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PHYSICS, M6 2016 HSC 8 MC

Which movement of the magnet(s) will produce the greatest deflection of the galvanometer?
 

 

 

Show Answers Only

`D`

Show Worked Solution

→ A larger induced current will produce greater deflection of the galvanometer.

→ The induced current is proportional to the total rate of change of magnetic flux through the coil.

→ The faster the movement of the magnet, the greater the rate of change of flux.

`=>D`

PHYSICS, M6 2016 HSC 5 MC

The diagram shows two parallel charged plates `5 × 10^(-3)  text{m}` apart.
 

What is the magnitude of the electric field between the plates in `text{V m}^(-1)` ?

  1. `3.3 × 10^(-4)`
  2. `0.33`
  3. `3`
  4. `3000`
Show Answers Only

`D`

Show Worked Solution
`E` `=(V)/(d)`  
  `=(15)/(0.005)`  
  `=3000\ text{V m}^(-1)`  

 
`=>D`

PHYSICS, M6 2016 HSC 1 MC

Some mobile phones are recharged at a power point using a charger that contains a transformer.

What is the purpose of the transformer?

  1. To convert AC at the power point to DC
  2. To convert DC at the power point to AC
  3. To increase the AC voltage at the power point
  4. To decrease the AC voltage at the power point
Show Answers Only

`D`

Show Worked Solution

→ Transformers cannot convert AC to DC or vice versa, they only change the voltage.

→ As transmission voltages are incredibly high in order to decrease power loss, transformers in the home decrease voltages to a level safe for devices to use.

`=>D`

PHYSICS, M5 2017 HSC 24

The escape velocity from a planet is given by  `v = sqrt((2GM)/(r))`.

  1. The radius of Mars is  `3.39 × 10^(6) \ text{m}`  and its mass is  `6.39 × 10^(23) \ text{kg}`.
  2. Calculate the escape velocity from the surface of Mars.   (2 marks)
  1. Using the law of conservation of energy, show that the escape velocity of an object is independent of its mass.   (3 marks)
Show Answers Only

a.   `v=5010  text{m s}^(-1)`

b.   Applying the law of conservation of energy:

→ The object’s initial mechanical energy must equal its final mechanical energy.

`KE_(i)+U_(i)=KE_(f)+U_(f)`

→ The escape velocity is the minimum velocity required for an object to escape from a central mass and never return.

→ As an object reaches an infinite distance away, `U_(f)=0`

→ When an object has just enough speed to escape, its final speed is zero, hence, `KE_(f)=0`.

→ It follows

`KE_(i)+U_(i)` `=0`  
`(1)/(2)mv^(2)-(GMm)/(r)` `=0`  
`mv^(2)` `=(2GMm)/(r)`  
`∴ v_(esc)` `=sqrt((2GM)/(r))`  

 
→ Which is independent of the object’s mass.

Show Worked Solution
a.    `v` `=sqrt((2xx6.67 xx10^(-11)xx6.39 xx10^(23))/(3.39 xx10^(6)))`
    `=5014.5  text{m s}^(-1)`
    `=5015  text{m s}^(-1)\ \ text{(to 0 d.p.)}`

 

b.   Applying the law of conservation of energy:

→ The object’s initial mechanical energy must equal its final mechanical energy.

`KE_(i)+U_(i)=KE_(f)+U_(f)`

→ The escape velocity is the minimum velocity required for an object to escape from a central mass and never return.

→ As an object reaches an infinite distance away, `U_(f)=0`

→ When an object has just enough speed to escape, its final speed is zero, hence, `KE_(f)=0`.

→ It follows

`KE_(i)+U_(i)` `=0`  
`(1)/(2)mv^(2)-(GMm)/(r)` `=0`  
`mv^(2)` `=(2GMm)/(r)`  
`∴ v_(esc)` `=sqrt((2GM)/(r))`  

 
→ Which is independent of the object’s mass.


Mean mark (b) 53%.

PHYSICS, M5 2017 HSC 12 MC

A satellite orbits Earth with an elliptical orbit that passes through positions `X` and `Y`.
 

Which row of the table correctly identifies the position at which the satellite has greater kinetic energy and the position at which it has greater potential energy?
 

