=ma\ \ \Rightarrow\ \ a=\dfrac{F}{m}\).

→ The blocks will experience the same acceleration.

→ Both blocks will have the same force to mass ratio. Since the 5 kg block is half the mass of the 10 kg block, it will experience half the magnitude of the net force as the 10 kg block.

\(\Rightarrow C\)

♦ Mean mark 49%.

 
\(\text{Question 10}\)

\(W\) \(=F_{\parallel}s\)  
  \(=250 \times 20\)  
  \(=5000\ \text{J}\)  
  \(=5\ \text{kJ}\)  

 
\(\Rightarrow A\)

PHYSICS, M3 2021 VCE 14

A distant fire truck travelling at 20 ms\(^{-1}\) to a fire has its siren emitting sound at a constant frequency of 500 Hz.

Chris is standing on the edge of the road. Assume that the fire truck is travelling directly towards him as it approaches and directly away from him as it goes past. The arrangement is shown in Figure 14.
 

  1. On the diagram below, sketch the frequency that Chris will hear as the truck moves towards him and then moves away from him. The \(500 \text{ Hz}\) siren signal is shown as a dotted line for reference. No calculations are required.  (2 marks)
     


 

  1. Name the physics principle involved in Chris’s experience.  (1 mark)
Show Answers Only

a.    

  
b.   The Doppler effect
Show Worked Solution

a.    

♦♦♦ Mean mark (a) 13%.
COMMENT: Poorly completed by most students, many incorrectly drew a smooth line that passed Chris at 500 Hz.
  
b.   The Doppler effect

PHYSICS, M3 2022 VCE 16

A small sodium lamp, emitting light of wavelength 589 nm, is viewed at night through two windows from across a street. The glass of one window has a fine steel mesh covering it and the other window is open, as shown in Figure 18. Assume that the sodium lamp is a point source at a distance.

A Physics student is surprised to see a pattern formed by the light passing through the steel mesh but no pattern for the light passing through the open window. She takes a photograph of the observed pattern to show her teacher, who assures her that it is a diffraction pattern.
 

  1. State the condition that the fine steel mesh must satisfy for a diffraction pattern to form.  (1 mark)
  1. Explain why the condition stated in part a. does not apply to the open window.  (2 marks)
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a.   Condition to satisfy:

→ The size of the gaps from the fine steel mesh must be of the same order of magnitude of the wavelengths of the light from the sodium lamp.

b.   Open window differences:

→ The width of the window is significantly greater than wavelength of the light from the sodium lamp and the width of the glass molecules that make up the glass window is significantly smaller than the light from the sodium lamp.

→ These widths contrast greatly to the size of the steel mesh and are not of the same order of magnitude of the light from the sodium lamp, hence no diffraction will occur.

Show Worked Solution

a.   Condition to satisfy:

→ The size of the gaps from the fine steel mesh must be of the same order of magnitude of the wavelengths of the light from the sodium lamp.


b.   Open window differences:

→ The width of the window is significantly greater than wavelength of the light from the sodium lamp and the width of the glass molecules that make up the glass window is significantly smaller than the light from the sodium lamp.

→ These widths contrast greatly to the size of the steel mesh and are not of the same order of magnitude of the light from the sodium lamp, hence no diffraction will occur.

♦ Mean mark (b) 40%.

PHYSICS, M2 2022 VCE 7

Kym and Kelly are experimenting with trolleys on a ramp inclined at 25°, as shown in Figure 7. They release a trolley with a mass of 2.0 kg from the top of the ramp. The trolley moves down the ramp, through two light gates and onto a horizontal, frictionless surface. Kym and Kelly calculate the acceleration of the trolley to be 3.2 m s\(^{-2}\) using the information from the light gates.
 
 

  1.  i. Show that the component of the gravitational force of the trolley down the slope is \(8.3 \text{ N}\). Use \(g=9.8 \text{ m s}^{-2}\).  (2 marks)
  1. ii. Assume that on the ramp there is a constant frictional force acting on the trolley and opposing its motion.
  2.      Calculate the magnitude of the constant frictional force acting on the trolley.  (2 marks)
  1. When it reaches the bottom of the ramp, the trolley travels along the horizontal, frictionless surface at a speed of 4.0 m s\(^{-1}\) until it collides with a stationary identical trolley. The two trolleys stick together and continue in the same direction as the first trolley.
    1. Calculate the speed of the two trolleys after the collision. Show your working and clearly state the physics principle that you have used.  (3 marks)
    1. Determine, with calculations, whether this collision is an elastic or inelastic collision. Show your working.  (3 marks)
Show Answers Only

a.i.   See Worked Solutions

a.ii.  \(F_f=1.9\ \text{N}\)

b.i.   \(2.0\ \text{ms}^{-1}\)

 b.ii.  For the collision to be elastic, the kinetic energy must be conserved.

