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v1 Algebra, STD2 A2 2007 HSC 27b

A cafe uses eight long-life light globes for 7 hours every day of the year. The purchase price of each light globe is $11.00 and they each cost  \($f\)  per hour to run.

  1. Write an equation for the total cost (\($c\)) of purchasing and running these eight light globes for one year in terms of  \(f\).    (2 marks)

  2. Find the value of  \(f\)  (correct to three decimal places) if the total cost of running these eight light globes for one year is $850.   (1 mark)

  3. If the use of the light globes increases to ten and a half hours per night every night of the year, does the total cost increase by one-and-a-half times? Justify your answer with appropriate calculations.   (1 mark)

Show Answers Only
  1. \($c=88+20\ 440f\)
  2. \(0.037\ $/\text{hr}\ \text{(3 d.p.)}\)
  3. \(\text{Proof:  (See Worked Solutions)}\)
Show Worked Solution

i.  \(\text{Purchase price}=8\times 11=$88\)

\(\text{Running cost}\) \(=\text{No. of  hours}\times \text{Cost per hour}\)
  \(=8\times 7\times 365\times f\)
  \(=20\ 440f\)

  
\(\therefore\ $c=88+20\ 440f\)
  

ii.  \(\text{Given}\ \ $c=$850\)

\(850\) \(=88+20\ 440f\)
\(20\ 440f\) \(=850-88\)
\(f\) \(=\dfrac{762}{20440}\)
  \(= 0.03727\dots\)
  \(=0.037\ $/\text{hr}\ \text{(3 d.p.)}\)

 

iii.  \(\text{If}\ f\ \text{is multiplied by  }1.5 =\dfrac{10.5}{7}\)

\(f=1.5\times0.037=0.0555\ \ $/\text{hr}\)

\(\therefore\ $c\) \(=88+20\ 440\times 0.0555\)
  \(=$1222.42\)

  
\(\text{Since }$1222.42\ \text{is less than}\ 1.5\times $850 = $1275,\)

\(\text{the total cost increases to less than 1.5 times the}\)

\(\text{the original cost.}\)

v1 Algebra, STD2 A2 2004 HSC 22 MC

Mary-Anne knows that

• one Australian dollar (AUD) is worth 0.64 euros, and
• one Canadian dollar (CAD) is worth 0.97 euros.

Mary-Anne changes 75 AUD to Canadian dollars.

How many Canadian dollars will she get?

  1. 46.56 CAD
  2. 49.48 CAD
  3. 113.67 CAD
  4. 120.75 CAD
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Mary-Anne has 75 AUD.}\)

\(\text{Converting to Euros}\)

\(25\ \text{AUD}\) \(=75\times 0.64\)
  \(=48\ \text{Euros}\)

 

\(\text{Converting to CAD}\)

\(48\ \text{euros}\) \(=\dfrac{48}{0.97}\)
  \(=49.484\dots\)
  \(=49.48\ \text{CAD}\)

\(\Rightarrow B\)

v1 Algebra, STD2 A2 2016 HSC 29e

The graph shows the life expectancy of people born between 1900 and 2010.
 


  1. According to the graph, what is the life expectancy of a person born in 1968?  (1 mark)

  2. With reference to the value of the gradient, explain the meaning of the gradient in this context.  (2 marks)

Show Answers Only
  1. \(\text{76 years}\)
  2. \(\text{After 1900, life expectancy increases by 0.38 years for}\)
    \(\text{each year later that someone is born.}\)
Show Worked Solution

i.    \(\text{76 years}\)

ii.    \(\text{Using (2000, 88) and (1900, 50):}\)

\(\text{Gradient}\) \(= \dfrac{y_2-y_1}{x_2-x_1}\)
  \(= \dfrac{88-50}{2000-1900}\)
  \(= 0.38\)

 
\(\text{After 1900, life expectancy increases by 0.38 years for}\)

\(\text{each year later that someone is born.}\)

♦♦ Mean mark (ii) 33%.

v1 Algebra, STD2 A2 SM-Bank 2

The cost of apples per kilogram, \(C\), varies directly with the weight of apples purchased, \(w\).

If 12 kilograms costs $56.64, calculate the cost of 4.5 kilograms of apples.  (2 marks)

Show Answers Only

\($21.24\)

Show Worked Solution

\(C\ \propto \ w\)

\(C=kw\)

\(\text{When}\ C=$56.64\ \text{kg},\ w=12\ \text{kg}\)

\(56.64\) \(=k\times 12\)
\(k\) \(=\dfrac{56.64}{12}\)
  \(=$4.72\)

 

\(\text{When}\ \ w=4.5\ \text{kg,}\)

\(C\) \(=4.72\times 4.5\)
  \(=$21.24\)

v1 Algebra, STD2 A2 SM-Bank 3

The average height, \(L\), in centimetres, of a boy between the ages of 7 years and 10 years can be represented by a line with equation

\(L=7A+85\)

where \(A\) is the age in years. For this line, the gradient is 7.

  1. What does this indicate about the heights of boys aged 7 to 10?   (1 mark)

  2. Give ONE reason why this equation is not suitable for predicting heights of boys older than 10.   (1 mark)

Show Answers Only
  1. \(\text{It indicates that 7-10 year old boys, on average, grow 7 cm per year.}\)
  2. \(\text{Boys eventually stop growing, and the equation doesn’t factor this in.}\)
Show Worked Solution

i.    \(\text{It indicates that 7-10 year old boys, on average, grow}\)

\(\text{7 cm per year.}\)
 

ii.   \(\text{Boys eventually stop growing, and the equation doesn’t}\)

\(\text{factor this in.}\)

v1 Algebra, STD2 A2 2019 HSC 34

The relationship between British pounds \((p)\) and Australian dollars \((d)\) on a particular day is shown in the graph.
 

  1. Write the direct variation equation relating British pounds to Australian dollars in the form  \(p=md\). Leave \(m\) as a fraction.  (1 mark)

  2. The relationship between Japanese yen \((y)\) and Australian dollars \((d)\) on the same day is given by the equation  \(y=84d\).

     

    Convert \(107\ 520\) Japanese yen to British pounds.  (2 marks)

Show Answers Only
  1. \(p=\dfrac{5}{8}d\)
  2. \(107\ 520\ \text{yen = 800 pounds}\)
Show Worked Solution

a.   \(m=\dfrac{\text{rise}}{\text{run}}=\dfrac{5}{8}\)

\(p=\dfrac{5}{8}d\)


♦ Mean mark 42%.

b.   \(\text{Yen to Australian dollars:}\)

\(y\) \(=84d\)
\(107\ 520\) \(=84d\)
\(d\) \(=\dfrac{107\ 520}{84}\)
  \(= 1280\ $\text{A}\)

 
\(\text{Australian dollars to pounds:}\)

\(p\) \(=\dfrac{5}{8}\times 1280\)
  \(=800\ \text{pounds}\)

  
\(\therefore\ 107\ 520\ \text{yen = 800 pounds}\)

v1 Algebra, STD1 A3 2021 HSC 25

The diagram shows a container which consists of a large hexagonal prism on top of a smaller hexagonal prism.
 

The container is filled with water at a constant rate into the top of the larger hexagonal prism.

The smaller prism is totally filled before the larger prism begins to fill.

It takes 5 minutes to fill the smaller cylinder.

Draw a possible graph of the water level in the container against time.  (2 marks)
 

Show Answers Only

Show Worked Solution


♦♦ Mean mark 38%.

v1 Algebra, STD2 A2 2020 HSC 10 MC

An electrician charges a call-out fee of $75 as well as $1.50 per minute while working.

Suppose the electrician works for \(t\) hours.

Which equation expresses the amount the plumber charges ($\(C\)) as a function of time (\(t\) hours)?

  1. \(C=75+1.50t\)
  2. \(C=150+75t\)
  3. \(C=75+90t\)
  4. \(C=90+75t\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Hourly rate}=60\times 1.50=$90\)

\(\therefore\ C=75+90t\)
  

\(\Rightarrow C\)


♦ Mean mark 42%.

v1 Algebra, STD1 A2 2020 HSC 20

The height of a bundle of photographic paper (\(H\) mm) varies directly with the number of sheets (\(N\)) of photographic paper that the bundle contains.

This relationship is modelled by the formula  \(H=kN\), where  \(k\)  is a constant.

The height of a bundle containing 150 sheets of photographic paper is 2.7 centimetres.

  1. Show that the value of  \(k\)  is 0.18.   (1 mark)

  2. A bundle of photographic paper has a height of 36 centimetres. Calculate the number of sheets of photographic paper in the bundle.   (2 marks)

Show Answers Only
  1. \(\text{See Worked Solutions}\)
  2. \(2000\ \text{sheets}\)
Show Worked Solution

a.    \(H=2.7\ \text{cm }=27\ \text{mm, when}\  N=150:\)

\(H\) \(=kN\)
\(2.7\) \(=k\times 150\)
\(\therefore\ k\) \(=\dfrac{2.7}{150}\)
  \(=0.18\)

  

b.     \(\text{Find}\ \ N \ \text{when} \ \ H=36\ \text{cm}=360\ \text{mm:}\)

\(360\) \(=0.18\times N\)
\(\therefore\ N\) \(=\dfrac{360}{0.18}\)
  \(=2000\ \text{sheets}\)

♦ Mean mark 50%.

v1 Algebra, STD2 A2 2012 HSC 5 MC

The line below has intercepts  \(m\)  and  \(n\),  where  \(m\) and  \(n\) are positive integers. 
  

What is the gradient of the line? 

