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CHEMISTRY, M6 2020 HSC 33

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

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

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

Show Answers Only

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

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

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

b.   `text{pH} = 11.35`

Show Worked Solution

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

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

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

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

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

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

 
`:.\ text{pH} = 14-2.650 = 11.35`


♦♦♦ Mean mark (b) 20%.

PHYSICS, M5 2018 HSC 21

  1. Compare the force of gravity exerted on the moon by Earth with the force of gravity exerted on Earth by the moon.    (2 marks)
  1. The acceleration due to gravity on the moon is `1.6 \ text{m s}^(-2)` and on Earth it is `9.8 \ text{m s}^(-2)`. Quantitatively compare the mass and weight of a 70 kg person on the moon and on Earth.   (2 marks)
Show Answers Only

a.   Using Newton’s Third Law:

  • The force of gravity of the Earth on the moon is equal in magnitude and opposite in direction to the force of gravity exerted on Earth by the moon.
     

b.   Comparison of mass:

  • The mass of the person on both Earth and the moon is 70 kg.

Comparison of weight:

  • The weight of the person on Earth is given by  `W_e=mg=70 xx9.8=686\ text{N.}`
  • The weight of the person on the moon is given by  `W_m=mg=70xx1.6=112\ text{N.}`
  • The persons weight on Earth is greater than it is on the moon.
Show Worked Solution

a.   Using Newton’s Third Law:

  • The force of gravity of the Earth on the moon is equal in magnitude and opposite in direction to the force of gravity exerted on Earth by the moon.

♦♦ Mean mark (a) 33%.

b.   Comparison of mass:

  • The mass of the person on both Earth and the moon is 70 kg.

Comparison of weight:

  • The weight of the person on Earth is given by  `W_e=mg=70 xx9.8=686\ text{N.}`
  • The weight of the person on the moon is given by  `W_m=mg=70xx1.6=112\ text{N.}`
  • The persons weight on Earth is greater than it is on the moon.

CHEMISTRY, M5 2020 HSC 26

Nitric oxide gas (`text{NO}`) can be produced from the direct combination of nitrogen gas and oxygen gas in a reversible reaction.

  1. Write the balanced chemical equation for this reaction.   (1 mark)

  2. The energy profile diagram for this reaction is shown. 
     

   

  1. Explain, using collision theory, how an increase in temperature would affect the value of `K_{eq}` for this system. Refer to the diagram in your answer.   (4 marks)

Show Answers Only

a.   \(\ce{N2(g) + O2(g) \rightleftharpoons 2NO(g) }\)
 

b.    Increased temperature’s effect on `K_{eq}`:

  • From the graph, the forward reaction is endothermic.
  • The activation energy of the forward endothermic reaction is greater than the activation energy of the reverse exothermic reaction.
  • An increase in temperature would cause the rates of both the forward and reverse reaction due to the higher average kinetic energy, resulting in a larger likelihood of a successful collisions.
  • However, the rate of the forward reaction would increase to a higher extent than the reverse reaction, since it is an endothermic reaction.
  • Using  `K_(eq) = ([text{NO}]^2)/([text{N}_2][text{O}_2])`, as the equilibrium shifts right, the equilibrium constant would increase.
Show Worked Solution

a.   \(\ce{N2(g) + O2(g) \rightleftharpoons 2NO(g) }\)
 

b.    Increased temperature’s effect on `K_{eq}`:

  • From the graph, the forward reaction is endothermic.
  • The activation energy of the forward endothermic reaction is greater than the activation energy of the reverse exothermic reaction.
  • An increase in temperature would cause the rates of both the forward and reverse reaction due to the higher average kinetic energy, resulting in a larger likelihood of a successful collisions.
  • However, the rate of the forward reaction would increase to a higher extent than the reverse reaction, since it is an endothermic reaction.
  • Using  `K_(eq) = ([text{NO}]^2)/([text{N}_2][text{O}_2])`, as the equilibrium shifts right, the equilibrium constant would increase.

BIOLOGY, M8 2020 HSC 24

An indicator of kidney function is the volume of filtrate formed at the glomerulus in 1 minute (GFR).
 

A patient's kidney function was monitored and the following data recorded.

  1. Plot the data on the grid.   (2 marks)
      

      
  2. Use the graph to show the year that the patient is predicted to require dialysis. Show your working and answer on the graph.   (2 marks)
  3. Explain how dialysis compensates for the loss of a function of the kidneys.   (3 marks)

Show Answers Only

a. & b.   

c.    Kidneys that lose function cannot remove urea from the blood.

  • In dialysis, blood is passed through a permeable dialysis tube in which fluid around the tube called dialysate, which has a similar composition to blood with no urea present, flows the opposite way. 
  • Through the concentration gradient, urea is removed from the blood (where it has high concentration) to the dialysate (low concentration).
Show Worked Solution

a. & b.    


♦♦ Mean mark (b) 36%.

c.    Kidneys that lose function cannot remove urea from the blood.

  • In dialysis, blood is passed through a permeable dialysis tube in which fluid around the tube called dialysate, which has a similar composition to blood with no urea present, flows the opposite way. 
  • Through the concentration gradient, urea is removed from the blood (where it has high concentration) to the dialysate (low concentration).

♦♦ Mean mark (c) 32%.

CHEMISTRY, M6 2020 HSC 2 MC

Which indicator in the table would be best for distinguishing between a face cleanser (pH = 5.0) and a soap (pH = 9.0)?
 

Show Answers Only

`C`

Show Worked Solution
  • The phenol red would be yellow for face cleanser (pH of 5.0) and red for soap (pH of 9.0).
  • The other indicators would give off the same colour for both.

`=> C`

CHEMISTRY, M7 2022 HSC 29

The enthalpies of combustion of four alcohols were determined in a school laboratory.

The results are shown in the table.
 

  1. Plot the results, including a curved line of best fit, to estimate the enthalpy of combustion of butan-1-ol.   (3 marks)

  1. The published value for the enthalpy of combustion of pentan-1-ol is closer to `-3331\ text{kJ}\ text{mol}^(-1)`.
  2. Justify ONE possible reason for the difference between the school's results and published values.   (2 marks)
Show Answers Only

a.    

From interpolating the graph, the enthalpy of combustion of butan-1-ol is –2120 kJ mol ¯1.

b.   Heat loss to the surroundings.

  • The school’s results are lower in magnitude than the published values because heat is lost to the surroundings, making the measured change in temperature smaller.

Other possible answers:

  • Incomplete combustion
  • Temperature change will be reduced if combustion is incomplete as less heat is released. Enthalpy of combustion will be lower as a result.
Show Worked Solution

a.    

From interpolating the graph, the enthalpy of combustion of butan-1-ol is –2120 kJ mol ¯1.
 

b.   Heat loss to the surroundings.

  • The school’s results are lower in magnitude than the published values because heat is lost to the surroundings, making the measured change in temperature smaller.