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex}\textit{} & \textit{} \\
\textit{}\rule[-1ex]{0pt}{0pt}& \textit{} \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Greater} & \textit{Greater} \\
\quad \textit{kinetic energy}\quad \rule[-1ex]{0pt}{0pt}& \quad \textit{potential energy} \quad \\
\hline
\rule{0pt}{2.5ex}X\rule[-1ex]{0pt}{0pt}&X\\
\hline
\rule{0pt}{2.5ex}X\rule[-1ex]{0pt}{0pt}& Y\\
\hline
\rule{0pt}{2.5ex}Y\rule[-1ex]{0pt}{0pt}& X \\
\hline
\rule{0pt}{2.5ex}Y\rule[-1ex]{0pt}{0pt}& Y \\
\hline
\end{array}
\end{align*}

Show Answers Only

`B`

Show Worked Solution

→ `Y`  is at a greater distance from the centre of Earth so it has a greater potential energy at `Y`.

→ As it moves towards `X` its potential energy is converted into kinetic energy, so it has a greater kinetic energy at `X`.

`=>B`

PHYSICS, M7 2017 HSC 2 MC

Which of the following is an inertial frame of reference?

  1. A rocket during launch
  2. A train travelling at a constant velocity
  3. A car turning a corner at a constant speed
  4. A lift slowing down as it approaches the ground floor
Show Answers Only

`B`

Show Worked Solution

→ An inertial frame of reference is one which is not accelerating.

`=>B`

PHYSICS, M5 2018 HSC 28

The radius of the moon is 1740 km. The moon's mass is `7.35 × 10^(22)` kg. In this question, ignore the moon's rotational and orbital motion.

A 20 kg mass is launched vertically from the moon's surface at a velocity of `1200 \ text{m s}^(-1)`.

  1. Show that the change in potential energy of the mass in moving from the surface to an altitude of 500 km is `1.26 × 10^7 \ text{J}`.   (2 marks)
  1. Calculate the velocity of the 20 kg mass at an altitude of 500 km.   (3 marks)
Show Answers Only

a.   Proof (See Worked Solutions)

b.   `424  text{m s}^-1`

Show Worked Solution
a.       `U_(i)` `=-(GMm)/(r)`
    `=(-6.67 xx10^(-11)xx7.35 xx10^(22)xx20)/(1.74 xx10^(6))`
    `=-5.64 xx10^(7)\ text{J}`
  `U_(f)` `=(-6.67 xx10^(-11)xx7.35 xx10^(22)xx20)/(2.24 xx10^(6))`
    `=-4.38 xx10^(7)\ text{J}`

 

`:.Delta U` `=U_(f)-U_(i)`  
  `=1.26 xx10^(7)\ text{J}`  

 

b.       `KE_(i)` `=(1)/(2)m u^(2)`  
    `=(1)/(2)xx20 xx1200^(2)`  
    `=1.44 xx10^(7)  text{J}`  
     `KE_(f)` `=KE_(i)-DeltaE_(p)\ \ text{(by LCE)}`  
    `=1.44 xx10^(7)-1.26 xx10^(7)`  
    `=1.8 xx10^(6) text{J}`  

 

`(1)/(2)mv^(2)` `=1.8 xx10^(6)`  
`v^2` `=(2xx1.8 xx10^(6))/(20)`  
`:.v` `=sqrt((2xx1.8 xx10^(6))/(20))`  
  `= 424  text{m s}^-1`  

♦♦ Mean mark (b) 32%.

ENGINEERING, AE 2022 HSC 5 MC

Contemporary aircraft construction includes the use of alloys and composite materials.

Which of the following statements is NOT true for composite materials used in aircraft?

  1. They are generally corrosion resistant.
  2. They can be made into complex shapes.
  3. They have a high strength-to-weight ratio.
  4. They generally cost less than other materials.
Show Answers Only

`D`

Show Worked Solution

→ Composite materials have high corrosion resistance and strength to weight ratios.

→ They can also be made into complex shapes, however, they are far more expensive than other materials.

`=>D`

Networks, STD1 N1 2022 HSC 20

The table below shows the distances, in kilometres, between a number of towns.
 