  1. \(\dfrac{m}{n}\)
  2. \(\dfrac{n}{m}\)
  3. \(-\dfrac{m}{n}\)  
  4. \(-\dfrac{n}{m}\)
Show Answers Only

\(C\)

Show Worked Solution
 
♦ Mean mark 45%
\(\text{Gradient}\) \(=\dfrac{\text{rise}}{\text{run}}\)
  \(=-\dfrac{m}{n}\)

\(\Rightarrow C\)

v1 Algebra, STD2 A2 2009 HSC 14 MC

If   \(C=5x+4\), and  \(x\)  is increased by  3, what will be the corresponding increase in \(C\) ?

  1. \(3\)
  2. \(15\)
  3. \(3x\)
  4. \(5x\)
Show Answers Only

\(B\)

Show Worked Solution

\(C=5x+4\)

\(\text{If}\ x\ \text{increases by 3}\)

\(C\ \text{increases by}\ 5\times 3=15\)

\(\Rightarrow B\)


♦ Mean mark 50%.
STRATEGY: Substituting real numbers into the equation can work well in these type of questions. eg. If \(x=0,\ C=4\) and when \(x=3,\ C=19\).

EXAMCOPY Algebra, STD2 A1 2009 HSC 16 MC

The time for a car to travel a certain distance varies inversely with its speed.

Which of the following graphs shows this relationship?
 

Show Answers Only

`A`

Show Worked Solution
`T` `prop 1/S`
`T` `= k/S`

 
`text{By elimination:}`

`text(As   S) uarr text(, T) darr => text(cannot be B or D)`

♦ Mean mark 38%

`text(C  is incorrect because it graphs a linear relationship)`

`=>  A`

v1 Algebra, STD2 A2 2011 HSC 23b

Sticks were used to create the following pattern. 
  

The number of sticks used is recorded in the table.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Shape $(S)$} \rule[-1ex]{0pt}{0pt} & \;\;\; 1 \;\;\; & \;\;\; 2 \;\;\; & \;\;\; 3 \;\;\; \\
\hline
\rule{0pt}{2.5ex} \text{Number of sticks $(N)$}\; \rule[-1ex]{0pt}{0pt} & \;\;\; 6 \;\;\; & \;\;\; 10 \;\;\; & \;\;\; 14 \;\;\; \\
\hline
\end{array}

  1. Draw Shape 4 of this pattern.  (1 mark)

  2. How many sticks would be required for Shape 128?    (1 mark)

  3. Is it possible to create a shape in this pattern using exactly 609 sticks?

     

    Show suitable calculations to support your answer.    (2 marks)

Show Answers Only
  1. \(\text{See Worked Solutions.}\)
  2. \(514\)
  3. \(\text{No (See worked solution)}\)
Show Worked Solution

  i.    \(\text{Shape 4 is shown below:}\)

ii.    \(\text{Since}\ \ N=2+4S\)

♦ Mean mark 48%.
MARKER’S COMMENT: Students should attempt to find a “rule” in such questions, and use this formula to solve the question, as per the Worked Solution.  
\(\text{If }S\) \(=128\)
\(N\) \(=2+(4\times 128)\)
  \(=514\)

 

iii.    \(609\) \(=2+4S\)
  \(4S\) \(=607\)
  \(S\) \(=151.75\)

    
\(\text{Since}\ S\ \text{is not a whole number, 609 sticks}\)

\(\text{will not create a shape in this pattern.}\)

v1 Algebra, STD2 A2 2005 HSC 17 MC

The total cost, \($C\), of a school excursion is given by  \(C=4n+9\), where \(n\) is the number of students.

If five extra students go on the excursion, by how much does the total cost increase?

  1. $4
  2. $20
  3. $18
  4. $29
Show Answers Only

\(B\)

Show Worked Solution

\(C=4n+9\)

\(\text{If}\ n\ \text{increases to}\ n+5\)

\(C\) \(=4(n+5)+9\)
  \(=4n+20+9\)
  \(=4n+29\)

 

\(\therefore \text{Total cost increases by }$20\)

\(\Rightarrow B\)

v1 Algebra, STD2 A2 2015 HSC 13 MC

What is the equation of the line \(l\)?
 

 

  1. \(y=-\dfrac{1}{5}x+5\)
  2. \(y=\dfrac{1}{5}x+5\)
  3. \(y=-5x+5\)
  4. \(y=5x+5\)
Show Answers Only

\(C\)

Show Worked Solution

\(l\ \ \text{passes through (0, 5) and (1, 0)}\)

\(\text{Gradient}\) \(=\dfrac{y_2-y_1}{x_2-x_1}\)
  \(=\dfrac{5-0}{0-1}\)
  \(=-5\)

 
\(y\ \text{-intercept}= 5\)

\(\therefore\ y=-5x+5\)

\(\Rightarrow C\)


♦ Mean mark 48%.

v1 Algebra, STD2 A2 2017 HSC 20 MC

A pentagon is created using matches.

By adding more matches, a row of two pentagons is formed.

Continuing to add matches, a row of three pentagons can be formed.

Continuing this pattern, what is the maximum number of complete pentagons that can be formed if 230 matches in total are available?

  1. 55
  2. 56
  3. 57
  4. 58
Show Answers Only

\(C\)

Show Worked Solution

\(\text{1 pentagon:}\ 5+4\times 0=5\)

\(\text{2 pentagons:}\ 5+4\times 1=9\)

\(\text{3 pentagons:}\ 5+4\times 2 = 13\)

\(\vdots\)

\(n\ \text{pentagons:}\ 5 + 4(n – 1)\)

\(5+4(n – 1)\) \(=230\)
\(4n-4\) \(=225\)
\(4n\) \(=229\)
\(n\) \(=57.25\)

 

\(\text{Complete pentagons possible}\ =57\)

\(\Rightarrow C\)

v1 Algebra, STD2 A1 2011 HSC 21 MC

A train departs from Town A at 4.00 pm to travel to Town B. Its average speed for the journey is 80 km/h, and it arrives at 6.00 pm. A second train departs from Town A at 4.30 pm and arrives at Town B at 6.10 pm.

What is the average speed of the second train?

  1. 96 km/h
  2. 114 km/h
  3. 224 km/h
  4. 280 km/h
Show Answers Only

\(A\)

Show Worked Solution

\(\text{1st train:}\)

\(\text{Travels 2hrs at 80km/h}\)

\(\text{Distance}\) \(=\text{Speed}\times\text{Time}\)
  \(=80\times 2\)
  \(=160\ \text{km}\)

 
\(\text{2nd train:}\)

\(\text{Travels 160 km in 1 hr 40 min}\ \rightarrow\ \dfrac{5}{3}\ \text{hrs}\)

\(\text{Speed}\) \(=\dfrac{\text{Distance}}{\text{Time}}\)
  \(=160\ ÷\ \dfrac{5}{3}\)
  \(=160\times \dfrac{3}{5}\)
  \(=96\ \text{km/h}\)

\(\Rightarrow A\)


♦♦ Mean mark 49%.

v1 Algebra, STD2 A1 2009 HSC 16 MC

The time for a train to travel a certain distance varies inversely with its speed.

Which of the following graphs shows this relationship?

 

 

Show Answers Only

\(C\)

Show Worked Solution
\(T\) \(\propto \dfrac{1}{S}\)
\(T\) \(=\dfrac{k}{S}\)

 
\(\text{By elimination:}\)

\(\text{As   Speed} \uparrow \ \text{, Time}\downarrow\ \Rightarrow\ \text{cannot be A or B}\)

\(\text{D  is incorrect because it graphs a linear relationship}\)

\(\Rightarrow C\)


♦♦ Mean mark 38%.

v1 Algebra, STD2 A1 2014 HSC 4 MC

Young’s formula below is used to calculate the required dosages of medicine for children aged 1–12 years.
  

 \(\text{Dosage}=\dfrac{\text{age of child (in years)}\ \times\ \text{adult dosage}}{\text{age of child (in years)}\ +\ 12}\)
  

How much of the medicine should be given to an 18-month-old child in a 24-hour period if each adult dosage is 27 mL? The medicine is to be taken every 8 hours by both adults and children.

  1. 3 mL
  2. 6 mL
  3. 9 mL
  4. 12 mL
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Age of child} = 18\ \text{months}=1.5\ \text{years}\)

\(\text{Dosage}\) \(=\dfrac{1.5\times 27}{1.5+12}\)
  \(=3\ \text{mL}\)

 
\(\text{Dosage every 8 hrs}\)

\(\therefore\ \text{In 24 hours, medicine given} = 3\times 3=9\ \text{mL}\)
  
\(\Rightarrow C\)


♦♦ Mean mark 42%.

v1 Algebra, STD2 A1 2015 HSC 30d

Monica is driving on a motorway at a speed of 105 kilometres per hour and has to brake suddenly. She has a reaction time of 1.3 seconds and a braking distance of 54.3 metres.

Stopping distance can be calculated using the following formula
 

\(\text{stopping distance = {reaction time distance} + {braking distance}}\)

 
What is Monica's stopping distance? Give your answer to 1 decimal place.  (2 marks)

Show Answers Only

\(92.2\ \text{metres  (to 1 d.p.)}\)

Show Worked Solution
\(105\ \text{km/hr}\) \(=105\ 000\ \text{m/hr}\)
  \(=\dfrac{105\ 000}{60\times 60}\ \text{m/sec}\)
  \(=29.166\dots\ \text{m/sec}\)

 

\(\text{Reaction time distance}\) \(=1.3\times 29.166\dots\)
  \(=37.916\dots\ \text{metres}\)

 

\(\text{Stopping distance}\)

\(\text{ = {Reaction time distance} + {braking distance}}\)

\(=37.916…+54.3\)

\(=92.216\dots\)

\(=92.2\ \text{metres  (to 1 d.p.)}\)


♦ Mean mark 34%.

v1 Algebra, STD2 A1 SM-Bank 4

Yuan is driving in a school zone at a speed of 30 kilometres per hour and needs to stop immediately to avoid an accident.

It takes him 1.4 seconds to react and his breaking distance is 6.2 metres.