\(KE_{\text{init}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 2 \times 4^2=16\ \text{J}\)

\(KE_{\text{final}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 4 \times 2^2=8\ \text{J}\)

→ As the kinetic energy of the system decreases after the collision, it is not an elastic collision.

Show Worked Solution

a.i.  The gravitational force down the slope:

\(F\) \(=mg\, \sin \theta\)  
  \(=2.0 \times 9.8 \times \sin 25\)  
  \(=8.3\ \text{N}\)   

 

a.ii.    \(F_{net}\) \(=ma\)
    \(=2.0 \times 3.2\)
    \(=6.4\)

 

\(6.4\) \(=F-F_f\)  
\(F_f\) \(=8.3-6.4\)  
  \(=1.9\ \text{N}\)  
♦ Mean mark (a.ii) 53%.

 
b.i.   By the conservation of momentum:

\(m_1u_1+m_2u_2\) \(=v(m_1+m_2)\)  
\(2 \times 4 + 2 \times 0\) \(=v(2 +2)\)  
\(8\) \(=4v\)  
\(v\) \(=2\ \text{ms}^{-1}\)  
 

 b.ii.  For the collision to be elastic, the kinetic energy must be conserved.

\(KE_{\text{init}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 2 \times 4^2=16\ \text{J}\)

\(KE_{\text{final}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 4 \times 2^2=8\ \text{J}\)

→ As the kinetic energy of the system decreases after the collision, it is not an elastic collision.

PHYSICS, M3 2022 VCE 11 MC

Which one of the following statements best describes transverse and longitudinal waves?

  1. Both transverse waves and longitudinal waves travel in a direction parallel to their vibrations.
  2. Both transverse waves and longitudinal waves travel in a direction perpendicular to their vibrations.
  3. Transverse waves travel in a direction perpendicular to their vibrations; longitudinal waves travel parallel to their vibrations.
  4. Transverse waves travel in a direction parallel to their vibrations; longitudinal waves travel perpendicular to their vibrations.
Show Answers Only

\(C\)

Show Worked Solution

→ The energy transfer through transverse waves is perpendicular to the direction of vibrations.

→ The energy transfer through longitudinal waves is parallel to the direction of vibrations.
 

\( \Rightarrow C\)

PHYSICS, M3 2022 VCE 10 MC

Which one of the following statements best describes an observation of the Doppler effect for sound?

  1. a decrease in frequency received when a source of sound moves towards you
  2. a decrease in frequency received when moving towards a stationary source of sound
  3. an increase in frequency received when moving towards a stationary source of sound
  4. a decrease in wavelength received when moving away from a stationary source of sound
Show Answers Only

\(C\)

Show Worked Solution

→ As a result of the Doppler effect, when there is relative movement towards a source of sound waves the frequency of the observed sound waves is increased.

\( \Rightarrow C\)

PHYSICS, M2 2022 VCE 6-7 MC

A railway truck \(\text{(X)}\) of mass 10 tonnes, moving at 3.0 m s\(^{-1}\), collides with a stationary railway truck \(\text{(Y)}\), as shown in the diagram below.

After the collision, they are joined together and move off at speed  \(v= 2.0\ \text{m s}^{-1}\).
 


 

\(\text{Question 6}\)

Which one of the following is closest to the mass of railway truck \(\text{Y}\)?

  1. 3 tonnes
  2. 5 tonnes
  3. 6.7 tonnes
  4. 15 tonnes

 
\(\text{Question 7}\)

Which one of the following best describes the force exerted by the railway truck \(\text{X}\) on the railway truck \(\text{Y} \left(F_{\text { X on Y}}\right)\) and the force exerted by the railway truck \(\text{Y}\) on the railway truck \(\text{X} \left(F_{\text {Y on X}}\right)\) at the instant of collision?