  1. Using the vertices given, draw a weighted network diagram to represent the information shown in the table.  (2 marks)
     

     
  2. A tourist wishes to visit each town.
  3. Draw the minimum spanning tree which will allow for this AND determine its length.  (3 marks)
     

Show Answers Only
  1.  
     
     
  2.   
     
  3. `1015\ text{km}`
Show Worked Solution

a. 

b.   `text{Using Prim’s algorithm (starting at}\ Y):`

`text{1st edge:}\ YC`

`text{2nd edge:}\ CB`

`text{3rd edge:}\ SB`

`text{4th edge:}\ YM`

`text{Length of minimum spanning tree}`

`=275 + 150+60+530`

`=1015\ text{km}`

Probability, STD1 S2 2022 HSC 17

Each number from 1 to 30 is written on a separate card. The 30 cards are shuffled. A game is played where one of these cards is selected at random. Each card is equally likely to be selected.

Ezra is playing the game, and wins if the card selected shows an odd number between 20 and 30.

  1. List the numbers which would result in Ezra winning the game.  (1 mark)

  2. What is the probability that Ezra does NOT win the game?  (2 marks)

Show Answers Only
  1. `21, 23, 25, 27, 29`
  2. `Ptext{(not win)} = 5/6`
Show Worked Solution

a.   `21, 23, 25, 27, 29`
 

b.    `Ptext{(not win)}` `=1-Ptext{(win)}`
    `=1-5/30`
    `=25/30`
    `=5/6`


♦ Mean mark 51%.

BIOLOGY, M5 2019 HSC 27

Yeast is a single-celled fungus that can reproduce by budding.

  1. What type of reproduction is budding?   (1 mark)

  2. Outline a procedure that could be used to test the effect of temperature on reproduction in yeast.   (4 marks)

Show Answers Only

a.   Asexual.

b.   Steps of procedure:

  1. Prepare 4 water baths of 20, 30, 40 and 50 degrees Celsius.
  2. Add 10mL of yeast suspension in 12 different test tubes.
  3. Place 3 test tubes into each of the 4 water baths.
  4. Incubate for 3 hours.
  5. Take a drop from each test tube and inspect under a microscope using a mm grid and recording yeast counts for each.
  6. Compare the number of cells present at each temperature to see which temperature is most optimal for yeast reproduction.
Show Worked Solution

a.   Asexual.

b.   Steps of procedure:

  1. Prepare 4 water baths of 20, 30, 40 and 50 degrees Celsius.
  2. Add 10mL of yeast suspension in 12 different test tubes.
  3. Place 3 test tubes into each of the 4 water baths.
  4. Incubate for 3 hours.
  5. Take a drop from each test tube and inspect under a microscope using a mm grid and recording yeast counts for each.
  6. Compare the number of cells present at each temperature to see which temperature is most optimal for yeast reproduction.

♦♦ Mean mark (b) 45%.

Measurement, STD1 M5 2022 HSC 11

The floor plan of a home unit has been drawn to scale.
 


 

  1. How many doors are shown on the plan?  (1 mark)

  2. What is the total floor area of the home unit in square metres?  (2 marks)

Show Answers Only
  1. `5`
  2. `text{55.5 m²}`
Show Worked Solution

a.    `5`

b.   `text{Conversion: 1000 mm = 1 metre}`

`text{Calculate area by splitting into 2 rectangles}`

`text{Area}`  `=  5.5 × 5 + (4.5 +2.5) xx 4`  
   `=  55.5\ text{m²}`  

♦♦ Mean mark part (b) 26%.

Algebra, STD1 A2 2022 HSC 6 MC

A water tank holds 6000 litres when full.

The tank is full when water starts to flow out of it at a constant rate of 3 litres per minute until the tank is empty.

Which expression represents the volume (`V` litres) of water in the tank after `t` minutes?