Other possible answers:

  • Incomplete combustion
  • Temperature change will be reduced if combustion is incomplete as less heat is released. Enthalpy of combustion will be lower as a result.

♦ Mean mark (b) 48%.

CHEMISTRY, M8 2022 HSC 27

A bottle labelled 'propanol' contains one of two isomers of propanol.

  1. Draw the TWO isomers of propanol.   (2 marks)
     

  1. Describe how \( \ce{^13C NMR}\) spectroscopy might be used to identify which isomer is in the bottle.   (2 marks)
  1. Each isomer produces a different product when oxidised.
  2. Write equations to represent the oxidation reactions of the two isomers. Include reaction conditions.   (3 marks)
Show Answers Only

a.    Isomer 1:

Isomer 2:

b.   Identifying isomers with \( \ce{^13C NMR} \) spectroscopy:

  • this can be used to identify the isomers in the bottle because they show a different number of signals which helps deduce the carbon environment.
  • Propan-1-ol contains 3 \( \ce{C}\) environments so it would have 3 peaks on a \( \ce{^13C NMR}\) spectrum whereas propan-2-ol only contains 2 \( \ce{C}\) environments (due to symmetry), so it would only have 2 signals on a \( \ce{^13C NMR}\) spectrum.

c.  


  

Show Worked Solution

a.    Isomer 1:

   

Isomer 2:

   

b.   Identifying isomers with \( \ce{^13C NMR} \) spectroscopy:

  • this can be used to identify the isomers in the bottle because they show a different number of signals which helps deduce the carbon environment.
  • Propan-1-ol contains 3 \( \ce{C}\) environments so it would have 3 peaks on a \( \ce{^13C NMR}\) spectrum whereas propan-2-ol only contains 2 \( \ce{C}\) environments (due to symmetry), so it would only have 2 signals on a \( \ce{^13C NMR}\) spectrum.

c.  


  


♦ Mean mark (c) 52%.

CHEMISTRY, M5 2022 HSC 23

Consider the following system which is at equilibrium in a rigid, sealed container.

\( \ce{4NH3(g) + 5O2(g) \rightleftharpoons 4NO(g) + 6H2O(g)} \ \ \ \ \ \  \Delta H = -950\ \text{kJ mol}^{-1} \) 

  1. Identify what would happen to the amount of \( \ce{NO(g)} \) if the temperature was increased.   (1 mark)
  1. Explain why a catalyst does not affect the equilibrium position of this system.   (2 marks)
  1. Using collision theory, explain what would happen to the concentration of \( \ce{NO(g)} \) if \( \ce{H2O(g)} \) was removed from the system.   (3 marks)
Show Answers Only

a.   The amount of \( \ce{NO2}\) decreases.

b.   Catalyst affect on equilibrium:

  • A catalyst lowers the activation energy and equally increases the rate of both the forward and reverse reactions.
  • As a result, there is no net change in equilibrium, and thus the equilibrium position remains unchanged.

c.   If \( \ce{H2O}\) is removed from the system:

  • This would cause the reverse reaction to decrease because there would be a lower likelihood of successful collisions between \( \ce{NO}\) and \( \ce{H2O}\) molecules.
  • As a result, the forward reaction rate is greater than the reverse reaction rate, thus, the equilibrium would shift to the right, causing \( \ce{[NO]}\) and \( \ce{[H2O]}\) to increase.
  • As the equilibrium is shifting to the right, the forward reaction rate decreases, whilst the reverse reaction rate increases, until they reach equilibrium.
  • At equilibrium, the concentration of all substances remain constant.
Show Worked Solution

a.   The amount of \( \ce{NO2}\) decreases. 

b.   Catalyst affect on equilibrium:

  • A catalyst lowers the activation energy and equally increases the rate of both the forward and reverse reactions.
  • As a result, there is no net change in equilibrium, and thus the equilibrium position remains unchanged.

c.   If \( \ce{H2O}\) is removed from the system:

  • This would cause the reverse reaction to decrease because there would be a lower likelihood of successful collisions between \( \ce{NO}\) and \( \ce{H2O}\) molecules.
  • As a result, the forward reaction rate is greater than the reverse reaction rate, thus, the equilibrium would shift to the right, causing \( \ce{[NO]}\) and \( \ce{[H2O]}\) to increase.
  • As the equilibrium is shifting to the right, the forward reaction rate decreases, whilst the reverse reaction rate increases, until they reach equilibrium.
  • At equilibrium, the concentration of all substances remain constant.

CHEMISTRY, M6 2022 HSC 22

The following equation describes an equilibrium reaction.

\( \ce{HF(aq) + PO4^3-(aq) \rightleftharpoons HPO4^2-(aq) + F-(aq)} \)

Identify ONE base and its conjugate acid in the above equation.   (2 marks)
 

Show Answers Only

Possible answers:

\begin{array} {ll}
\text{Base} & \text{Conjugate Acid} \\
\hline \ce{PO4^3-(aq)} & \ce{HPO4^2-(aq)} \\
\ce{F-(aq)} & \ce{HF(aq)} \\
  \end{array}

Show Worked Solution

Possible answers:

\begin{array} {ll}
\text{Base} & \text{Conjugate Acid} \\
\hline \ce{PO4^3-(aq)} & \ce{HPO4^2-(aq)} \\
\ce{F-(aq)} & \ce{HF(aq)} \\
  \end{array}

Complex Numbers, EXT2 N2 2022 HSC 13c

Consider the equation  `z^5+1=0`, where `z` is a complex number.

  1. Solve the equation  `z^5+1=0`  by finding the 5th roots of `-1`.  (2 marks)

  2. Show that if `z` is a solution of  `z^5+1=0`  and  `z !=-1`, then  `u=z+(1)/(z)`  is a solution of  `u^2-u-1=0`.  (2 marks)

  3. Hence find the exact value of  `cos\ (3pi)/(5)`.  (3 marks)

Show Answers Only
  1. `z=e^(i(pi)/5), e^(i(3pi)/5), e^(-i(pi)/5), -1, e^(-i(3pi)/5)`
  2. `text{Proof (See Worked Solutions)}`
  3. `(1-sqrt5)/4`
Show Worked Solution

i.   `z^5+1=0\ \ =>\ \ z^5=-1`

`z=e^(i((2k+1)/5)),\ \ kin{0,1,-1,2,-2}`

`:.z=e^(i(pi)/5), e^(i(3pi)/5), e^(-i(pi)/5), -1, e^(-i(3pi)/5)`
  

ii.   `z^5+1=(z+1)(z^4-z^3+z^2-z+1)`

`text{Given}\ \ z!=-1,`

`z^4-z^3+z^2-z+1=0`
 

`text{Divide by}\ z^2\ \ (z!=0)`

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

 
`text{Let}\ \ u=z+1/z:`

`:.u^2-u-1=0`
 


Mean mark (ii) 53%.

iii.  `u^2-u-1=0`

`text{By quadratic formula:}`

`u` `=(1+-sqrt(1-4xx1xx(-1)))/(2)`  
  `=(1+-sqrt5)/2`  

♦ Mean mark (iii) 43%.
`z+1/z` `=(1+-sqrt5)/2`  
`e^(i(3pi)/5)+e^(-i(3pi)/5)` `=(1-sqrt5)/2,\ \ (cos\ (3pi)/5 <0)`  
`2cos((3pi)/5)` `=(1-sqrt5)/2`  
`cos((3pi)/5)` `=(1-sqrt5)/4`  

CHEMISTRY, M8 2022 HSC 5 MC

Which pair of ions can be distinguished using a flame test in the school laboratory?