  1. \(F_{  \text { X on Y }  }<F_{ \text {Y on X}  }\)
  2. \(F_{ \text { X on Y }  }=F_{ \text { Y on X}  }\)
  3. \(F_{ \text { X on Y }  }=-F_{  \text { Y on X}  }\)
  4. \(F_{  \text {X on Y } }>F_{ \text {Y on X}  }\)
Show Answers Only

\(\text{Question 6:}\ B\)

\(\text{Question 7:}\ C\)

Show Worked Solution

\(\text{Question 6}\)

→ By the law of conservation of momentum, the momentum before the collision is equal to the momentum after the collision.

\(m_Xu_X+m_Yu_Y\) \(=m_Xv_X+m_Yv_Y\)  
  \(=v(m_X + m_Y)\ \ \ (v_X=v_Y) \)  
\(10000 \times 3 +0\) \(= 2(10000 + m_Y)\)  
\(15000\) \(=10000 +m_Y\)  
\(m_Y\) \(=5000\ \text{kg}\)   
  \(=5\ \text{tonnes}\)   

\( \Rightarrow B\)

 
\(\text{Question 7}\)

→ By Newton’s 3rd law of motion, each action has an equal and opposite reaction. 

→ Hence, the force of \(F_{ \text { X on Y }  }\) is equal in magnitude to \(F_{  \text { Y on X}  }\) but opposite in direction which is indicated by the negative sign (–).

\( \Rightarrow C\)

♦ Mean mark (Q7) 47%.

BIOLOGY, M3 2022 VCE 32-33 MC

Victoria’s fossil emblem Koolasuchus cleelandi, shown on the stamp above, was chosen in January 2022.

K. cleelandi fossils have been found in Boonwurrung country in Gippsland. Studies of K. cleelandi fossils have revealed many interesting findings. Several of these findings are summarised below.

K. cleelandi findings

    1. It was an amphibian that lived 125 million years ago.
    2. It had a jaw, a skull, vertebrae and ribs made of bone.
    3. It was covered in scales and had fangs on the roof of its mouth.
    4. It lived in rivers carrying many sediments.
    5. It lived in Victoria for 50 million years after its closest relatives in the rest of the world became extinct.
    6. It fed on small dinosaurs, turtles and fish.
    7. It was the size of a small car, 4 m long and weighed 500 kg.
       

Part 1

Which combination of findings from the list above is most likely to be associated with an increased chance of formation of K. cleelandi fossils?

  1. findings 1 and 6
  2. findings 2 and 4
  3. findings 3 and 5
  4. findings 2 and 7

 
Part 2

It is most likely that scientists determined the absolute age of K. cleelandi fossils

  1. through mtDNA analysis of K. cleelandi scales and fangs.
  2. through carbon-14 dating of jaw bones, skull, vertebrae and ribs.
  3. by finding index fossils nearby, such as those of small dinosaurs and turtles.
  4. through radioisotope dating of rock surrounding the fossils.
Show Answers Only

\(\text{Part 1:}\ B\)

\(\text{Part 2:}\ D\)

Show Worked Solution

\(\text{Part 1}\)

→ K. cleelandi has large, dense bones (finding 2) which are less likely to decay overtime and hence have a higher chnace of being fossilised for a long period of time.

→ Because these bones are remaining in a river carrying many sediments (finding 4) the bones are more likely to be fossilised as they are covered quicker, reducing disturbance and hiding them from scavnegers.

\(\Rightarrow B\)
  

\(\text{Part 2}\)

→ Radioisotope dating of rocks is extremely accurate and reliable source of dating. Therefore it ios likely that scientists dated K. cleelandi fossils by dating the surrounding rock.

\(\Rightarrow D\)

♦ Mean mark (part 2) 52%.

BIOLOGY, M1 2022 VCE 19 MC

Enzymes are associated with the biochemical pathways of cellular respiration and photosynthesis.

Consider a cell that carries out both cellular respiration and photosynthesis.

Which one of the following is a correct statement about the enzymes associated with these pathways?