  1. `V=6000-3t`
  2. `V=6000 t-3`
  3. `V=3-6000 t`
  4. `V=3t-6000`
Show Answers Only

`A`

Show Worked Solution

`text{When}\ \ t=0,  \ V=6000-0 xx 3 =6000`
 
`text{When}\ \ t=1,  \ V=6000-1 xx 3 =5997`

`vdots`

`text{After}\ t\ text{minutes}, \ V=6000-t xx 3 =6000-3t`

`=> A`

Calculus, EXT1 C3 2022 HSC 12a

A direction field is to be drawn for the differential equation

`(dy)/(dx)=(x-2y)/(x^(2)+y^(2)). `

On the diagram, clearly draw the correct slopes of the direction field at the points `P, Q` and `R`.  (2 marks)
  

           

Show Answers Only

Show Worked Solution

`text{At (–1, 1):}\ \ dy/dx=(-1-2)/(1+1)=-3/2`

`text{At (1, 1):}\ \ dy/dx=(1-2)/(1+1)=-1/2`

`text{At (2, 1):}\ \ dy/dx=(2-2)/(4+1)=0`
 

Calculus, EXT2 C1 2022 HSC 14b

Let  `J_(n)=int_(0)^(1)x^(n)e^(-x)\ dx`, where "n" is a non-negative integer.

  1. Show that  `J_(0)=1-(1)/(e)`.  (1  mark)

  2. Show that  `J_(n) <= (1)/(n+1)`.  (2 marks)

  3. Show that  `J_(n)=nJ_(n-1)-(1)/(e)`.  (2 marks)

  4. Using parts (i) and (iii), show by mathematical induction, or otherwise, that for all `n >= 1`,
  5.        `J_(n)=n!-(n!)/(e)sum_(r=0)^(n)(1)/(r!)`    (2 marks)

  6. Using parts (ii) and (iv) prove that  `e=lim_(n rarr oo)sum_(r=0)^(n)(1)/(r!)`.  (1  mark)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `text{Proof (See Worked Solutions)}`
  4. `text{Proof (See Worked Solutions)}`
  5. `text{Proof (See Worked Solutions)}`
Show Worked Solution
i.    `J_0` `=int_0^1 e^(-x)\ dx`
    `=[-e^(-x)]_0^1`
    `=-e^(-1)+1`
    `=1-1/e`

 


Mean mark (i) 93%.

ii.  `text{Show}\ \ J_n<=1/(n+1)`

`text{Note:}\ e^(-x)<1\ \ text{for}\ \ x in [0,1]`

`J_n` `=int_0^1 x^n e^(-x)\ dx`  
  `leq int_0^1 x^n \ dx`  
  `leq 1/(n+1)[x^(n+1)]_0^1`  
  `leq 1/(n+1)(1^(n+1)-0)`  
  `leq 1/(n+1)\ \ text{… as required}`  

 


♦♦ Mean mark (ii) 28%.
 

iii.  `text{Show}\ \ J_n=nJ_(n-1)-1/e`

`u` `=x^n` `v′` `=e^(-x)`
`u′` `=nx^(n-1)` `v` `=-e^(-x)`
`J_n` `=[-x^n * e^(-x)]_0^1-int_0^1 nx^(n-1)*-e^(-x)\ dx`  
  `=(-1^n * e^(-1)+0^n e^0)+nint_0^1 x^(n-1)*e^(-x)\ dx`  
  `=nJ_(n-1)-1/e`  

 
iv.   `text{Prove}\ \ J_(n)=n!-(n!)/(e)sum_(r=0)^(n)(1)/(r!)\ \ text{for}\ \ n >= 0`

`text{If}\ \ n=0:`

`text{LHS} = 1-1/e\ \ text{(see part (i))}`

`text{RHS} = 0!-0!/e (1/(0!)) = 1-1/e(1)=\ text{LHS}`

`:.\ text{True for}\ \ n=0.`
 

`text{Assume true for}\ \ n=k:`

`J_(k)=k!-(k!)/(e)sum_(r=0)^(k)(1)/(r!)`
   


♦ Mean mark (iv) 50%.