  1. `\text{Ag}^(+) and \text{Mg}^(2+)`
  2. `\text{Ba}^(2+) and \text{Ca}^(2+)`
  3. `\text{Br}^(-) and \text{Cl}^(-)`
  4. `\text{Fe}^(2+) and \text{Fe}^(3+)`
Show Answers Only

`B`

Show Worked Solution

By elimination:

  • Silver and Magnesium do not emit visible wavelengths of light. (A is incorrect)}
  • The flame test only works on metals (C is incorrect)
  • Ions of the same element but different oxidation states cannot be distinguished using the flame test (D is incorrect)
  • In a flame test `text{Ba}^(2+)` has an apple-green flame colour whilst `text{Ca}^(2+)` has a brick-red flame colour, and thus can be distinguished. (B is correct)

`=> B`

CHEMISTRY, M5 2022 HSC 3 MC

Which of the following features is NOT a characteristic of a state of equilibrium?

  1. Equilibrium is achieved in a closed system.
  2. Equilibrium position depends on temperature.
  3. Equilibrium can be reached from either direction.
  4. Equilibrium concentrations of reactants and products are equal.
Show Answers Only

`D`

Show Worked Solution
  • The concentration of the reactants and products remains constant but is not required to be equal at equilibrium.

`=> D`

Calculus, EXT2 C1 2022 HSC 12d

Using partial fractions, evaluate  `int_(2)^(n)(4+x)/((1-x)(4+x^(2))) dx`, giving your answer in the form  `(1)/(2)ln((f(n))/(8(n-1)^(2)))`, where `f(n)` is a function of `n`.  (4 marks)

Show Answers Only

`1/2ln((4+n^2)/(8(1-n^2)))`

Show Worked Solution
`(4+x)/((1-x)(4+x^(2)))` `≡ A/(1-x) + (Bx+C)/(4+x^2)`  
`4+x` `≡A(4+x^2)+(Bx+C)(1-x)`  

 
`text{If}\ \ x=1, \ 5=5A\ \ =>\ \ A=1`

`(4+x)` `≡ 4+x^2+Bx-Bx^2+C-Cx`  
`4+x` `≡ (1-B)x^2+(B-C)x+C+4`  

 
`=>\ B=1, \ C=0`
 


Mean mark 85%.

`:.int_(2)^(n)(4+x)/((1-x)(4+x^(2))) dx`

`=int_2^n 1/(1-x) +x/(4+x^2)\ dx`

`=[-ln abs(1-x)+1/2ln(4+x^2)]_2^n`

`=-ln abs(1-n)+1/2ln(4+n^2)+lnabs(1-2)-1/2ln(4+2^2)`

`=-1/2ln(1-n)^2+1/2ln(4+n^2)-1/2ln(8)`

`=1/2ln((4+n^2)/(8(1-n^2)))`

Mechanics, EXT2 M1 2022 HSC 12c

A particle with mass 1 kg is moving along the `x`-axis. Initially, the particle is at the origin and has speed `u` m s-1 to the right. The particle experiences a resistive force of magnitude  `v+3 v^2`  newtons, where `v` m s-1 is the speed of the particle after `t` seconds. The particle is never at rest.

  1. Show that  `(dv)/(dx)=-(1+3v)`  (1 mark)

  2. Hence, or otherwise, find `x` as a function of `v`.  (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `x=1/3 ln((1+3u)/(1+3v))`
Show Worked Solution

i.   `F=m ddotx=ddotx\ \ (m=1)`

`ddotx` `=-(v+3v^2)`  
`v*(dv)/(dx)` `=-(v+3v^2)`  
`:. (dv)/(dx)` `=-(1+3v)\ \ text{… as required}`  

  

ii.   `(dx)/(dv)` `=- 1/(1+3v)`
  `x` `=-int1/(1+3v)\ dv`
    `=-1/3 ln(1+3v)+c`

 
`text{When}\ \ t=0, \ v=u\ \ =>\ \ c=1/3 ln(1+3u)`

`:.x` `=1/3 ln(1+3u)-1/3 ln(1+3v)`  
  `=1/3 ln((1+3u)/(1+3v))`  

Mechanics, EXT2 M1 2022 HSC 12b

A particle is moving in a straight line with acceleration  `a=12-6 t`. The particle starts from rest at the origin.

What is the position of the particle when it reaches its maximum velocity?  (3 marks)

Show Answers Only

`x=16`

Show Worked Solution
`a` `=12-6t`  
`v` `=int 12-6t\ dt`  
  `=12t-3t^2+c`  

 
`text{When}\ \ t=0, v=0\ \ =>\ \ c=0`

`x` `=int 12t-3t^2\ dt`  
  `=6t^2-t^3+c`  

 
`text{When}\ \ t=0, x=0\ \ =>\ \ c=0`

 
`v_max\ \ text{occurs when}\ \ a=0:`

`12-6t=0\ \ =>\ \ t=2`

`:.x|_(t=2)` `=6(2^2)-2^3`  
  `=16`  

Calculus, EXT2 C1 2022 HSC 11f

Using the substitution  `t=tan\ x/2`, find

`int(dx)/(1+cos x-sin x)`  (3 marks)

 

Show Answers Only

`-lnabs(1-tan(x/2))+C`

Show Worked Solution

`text(Let)\ \ t = tan\ x/2, \ cos\ x = (1-t^2)/(1 + t^2), \ sin\ x=(2t)/(1+t^2)`

`dt = 1/2 sec^2\ x/2\ dx \ => \ d x = (2\ dt)/(sec^2\ x/2) = 2/(1 + t^2)\ dt`

`text{I}` `= int(dx)/(1+cos x-sin x)`  
  `=int 1/(1+(1-t^2)/(1 + t^2)-(2t)/(1 + t^2)) *2/(1 + t^2)\ dt`  
  `=int 2/(1+t^2+1-t^2-2t)\ dt`  
  `=int 1/(1-t)\ dt`  
  `=-ln abs(1-t)+C`  
  `=-lnabs(1-tan(x/2))+C`  

Vectors, EXT2 V1 2022 HSC 11e

Let `ℓ_(1)` be the line with equation `([x],[y])=([-1],[7])+lambda([3],[2]),lambda inRR`.

The line `ℓ_(2)` passes through the point  `A(-6,5)`  and is parallel to `ℓ_(1)`.