  1. Both pathways are catalysed by the same enzymes.
  2. An increase in pH above the optimal pH for cellular respiration may cause the enzymes to denature.
  3. The rates of these biochemical pathway reactions are not affected by changes in the concentrations of the enzymes.
  4. A drastic increase in temperature above the optimal temperature for photosynthesis will cause an increase in the rate of photosynthesis.
Show Answers Only

\(B\)

Show Worked Solution

By Elimination

→ While the net chemical equations for photosynthesis and cellular respiration appear to be the reverse of each other, each reaction is made up of various processes which require different reaction conditions, including different enzymes (Eliminate A).

→ Enzyme concentrations can increase/decrease the amount of available active sites which can change the rate of reaction rate (Eliminate C).

→ A large increase in temperature would not increase reaction rate but would denature the enzymes (Eliminate D).

→ Changes in various factors, including pH, can denature an enzyme if the change is outside the optimal range for that factor.

\(\Rightarrow B\)

CHEMISTRY, M4 2014 VCE 1

The decomposition of ammonia is represented by the following equation.

\(\ce{2NH3(g)\rightleftharpoons N2(g) + 3H2(g) \quad \quad \Delta H = 92.4 \text{kJ mol}^{–1}}\)

  1. The activation energy for the uncatalysed reaction is 335 kJ mol\(^{-1}\).
  2. The activation energy for the reaction when tungsten is used as a catalyst is 163 kJ mol\(^{-1}\).
  3. On the grid provided below, draw a labelled energy profile diagram for the uncatalysed and catalysed reactions.  (3 marks)
     

  1. When osmium is used as a catalyst, the activation energy is 197 kJ mol\(^{-1}\).
  2. Which catalyst – osmium or tungsten – will cause ammonia to decompose at a faster rate? Justify your answer in terms of the chemical principles you have studied this year.  (2 marks)

Show Answers Only

a.    

 

b.    → Tungsten will cause ammonia to decompose faster.

→ Faster rate due to lower activation energy.

→ Lower activation energy results in a greater proportion of successful collisions.

Show Worked Solution

a.    

     

b.   → Tungsten will cause ammonia to decompose faster.

→ Faster rate due to lower activation energy.

→ Lower activation energy results in a greater proportion of successful collisions.

CHEMISTRY, M4 2019 VCE 9 MC

A reaction has the energy profile diagram shown below.
 

Which of the following represents the energy profile of the reverse reaction.
 

  \(\text{Final product energy}\)
\(\text{(kJ mol}^{−1})\)		 \(\Delta H\)
\(\text{(kJ mol}^{−1})\)
A.   \(40\) \(+10\)
B. \(50\) \(+10\)
C. \(50\) \(-10\)
D. \(40\) \(-10\)
 
Show Answers Only

\(B\)

Show Worked Solution

Reverse reaction  \(\Rightarrow\)  Endothermic

\(\Delta H \) \(=\ \text{E(products) – E(reactants)}\)  
  \(= 50-40\)  
  \(=10\ \text{kJ mol}^{-1}\)  

 
\(\Rightarrow B\)

CHEMISTRY, M4 2018 VCE 13 MC

The energy profile diagram below represents a particular reaction. One graph represents the uncatalysed reaction and
the other graph represents the catalysed reaction.
 

Which of the following best matches the energy profile diagram?
 

  \(E_\text{a}\)
\(\text{uncatalysed reaction}\)
\(\text{(kJ mol}^{-1})\)		 \(\Delta H\)
\(\text{catalysed reaction}\)
\(\text{(kJ mol}^{-1}) \)
A.   \(40\) \(-140\)
B. \(90\) \(-140\)
C. \(40\) \(-50\)
D. \(90\) \(-50\)
Show Answers Only

\(D \)

Show Worked Solution

→ The \(\text{E}_{a}\) of the uncatalysed reaction is higher (90) than the catalysed reaction (40).

→ The \(\Delta H\) energy changes are based off the zero reading (pre-reaction) to the final energy. 

\(\Rightarrow D \)

CHEMISTRY, M4 2017 VCE 29 MC

The following energy profile shows the results obtained during an enzyme-catalysed reaction. Each stage of the reaction is labelled: M represents the initial reactants, N represents a stable intermediate and P represents the final products.
 

Which one of the following statements is correct?

  1. The energy change from M to N is exothermic and the energy change from N to P is exothermic.
  2. The energy change from M to P is exothermic and the energy change from N to P is endothermic.
  3. The energy change from M to N is endothermic and the energy change from N to P is endothermic.
  4. The energy change from M to N is endothermic and the energy change from M to P is endothermic.
Show Answers Only

\(D\)

Show Worked Solution

\(\text{M → P}\) is endothermic since energy of products > energy of reactants.