`text{Prove true for}\ \ n=k+1:`

`text{i.e.}\ \ J_(k+1)=(k+1)!-((k+1!))/(e)sum_(r=0)^(k+1)(1)/(r!)`

`J_(k+1)` `=(k+1)J_k-1/e\ \ text{(using part (iii))}`  
  `=(k+1)(k!-(k!)/(e)sum_(r=0)^(k)(1)/(r!))-1/e`  
  `=(k+1)!-((k+1)!)/(e)sum_(r=0)^(k)(1)/(r!)-1/e xx ((k+1)!)/((k+1)!)`  
  `=(k+1)!-((k+1)!)/e(\ sum_(\ r=0)^(k)(1)/(r!)+1/((k+1)!))`  
  `=(k+1)!-((k+1)!)/e(\ sum_(\ r=0)^(k+1)(1)/(r!))`  

 
`=>\ text{True for}\ \ n=k+1`

`:.\ text{S}text{ince true for}\ n=1,\ text{by PMI, true for integers}\ n>=1`
 

v.   `0<=J_n<= 1/(n+1)\ \ \ text{(part (ii))}`

`lim_(n->oo) 1/(n+1)=0\ \ => \ lim_(n->oo) J_n=0`

  
`text{Using part (iv):}`

`J_n/(n!)` `=1-1/e sum_(r=0)^(n)(1)/(r!)`  
`1/e sum_(r=0)^(n)(1)/(r!)` `=1-J_n/(n!)`  
`sum_(r=0)^(n)(1)/(r!)` `=e-(eJ_n)/(n!)`  
`lim_(n->oo)(\ sum_(\ r=0)^(n)(1)/(r!))`  `=lim_(n->oo)(e-(eJ_n)/(n!))`  
  `=e-0`  
  `=e`  

♦♦ Mean mark (v) 34%.

BIOLOGY, M5 2019 HSC 14 MC

The following DNA base sequence is used to code for a sequence of four amino acids.

`text{CGC  ATC  ATG  CTA}`

Which of the following correctly represents the anticodons on the transfer RNA during synthesis of this string of amino acids?

  1. `text{GCG  UAG  UAC  GAU}`
  2. `text{CGC  AUC  AUG  CUA}`
  3. `text{CGC  ATC  ATG  CTA}`
  4. `text{GCG  TAG  TAC  GAT}`
Show Answers Only

`B` or `A`

Show Worked Solution

By Elimination:

→ RNA has Uracil (U) in replace of Thymine (T) (eliminate C and D).

→ The DNA segment above must be copied into mRNA with complementary bases (with U replacing T), then anticodons on tRNA have complementary bases again. Therefore the tRNA associated with he DNA segment above will be the same with U instead of T.

`=>B`

→ `A` –  also considered an acceptable answer due to confusion as to whether the segment above was DNA or RNA.

CHEMISTRY, M7 2019 HSC 23

The following apparatus was used in an experiment to determine the molar enthalpy of combustion of ethanol.
 


 

  1. Calculate the experimental molar enthalpy of combustion `(Delta_(c) H)` of ethanol when 0.370 g ethanol was used to raise the water temperature from 18.5°C to 30.0°C.   (4 marks)

  2. Upon replication, the molar enthalpy of combustion obtained in the experiment was consistently much lower than the accepted value.
  3. Explain ONE change that could be made to the experiment that would improve the accuracy of the obtained value.   (2 marks)

Show Answers Only

a.  `-628\ text{kJ mol}^(-1)`

b.   Improvement to experiment accuracy (one of many possible answers):

→ The molar enthalpy of combustion measured in the experiment was not very accurate due to heat being lost to the surroundings.

→ To improve the accuracy, it would be helpful to move the spirit burner closer to the beaker to reduce heat loss.

Show Worked Solution
a.    `text{n(ethanol)}` `= text{m}/text{MM}`
    `=0.370/46.068`
    `=0.008032\ text{mol}`

 
`q=mC DeltaT = 105\ text{g} xx 4.18\ text{J g}^(-1)\ text{K}^(-1) xx (30-18.5)\ text{K} = 5047.35\ text{J}`

`(Delta_(c) H)` `=-q/n`  
  `=- 5047.35/0.008032`  
  `=-628\ 405\ text{J mol}^(-1)`  
  `=-628\ text{kJ mol}^(-1)\ \ text{(3 sig fig)}`  

 

b.   Improvement to experiment accuracy (one of many possible answers):

→ The molar enthalpy of combustion measured in the experiment was not very accurate due to heat being lost to the surroundings.