Stopping distance can be calculated using the following formula
 

\(\text{stopping distance = {reaction time distance} + {braking distance}}\)
 

What is Yuan's total stopping distance? Give your answer to 1 decimal place.  (2 marks)

Show Answers Only

\(17.9\ \text{metres  (to 1 d.p.)}\)

Show Worked Solution
\(30\ \text{km/hr}\) \(=30\ 000\ \text{m/hr}\)
  \(=\dfrac{30\ 000}{60\times 60}\ \text{m/sec}\)
  \(=8.33\dots\ \text{m/sec}\)

 

\(\therefore\ \text{Total stopping distance}\)

\(\text{ = {Reaction time distance} + {braking distance}}\)

\(=1.4\times 8.33…+6.2\)

\(=17.866\dots\)

\(=17.9\ \text{metres  (to 1 d.p.)}\)

v1 Algebra, STD2 A1 2018 HSC 28e

Drake is driving at 80 km/h. He notices a branch on the road ahead and decides to apply the brakes. His reaction time is 1.2 seconds. His braking distance (\(D\) metres) is given by  \(D=0.01v^2\), where  \(v\) is speed in km/h.

Stopping distance can be calculated using the following formula
 

\(\text{stopping distance = {reaction time distance} + {braking distance}}\)
 

What is Drake’s stopping distance, to the nearest metre?  (3 marks)

Show Answers Only

\(91\ \text{m  (nearest m)}\)

Show Worked Solution
\(\text{80 km/hr}\) \(=80\ 000\ \text{m/hr}\)
  \(=\dfrac{80\ 000}{60\times 60}\ \text{m/sec}\)
  \(=22.22\dots\ \text{m/sec}\)

  
\(\text{Total stopping distance}\)

\(\text{ = {reaction time distance} + {braking distance}}\)

\(=1.2\times 22.22\dots + 0.01\times 80^2\)

\(=90.66\dots\)

\(=91\ \text{m  (nearest m)}\)


♦ Mean mark 46%.

v1 Algebra, STD2 A1 2013 HSC 29a

Jeremy tried to solve this equation and made a mistake in Line 2. 

\(\dfrac{M+3}{2}-\dfrac{2M-1}{5}\) \(=1\) \(\text{... Line 1}\)
\(5M+15-4M-2\) \(=10\) \(\text{... Line 2}\)
\(M+13\) \(=10\) \(\text{... Line 3}\)
\(M\) \(=-3\) \(\text{... Line 4}\)

  
Copy the equation in Line 1 and continue your solution to solve this equation for \(M\).

Show all lines of working.   (2 marks)

Show Answers Only
\(\dfrac{M+3}{2}-\dfrac{2M-1}{5}\) \(=1\) `text(… Line 1)`
\(5M+15-4M+2\) \(=10\) `text(… Line 2)`
\(M+17\) \(=10\) `text(… Line 3)`
\(M\) \(=-7\) `text(… Line 4)`
Show Worked Solution
♦♦ Mean mark 27%
STRATEGY: The RHS of the equation increases from 1 to 10 (from Line 1 to Line 2), indicating both sides must have been multiplied by 10.
\(\dfrac{M+3}{2}-\dfrac{2M-1}{5}\) \(=1\) `text(… Line 1)`
\(5M+15-4M+2\) \(=10\) `text(… Line 2)`
\(M+17\) \(=10\) `text(… Line 3)`
\(M\) \(=-7\) `text(… Line 4)`

v1 Algebra, STD2 A1 2018 HSC 28b

Solve the equation  \(\dfrac{3x}{4}+1=\dfrac{5x+1}{3}\), leaving your answer as a fraction.  (3 marks)

Show Answers Only

\(\dfrac{8}{11}\)

Show Worked Solution

♦ Mean mark 35%.

\(\underbrace{\dfrac{3x}{4} + 1}_\text{multiply x 12}\) \(=\underbrace{\dfrac{5x+1}{3}}_\text{multiply x 12}\)
\(9x+12\) \(=20x+4\)
\(11x\) \(=8\)
\(x\) \(=\dfrac{8}{11}\)

v1 Algebra, STD2 A1 2013 HSC 21 MC

Which equation correctly shows  \(n\)  as the subject of  \(V=600(1-n)\)?

  1. \(n=\dfrac{V-600}{600}\)
  2. \(n=\dfrac{600-V}{600}\)
  3. \(n=V-600\)
  4. \(n=600-V\)
Show Answers Only

\(B\)

Show Worked Solution
♦♦♦ Mean mark 27%
\(V\) \(=600(1-n)\)
\(1-n\) \(=\dfrac{V}{600}\)
\(n\) \(=1-\dfrac{V}{600}\)
  \(=\dfrac{600-V}{600}\)

 
\(\Rightarrow B\)

v1 Algebra, STD2 A1 2011 HSC 18 MC

Which of the following correctly expresses  \(b\)  as the subject of  \(y= ax+\dfrac{1}{4}bx^2\)?

  1. \(b=\dfrac{4y-ax}{x^2}\)
  2. \(b=\dfrac{4(y-ax)}{x^2}\)
  3. \(b=\dfrac{\dfrac{1}{4}y-ax}{x^2}\)
  4. \(b=\dfrac{\dfrac{1}{4}(y-ax)}{x^2}\)
Show Answers Only

\(B\)

Show Worked Solution
\(y\) \(= ax+\dfrac{1}{4}bx^2\)
\(\dfrac{1}{4}bx^2\) \(=y-ax\)
\(bx^2\)  \(=4(y-ax)\)
\(b\) \(=\dfrac{4(y-ax)}{x^2}\)

 
\(\Rightarrow B\)

v1 Algebra, STD2 A1 2006 HSC 18 MC

What is the formula for \(g\) as the subject of \(7d=8e+5g^2\)?

  1. \(g =\pm\sqrt{\dfrac{8e-7d}{5}}\)
  2. \(g =\pm\sqrt{\dfrac{7d-8e}{5}}\)
  3. \(g =\pm\dfrac{\sqrt{7d+8e}}{5}\)
  4. \(g =\pm\dfrac{\sqrt{8e-7d}}{5}\)
Show Answers Only

\(B\)

Show Worked Solution
\(7d\) \(=8e+5g^2\)
\(5g^2\) \(=7d-8e\)
\(g^2\) \(=\dfrac{7d-8e}{5}\)
\(g\) \(=\pm\sqrt{\dfrac{7d-8e}{5}}\)

\(\Rightarrow B\)

v1 Algebra, STD2 A1 2007 HSC 19 MC

Which of the following correctly expresses  \(X\)  as the subject of  \(Y=4\pi\Bigg(\dfrac{X}{4}+L\Bigg)\)?

  1. \(X=\dfrac{Y}{\pi}-L\)
  2. \(X=\dfrac{Y}{\pi}-4L\)
  3. \(X=4L-\dfrac{Y}{2\pi}\)
  4. \(X=\dfrac{Y}{8\pi}-\dfrac{L}{4}\)
Show Answers Only

\(B\)

Show Worked Solution
\(Y\) \(=4\pi\Bigg(\dfrac{X}{4}+L\Bigg)\)
\(\dfrac{Y}{4\pi}\) \(=\dfrac{X}{4}+L\)
\(\dfrac{X}{4}\) \(=\dfrac{Y}{4\pi}-L\)
\(X\) \(=4\Bigg(\dfrac{Y}{4\pi}-L\Bigg)\)
\(X\) \(=\dfrac{Y}{\pi}-4L\)

 
\(\Rightarrow B\)

v1 Algebra, STD2 A1 2019 HSC 11 MC

Which of the following correctly expresses \(y\) as the subject of the formula  \(5x-2y-9=0\)?

  1.  \(y=\dfrac{5}{2}x-9\)
  2.  \(y=\dfrac{5}{2}x+9\)
  3.  \(y=\dfrac{5x+9}{2}\)
  4.  \(y=\dfrac{5x-9}{2}\)
Show Answers Only

\(D\)

Show Worked Solution

♦ Mean mark 50%.

\(5x-2y-9\) \(=0\)
\(2y\) \(=5x-9\)
\(\therefore\ y\) \(=\dfrac{5x-9}{2}\)

 
\(\Rightarrow D\)

CHEMISTRY, M8 2022 VCE 5*

A chemist uses spectroscopy to identify an unknown organic molecule, Molecule \(\text{J}\), that contains chlorine.

The \({}^{13}\text{C NMR}\) spectrum of Molecule \(\text{J}\) is shown below.
 

The infra-red (IR) spectrum of Molecule \(\text{J}\) is shown below.
 

  1. Name the functional group that produces the peak at 168 ppm in the \({}^{13}\text{C NMR}\) spectrum on the first image, which is consistent with the IR spectrum shown above. Justify your answer with reference to the IR spectrum.   (2 marks)

The mass spectrum of Molecule \(\text{J}\) is shown below
 

  1. The molecular mass of Molecule \(\text{J}\) is 108.5
  2.  Explain the presence of the peak at 110 m/z.  (1 mark)

The \({ }^1 \text{H NMR}\) spectrum of Molecule \(\text{J}\) is shown below.
 

  1. The \({ }^1 \text{H NMR}\) spectrum consists of two singlet peaks.
  2. What information does this give about the molecule?   (2 marks)

  3. Draw a structural formula for Molecule \(\text{J}\) that is consistent with the information provided in parts a–c.   (2 marks)

Show Answers Only

a.    → Absence of a very broad \(\ce{OH}\) acid peak between 2500–3000. 

→ Molecule \(\text{J}\) must be an ester.
 

b.    The peak at 110 m/z:

→ due to the Chlorine-37 isotope which is slightly heavier than the more abundant Chlorine-35.
 

c.    The two singlet peaks indicate:

→ two different hydrogen environments within the molecule.