\(\text{M → N}\) is endothermic

\(\text{N → P}\) is exothermic

\(\Rightarrow D\)

CHEMISTRY, M4 2005 HSC 17

  1. The heat of combustion of ethanol is 1367 kJ mol\(^{-1}\). In a first-hand investigation to determine the heat of combustion of ethanol, the experimental value differed from the theoretical value.
  2. Identify a reason for this difference.   (1 mark)

  3. Calculate the theoretical mass of ethanol \(\ce{(C2H6O)}\) required to heat 200 mL of water from 21.0°C to 45.0°C.   (2 marks)
Show Answers Only

a.    Answers could include one of the following:

→ Heat loss into the surroundings from combustion process.

→ Specific examples of heat loss such as inefficient heating of water caused by proximity of the heat or not using a beaker lid.

b.    \(0.68\ \text{g} \)

Show Worked Solution

a.    Answers could include one of the following:

→ Heat loss into the surroundings from combustion process.

→ Specific examples of heat loss such as inefficient heating of water caused by proximity of the heat or not using a beaker lid.
 

b.    \(\Delta H = mC \Delta T = 0.200 \times 4.18 \times 10^{3} \times (45-21) = 20\ 064\ \text{J}\)

\(\ce{n(C2H6O)_{\text{req}} = \dfrac{20.064}{1367} = 0.014677\ \text{mol}} \)

\(\ce{MM(C2H6O) = 2 \times 12.01 + 6 \times 1.008 + 16 = 46.068\ \text{g mol}^{-1}} \)

\(\ce{m(C2H6O)_{\text{req}} = n \times MM = 0.014677 \times 46.068 = 0.68\ \text{g}} \)

CHEMISTRY, M4 EQ-Bank 10 MC

Which statement explains why catalysts are often used in chemical reactions?

  1. Catalysts increase the activation energies of reactions.
  2. Catalysts increase the rate of reactions.
  3. Catalysts increase the yield of products of reactions.
  4. Catalysts increase the purity of products of reactions.
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\(B\)

Show Worked Solution

→ Catalysts increase the rate of reactions.

\(\Rightarrow B\)

CHEMISTRY, M3 2018 VCE 2

Hydrogen peroxide, \(\ce{H2O2}\), in aqueous solution at room temperature decomposes slowly and irreversibly to form water, \(\ce{H2O}\), and oxygen, \(\ce{O2}\), according to the following equation.

\(\ce{2H2O2(aq)\rightarrow 2H2O(l) + O2(g)}\)    \(\Delta H < 0\)

  1. What effect will increasing the temperature have on the rate of \(\ce{O2}\) production? Use collision theory to explain your answer.  (3 marks)

  2. When a small lump of manganese(\(\text{IV}\)) dioxide, \(\ce{MnO2}\), is added to the \(\ce{H2O2}\) solution, the rate of \(\ce{O2}\) production increases, but when powdered \(\ce{MnO2}\) is added instead, the rate of \(\ce{O2}\) production is greatly increased. The \(\ce{MnO2}\) is recovered at the end of the reaction.
  3. State the function of \(\ce{MnO2}\) in this reaction.  (1 mark)

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a.    → Increasing the temperature will increase the rate of \(\ce{O2}\) production.

→ An increase in temperature increases the kinetic energy of all molecules which causes both the frequency of collisions and the energy of each collision to increase.

→ The temperature increase results in more collisions, of which a higher proportion are successful (i.e. occur with greater energy than the activation energy threshold required to react).
 

b.    \(\ce{MnO2}\) acts as a catalyst in this reaction.

Show Worked Solution

a.    → Increasing the temperature will increase the rate of \(\ce{O2}\) production.

→ An increase in temperature increases the kinetic energy of all molecules which causes both the frequency of collisions and the energy of each collision to increase.

→ The temperature increase results in more collisions, of which a higher proportion are successful (i.e. occur with greater energy than the activation energy threshold required to react).
 

b.    \(\ce{MnO2}\) acts as a catalyst in this reaction.