→ To improve the accuracy, it would be helpful to move the spirit burner closer to the beaker to reduce heat loss.

CHEMISTRY, M7 2019 HSC 21

  1. The structural formula for 2-methylpropan-2-ol is shown in the table.
  2. Draw one structural isomer of this alcohol and state its name.   (2 marks)

     

     

  3. The structural formulae for two compounds are shown below.
     

   
 

  1. Why are these two compounds classed as functional group isomers?   (2 marks)

  2. A chemical test is required to distinguish between the isomers in part (b).
  3. Identify a suitable test and explain the expected observations.   (3 marks)

Show Answers Only

a.   Successful answers should have one of the following:

 

b.  Functional Group isomers

→ Both isomers have the same number and type of atoms, but they have different arrangements of those atoms and therefore have different functional groups.

→ Isomer A has a ketone functional group, while isomer B has an aldehyde functional group.
 

c.   Tollens’ Test:

→ The Tollens’ test can be used to differentiate between Isomer A (a ketone) and Isomer B (an aldehyde).

→ Isomer B will be readily oxidised to a carboxylic acid, whereas isomer A will not.

→ As a result, Isomer B will reduce the silver-ions in Tollens’ reagent to form a silver mirror inside the test tube, while Isomer A will not react.

Show Worked Solution

a.   Successful answers should have one of the following:

b.  Functional Group isomers

→ Both isomers have the same number and type of atoms, but they have different arrangements of those atoms and therefore have different functional groups.

→ Isomer A has a ketone functional group, while isomer B has an aldehyde functional group.
 

c.   Tollens’ Test:

→ The Tollens’ test can be used to differentiate between Isomer A (a ketone) and Isomer B (an aldehyde).

→ Isomer B will be readily oxidised to a carboxylic acid, whereas isomer A will not.

→ As a result, Isomer B will reduce the silver-ions in Tollens’ reagent to form a silver mirror inside the test tube, while Isomer A will not react.


♦♦ Mean mark (c) 39%.

CHEMISTRY, M5 2019 HSC 16 MC

At equilibrium, a 1.00 L vessel contains 0.0430 mol of \(\ce{H2}\), 0.0620 mol of \(\ce{I2}\), and 0.358 mole of \(\ce{HI}\). The system is represented by the following equation:

\(\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)}\)

Which of the following is closest to the value of the equilibrium constant, \(K_{eq}\), for this reaction?

  1. 0.0208
  2. 48.1
  3. 134
  4. 269
Show Answers Only

\(B\)

Show Worked Solution
\(K_{eq}\) \(= \dfrac{\ce{[HI]^2}}{\ce{[H_2][I_2]}} \)  
  \(=\dfrac{0.358^2}{(0.0430)(0.0620)}\)  
  \(=48.1\)  

 
\(\Rightarrow B\)

BIOLOGY, M6 2019 HSC 8 MC

A group of four students set out to determine the animal species diversity over an area of one hectare in each of five different habitats. Each student graphed their data as shown.

Which of the graphs produced is the most suitable to represent animal species diversity in the different habitats?
 

Show Answers Only

`B`

Show Worked Solution

By Elimination:

→ The data type is categorical, and a line graph / area under line graph would not be suited (eliminate C and D).

→ A (and D) refers to the number of animals rather than number of species (eliminate A)

→ A column graph is the best graph to show the numerical value of species counts in each category, and would therefore be the most effective way to show this data.

`=>B`

CHEMISTRY, M7 2019 HSC 10 MC

Which class of organic compound must contain at least three carbon atoms?

  1. Aldehydes
  2. Alkenes
  3. Carboxylic acids
  4. Ketones
Show Answers Only

`D`

Show Worked Solution

Consider each option

→ Aldehydes can form methanal (1 carbon atom – eliminate)

→ Alkenes can form ethene (2 carbon atoms – eliminate)

→ Carboxylic acids can form methanoic acid (1 carbon atom – eliminate)

→ Ketones can for propanone (3 carbon atoms)

`=>D`

CHEMISTRY, M6 2019 HSC 5 MC

The diagram represents the titration curve for a reaction between a particular acid and a particular base. 
 