→ there are no adjacent hydrogen environments.

→ The relative heights of the peaks show the ratios of the hydrogens in the environments are 2 : 3.
 

d.    Either of the two molecules show below are correct:

Show Worked Solution

a.    → Absence of a very broad \(\ce{OH}\) acid peak between 2500–3000. 

→ Molecule \(\text{J}\) must be an ester.
 

♦ Mean mark (a) 41%.

b.    The peak at 110 m/z:

→ due to the Chlorine-37 isotope which is slightly heavier than the more abundant Chlorine-35.
 

c.    The two singlet peaks indicate:

→ two different hydrogen environments within the molecule.

→ there are no adjacent hydrogen environments.

→ The relative heights of the peaks show the ratios of the hydrogens in the environments are 2 : 3.

♦♦♦ Mean mark (b) 15%.
COMMENT: Know the masses of common isotopes.

d.    Either of the two molecules show below are correct:
 

♦ Mean mark (d) 40%.

CHEMISTRY, M8 2021 VCE 7*

Two students are given a homework assignment that involves analysing a set of spectra and identifying an unknown compound. The unknown compound is one of the molecules shown below.
 

The \(^{13}\text{C NMR}\) spectrum of the unknown compound is shown below.
 

  1. Based on the number of peaks in the \(^{13}\text{C NMR}\) spectrum above, which compound – \(\text{P, Q, R, S}\) or \(\text{T}\) – could be eliminated as the unknown compound?   (1 mark)

  2. The infra-red (IR) spectrum of the unknown compound is shown below.
     

  1. Identify which of the five compounds (1 or more) can be eliminated on the basis of the IR spectrum. Justify your answer using data from the IR spectrum.   (3 marks)

  2. The mass spectrum of the unknown compound is shown below.
     

  1. Write the chemical formula of the species that produces a peak at m/z = 43.   (1 mark)

  2. Explain why one molecule can produce multiple peaks on a mass spectrum.   (2 mark)

Show Answers Only

a.    → The \(^{13}\text{C NMR}\) shows 4 different carbon environments.

→ Compound \(T\) has 5 unique carbon environments and can be eliminated.
 

b.    → Compounds \(\text{P, Q}\) and \(\text{S}\) can be eliminated.

→ Compounds \(\text{S}\) and \(\text{P}\) both don’t have an \(\ce{OH}\) alcohol group, however the IR spectrum clearly shows an \(\ce{OH}\) alcohol group with an absorbance at 3500 cm\(^{-1}\).

→ Compound \(\text{Q}\) contains an \(\ce{OH}\) acid group whereas the IR spectrum shows an \(\ce{OH}\) alcohol group at 3500 cm\(^{-1}\). There is no evidence of a broad \(\ce{OH}\) acid group between 2500–3000 cm\(^{-1}\).
 

c.i.  → The unknown compound is compound \(\text{R}\).

→ The chemical species are fragments of the original compound with a positive charge.

→ The chemical formulas could include \(\ce{[CH3CO]^+}\) or \(\ce{[C2H3O]^+}\)
 

 ii.   → Ions of that molecule must be formed to produce peaks on the mass spectrum.

→ Organic compounds can be split up into numerous different ions when producing fragment patterns leading to multiple different peaks on the mass spectrum.

Show Worked Solution

a.    → The \(^{13}\text{C NMR}\) shows 4 different carbon environments.

→ Compound \(T\) has 5 unique carbon environments and can be eliminated.
 

b.    → Compounds \(\text{P, Q}\) and \(\text{S}\) can be eliminated.

→ Compounds \(\text{S}\) and \(\text{P}\) both don’t have an \(\ce{OH}\) alcohol group, however the IR spectrum clearly shows an \(\ce{OH}\) alcohol group with an absorbance at 3500 cm\(^{-1}\).

→ Compound \(\text{Q}\) contains an \(\ce{OH}\) acid group whereas the IR spectrum shows an \(\ce{OH}\) alcohol group at 3500 cm\(^{-1}\). There is no evidence of a broad \(\ce{OH}\) acid group between 2500–3000 cm\(^{-1}\).
 

♦ Mean mark (b) 43%.
COMMENT: Recognise the difference between OH alcohols and OH acid groups.

c.i.  → The unknown compound is compound \(\text{R}\).

→ The chemical species are fragments of the original compound with a positive charge.

→ The chemical formulas could include \(\ce{[CH3CO]^+}\) or \(\ce{[C2H3O]^+}\)

♦ Mean mark (c.i.) 50%.
COMMENT: Ions require a positive charge.

 ii.   → Ions of that molecule must be formed to produce peaks on the mass spectrum.

→ Organic compounds can be split up into numerous different ions when producing fragment patterns leading to multiple different peaks on the mass spectrum.

♦ Mean mark (c.ii.) 39%.

CHEMISTRY, M8 2023 VCE 7-1*

Molecule \(\text{V}\) contains only carbon atoms, hydrogen atoms and one oxygen atom.

The mass spectrum of molecule \(\text{V}\) is shown below.
 

  1.  i.  State the molecular formula of molecule \(\text{V}\). Justify your answer by using the information in the mass spectrum.   (2 marks)

  2. ii.  State why there is a small peak at \(\text{m/z} =87\).   (1 mark)

The \({ }^1 \text{H NMR}\) spectrum of molecule \(\text{V}\) is shown below.
 

  1. State what information the doublet at 1.1 ppm provides about the structure of the molecule.   (1 mark)

The \({ }^{13} \text{C NMR}\) spectrum of molecule \(\text{V}\) is shown below.
 

  1. In the space below, draw a structural formula of molecule \(\text{V}\) that is consistent with the information provided in parts a.-c.   (3 marks)

Show Answers Only

a.i.   Molar mass of \(\text{V}\ = 86\ \text{g mol}^{-1}\)

→ Indicated by the parent ion peak at 86 m/z.

→ Molecular formula: \(\ce{C5H10O}\)

a.ii.  Small peak at m/z = 87:

→ A carbon-13 isotope being present in the molecule.

b.    The doublet peak at 1.1 ppm:

→ Indicates there is a single hydrogen with a different hydrogen environment bonded to an adjacent carbon atom. 

c.   
       

Show Worked Solution

a.i.   Molar mass of \(\text{V}\ = 86\ \text{g mol}^{-1}\)

 → Indicated by the parent ion peak at 86 m/z.

 → Molecular formula: \(\ce{C5H10O}\)
 

a.ii.  Small peak at m/z = 87:

→ A carbon-13 isotope being present in the molecule.
 

b.    The doublet peak at 1.1 ppm:

→ Indicates there is a single hydrogen with a different hydrogen environment bonded to an adjacent carbon atom.
 

♦ Mean mark (b) 50%.

c.    From the carbon NMR graph:

→ There are 4 carbon environments, one shifted above 200 ppm indicating a ketone or aldehyde.

→ As there are 5 carbons, 2 of the carbons must have the same environment.
 

From the hydrogen NMR graph:

→ There are 3 hydrogen environments.

→ The septet peak indicates there are 6 hydrogens with the same chemical environment on adjacent carbon atoms.
 

♦ Mean mark (c) 43%.

CHEMISTRY, M8 2022 VCE 28 MC

The \({ }^{13}\text{C NMR}\) spectrum of an organic compound is shown below.

The organic compound could be

Show Answers Only

\(D\)

Show Worked Solution

→ The \({ }^{13}\text{C NMR}\) has five peaks indicating 5 different carbon environments within the molecule.

→ The peak at 140 indicates the presence of the \(\ce{C=C}\).

\(\Rightarrow D\)

♦ Mean mark 42%.
COMMENT: Solving by elimination is an effective strategy here.

CHEMISTRY, M8 2023 VCE 29 MC

Which one of the following statements about mass spectrometry is always correct?

  1. The relative molecular mass of a molecule is determined from the base peak.
  2. The peaks in a mass spectrum are caused by the presence of different isotopes.
  3. The base peak is formed when an uncharged species is removed from the molecule.
  4. The height of each peak in the mass spectrum is measured relative to the height of the base peak.
Show Answers Only

\(D\)

Show Worked Solution

→ The relative molecular mass of a molecule is determined from the peak with the largest m/z ratio (eliminate \(A\)).

→ The peaks in a mass spectrum is caused by the molecules splitting up into different ions as they are bombarded with electrons. Some peaks that differ by a value of 1 are caused by the presence of different isotopes but this is not for all peaks (eliminate \(B\)).

→ All peaks on the mass spectrometry graph are charged ions (eliminate \(C\)).

→ The base peak is the largest peak on a mass spectrometry graph and therefore the abundances of each peak are relative to the height of the base peak.

\(\Rightarrow D\)

♦ Mean mark 46%.

CHEMISTRY, M8 2023 VCE 16 MC

Consider the following molecule.
 

How many peaks will be observed in a \({ }^{13} \text{C NMR}\) spectrum of this molecule

  1. 5
  2. 6
  3. 7
  4. 8
Show Answers Only

\(C\)

Show Worked Solution

→ There are 7 different carbon environments in the molecule:

\(\Rightarrow C\)

♦ Mean mark 43%.

CHEMISTRY, M8 2014 VCE 16-17 MC

An atomic absorption spectrometer can be used to determine the level of copper in soils. The calibration curve below plots the absorbance of four standard copper solutions against the concentration of copper ions in ppm.

The concentrations of copper ions in the standard solutions were 1.0, 2.0, 3.0 and 4.0 mg L\(^{-1}\). (1 mg L\(^{-1}\) = 1 ppm)
 

Question 16

The concentration of copper in a test solution can be determined most accurately from the calibration curve if it is between

  1. 0.0 ppm and 5.0 ppm.
  2. 0.0 ppm and 4.0 ppm.
  3. 1.0 ppm and 4.0 ppm.
  4. 1.0 ppm and 5.0 ppm.