Find the equation of the line `ℓ_(2)` in the form  `y=mx+c`.  (2 marks)

Show Answers Only

`y=2/3x+9`

Show Worked Solution

`m_(ℓ_(1))=2/3`

`text{Equation of}\ ℓ_(2)\ text{has}\ m=2/3\ text{and passes through}\ (-6,5):`

`y-5` `=2/3(x+6)`  
`y` `=2/3x+9`  

CHEMISTRY, M5 2021 HSC 22

Consider the following equilibrium system.

The solution is orange.

Justify TWO ways to shift the equilibrium to the left to change the colour of the solution.   (3 marks)

Show Answers Only

Answers could include two of the following methods.

Add `text{Cr}_2 text{O}_7 \ ^(2-)` ions:

  • This would increase the concentration of `text{Cr}_2 text{O}_7\ ^(2-)` ions.
  • According to Le Chatelier’s Principle, this would cause the equilibrium to shift left to counteract the change and decrease the concentration of `text{Cr}_2 text{O}_7\ ^(2-)` ions.

Decrease the concentration of `text{H}^+` ions by adding a base:

  • This would decrease the concentration of `text{H}^+` ions through an acid-base reaction.
  • According to Le Chatelier’s Principle, this would cause the equilibrium to shift left to counteract the change and increase the concentration of `text{H}^+` ions.

Increase the temperature:

  • According to Le Chatelier’s Principle, this would cause the equilibrium to shift left towards the endothermic side to counteract the change and decrease the temperature. 
Show Worked Solution

Answers could include two of the following methods.

Add `text{Cr}_2 text{O}_7 \ ^(2-)` ions:

  • This would increase the concentration of `text{Cr}_2 text{O}_7\ ^(2-)` ions.
  • According to Le Chatelier’s Principle, this would cause the equilibrium to shift left to counteract the change and decrease the concentration of `text{Cr}_2 text{O}_7\ ^(2-)` ions.

Decrease the concentration of `text{H}^+` ions by adding a base:

  • This would decrease the concentration of `text{H}^+` ions through an acid-base reaction.
  • According to Le Chatelier’s Principle, this would cause the equilibrium to shift left to counteract the change and increase the concentration of `text{H}^+` ions.

Increase the temperature:

  • According to Le Chatelier’s Principle, this would cause the equilibrium to shift left towards the endothermic side to counteract the change and decrease the temperature. 

CHEMISTRY, M8 2021 HSC 21

Four organic liquids are used in an experiment. The four liquids are

    • hexane
    • hex-1-ene
    • propan-1-ol
    • propanoic acid
  1. State ONE safety concern associated with organic liquids and suggest ONE way to address this safety concern.   (2 marks)

  2. The organic liquids are held separately in four flasks but the flasks are not labelled. Tests were conducted to identify these liquids. The outcomes of the tests are summarised below.   (2 marks)
     
           
    Identify the FOUR liquids.
     
           
  3. What chemical test, other than those used in part (b), could be used to confirm the identification of ONE of the liquids? Include the expected observation in your answer.   (2 marks)

Show Answers Only

a.    A safety concern is that the organic liquids are flammable.

To address this, keep substance away from open flames and keep away from ignition sources.
 

b.    Flask 1: propanoic acid  (carboxylic acids can’t be oxidised and are polar)

Flask 2: hex-1-ene (alkenes can be oxidised and are non-polar)

Flask 3: propan-1-ol (primary alcohols can be oxidised and are polar)

Flask 4: hexane (alkanes don’t react with acidified oxidants and are non-polar)
 

c.   Hex-1-ene

  • Could be identified using the bromine water test.
  • The addition of brown bromine water to an alkene causes an addition reaction where the solution changes colours from brown to colourless.

Propanoic acid

  • Could be identified through a neutralisation reaction using `text{Na}_2text{CO}_3`.
  • Effervescent reaction will result.

Propan-1-ol

  • Could be identified through an oxidation reaction using acidified dichromate.
  • The reaction would cause the solution to change from green to orange.
Show Worked Solution

a.    A safety concern is that the organic liquids are flammable.

To address this, keep substance away from open flames and keep away from ignition sources.
 

b.    Flask 1: propanoic acid  (carboxylic acids can’t be oxidised and are polar)

Flask 2: hex-1-ene (alkenes can be oxidised and are non-polar)

Flask 3: propan-1-ol (primary alcohols can be oxidised and are polar)

Flask 4: hexane (alkanes don’t react with acidified oxidants and are non-polar)
 

c.   Hex-1-ene

  • Could be identified using the bromine water test.
  • The addition of brown bromine water to an alkene causes an addition reaction where the solution changes colours from brown to colourless.

Propanoic acid

  • Could be identified through a neutralisation reaction using `text{Na}_2text{CO}_3`.
  • Effervescent reaction will result.

Propan-1-ol

  • Could be identified through an oxidation reaction using acidified dichromate.
  • The reaction would cause the solution to change from green to orange.

Vectors, EXT2 V1 2022 HSC 11d

A triangle is formed in three-dimensional space with vertices  `A(1,-1,2)`, `B(0,2,-1)`  and  `C(2,1,1)`.

Find the size of `/_ABC`, giving your answer to the nearest degree.  (3 marks)

Show Answers Only

`33°`

Show Worked Solution

`vec(BA)=((1),(-1),(2))-((0),(2),(-1))=((1),(-3),(3))`

`abs(vec(BA))=sqrt(1^2+3^2+3^2)=sqrt19`
 

`vec(BC)=((2),(1),(1))-((0),(2),(-1))=((2),(-1),(2))`

`abs(vec(BC))=sqrt(2^2+1^2+2^2)=sqrt9=3`
 

`vec(BA)*vec(BC)=1xx2+ -3xx-1+3xx2=11`

`cos/_ABC` `=(vec(BA)*vec(BC))/(abs(vec(BA)abs(vec(BC))`  
  `=11/(3sqrt19)`  
`:./_ABC` `=cos^(-1)(11/(3sqrt19))`  
  `=32.733…`  
  `=33°\ \ text{(nearest degree)}`  

Complex Numbers, EXT2 N1 2022 HSC 11c

  1. Write the complex number  \(-\sqrt{3}+i\) in exponential form.  (2 marks)

  2. Hence, find the exact value of  \((-\sqrt{3}+i)^{10}\) giving your answer in the form  \(x+i y\).  (2 marks)

Show Answers Only
  1. \(2 e^{\small{\dfrac{5 \pi}{6}} i}\)
  2. \(512+512 \sqrt{3} i\)
Show Worked Solution

i.   