CHEMISTRY, M2 2022 VCE 3* MC

The correct equation for the incomplete combustion of ethanol is

  1. \(\ce{C2H5OH(l) + O2(g)\rightarrow CO(g) + H2O(l)}\)
  2. \(\ce{C2H5OH(l) + 2O2(g)\rightarrow 2CO(g) + 4H2O(l)}\)
  3. \(\ce{C2H5OH(l) + 2O2(g)\rightarrow 2CO(g) + 3H2O(l)}\)
  4. \(\ce{2C2H5OH(l) + 4O2(g)\rightarrow 4CO(g) + 5H2O(l)}\)
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\(C\)

Show Worked Solution

→ The products of incomplete combustion are \(\ce{CO}\) and \(\ce{H2O}\) and correctly balanced as shown in \(C\).

\(\Rightarrow C\)

CHEMISTRY, M2 2020 VCE 6a*

Methane gas, \(\ce{CH4}\), can be captured from the breakdown of waste in landfills. \(\ce{CH4}\) is also a primary component of natural gas. \(\ce{CH4}\) can be used to produce energy through combustion.

Write the balanced equation for the complete combustion of \(\ce{CH4}\) to produce carbon dioxide, \(\ce{CO2}\) given the unbalanced equation below.  (1 mark)

\(\ce{CH4(g) + O2 \rightarrow CO2(g) + H2O(l)}\)

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\(\ce{2CH4(g) + 3O2 \rightarrow 2CO2(g) + 4H2O(l)\ \ \text{or}}\)

\(\ce{CH4(g) + \dfrac{3}{2}O2 \rightarrow CO2(g) + 2H2O(l)}\)

Show Worked Solution

\(\ce{2CH4(g) + 3O2 \rightarrow 2CO2(g) + 4H2O(l)\ \ \text{or}}\)

\(\ce{CH4(g) + \dfrac{3}{2}O2 \rightarrow CO2(g) + 2H2O(l)}\)

CHEMISTRY, M2 2018 VCE 10 MC

Bioethanol, \(\ce{C2H5OH}\), is produced by the fermentation of glucose, \(\ce{C6H12O6}\), according to the following equation.

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

The mass of \(\ce{C2H5OH}\) obtained when 5.68 g of carbon dioxide, \(\ce{CO2}\), is produced is

  1. 0.168 g
  2. 0.337 g
  3. 2.97 g
  4. 5.94 g
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\(D\)

Show Worked Solution

\(\ce{MM(CO2) = 12.01 + 2 \times 16 = 44.0\ \text{g}}\)

\(\ce{n(CO2) = \dfrac{5.68}{44.0} = 0.129\ \text{mol}} \)

\(\ce{n(C2H5OH) = n(CO2) = 0.129\ \text{mol}}\)

\(\ce{MM(C2H5OH) = 2 \times 12.01 + 6 \times 1.008 + 16 = 46.07\ \text{g}}\)

\(\ce{m(C2H5OH) = 0.129 \times 46.07 = 5.94\ \text{g}} \)

\(\Rightarrow D\)

PHYSICS, M1 2012 HSC 21

  1. Outline a first-hand investigation that could be performed to measure a value for acceleration due to gravity.   (3 marks)

  2. How would you assess the accuracy of the result of the investigation?   (1 mark)

  3. How would you increase the reliability of the data collected?   (1 mark)

  4. How would you assess the reliability of the data collected?   (1 mark)

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a.    Timing of a falling mass.

→ Set up an electronic and automatic timing system with sensors to detect the presence of a small falling metal ball.

→ Heights for the ball should be set up between 0.2 m to 1 m with intervals every 0.2 m. To increase the reliability of the results, multiple trials should be conducted at each height and the average falling time for each height should be calculated which can then be used to graph the data.

→ The results should be plotted on a graph of height vs time\(^2\). This uses the equation  \(s=ut +\dfrac{1}{2}at^2\)  where  \(u=0\)  which becomes  \(s=\dfrac{1}{2}at^2\).

→ After plotting the data, the acceleration due to gravity, \(a\), can be calculated using  \(a=\dfrac{2s}{t^2}\), which will make it equal to 2 × the gradient of the line of best fit.
 

b.    → Look up known value on a reliable website (e.g. National Measurement Institute).

→ Ensure the value is for the location of the experiment (it can differ slightly).

→ Compare the known value to the value determined experimentally and the closer they are, the greater the accuracy of the experiment.
 

c.    → Conduct multiple trials at each height.