Question 17

If the test solution gave an absorbance reading of 0.40, what would be the concentration of copper ions in the solution in mol L\(^{-1}\)?

  1. 2.5
  2. 3.9 × 10\(^{-2}\)
  3. 3.9 × 10\(^{-5}\)
  4. 2.5 × 10\(^{-6}\)
Show Answers Only

\(\text{Question 16:}\ C\)

\(\text{Question 17:}\ C\)

Show Worked Solution

Question 16

→ The concentration of copper ions with be most accurate on the calibration curve in between the limits of the concentration of standard solutions used to produce the calibration curve.

→ Thus it will be most accurate between 1.0 ppm and 4.0 ppm.

\(\Rightarrow C\)
 

Question 17

From the graph: Absorbance of 0.4 → 2.5 ppm (2.5 mg L\(^{-1}\))

\(\ce{M(copper ions) = 63.55\ \text{g mol}^{-1}}\)

\(\ce{c(copper ions) = 2.5 \times 10^{-3}\ \text{g L}^{-1} = \dfrac{2.5 \times 10^{-3}}{63.55}=3.9 \times 10^{-5}\ \text{mol L}^{-1}}\)

\(\Rightarrow C\)

♦ Mean mark (Q17) 49%.

CHEMISTRY, M8 2015 VCE 4

UV-visible spectroscopy was used to measure the spectra of two solutions, \(\text{A}\) and \(\text{B}\). Solution \(\text{A}\) was a pink colour, while Solution \(\text{B}\) was a green colour.

The analyst recorded the absorbance of each solution over a range of wavelengths on the same axes. The resultant absorbance spectrum is shown below.
 

  1. If 10.00 mL of Solution \(\text{A}\) was mixed with 10.00 mL of Solution \(\text{B}\), which wavelength should be used to measure the absorbance of Solution \(\text{B}\) in this mixture? Justify your answer.   (2 marks)

The analyst used two sets of standard solutions and blanks to determine the calibration curves for the two solutions. The absorbances were plotted on the same axes. The graph is shown below.
 

  1. The analyst found that, when it was measured at the appropriate wavelength, Solution \(\text{A}\) had an absorbance of 0.2
  2. If Solution \(\text{A}\) was cobalt\(\text{(II)}\) nitrate, \(\ce{Co(NO3)2}\), determine its concentration in mg L\(^{–1}\)   (2 marks)
  3.    \(\ce{ M(Co(NO3)2) = 182.9 g mol^{–1} \quad \quad 1 mM = 10^{–3} M}\)
  4. In another mixture, the pink compound in Solution \(\text{A}\) and the green compound in Solution \(\text{B}\) each have a concentration of approximately 1.5 × 10\(^{-2}\) M.
  5. Could the analyst reliably use both of the calibration curves to determine the concentrations for Solution \(\text{A}\) and Solution \(\text{B}\) by UV-visible spectroscopy? Justify your answer.   (2 marks)

Show Answers Only

a.    → Wavelength: 625 nm (600 to 650 was accepted)

→ At this wavelength there is the maximum absorbance of solution \(\text{B}\) with no significant absorbance (interference) from solution \(\text{A}\).

b.    \(1.8 \times 10^3\ \text{mgL}^{-1}\)

c.    → \(1.5 \times 10^{-2}\ \text{M} = 15\ \text{mM}\).

→ This value lies within the calibration curve for solution \(\text{A}\), so the calibration curve can be used to determine the concentration of solution \(\text{A}\).

→ This value lies beyond the calibration curve for solution \(\text{B}\), therefore the curve can’t be used to determine the concentration of solution \(\text{B}\). The analyst would need to extrapolate the curve but this would not be reliable.

Show Worked Solution

a.    → Wavelength: 625 nm (600 to 650 was accepted)

→ At this wavelength there is the maximum absorbance of solution \(\text{B}\) with no significant absorbance (interference) from solution \(\text{A}\).
 

b.    From the graph:

\(\text{Absorbance of 0.2} \Rightarrow 10 \times 10^{-3}\ \text{mol L}^{-1}\)

\(\ce{n(Co(NO3)2) = 10 \times 10^{-3} \times 182.9 = 1.829\ \text{g L}^{-1} = 1.8 \times 10^3\ \text{mg L}^{-1}}\)
 

c.    → \(1.5 \times 10^{-2}\ \text{M} = 15\ \text{mM}\).

→ This value lies within the calibration curve for solution \(\text{A}\), so the calibration curve can be used to determine the concentration of solution \(\text{A}\).

→ This value lies beyond the calibration curve for solution \(\text{B}\), therefore the curve can’t be used to determine the concentration of solution \(\text{B}\). The analyst would need to extrapolate the curve but this would not be reliable.

♦♦ Mean mark (c) 30%.

CHEMISTRY, M8 2016 VCE 2*

A common iron ore, fool’s gold, contains the mineral iron pyrite, \(\ce{FeS2}\).

Typically, the percentage by mass of \(\ce{FeS2}\) in a sample of fool’s gold is between 90% and 95%. The actual percentage in a sample can be determined by gravimetric analysis.

The sulfur in \(\ce{FeS2}\) is converted to sulfate ions, \(\ce{SO4^2–}\) as seen below:

\(\ce{4FeS2 + 11O2 \rightarrow 2Fe2O3 + 8SO4^2-}\)

This is then mixed with an excess of barium chloride, \(\ce{BaCl2}\), to form barium sulfate, \(\ce{BaSO4}\), according to the equation

\(\ce{Ba^2+(aq) + SO4^2–(aq)\rightarrow BaSO4(s)}\)

When the reaction has gone to completion, the \(\ce{BaSO4}\) precipitate is collected in a filter paper and carefully washed. The filter paper and its contents are then transferred to a crucible. The crucible and its contents are heated until constant mass is achieved.

The data for an analysis of a mineral sample is as follows.
 

\(\text{initial mass of mineral sample}\) \(\text{14.3 g}\)
\(\text{mass of crucible and filter paper}\) \(\text{123.40 g}\)
\(\text{mass of crucible, filter paper and dry}\ \ce{BaSO4}\) \(\text{174.99 g}\)
\(\ce{M(FeS2)}\) \(\text{120.0 g mol}^{-1}\)
\(\ce{M(BaCl2)}\) \(\text{208.3 g mol}^{-1}\)
\(\ce{M(BaSO4)}\) \(\text{233.4 g mol}^{-1}\)
     
  1. Calculate the percentage by mass of \(\ce{FeS2}\) in this mineral sample.  (4 marks)

  2. State one assumption that was made in completing the calculations for this analysis.  (1 mark)

Show Answers Only

a.  \(\ce{\% FeS2} =92.7\%\ \text{(3 sig.fig.)}\)

b.    Answers could have included one of the following:

→ All the sulfur is converted to \(\ce{SO4^2-}\).

→ Precipitate was pure.

→ No \(\ce{BaSO4}\) was lost in washing the sample.

→ \(\ce{BaSO4}\) was the only precipitate.

→ The sample was fully dehydrated.

Show Worked Solution

a.    \(\ce{m(BaSO4 \text{final})} = 174.99-123.40=51.59\ \text{g}\)

\(\ce{n(BaSO4(s))}=\dfrac{51.59}{233.4}=0.221\ \text{mol}\)

\(\Rightarrow \ce{n(SO4^2-(aq))}=0.221\ \text{mol}\)
 

Molar ratio  \(\ce{FeS2 : SO4^2-} = 1:2\)

\(\Rightarrow \ce{n(FeS2)}=0.221 \times \dfrac{1}{2}=0.1105\ \text{mol}\)

\(\ce{m(FeS2)}=0.1105 \times 120.0 =13.26\ \text{g}\)

\(\ce{\% FeS2} = \dfrac{13.26}{14.3} \times 100=92.7\%\ \text{(3 sig.fig.)}\)
 

b.    Answers could have included one of the following:

→ All the sulfur is converted to \(\ce{SO4^2-}\).

→ Precipitate was pure.

→ No \(\ce{BaSO4}\) was lost in washing the sample.

→ \(\ce{BaSO4}\) was the only precipitate.

→ The sample was fully dehydrated.

♦ Mean mark (b) 39%.

CHEMISTRY, M3 2013 VCE 11*

The following is a student’s summary of catalysts. It contains some correct and incorrect statements.

    1. A catalyst increases the rate of a reaction.
    2. All catalysts are solids.
    3. The mass of a catalyst is the same before and after the reaction.
    4. All catalysts align the reactant particles in an orientation that is favourable for a reaction to occur.
    5. A catalyst lowers the enthalpy change of a reaction, enabling more particles to have sufficient energy to successfully react.
  1. Identify two correct statements.   (1 mark)

  2. Evaluate the student’s summary by identifying two incorrect statements. In each case, explain why it is incorrect.   (4 marks)

Show Answers Only

a.    Statements 1 and 3 are correct.

b.   Answers could include two of the following:

Incorrect statement: All catalysts are solids

→ Catalysts can be solids, liquids, gases or within aqueous solutions.
 

Incorrect statement: All catalysts align the reactant particles in an orientation that is favourable for a reaction to occur.

→ Although solid catalysts orient reactant particles to favor reactions, liquid and gaseous catalysts do not arrange reactants in this manner.
 

Incorrect statement: A catalyst lowers the enthalpy change of a reaction, enabling more particles to have sufficient energy to successfully react.

→ A catalyst does not change the relative energy contents of reactants and products.

→ It does, however, lower the activation energy/increases the proportion of successful collisions/provides an alternative reaction pathway with the same overall \(\Delta H.\)

Show Worked Solution

a.    Statements 1 and 3 are correct.

b.   Answers could include two of the following:

Incorrect statement: All catalysts are solids

→ Catalysts can be solids, liquids, gases or within aqueous solutions.
 