\(\text {Let}\ \ z=-\sqrt{3}+i\)

\(\abs{z}=\sqrt{(-\sqrt{3})^2+1^2}=2\)

\(\text{Find}\ \ \arg (z):\)

\(\tan \theta=\dfrac{1}{\sqrt{3}} \Rightarrow \theta=\dfrac{\pi}{6}\)

\(\Rightarrow \arg (z)=\dfrac{5 \pi}{6}\)

\(\therefore z\) \(=2\left(\dfrac{\cos (5 \pi)}{6}+\dfrac{\sin (5 \pi)}{6} i\right)\)  
  \(=2 e^{\small{\dfrac{5 \pi}{6}} i}\)  

 

ii.    \((-\sqrt{3}+i)^{10}\) \(=\left(2 e^{\small{\dfrac{5 \pi}{6}} i}\right)^{10}\)
    \(=2^{10} e^{\small{\dfrac{50 \pi}{6}} i}\)
    \(=1024 e^{\small{\dfrac{\pi}{3}} i}\)
    \(=1024\left(\cos \left(\dfrac{\pi}{3}\right)+\sin \left(\dfrac{\pi}{3}\right) i\right)\)
    \(=1024\left(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2} i\right)\)
    \(=512+512 \sqrt{3} i\)

ENGINEERING, AE 2021 HSC 23a

Due to a reduction in air travel, an airline company has had to store some of its unused aircraft.

Explain why the desert is a suitable place to store the unused aircraft.   (3 marks)

Show Answers Only
  • Aircraft are susceptible to corrosion as they are constructed from aluminium. Specifically, crevice corrosion in the joints between the aluminium sheets.
  • While the aircraft are in storage they should be kept in a dry area to prevent any moisture from entering the crevices.
  • A desert provides this environment at a low cost. 

Successful answers could also include:

  • Deserts also have fewer insects/wildlife which could nest in the aircraft.
  • The surface of the desert is also hard and dry so paving is not required.
Show Worked Solution
  • Aircraft are susceptible to corrosion as they are constructed from aluminium. Specifically, crevice corrosion in the joints between the aluminium sheets.
  • While the aircraft are in storage they should be kept in a dry area to prevent any moisture from entering the crevices.
  • A desert provides this environment at a low cost. 

Successful answers could also include:

  • Deserts also have fewer insects/wildlife which could nest in the aircraft.
  • The surface of the desert is also hard and dry so paving is not required.

BIOLOGY, M6 2020 HSC 12 MC

What is the purpose of cloning in agriculture?

  1. Increasing the frequency of recessive traits
  2. Preserving favourable traits in the offspring
  3. Preserving genetic variability in a population
  4. Increasing combinations of alleles in a population
Show Answers Only

`B`

Show Worked Solution
  • Cloning is often used in agricultural practises within both flora and fauna to preserve favourable market characteristics, such as muscle size in cows.

`=>B`

BIOLOGY, M8 2020 HSC 1 MC

In maintaining homeostasis, which of the following is a behavioural adaptation?

  1. Sweating to cool down
  2. Curling up in a ball to keep warm
  3. Speeding up or slowing down cell metabolism
  4. Skin going red as more blood flows to surface
Show Answers Only

`B`

Show Worked Solution
  • Curling into a ball to keep warm is a behavioural response of an organism that helps it maintain body temperature.

`=>B`

BIOLOGY, M7 2021 HSC 16 MC

Scientists conducted an experiment to investigate the effectiveness of treating water from storage dams with UV radiation.

The experiment was conducted more than three times. The results are shown in the table.
 

What conclusion may be drawn from the data obtained?

  1. The control plates are contaminated.
  2. High doses of UV eliminate all pathogens.
  3. Exposure to UV inhibits reproduction of bacteria.
  4. The presence of bacteria reduces the amount of UV.
Show Answers Only

`C`

Show Worked Solution
  • As the UV dose increased, the number of bacteria decreased.

`=>C`

♦ Mean mark 79%.

PHYSICS, M7 2017 HSC 21

A laser emits light of wavelength 550 nm.

  1. Calculate the frequency of this light.   (2 marks)
  1. The electrons in a specific metal must absorb a minimum of  `5 × 10^(-19)\ text{J}`  in order to be ejected from its surface.
  2. Explain why electrons will not be ejected from this metal when photons of wavelength 550 nm strike its surface. Support your answer with relevant calculations.   (3 marks)
Show Answers Only

a.   `f=5.45 xx10^(14)  text{Hz}`

b.    `E` `=3.63 xx10^(-19)  text{J}`
    `<5 xx10^(-19)  text{J}`

 
`text{So, no electrons are ejected.}`

Show Worked Solution
a. `v` `=flambda`
  `f` `=(v)/(lambda)=(3xx10^(8))/(550 xx10^(-9))=(3xx10^(8))/(550 xx10^(-9))`

 

b. `E` `=hf`
    `=6.62 xx10^(-34)xx5.46 xx10^(14)`
    `=3.63 xx10^(-19)  text{J}`
     
  •  The work function of the metal sample is  `5 xx10^(-19)\ text{J}`.
  • Since the incident photon energy of  `3.63 xx10^(-19)  text{J}`  is less than  `5 xx10^(-19)\ text{J}`, they are unable to eject electrons from the metal.

PHYSICS, M5 2022 HSC 25

A rocket is launched vertically from a planet of mass \(M\). After it leaves the atmosphere, the rocket's engine is turned off and it continues to move away from the planet. From this time the rocket's mass is 200 kg. The rocket's speed, \(v\), at two different distances from the planet's centre, \(R\), is shown.

\begin{array}{|c|c|c|}
\hline \rule{0pt}{2.5ex} \text {Point } \rule[-1ex]{0pt}{0pt}& R\ \text{(m)} & \quad v\left(\text{m s}^{-1}\right) \quad \\
\hline \rule{0pt}{2.5ex} 1 \rule[-1ex]{0pt}{0pt}& \quad 4.3 \times 10^6 \quad & 5500 \\
\hline \rule{0pt}{2.5ex} 2 \rule[-1ex]{0pt}{0pt}& 2.5 \times 10^7 & 2900 \\
\hline
\end{array}

  1. Show that the magnitude of the change in kinetic energy from point 1 to point 2 is  \(2.2 \times 10^9 \ \text{J}\).  (2 marks)

  1. Determine the mass \(M\) of the planet using the law of conservation of energy.  (3 marks)

Show Answers Only

a.  \(\Delta K=K_f-K_i\)

\(K_f=\dfrac{1}{2} m v_2^2=\dfrac{1}{2} 200 \times 2900^2=8.41 \times 10^8 \ \text{J}\)

\(K_i=\dfrac{1}{2} m v_1^2=\dfrac{1}{2} 200 \times 5500^2=3.025 \times 10^9 \ \text{J}\)

  \(\therefore \Delta K\) \(=8.41 \times 10^8-3.025 \times 10^9\)
    \(=-2.2 \times 10^9 \ \text{J}\)

 
\(\therefore\) This change has a magnitude of \(2.2 \times 10^9 \ \text{J}\).
  

b.   Applying the law of conservation of energy:

  • The magnitude of kinetic energy lost is equal to the magnitude of potential energy gained by the rocket:
  •    \(\Delta U=U_f-U_i=-\dfrac{G M m}{r_f}-\left(-\dfrac{G M m}{r_i}\right)\)

