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v1 Algebra, STD2 A1 2015 HSC 24 MC

Consider the equation  \(\dfrac{5x}{2}-3=\dfrac{3x}{5}+1\).

Which of the following would be a correct step in solving this equation?

  1. \(\dfrac{5x}{2}-2=\dfrac{3x}{5}\)
  2. \(\dfrac{10x}{4}-4=\dfrac{6x}{5}\)
  3. \(\dfrac{5x}{2}=\dfrac{3x}{5}+4\)
  4. \(5x-3=\dfrac{6x}{5}+2\)
Show Answers Only

\(C\)

Show Worked Solution
\(\dfrac{5x}{2}-3\) \(=\dfrac{3x}{5}+1\)
\(\dfrac{5x}{2}-3+3\) \(=\dfrac{3x}{5}+1+3\)
\(\dfrac{5x}{2}\) \(=\dfrac{3x}{5}+4\)

 
\(\Rightarrow C\)

v1 Algebra, STD2 A1 2015 HSC 28d

The formula  \(C=\dfrac{5}{9}(F-32)\)  is used to convert temperatures between degrees Fahrenheit \((F)\) and degrees Celsius \((C)\).

Convert 18°C to the equivalent temperature in Fahrenheit.  (2 marks)

Show Answers Only

\(64.4\ \text{degrees}\ F\)

Show Worked Solution
\(C\) \(=\dfrac{5}{9}(F-32)\)
\(F-32\) \(=\dfrac{9}{5}C\)
\(F\)  \(=\dfrac{9}{5}C+32\)

 
\(\text{When}\ \ C = 18,\)

\(F\)  \(=\dfrac{9}{5}\times 18+32\)
  \(=64.4\ \text{degrees}\ F\)

EXAMCOPY Algebra, STD2 A1 SM-Bank 9

The volume of a sphere is given by  `V = 4/3 pi r^3`  where  `r`  is the radius of the sphere.

If the volume of a sphere is  `220\ text(cm)^3`, find the radius, to 1 decimal place.  (3 marks)

Show Answers Only

`3.7\ \ text{cm  (to 1 d.p.)}`

Show Worked Solution
`V` `= 4/3 pi r^3`
`3V` `= 4 pi r^3`
`r^3` `= (3V)/(4 pi)`

 

`text(When)\ \ V = 220`

`r^3` `= (3 xx 220)/(4 pi)`
  `= 52.521…`
`:. r` `=root3 (52.521…)`
  `= 3.744…\ \ \ text{(by calc)}`
  `= 3.7\ \ text{cm   (to 1 d.p.)}`

v1 Algebra, STD2 A1 SM-Bank 10

For adults (18 years and older), the Body Mass Index is given by:
  

\(B=\dfrac{m}{h^2}\),    where  \(m=\) mass in kilograms and  \(h=\) height in metres.
  

The medically accepted healthy range for  \(B\)  is  \(21\leq B\leq 25\).

What is the minimum weight for a 172 cm adult female to be considered healthy, correct to 1 decimal place? (2 marks)

Show Answers Only

\(62.1\ \text{kg}\)

Show Worked Solution

\(B=\dfrac{m}{h^2}\)

\(h=172\ \text{cm} =1.72\ \text{m}\)
 

\(\text{Given}\ \ 21\leq B\leq 25,\)

\(\rightarrow\ B = 21\ \text{for minimum healthy weight.}\)

\(21\) \(=\dfrac{m}{1.72^2}\)
\(\therefore\ m\) \(=21\times 1.72^2\)
  \(=62.1264\)
  \(=62.1\ \text{kg}\ \text{(1 d.p.)}\)

EXAMCOPY Algebra, STD1 A1 2019 HSC 34

Given the formula  `C = (A(y + 1))/24`, calculate the value of  `y`  when  `C = 120`  and  `A = 500`.  (3 marks)

Show Answers Only

`4.76`

Show Worked Solution

`text(Make)\ \ y\ \ text(the subject:)`

`C` `= (A(y + 1))/24`
`24C` `= A(y + 1)`
`y + 1` `= (24C)/A`
`y` `= (24C)/A-1`
  `= (24 xx 120)/500-1`
  `= 4.76`

v1 Algebra, STD1 A1 2020 HSC 18

The distance, \(d\) metres, travelled by a car slowing down from \(u\) km/h to \(v\) km/h can be obtained using the formula

\(v^2=u^2-100 d\)

What distance does a car travel while slowing down from 100 km/h to 70 km/h?   (2 marks)

Show Answers Only

\(51\ \text{metres}\)

Show Worked Solution

\(u=100 \ , \ v=70\)

\(v^2\) \(=u^2-100d\)
\(70^2\) \(=100^2-100d\)
\(100d\) \(=100^2-70^2\)
\(\therefore\ d\) \(=\dfrac{100^2-70^2}{100}\)
  \(=51\ \text{metres}\)

v1 Algebra, STD2 A1 2012 HSC 21 MC

Which of the following correctly expresses  \(r\)  as the subject of  \(V=\pi r^2+x\) ?