Incorrect statement: All catalysts align the reactant particles in an orientation that is favourable for a reaction to occur.

→ Although solid catalysts orient reactant particles to favor reactions, liquid and gaseous catalysts do not arrange reactants in this manner.
 

Incorrect statement: A catalyst lowers the enthalpy change of a reaction, enabling more particles to have sufficient energy to successfully react.

→ A catalyst does not change the relative energy contents of reactants and products.

→ It does, however, lower the activation energy/increases the proportion of successful collisions/provides an alternative reaction pathway with the same overall \(\Delta H.\)

♦ Mean mark (b) 51%.

CHEMISTRY, M4 2016 VCE 10a*

A senior Chemistry student created an experiment to calculate the molar heat of combustion of butane.

The experimental steps are as follows:

    1. Measure the initial mass of a butane canister
    2. Measure the mass of a metal can, add 250 mL of water and re-weigh.
    3. Set up the apparatus as in the diagram and measure the initial temperature of the water.
    4. Burn the butane gas for five minutes.
    5. Immediately measure the final temperature of the water.
    6. Measure the final mass of the butane canister when cool.
       

Results 

Quantity Measurement
mass of empty can 52.14 g
mass of can + water before combustion 303.37 g
mass of butane canister before heating 260.15 g
mass of butane canister after heating 259.79 g
initial temperature of water 22.1 °C
final temperature of water 32.7 °C

 

The balanced thermochemical equation for the complete combustion of butane is

\(\ce{2C4H10(g) + 13O2(g) \rightarrow 8CO2(g) + 10H2O(l)},\ \ \Delta H=-5748\ \text{kJ mol}^{-1}\)

  1. Calculate the amount of heat energy absorbed by the water when it was heated by burning the butane. Give your answer in kilojoules.   (2 marks)

  2. Calculate the experimental value of the molar heat of combustion of butane. Give your answer in kJ mol\(^{–1}\).   (2 marks)
  3. Use the known enthalpy change for butane to calculate the percentage energy loss to the environment.   (2 marks) 
Show Answers Only

i.    \(11.13\ \text{kJ}\)

ii.   \(-1.8 \times 10^{3}\ \text{kJ mol}^{-1} \)

iii.   \(37.4\% \)

Show Worked Solution

i.    \(\ce{m(water) = 303.37-52.14=251.23\ \text{g}}\)

\(\Delta T = 32.7-22.1=10.6^{\circ}\text{C}\)

\(\text{Energy absorbed}\) \(=4.18 \times 251.23 \times 10.6\)  
  \(=1.113 \times 10^{4}\ \text{J}\)  
  \(=11.13\ \text{kJ}\)  

 

ii.   \(\ce{m(C4H10 reacted) = 260.15-259.79=0.36\ \text{g}}\)

\(\ce{n(C4H10 reacted) = \dfrac{0.36}{\text{MM}} = \dfrac{0.36}{58.0} = 0.0062\ \text{mol}}\)

\(\text{Energy released (per mol}\ \ce{C4H10}\text{)} = \dfrac{11.13}{0.0062}=1.8 \times 10^{3}\ \text{mol}\)

\(\text{Experimental molar heat of combustion} = -1.8 \times 10^{3}\ \text{kJ mol}^{-1} \)
 

♦ Mean mark (ii) 42%.

iii.   \(\text{2 moles}\ \ce{C4H10},\ \Delta H = -5748\ \ \Rightarrow\ \ \text{1 mole}\ \ce{C4H10}, \Delta H = -2874\) 

\(\text{% Energy loss}\ = \dfrac{(2874-1.8 \times 10^{3})}{2874} \times 100 = 37.4\% \)

CHEMISTRY, M4 2014 VCE 3a*

The combustion of ethanol is represented by the following equation.

\(\ce{C2H5OH(l) + 3O2(g)\rightarrow 2CO2(g) + 3H2O(l)}\ \ \ \ \ \ \Delta H=-1364\ \text{kJ mol}^{-1}\)

A spirit burner used 1.80 g of ethanol to raise the temperature of 100.0 g of water in a metal can from 25.0 °C to 40.0 °C.
 

Calculate the percentage of heat lost to the environment and to the apparatus.   (5 marks)

Show Answers Only

\(\text{% heat lost}\ = 88.3\% \)

Show Worked Solution

\(\ce{n(CH3CH2OH) = \dfrac{1.80}{46.0} = 0.0391\ \text{mol}}\)

\(\Delta H \ce{(CH3CH2OH) =-1364\ \text{kJ mol}^{-1}}\)

\(\text{Energy released}\ = 0.0391\ \text{mol}\ \times 1364\ \text{kJ mol}^{-1} = 53.4\ \text{kJ}\)

\(\text{Energy absorbed by water}\ = 4.18 \times 100 \times 15.0 = 6.27\ \text{kJ}\)

\(\text{Energy not absorbed by water}\ = 53.4-6.27 = 47.13\ \text{kJ}\)

\(\text{% heat lost}\ = \dfrac{47.13}{53.4} \times 100 = 88.3\% \)

PHYSICS, M8 2019 VCE 17

Students are comparing the diffraction patterns produced by electrons and X-rays, in which the same spacing of bands is observed in the patterns, as shown schematically in Figure 18. Note that both patterns shown are to the same scale.
 

The electron diffraction pattern is produced by 3.0 × 10\(^3\) eV electrons.

  1. Explain why electrons can produce the same spacing of bands in a diffraction pattern as X-rays.  (3 marks)
  1. Calculate the frequency of X-rays that would produce the same spacing of bands in a diffraction pattern as for the electrons. Show your working.  (4 marks)
Show Answers Only

a.    → Electrons with momentum exhibit wave-like properties.

→  In this way, moving electrons produce a de Broglie wavelength.

→  As diffraction is a wave phenomenon and is dependent on wavelengths, if the de Broglie wavelength of an electron matches the wavelength of an X-ray then spacing of the bands will be the same.
 

b.    \(1.34 \times 10^{19}\ \text{Hz}\)

Show Worked Solution

a.    → Electrons with momentum exhibit wave-like properties.

→  In this way, moving electrons produce a de Broglie wavelength.

→  As diffraction is a wave phenomenon and is dependent on wavelengths, if the de Broglie wavelength of an electron matches the wavelength of an X-ray then spacing of the bands will be the same.
 

♦ Mean mark (a) 49%.

b.    Find velocity of the electrons using  \(E=\dfrac{1}{2}mv^2 :\)

\(3.0 \times 10^3 \times 1.602 \times 10^{-19}=\dfrac{1}{2} \times 9.109 \times 10^{-31} \times v^2\)

\(v^2\) \(=\dfrac{4.806 \times 10^{-16}}{4.5545 \times 10^{-31}}\)  
\(v\) \(=\sqrt{1.055 \times 10^{15}}\)  
  \(=3.25 \times 10^7\ \text{ms}^{-1}\)  

 

The de Broglie wavelength of the electron is:

\(\lambda\) \(=\dfrac{h}{mv}\)  
  \(=\dfrac{6.626 \times 10^{-34}}{3.25 \times 10^7 \times 9.109 \times 10^{-31}}\)  
  \(=2.24 \times 10^{-11}\ \text{m}\)  

 
Frequency of the X-ray:

\(f=\dfrac{c}{\lambda}=\dfrac{3 \times 10^8}{2.24 \times 10^{-11}}=1.34 \times 10^{19}\ \text{Hz}\)

♦♦♦ Mean mark (b) 27%.
COMMENT: Multi-step solutions require clear and logical working.

PHYSICS, M7 2019 VCE 16

Students are studying the photoelectric effect using the apparatus shown in Figure 15.
 

     

Figure 16 shows the results the students obtained for the maximum kinetic energy \((E_{\text{k max }})\) of the emitted photoelectrons versus the frequency of the incoming light.
 

  1. Using only data from the graph, determine the values the students would have obtained for
    1. Planck's constant, \(h\). Include a unit in your answer  (2 marks)
    1. the maximum wavelength of light that would cause the emission of photoelectrons   (1 mark)
    1. the work function of the metal of the photocell.   (1 mark)
  1. The work function for the original metal used in the photocell is \(\phi\).
    On Figure 17, draw the line that would be obtained if a different metal, with a work function of \(\dfrac{1}{2} \phi\), were used in the photocell. The original graph is shown as a dashed line.   (2 marks)
     

Show Answers Only

a. i.  \(5.4 \times 10^{-15}\ \text{eV s}\)

    ii.  \(811\ \text{nm}\)

   iii.  \(1.9\ \text{eV}\).

b.  

Show Worked Solution

a.i.  Planck’s constant \((h)\):

→ Equal to the gradient of the line when \(E_{\text{k max}}\) is graphed against frequency.

\(\therefore h=\dfrac{\text{rise}}{\text{run}}=\dfrac{1.25-0}{6 \times 10^{14}-3.7\times 10^{-14}}=5.4 \times 10^{-15}\ \text{eV s}\)
 

a.ii.  → Max wavelength = minimum frequency of emitted photoelectron.

\(\lambda=\dfrac{c}{f}=\dfrac{3 \times 10^8}{3.7 \times 10^{14}}=811\ \text{nm}\)
 

♦ Mean mark (a)(ii) 44%.

a.iii.  

   

→ The work function is the y-intercept of the graph, so by extending the graph as shown above, the work function is \(1.9\ \text{eV}\).
 

b.    → The new y-intercept for the graph will be \(-0.95\ \text{eV}\)

→ The gradient of the graph will remain the same (Planck’s constant)
 

PHYSICS, M7 2019 VCE 14*

Students have set up a double-slit experiment using microwaves. The beam of microwaves passes through a metal barrier with two slits, shown as \(\text{S}_1\) and \(\text{S}_2\) in Figure 13. The students measure the intensity of the resulting beam at points along the line shown. They determine the positions of maximum intensity to be at the points labelled \(\text{P}_0,\) \(\text{P}_1\), \(\text{P}_2\) and \(\text{P}_3\). 
 