\(\left(2.2 \times 10^9\right)=-\dfrac{\left(6.67 \times 10^{-11}\right)(200) M}{2.5 \times 10^7}+\dfrac{\left(6.67 \times 10^{-11}\right)(200) M}{4.3 \times 10^6}\)
 

\(\left(2.2 \times 10^9\right)\left(2.5 \times 10^7\right)\left(4.3 \times 10^6\right)\)

\(=\left(6.67 \times 10^{-11}\right) 200 M\left[\left(2.5 \times 10^7\right)-\left(4.3 \times 10^7\right)\right]\)
 

\(\therefore M\) \(=\dfrac{\left(2.2 \times 10^9\right)\left(2.5 \times 10^7\right)\left(4.3 \times 10^6\right)}{200\left(6.67 \times 10^{-11}\right)\left[\left(2.5 \times 10^7\right)-\left(4.3 \times 10^6\right)\right]}\)  
  \(=8.56 \times 10^{23} \ \text{kg}\)  
Show Worked Solution

a.  \(\Delta K=K_f-K_i\)

\(K_f=\dfrac{1}{2} m v_2^2=\dfrac{1}{2} 200 \times 2900^2=8.41 \times 10^8 \ \text{J}\)

\(K_i=\dfrac{1}{2} m v_1^2=\dfrac{1}{2} 200 \times 5500^2=3.025 \times 10^9 \ \text{J}\)

  \(\therefore \Delta K\) \(=8.41 \times 10^8-3.025 \times 10^9\)
    \(=-2.2 \times 10^9 \ \text{J}\)

 
\(\therefore\) This change has a magnitude of \(2.2 \times 10^9 \ \text{J}\).
 

b.   Applying the law of conservation of energy:

  • The magnitude of kinetic energy lost is equal to the magnitude of potential energy gained by the rocket:
  •    \(\Delta U=U_f-U_i=-\dfrac{G M m}{r_f}-\left(-\dfrac{G M m}{r_i}\right)\)

\(\left(2.2 \times 10^9\right)=-\dfrac{\left(6.67 \times 10^{-11}\right)(200) M}{2.5 \times 10^7}+\dfrac{\left(6.67 \times 10^{-11}\right)(200) M}{4.3 \times 10^6}\)
 

\(\left(2.2 \times 10^9\right)\left(2.5 \times 10^7\right)\left(4.3 \times 10^6\right)\)

\(=\left(6.67 \times 10^{-11}\right) 200 M\left[\left(2.5 \times 10^7\right)-\left(4.3 \times 10^7\right)\right]\)

 

\(\therefore M\) \(=\dfrac{\left(2.2 \times 10^9\right)\left(2.5 \times 10^7\right)\left(4.3 \times 10^6\right)}{200\left(6.67 \times 10^{-11}\right)\left[\left(2.5 \times 10^7\right)-\left(4.3 \times 10^6\right)\right]}\)  
  \(=8.56 \times 10^{23} \ \text{kg}\)

♦ Mean mark (b) 47%.

PHYSICS M8 2022 HSC 24

The radioactive decay curve for americium-242 is shown.
 


 

  1. Use the graph to find the half-life of Am-242 and hence show that the decay constant, `\lambda`, is `0.043` `text{h}^(-1).`   (2 marks)
  1. Calculate how long it takes until the mass of Am-242 is 8 micrograms.   (2 marks)
Show Answers Only

a.   `0.043  text{h}^(-1)`

b.  `53.55\ text{h}`

Show Worked Solution

a.   The half life of Am-242 is 16 hours.

`lambda=(ln 2)/(t_(1//2))=(ln 2)/(16)=0.043  text{h}^(-1)`
 

b.    `N` `=N_(0)e^(-lambda t)`
     `8` `=80e^(-0.043 t)`
  `e^(-0.043 t)` `=8/80`
  `-0.043t` `=ln(1/10)`
     `t` `=((-1)/(0.043))ln ((1)/(10))`
    `=53.55\ text{h}`

PHYSICS M8 2022 HSC 21

The positions of two stars, `X` and `Y`, are shown in the Hertzsprung-Russell diagram.
 

  1. Compare qualitatively the surface temperature and luminosity of `X` and `Y`.   (2 marks)

  1. Identify the elements undergoing fusion in the core of each star, `X` and `Y`.   (2 marks)
Show Answers Only

a.   Surface Temperature:

  • The surface temperature of `X` is greater than the surface temperature of `Y.`

Luminosity:

  • The luminosity of `Y` is greater than the luminosity of `X.`

b.   Elements undergoing fusion:

  • `X` is a main sequence star so in its core hydrogen is being fused into helium.
  • `Y` is a red giant so in its core helium is being fused into carbon.
Show Worked Solution

a.   Surface Temperature:

  • The surface temperature of `X` is greater than the surface temperature of `Y.`

Luminosity:

  • The luminosity of `Y` is greater than the luminosity of `X.`

b.   Elements undergoing fusion:

  • `X` is a main sequence star so in its core hydrogen is being fused into helium.
  • `Y` is a red giant so in its core helium is being fused into carbon.

BIOLOGY, M8 2021 HSC 25

A patient visited an audiologist for a hearing test. The audiologist tested both ears at specific frequencies. The volumes at which each frequency could be heard are shown.
 

  1. Plot the data on the grid provided and include a key.   (3 marks)
     

  1. What conclusions can be drawn about the patient's hearing?   (2 marks)

  2. It is discovered that there is a complete and permanent blockage of the outer ear, but the cochlea is still fully functional.
  3. Justify the use of a suitable technology to assist the patient's hearing.   (3 marks)

Show Answers Only

a.  

b.   Right ear is within normal hearing range.

Left ear has a deficit and cannot hear at a normal level.
 

c.    Effective technology: Bone Conduction Implants

Bone conduction implants would prove to be the most effective technology to restore hearing to this patient. 

Bone conduction implants detect sound waves via a microphone, relaying them to a sound processor that converts the waves into vibrations which are then directly transferred to the cochlea. This process bypasses the ear blockage, therefore restoring hearing to the patient.

Show Worked Solution

a.   


 

b.   Right ear is within normal hearing range.

  • Left ear has a deficit and cannot hear at a normal level.

c.    Effective technology: Bone Conduction Implants

  • Bone conduction implants would prove to be the most effective technology to restore hearing to this patient. 
  • Bone conduction implants detect sound waves via a microphone, relaying them to a sound processor that converts the waves into vibrations which are then directly transferred to the cochlea.
  • This process bypasses the ear blockage, therefore restoring hearing to the patient.

♦ Mean mark (c) 43%.

BIOLOGY, M5 2021 HSC 23

Ovulation in women is associated with a rapid increase in luteinising hormone (LH). Test strips can be used to detect high levels of LH in urine. Once a test strip is used, a control line should appear and the presence of a test line indicates high levels of LH in urine. The image below represents four different results that were obtained.
 