→ Use the average of the calculations as stated in the method above.
 

d.   → Compare the values obtained at a single height.

→ If there is a large variation in the calculations conducted at the same height, the data collected is less reliable.

Show Worked Solution

a.    Timing of a falling mass.

→ Set up an electronic and automatic timing system with sensors to detect the presence of a small falling metal ball.

→ Heights for the ball should be set up between 0.2 m to 1 m with intervals every 0.2 m. To increase the reliability of the results, multiple trials should be conducted at each height and the average falling time for each height should be calculated which can then be used to graph the data.

→ The results should be plotted on a graph of height vs time\(^2\). This uses the equation  \(s=ut +\dfrac{1}{2}at^2\)  where  \(u=0\)  which becomes  \(s=\dfrac{1}{2}at^2\).

→ After plotting the data, the acceleration due to gravity, \(a\), can be calculated using  \(a=\dfrac{2s}{t^2}\), which will make it equal to 2 × the gradient of the line of best fit.

♦ Mean mark (a) 55%.

b.    → Look up known value on a reliable website (e.g. National Measurement Institute).

→ Ensure the value is for the location of the experiment (it can differ slightly).

→ Compare the known value to the value determined experimentally and the closer they are, the greater the accuracy of the experiment.

♦♦♦ Mean mark (b) 26%.

c.    → Conduct multiple trials at each height.

→ Use the average of the calculations as stated in the method above.

 
d.   → Compare the values obtained at a single height.

→ If there is a large variation in the calculations conducted at the same height, the data collected is less reliable.

♦♦♦ Mean mark (d) 7%.

PHYSICS, M4 2012 HSC 6 MC

The diagram represents the electric field around a negative charge.
 

   

If the magnitude of the charge were doubled, which diagram would best represent the new electric field?

 


 

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\(C\)

Show Worked Solution

→ The density of the field lines corresponds to the strength of the electric field surrounding the point charge.

→ Hence a stronger charge produces a stronger field and so the field lines become denser.

\(\Rightarrow C\)

CHEMISTRY, M5 2023 HSC 12 MC

The industrial production of ammonia is represented by the Haber process reaction shown.

\( \ce{N2(g)} + 3 \ce{H2(g)} \rightleftharpoons \ce{2NH3(g)} \quad \Delta H=-91.8 \ \text{kJ} \ \text{mol}^{-1}\)

Factors such as temperature and pressure need to be considered in order to maximise yield.

Which of the following is correct?

  1. A lower pressure would result in a higher yield.
  2. A higher pressure would result in a higher yield.
  3. A lower temperature would result in a lower yield.
  4. A higher temperature would result in a higher yield.
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\(B\)

Show Worked Solution

→ Increasing the pressure would shift the position of equilibrium to the side of the equation with less moles (as per Le Chatelier’s principle).

→ This would increase the yield of \(\ce{NH3}\)

\(\Rightarrow B\)

CHEMISTRY, M5 2023 HSC 4 MC

Sodium chloride dissolves in water according to the following equation.

\( \ce{NaCl}(s) \rightleftharpoons \ce{Na^{+}} (aq) + \ce{Cl^{-}}(aq) \)

A saturated solution of \(\ce{NaCl}\) in water contains sodium and chloride ions at the following concentrations.
 

\( Ion \) \( Concentration \) \( \left(\text{mol L}^{-1}\right)  \)
\( \ce{Na}^{+}\)   \(6.13\)
\( \ce{Cl}^{-}\) \(6.13\)
  
What is the \(K_{s p}\) of sodium chloride?
 
  1. \(2.65 \times 10^{-2}\)
  2. \(8.16 \times 10^{-2}\)
  3. \(12.26\)
  4. \(37.6\)
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\(D\)

Show Worked Solution

\( \ce{NaCl}(s) \rightleftharpoons \ce{Na^{+}} (aq) + \ce{Cl^{-}}(aq) \)

\(K_{sp} = \dfrac{\ce{[Na+][Cl-]}}{\ce{[NaCl]}} \)

\(\ce{ \text{Since}\ NaCl\ \text{is a solid, its concentration is assumed to be 1}}\)

\(K_{sp}\) \(= 6.13 \times 6.13\)  
  \(= 37.6\)  

 
\(\Rightarrow D\)