The distance from \(\text{S}_1\) to \(\text{P}_3\) is 72.3 cm and the distance from \(\text{S}_2\) to \(\text{P}_3\) is 80.6 cm.

  1. What is the frequency of the microwaves transmitted through the slits? Show your working.   (2 marks)
  2. The signal strength is at a minimum approximately midway between points \(\text{P}_0\) and \(\text{P}_1\).
  3. Explain the reason why the signal strength would be a minimum at this location.   (2 marks)

  4. The microwaves from the source are polarised.
  5. Explain what is meant by the term 'polarised'. You may use a diagram in your answer.   (2 marks)

Show Answers Only

a.    \(1.08 \times 10^{10}\ \text{Hz}\)

b.    → Halfway between \(\text{P}_0\) and \(\text{P}_1\ \ \Rightarrow\)  path difference = \(\frac{\lambda}{2}\).

→ Therefore, destructive interference will occur and the signal strength will be a minimum.

c.   → Light is a transverse wave that can oscillate in any direction.

→ Light becomes polarised when its plane of oscillation is only in a single direction, as seen in the diagram below.
  

Show Worked Solution

a.    The path difference to \(P_3\) is equal to 3 wavelengths of the microwaves.

\(3\lambda\) \(=0.806-0.723\)  
  \(=0.083\)  
\(\lambda\) \(=0.0277\ \text{m}\)  

 
\(f=\dfrac{c}{\lambda}=\dfrac{3 \times 10^8}{0.0277}=1.08 \times 10^{10}\ \text{Hz}\)
 

♦♦ Mean mark (a) 35%.
COMMENT: Common error was not converting cm to m.

b.    → Halfway between \(\text{P}_0\) and \(\text{P}_1\ \ \Rightarrow\)  path difference = \(\frac{\lambda}{2}\).

→ Therefore, destructive interference will occur and the signal strength will be a minimum.
 

♦♦ Mean mark (b) 42%.

c.   → Light is a transverse wave that can oscillate in any direction.

→ Light becomes polarised when its plane of oscillation is only in a single direction, as seen in the diagram below.
 

♦ Mean mark (c) 50%.
Comment: Students are encouraged to use diagrams when possible.

PHYSICS, M7 2019 VCE 11

What is the second postulate of Einstein's theory of special relativity regarding the speed of light? Explain how the second postulate differs from the concept of the speed of light in classical physics.   (3 marks)

Show Answers Only

→ Einstein’s second postulate states that the speed of light \((c)\) is constant and independent of the motion of the source and motion of the observer.

→ The speed of light in classical physics was dependant on relative motion between the source and observer leading to different measurements for \((c)\),

→ For example, if the source and observer were approaching one another then the speed would be greater than \(3 \times 10^8\ \text{ms}^{-1}\) and if the source and observer were moving away from one another the speed would be less than \(3 \times 10^8\ \text{ms}^{-1}\).

Show Worked Solution

→ Einstein’s second postulate states that the speed of light \((c)\) is constant and independent of the motion of the source and motion of the observer.

→ The speed of light in classical physics was dependant on relative motion between the source and observer leading to different measurements for \((c)\),

→ For example, if the source and observer were approaching one another then the speed would be greater than \(3 \times 10^8\ \text{ms}^{-1}\) and if the source and observer were moving away from one another the speed would be less than \(3 \times 10^8\ \text{ms}^{-1}\).

♦ Mean mark 40%.
COMMENT: Many students failed to properly explain the classical predictions for c.

PHYSICS, M5 2019 VCE 8

A 250 g toy car performs a loop in the apparatus shown in Figure 8.
 

The car starts from rest at point \(\text{A}\) and travels along the track without any air resistance or retarding frictional forces. The radius of the car's path in the loop is 0.20 m. When the car reaches point \(\text{B}\) it is travelling at a speed of 3.0 m s\(^{-1}\).

  1. Calculate the value of \(h\). Show your working.   (3 marks)

  2. Calculate the magnitude of the normal reaction force on the car by the track when it is at point \(\text{B}\). Show your working.   (3 marks)

  3. Explain why the car does not fall from the track at point \(\text{B}\), when it is upside down.   (2 marks)

Show Answers Only

a.    \(0.86\ \text{m}\)

b.    \(8.8\ \text{N}\)

c.    Method 1

→ During the car’s circular motion around the loop, the magnitude of the centripetal force exceeds the magnitude of the weight force.

→ This produces a normal reaction force acting downwards which enables the car to travel around the loop at 3 ms\(^{-1}\).
 

Method 2

→ The minimum speed required for the car so that it stays on the track will be when the centripetal force is equal to the weight force. 

\(\dfrac{mv^2}{r}\) \(=mg\)  
\(v\) \(=\sqrt{gr}\)  
  \(=\sqrt{9.8 \times 0.2}\)  
  \(=1.4\ \text{ms}^{-1}\)  

 
→ As 3 ms\(^{-1}\) is greater than the minimum speed required, a normal force downwards will be present at point \(\text{B}\).

→ Therefore the car will not fall from the track at point \(\text{B}\).

Show Worked Solution

a.    Using the law of the conservation of energy:

\(mgh_A\) \(=mgh_B +\dfrac{1}{2}mv^2_b\)  
\(0.25 \times 9.8 \times h_A\) \(=0.25 \times 9.8 \times 0.4 + \dfrac{1}{2} \times 0.25 \times 3^2\)  
\(2.45h_A\) \(=2.105\)  
\(h_A\) \(=\dfrac{2.105}{2.45}\)  
  \(=0.86\ \text{m}\)  

 

b.     \(N + mg\) \(=\dfrac{mv^2}{r}\)
  \(N\) \(=\dfrac{mv^2}{r}-mg\)
    \(=\dfrac{0.25 \times 3^2}{0.2}- 0.25 \times 9.8\) 
    \(=8.8\ \text{N}\)

♦ Mean mark (b) 53%.

c.    Method 1

→ During the car’s circular motion around the loop, the magnitude of the centripetal force exceeds the magnitude of the weight force.

→ This produces a normal reaction force acting downwards which enables the car to travel around the loop at 3 ms\(^{-1}\).
 

Method 2

→ The minimum speed required for the car so that it stays on the track will be when the centripetal force is equal to the weight force. 

\(\dfrac{mv^2}{r}\) \(=mg\)  
\(v\) \(=\sqrt{gr}\)  
  \(=\sqrt{9.8 \times 0.2}\)  
  \(=1.4\ \text{ms}^{-1}\)  

 
→ As 3 ms\(^{-1}\) is greater than the minimum speed required, a normal force downwards will be present at point \(\text{B}\).

→ Therefore the car will not fall from the track at point \(\text{B}\).

♦♦♦ Mean mark (c) 23%.
COMMENT: Normal force is poorly understood here..

PHYSICS, M6 2019 VCE 7*

Students in a Physics practical class investigate the piece of electrical equipment shown in Figure 5. It consists of a single rectangular loop of wire that can be rotated within a uniform magnetic field. The loop has dimensions 0.50 m × 0.25 m and is connected to the output terminals with slip rings. The loop is in a uniform magnetic field of strength 0.40 T.
 

  1. What name best describes the piece of electrical equipment shown in Figure 5.   (1 mark)

  2. What is the magnitude of the flux through the loop when it is in the position shown in Figure 5? Explain your answer.   (2 marks)

The students connect the output terminals of the piece of electrical equipment to an oscilloscope. One student rotates the loop at a constant rate of 20 revolutions per second.

  1. Calculate the period of the rotation of the loop.  (1 mark)

  2. Calculate the maximum flux through the loop. Show your working.  (1 mark)

  3. The loop starts in the position shown in Figure 5.
  4. What is the average voltage measured across the output terminals for the first quarter turn? Show your working.  (2 marks)

Show Answers Only

a.    Alternator.

b.   → \(0\ \text{Wb}\)

→ This is because the plane of the area of the coil is parallel to the direction of the magnetic field, not perpendicular.

c.    \(0.05\ \text{s}\)

d.    \(0.05\ \text{Wb}\)

e.    \(4\ \text{V}\)

Show Worked Solution

a.    → The device is an alternator.

→ This is due to the input being mechanical motion and the output being AC current, whereas an AC motor is the opposite.

♦ Mean mark (a) 44%.
COMMENT: Many students incorrectly answered AC motor.

b.   → \(0\ \text{Wb}\)

→ This is because the plane of the area of the coil is parallel to the direction of the magnetic field, not perpendicular.
 

c.    \(T=\dfrac{1}{f}=\dfrac{1}{20}=0.05\ \text{s}\)
 

d.    \(\Phi=BA=0.4 \times (0.5 \times 0.25)=0.05\ \text{Wb}\)
 

e.     \(\varepsilon\) \(=\dfrac{\Delta \Phi}{\Delta t_{\text{1/4 rotation}}}\)
    \(=\dfrac{0.05}{0.0125}\)
    \(=4\ \text{V}\)
♦ Mean mark (e) 51%.

PHYSICS, M5 2019 VCE 5*

Navigation in vehicles or on mobile phones uses a network of global positioning system (GPS) satellites. The GPS consists of 31 satellites that orbit Earth.

In December 2018, one satellite of mass 2270 kg, from the GPS Block \(\text{IIIA}\) series, was launched into a circular orbit at an altitude of \(20\ 000\) km above Earth's surface.