Which of the results indicates a valid test which shows that ovulation is NOT occurring? Justify your answer.   (3 marks)

Show Answers Only
  • Test strips 3 and 4 do not contain a control line, therefore are invalid.
  • Test strip 2 contains a control line but no test line, indicating that there was no LH surge.
  • Therefore, test strip 2 indicates a valid negative result that shows ovulation is not occurring.
Show Worked Solution
  • Test strips 3 and 4 do not contain a control line, therefore are invalid.
  • Test strip 2 contains a control line but no test line, indicating that there was no LH surge.
  • Therefore, test strip 2 indicates a valid negative result that shows ovulation is not occurring.

PHYSICS, M7 2022 HSC 9 MC

The radiation emitted by a black body has a peak wavelength of  `5.8 xx10^(-7) \ text{m}`.

What is its temperature?

  1. `3000 \ text{K}`
  2. `4500 \ text{K}`
  3. `5000 \ text{K}`
  4. `5500 \ text{K}`
Show Answers Only

`C`

Show Worked Solution
`lambda_(max)` `=(b)/(T)`  
`T` `=(b)/(lambda_(max))=(2.898 xx10^(-3))/(5.8 xx10^(-7))~~5000  text{K}`  

 
`=>C`

PHYSICS M5 2022 HSC 8 MC

An object is launched with an initial velocity, `u`, and hits a wall with a final velocity, `v`.
 


 

Which statement correctly compares components of `u` and `v` ?

  1. The vertical component of `v` is less than the vertical component of `u`.
  2. The vertical component of `v` is greater than the vertical component of `u`.
  3. The horizontal component of `v` is less than the horizontal component of `u`.
  4. The horizontal component of `v` is greater than the horizontal component of `u`.
Show Answers Only

`A`

Show Worked Solution
  • The vertical velocity of the projectile decreases under the influence of gravity. The horizontal velocity of the projectile remains constant.

`=>A`

PHYSICS M7 2022 HSC 7 MC

A photon has an energy of  `9.0 xx10^(-24)\ text{J}`.

What is the frequency of this radiation?

  1. `1.00 xx10^(-40) \ text{Hz}`
  2. `7.36 xx10^(-11) \ text{Hz}`
  3. `1.36 xx10^(10) \ text{Hz}`
  4. `5.97 xx10^(11) \ text{Hz}`
Show Answers Only

`C`

Show Worked Solution
`E` `=hf`  
`f` `=(E)/(h)=(9.0 xx10^(-24))/(6.626 xx10^(-34))=1.36 xx10^(10)  text{Hz}`  

 
`=>C`

PHYSICS M8 2022 HSC 5 MC

Protons and neutrons are made up of quarks. The table shows the charges of these quarks.
 

What combination of quarks forms a neutron?

  1. 1 up, 1 down
  2. 1 up, 2 down
  3. 2 up, 1 down
  4. 2 up, 2 down
Show Answers Only

`B`

Show Worked Solution
  • Baryons such as neutrons are always are made up of three quarks.
  • Neutrons are electrically neutral and are comprised of one up quark and two down quarks.

`=>B`

PHYSICS M6 2022 HSC 3 MC

A radioisotope emits radiation which is deflected by an electric field, as shown.
 


 

What type of radiation is this?

  1. Alpha
  2. Gamma
  3. Beta positive (positron)
  4. Beta negative (electron)
Show Answers Only

`D`

Show Worked Solution
  • The radiation experiences a force of attraction towards the positive plate and is therefore negatively charged.

`=>D`

PHYSICS M8 2022 HSC 2 MC

The absorption lines in a star's spectrum are shown.
 

What feature of the star is directly responsible for these absorption lines?

  1. Size
  2. Colour
  3. Distance from Earth
  4. Chemical composition
Show Answers Only

`D`

Show Worked Solution
  • Elements and compounds in the stars’ atmosphere will absorb specific wavelengths of light which correspond to these absorption lines.
  • Chemical composition is responsible.

`=>D`

PHYSICS M6 2022 HSC 1 MC

An ideal transformer has 20 turns on the primary coil and an input voltage of 100 V.

How many turns are there on the secondary coil if the output voltage is 400 V ?

  1. 4
  2. 5
  3. 80
  4. 400
Show Answers Only

`C`

Show Worked Solution
`(V_(p))/(V_(s))` `=(N_(p))/(N_(s))`  
`(100)/(400)` `=(20)/(N_(s))`  
`N_(s)` `=(20 xx400)/(100)=80`  

 
`=>C`

BIOLOGY, M7 2021 HSC 21

  1. Label TWO features on the diagram below that would help to classify this pathogen as a bacterium.   (2 marks)
     
     

     
  2. A scientist followed Koch's postulates to confirm that this bacterium was causing diarrhoea in pigs on a local farm. 
  3. Complete the boxes in the flowchart provided to show the steps taken by the scientist.    (2 marks)
     
     

  4.  
  5. Two pig farmers on neighbouring farms noticed that their pigs were suffering from diarrhoea and gradually losing weight. The farmers each adopted a different strategy to deal with this disease, as shown in the table.  

\begin{array} {|c|l|l|}
\hline
\rule{0pt}{2.5ex} \quad \textit{Farm} \quad \rule[-1ex]{0pt}{0pt} & \quad\quad\quad \quad \textit{Strategy} & \quad\quad\quad \quad \textit{Result}\\
\hline
\rule{0pt}{2.5ex} 1  & \text{Treatment with antibiotics} & \text{All pigs recovered after two}\\
\rule[-1ex]{0pt}{0pt} & \text{} & \text{weeks}\\
\hline
\rule{0pt}{2.5ex} 2  & \text{Elimination of rats and mice} & \text{Decrease in number of sick}\\
& \text{from pig sheds to improve} & \text{animals over three months}\\
\rule[-1ex]{0pt}{0pt} & \text{hygiene} & \\
\hline
\end{array}

Outline ONE benefit and ONE limitation of the strategies used on each farm.   (3 marks)

Show Answers Only

a.    Include two of the following labels:
 
       

b.    Box 2: Bacteria grown in pure culture and identified.

Box 4: Healthy pig became ill with diarrhoea. 

c.   Benefits and Limitations of the strategies used on each farm.

  • The use of antibiotics on farm 1 has resulted in a rapid elimination of diarrhoea cases, however may induce antibiotic resistance in the
    future, rendering the strategy less effective.
  • The removal of rats and mice from pig sheds to increase hygiene on farm 2 is slow to eliminate diarrhoea cases, however provides reassurance to prevent future outbreaks. 

Other correct answers:

  • Proper hygiene practices can reduce incidence of other diseases, not just diarrhoea.
Show Worked Solution

a.    Include two of the following labels:
 
       
 

b.    Box 2: Bacteria grown in pure culture and identified.

Box 4: Healthy pig became ill with diarrhoea. 
 

c.   Benefits and Limitations of the strategies used on each farm.