Which indicator would be best for this titration?

Show Answers Only

`C`

Show Worked Solution

→ The pH range at which isopicramic acid exhibits a colour change includes the point at which the acid and base react in equal amounts (equivalence point), which is at approximately pH 5.

→ The colour change can be used to identify when the equivalence point has been reached in a titration.

`=>C`

CHEMISTRY, M5 2019 HSC 3 MC

Which of the following metal carbonates has the highest molar solubility?

  1. Calcium carbonate
  2. Copper`text{(II)}` carbonate
  3. Iron`text{(II)}` carbonate
  4. Lead`text{(II)}` carbonate
Show Answers Only

`A`

Show Worked Solution

→ Calcium carbonate has the greatest solubility product among the listed metal carbonates.

→ Since all of these metal carbonates release the same amount of ions in solutions, their solubility products can be directly compared to determine which substance is most soluble.

`=>A`

ENGINEERING, PPT 2020 HSC 3 MC

Which of the following manufacturing methods can be used to produce complex polymer components?

  1. Sand casting
  2. Blow moulding
  3. Lost wax casting
  4. Injection moulding
Show Answers Only

`D`

Show Worked Solution

By Elimination:

→ Sand casting and lost wax casting are for metals, so not `A` or `C`.

→ Blow moulding can only make simple hollow shapes (eliminate `B`).

`=>D`

ENGINEERING, AE 2020 HSC 2 MC

What is the main purpose of the flaps on the wings of an aircraft?

  1. To make the aircraft safer at higher speeds
  2. To decrease lift to allow faster take-off speeds
  3. To increase lift at lower speeds for take-off and landing
  4. To decrease drag to allow higher take-off and landing speeds
Show Answers Only

`C`

Show Worked Solution

→ The purpose of the flaps is to generate more lift at lower speeds.

→ This enables aircraft to fly at greatly reduced speeds with a lower risk of stalling.

`=>C`

BIOLOGY, M7 2022 HSC 9 MC

A new vaccine against an infectious disease was developed. The effectiveness of the vaccine in preventing infection in humans was plotted over time in three different age groups.
 

Which of the following is a valid conclusion that can be drawn from the data in the graph?

  1. The vaccine is ineffective after six months.
  2. The younger the person the more effective the vaccine.
  3. The younger the person, the faster the immunity to infection declines.
  4. The vaccine offers more protection to younger people than those over 65 .
Show Answers Only

`D`

Show Worked Solution

The graph shows that the vaccine has a lower effectiveness against infection for people aged over 65 than for the younger age groups over the whole period.

`=>D`

BIOLOGY, M8 2022 HSC 5 MC

Data for incidence of notifiable infectious diseases in Australia are shown in the table.

\begin{array}{|l|c|}
\hline
\rule{0pt}{2.5ex}\textit{Disease} & \textit{Percent of all cases of infectious} \\
\textit{} & \textit{disease notified for 2019} \\
\hline
\rule{0pt}{2.5ex}\text{Influenza}\rule[-1ex]{0pt}{0pt}&\text{53}\\
\hline
\rule{0pt}{2.5ex}\text{Sexually transmitted infections}\rule[-1ex]{0pt}{0pt}& \text{25}\\
\hline
\rule{0pt}{2.5ex}\text{Gastrointestinal diseases}\rule[-1ex]{0pt}{0pt}& \text{10} \\
\hline
\rule{0pt}{2.5ex}\text{Vaccine preventable diseases}\rule[-1ex]{0pt}{0pt}& \text{9} \\
\text{excluding influenza}\rule[-1ex]{0pt}{0pt}& \text{} \\
\hline
\rule{0pt}{2.5ex}\text{Other}\rule[-1ex]{0pt}{0pt}& \text{3} \\
\hline
\end{array}

Which type of graph should be chosen to best represent these data?

  1. A column graph because percentages are continuous.
  2. A column graph because diseases are categorical.
  3. A line graph because percentages are continuous.
  4. A line graph because diseases are categorical.
Show Answers Only

`B`

Show Worked Solution

A column graph would be the most appropriate as the diseases are divided into categories.

`=>B`

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