  1. Identify the type(s) of force(s) acting on the satellite and the direction(s) in which the force(s) must act to keep the satellite orbiting Earth.   (2 marks)
  1. Calculate the period of the satellite to three significant figures. You may use data from the table below in your calculations. Show your working.   (3 marks)

\begin{array}{|l|l|}
\hline \rule{0pt}{2.5ex}\text{mass of satellite} \rule[-1ex]{0pt}{0pt}& 2.27 \times 10^3 \ \text{kg} \\
\hline \rule{0pt}{2.5ex}\text{altitude of satellite above Earth's surface} \rule[-1ex]{0pt}{0pt}& 2.00 \times 10^7 \ \text{m} \\
\hline
\end{array}

Show Answers Only

a.    → Only force acting on a satellite is \(F_g\), the force due to gravity.

→ This force acts on the satellite directly towards the centre of the Earth.v

b.    \(4.25 \times 10^4\ \text{s}\)

Show Worked Solution


a.    → Only force acting on a satellite is \(F_g\), the force due to gravity.

→ This force acts on the satellite directly towards the centre of the Earth.
 



♦ Mean mark (a) 46%.
COMMENT: Thrust force is not a requirement for orbit. 


b.    \(\dfrac{r^3}{T^2}=\dfrac{GM}{4\pi^2}\ \ \Rightarrow \ \ T= \sqrt{\dfrac{4 \pi^2r^3}{GM}}\)

\(T=\sqrt{\dfrac{4\pi^2 \times (6.371 \times 10^6 + 2 \times 10^7)^3}{6.67 \times 10^{-11} \times 6 \times 10^{24}}}=42\ 533=4.25\times 10^4\ \text{s}\)

PHYSICS, M6 2019 VCE 1

A particle of mass \(m\) and charge \(q\) travelling at velocity \(v\) enters a uniform magnetic field \(\text{B}\), as shown in Figure 1.
 

  1. Is the charge \(q\) positive or negative? Give a reason for your answer.   (1 mark)
  1. Explain why the path of the particle is an arc of a circle while the particle is in the magnetic field.   (2 marks)

Show Answers Only

a.    Using the right-hand rule:

→ Thumb is in direction of velocity (right) and fingers are in direction of B-field (Into the page).

→ As the force on the charge is down the page (back of the hand), the charge must be negative.
 

b.    For circular motion to occur:

→ The force must be at right-angles to the velocity of the particle.

→ The force must be of constant magnitude \((F=qvB)\).

→ In the given example, the force on the charged particle due to the magnetic field is perpendicular to its velocity and the magnitude of the force remains constant.

→ The particle will therefore follow the arc of a circle while in the magnetic field.

Show Worked Solution

a.    Using the right-hand rule:

→ Thumb is in direction of velocity (right) and fingers are in direction of B-field (Into the page).

→ As the force on the charge is down the page (back of the hand), the charge must be negative.

♦ Mean mark (a) 44%.

b.    For circular motion to occur:

→ The force must be at right-angles to the velocity of the particle.

→ The force must be of constant magnitude \((F=qvB)\).

→ In the given example, the force on the charged particle due to the magnetic field is perpendicular to its velocity and the magnitude of the force remains constant.

→ The particle will therefore follow the arc of a circle while in the magnetic field.

♦♦♦ Mean mark (b) 27%.

PHYSICS, M8 2019 VCE 15 MC

Electrons pass through a fine metal grid, forming a diffraction pattern.

If the speed of the electrons was doubled using the same metal grid, what would be the effect on the fringe spacing?

  1. The fringe spacing would increase.
  2. The fringe spacing would decrease.
  3. The fringe spacing would not change.
  4. The fringe spacing cannot be determined from the information given.
Show Answers Only

\(B\)

Show Worked Solution

→ \(\lambda=\dfrac{h}{mv} \ \Rightarrow \  \lambda \propto \dfrac{1}{v}\)

→ If the velocity of the electron is doubled, the wavelength of the electron will halve.

→ Using Young’s double slit calculations:

\(\dfrac{d\Delta x}{D}=m \lambda\ \Rightarrow\  \Delta x \propto \lambda\)

→ Therefore, if lambda decreases so will the fringe spacing \((\Delta x)\).

\(\Rightarrow B\)

♦ Mean mark 40%.
COMMENT: Formulas from different syllabus areas required here.

PHYSICS, M8 2023 HSC 27b

The diagram represents one hydrogen emission line from the spectrum of a star.

Explain the changes to this spectral line that would be observed as a result of the star’s rotational velocity. Modify the diagram to support your answer.   (4 marks)

Show Answers Only

→ When a star rotates, one side of the star is moving towards the Earth and one side of the star is moving away from the Earth.

→ As a result of the Doppler Effect, the wavelengths of the light moving towards the earth are shortened and so are blueshifted while the wavelengths of the light moving away from the earth and lengthened and so are red shifted. 

→ The light from the centre of the star is not moving towards or away from the Earth so it is not shifted in any direction.

→ As a result of the simultaneous blue and red shifting of the 656 nm absorption line, the spectral line will become broader as seen in the modified diagram below.

Show Worked Solution

→ When a star rotates, one side of the star is moving towards the Earth and one side of the star is moving away from the Earth.

→ As a result of the Doppler Effect, the wavelengths of the light moving towards the earth are shortened and so are blueshifted while the wavelengths of the light moving away from the earth and lengthened and so are red shifted. 

→ The light from the centre of the star is not moving towards or away from the Earth so it is not shifted in any direction.

→ As a result of the simultaneous blue and red shifting of the 656 nm absorption line, the spectral line will become broader as seen in the modified diagram below.

♦ Mean mark 55%.

Calculus, MET1 2023 VCAA SM-Bank 5

Let  \(f: R \rightarrow R\),  where  \(f(x)=2-x^2\).

  1. Calculate the average rate of change of \(f\) between \(x=-1\) and \(x=1\).  (1 mark)

  2. Calculate the average value of \(f\) between \(x=-1\) and \(x=1\).  (2 marks)

  3. Four trapeziums of equal width are used to approximate the area between the functions  \(f(x)=2-x^2\)  and the \(x\)-axis from \(x=-1\) to \(x=1\).
  4. The heights of the left and right edges of each trapezium are the values of \(y=f(x)\), as shown in the graph below.

  1. Find the total area of the four trapeziums.  (2 marks)

Show Answers Only

a.    \(0\)

b.    \(\dfrac{5}{3}\)

c.    \(\dfrac{13}{4}\)

Show Worked Solution
a.     \(\text{Average rate of change}\) \(=\dfrac{f(1)-f(-1)}{1-(-1)}\)
    \(\dfrac{1-1}{2}\)
    \(=0\)

 

b.    \(\text{Avg value}\) \(=\dfrac{1}{1-(-1)}\displaystyle\int_{-1}^{1} \Big(2-x^2\Big)\,dx\)
    \(=\dfrac{1}{2}\displaystyle\Bigg[2x-\dfrac{x^3}{3}\displaystyle\Bigg]_{-1}^{1}\)
    \(=\dfrac{1}{2}\displaystyle\Bigg[\Bigg(2.(1)-\dfrac{(1)^3}{3}\Bigg)-\Bigg(2.(-1)-\dfrac{(-1)^3}{3}\Bigg)\Bigg]\)
    \(=\dfrac{1}{2}\times\dfrac{10}{3}=\dfrac{5}{3}\)

 

c.     \(\text{Total Area}\) \(=2\times\ \text{Area from 0 to 1}\)
    \(=2\times \dfrac{h}{2}\Big(f(0)+2.f(0.5)+f(1)\Big)\quad \text{where }h=\dfrac{1}{2}\)
    \(=\dfrac{1}{2}\Big(2+2\times\dfrac{7}{4}+1\Big)=\dfrac{13}{4}\)

CHEMISTRY, M6 2009 HSC 14 MC

Citric acid, the predominant acid in lemon juice, is a triprotic acid. A student titrated 25.0 mL samples of lemon juice with 0.550 mol L\(^{-1}\ \ce{NaOH}\). The mean titration volume was 29.50 mL. The molar mass of citric acid is 192.12 g mol\(^{-1}\).

What was the concentration of citric acid in the lemon juice?

  1. 1.04 g L\(^{-1}\)
  2. 41.6 g L\(^{-1}\)
  3. 125 g L\(^{-1}\)
  4. 374 g L\(^{-1}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\ce{H3A(aq) + 3NaOH(aq) → Na3A(aq) + 3H2O(l)}\)

\(\ce{n(NaOH)=c \times v=0.550 \times 0.0295=0.016225\ \text{mol}}\)

\(\ce{n(citric acid)= \dfrac{0.016225}{3}=0.00541\ \text{mol}}\)

\(\ce{[citric acid]=\dfrac{0.00541}{0.025}=0.216\ \text{molL}^{-1}}\)

Multiply by molar mass:.

\(\text{Concentration (citric acid)}\ = 0.216 \times 192.12 = 41.5\ \text{g L}^{-1}\)

\(\Rightarrow B\)

Functions, MET1 2023 VCAA SM-Bank 3

Find the general solution for  \(2 \sin (x)=\tan (x)\) for \(x \in R\).   (3 marks)

Show Answers Only

\(x=n\pi\ ,\ \Big(2n\pm \dfrac{\pi}{3}\Big)\pi\quad n\in \mathbb{Z}\)

Show Worked Solution

\(2\sin(x)\) \(=\tan(x)\)
\(2\sin(x)\) \(=\dfrac{\sin(x)}{\cos(x)}\quad \cos(x)\neq 0\)
\(2\sin(x)\cos(x)-\sin(x)\) \(=0\)
\(\sin(x)\Big(2\cos(x)-1\Big)\) \(=0\)
\(\therefore \sin(x)=0\quad\) \(\text{or}\) \(\quad 2\cos(x)-1=0\)
\(\therefore x=n\pi\quad\) \(\text{or}\) \(\quad \cos(x)=\dfrac{1}{2}\)
    \(\quad x=2n\pi\pm \dfrac{\pi}{3}=\Bigg(2n\pm \dfrac{1}{3}\Bigg)\pi\quad n\in\mathbb{Z}\)