  • The use of antibiotics on farm 1 has resulted in a rapid elimination of diarrhoea cases, however may induce antibiotic resistance in the
    future, rendering the strategy less effective.
  • The removal of rats and mice from pig sheds to increase hygiene on farm 2 is slow to eliminate diarrhoea cases, however provides reassurance to prevent future outbreaks. 

Other correct answers:

  • Proper hygiene practices can reduce incidence of other diseases, not just diarrhoea.

Calculus, EXT1 C1 2022 HSC 12d

In a room with temperature 12°C, coffee is poured into a cup. The temperature of the coffee when it is poured into the cup is 92°C, and it is far too hot to drink.

The temperature, `T`, in degrees Celsius, of the coffee, `t` minutes after it is made, can be modelled using the differential equation  `(dT)/(dt)=k(T-T_(1))`, where `k` is the constant of proportionality and `T_1` is a constant.

  1. It takes 5 minutes for the coffee to cool to a temperature of 76°C.
  2. Using separation of variables, solve the given differential equation to show that  `T=12+80e^((t)/(5)ln((4)/(5)))`.  (3 marks)

  3. The optimal drinking temperature for a hot beverage is 57°C.
  4. Find the value of `t` when the coffee reaches this temperature, giving your answer to the nearest minute.  (1 mark)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{13 minutes}`
Show Worked Solution

i.   `T_1=12\ \ text{(Room temperature = 12°C)}`

`(dT)/(dt)` `=k(T-12)`  
`(dt)/(dT)` `=1/(k(T-12))`  
`int dt` `=1/k int(1/(T-12))\ dT`  
`t` `=1/k ln|T-12|+c`  

 
`text{When}\ \ t=0,\ \ T=92:`

`0` `=1/k ln|92-12|+c`  
`c` `=-1/k ln(80)`  

 

`t` `=1/k ln|T-12|-1/k ln(80)`  
`kt` `=ln((|T-12|)/80)`  
`T-12` `=80e^(kt)`  
`T` `=12+80e^(kt)`  

 
`text{When}\ \ t=5,\ \ T=76:`

`76` `=12+80e^(5k)`  
`e^(5k)` `=64/80`  
`5k` `=ln(4/5)`  
`k` `=1/5 ln(4/5)`  

  
`:.T=12+80e^((t)/(5)ln((4)/(5)))\ \ \ text{… as required}`
 

ii.   `text{Find}\ \ t\ \ text{when}\ \ T=57:`

`57` `=12+80e^((t)/(5)ln((4)/(5)))`  
`e^((t)/(5)ln((4)/(5)))` `=45/80`  
`t/5ln(4/5)` `=ln(9/16)`  
`t` `=(5ln(9/16))/ln(4/5)`  
  `=12.89…`  
  `~~13\ text{minutes}`  

Combinatorics, EXT1 A1 2022 HSC 12b

A sports association manages 13 junior teams. It decides to check the age of all players. Any team that has more than 3 players above the age limit will be penalised.

A total of 41 players are found to be above the age limit.

Will any team be penalised? Justify your answer.   (2 marks)

Show Answers Only

`text{Yes. By PHP, at least one team will have at least}`

`text{4 players above the limit.}`

Show Worked Solution

`text{Pigeonholes}\ (k)=13`

`text{Pigeons}\ (n)=41`

`n/k=41/13=3\ text{remainder 2}`

`:.\ text{By PHP, at least one team must have 4 players above}`

`text{the age limit and therefore at least one team will be}`

`text{penalised.}`

Functions, EXT1 F1 2022 HSC 11f

Solve  `(x)/(2-x) >= 5`.  (3 marks)

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`5/3<=x<2`

Show Worked Solution

`(x)/(2-x) >= 5`

`text{Multiply b.s. by}\ \ (2-x)^2\ \ (>0):`

`x(2-x)` `>=5(2-x)^2`  
`2x-x^2` `>=5(x^2-4x+4)`  
`6x^2-22x+20` `<=0`  
`2(3x^2-11x+10)` `<=0`  
`(3x-5)(x-2)` `<=0`  

 
`text{Test}\ \ x=0:`

`(-5)(-2)=10>0`

`:. 5/3<=x<2\ \ (x!=2)`

Vectors, EXT1 V1 2022 HSC 11d

The vectors  `underset~u=([a],[2])`  and  `underset~v=([a-7],[4a-1])`  are perpendicular.

What are the possible values of `a`?  (2 marks)

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`a=1, -2`

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`text{If}\ \ underset~u ⊥ underset~v:`

`([a],[2])*([a-7],[4a-1])` `=0`  
`a(a-7)+2(4a-1)` `=0`  
`a^2-7a+8a-2` `=0`  
`a^2+a-2` `=0`  
`(a+2)(a-1)` `=0`  

 
`:.a=1\ \ text{or}\ \ -2`

Calculus, EXT1 C2 2022 HSC 11b

Find the exact value of  `int_(0)^(1)(x)/(sqrt(x^(2)+4))\ dx`  using the substitution `u=x^(2)+4`.  (3 marks)

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`sqrt5-2`

Show Worked Solution

`u=x^(2)+4`

`(du)/dx=2x\ \ =>\ \ du=2x\ dx`

`text{At}\ \ x=1,\ \ u=5`

`text{At}\ \ x=0,\ \ u=4`

`int_(0)^(1)(x)/(sqrt(x^(2)+4))\ dx` `=1/2 int_(4)^(5)(1)/(sqrt(u))\ du`  
  `=1/2[2sqrtu]_4^5`  
  `=[sqrtu]_4^5`  
  `=sqrt5-2`  

Vectors, EXT1 V1 2022 HSC 11a

For the vectors  `underset~u= underset~i- underset~j`  and  `underset~v=2 underset~i+ underset~j`, evaluate each of the following. 

  1. `underset~u+3 underset~v`   (1 mark)

  2. `underset~u * underset~v`   (1 mark)

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  1. `((7),(2))`
  2. `1`
Show Worked Solution

i.  `underset~u= ((1),(-1)),\ \ underset~v= ((2),(1))`

`underset~u+3 underset~v` `=((1),(-1))+3((2),(1))`  
  `=((1+3xx2),(-1+3xx1))`  
  `=((7),(2))`  

 

ii.    `underset~u * underset~v` `=((1),(-1))*((2),(1))`
    `=1xx2+(-1)xx1`
    `=1`

Functions, EXT1 F1 2022 HSC 4 MC

The diagram shows the graph of the sum of the functions  `f(x)`  and  `g(x)`.
 

Which of the following best represents the graphs of both  `f(x)`  and  `g(x)`?
 


 

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`A`

Show Worked Solution

`text{By Elimination,}`

`text{Consider the}\ ytext{-intersection of both graphs in each option:}`

`B and C\ text{will have result in a positive}\ ytext{-intersection (Eliminate)}`
 

`y=f(x)+g(x)=0\ \ text{when the two graphs are equidistant from}`

`text{the}\ xtext{-axis (Eliminate}\ D).`

`=>A`