  1. \(r=\pm\sqrt{\dfrac{V}{\pi}}-x\)
  2. \(r=\pm\sqrt{\dfrac{V}{\pi}-x}\)
  3. \(r=\pm\sqrt{\dfrac{V-x}{\pi}}\)
  4. \(r=\pm\dfrac{\sqrt{V-x}}{\pi}\)
Show Answers Only

\(C\)

Show Worked Solution
\(V\) \(=\pi r^2+x\)
\(\pi r^2\) \(=V-x\)
\(r^2\) \(=\dfrac{V-x}{\pi}\)
\(\therefore\ r\) \(=\pm\sqrt{\dfrac{V-x}{\pi}}\)

\(\Rightarrow C\)

v1 Algebra, STD2 A1 2004 HSC 11 MC

If  \(m = 8n^2\), what is a possible value of  \(n\)  when  \(m=7200\)?

  1. \(0.0\dot{3}\)
  2. \(30\)
  3. \(240\)
  4. \(900\)
Show Answers Only

\(B\)

Show Worked Solution
\(m\) \(=8n^2\)
\(n^2\) \(=\dfrac{m}{8}\)
\(n\) \(=\pm\sqrt{\dfrac{m}{8}}\)

 
\(\text{When}\ m=7200:\)

\(n\) \(=\pm\sqrt{\dfrac{7200}{8}}\)
  \(=\pm 30\)

 
\(\Rightarrow B\)

v1 Algebra, STD2 A1 2016 HSC 24 MC

Which of the following correctly expresses \(M\) as the subject of  \(y=\dfrac{M}{V}+cX\)?

  1. \(M=Vy-VcX\)
  2. \(M=Vy+VcX\)
  3. \(M=\dfrac{y-cX}{V}\)
  4. \(M=\dfrac{y+cX}{V}\)
Show Answers Only

\(A\)

Show Worked Solution
\(y\) \(=\dfrac{M}{V}+cX\)
\(\dfrac{M}{V}\) \(=y-cX\)
\(\therefore\ M\) \(=V(y-cX)\)
  \(=Vy-VcX\)

 
\(\Rightarrow A\)

v1 Algebra, STD2 A1 SM-Bank 11

Make  \(V\)  the subject of the equation  \(E=\dfrac{3}{2}mV^3\).  (3 marks)

Show Answers Only

\(v=\sqrt[3]{\dfrac{2E}{3m}}\)

Show Worked Solution
\(E\) \(=\dfrac{3}{2}mV^3\)
\(2E\) \(=3mV^3\)
\(\dfrac{2E}{3}\) \(=mV^3\)
\(V^3\) \(=\dfrac{2E}{3m}\)
\(V\) \(=\sqrt[3]{\dfrac{2E}{3m}}\)

v1 Algebra, STD1 A1 2019 HSC 34

Given the formula  \(D=\dfrac{B(x+1)}{18}\), calculate the value of  \(x\)  when  \(D=90\)  and  \(B=400\).  (3 marks)

Show Answers Only

\(3.05\)

Show Worked Solution

\(\text{Make}\ x\ \text{the subject:}\)

\(D\) \(=\dfrac{B(x+1)}{18}\)
\(18D\) \(=B(x+1)\)
\(x+1\) \(=\dfrac{18D}{B}\)
\(x\) \(=\dfrac{18D}{B}-1\)
\(\text{When }\) \(D=90, B=400\)
\(\therefore\ x\) \(=\dfrac{18\times 90}{400}-1=3.05\)

v1 Algebra, STD2 A2 2022 HSC 14 MC

Which of the following correctly expresses \(x\) as the subject of  \(y=\dfrac{mx-c}{3}\) ?

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

\(D\)

Show Worked Solution
\(y\) \(=\dfrac{mx-c}{3}\)
\(3y\) \(=mx-c\)
\(mx\) \(=3y+c\)
\(\therefore\ x\) \(=\dfrac{3y+c}{m}\)

 
\(\Rightarrow D\)

v1 Algebra, STD2 A1 SM-Bank 9

The volume of a sphere is given by  \(V=\dfrac{4}{3}\pi r^3\)  where  \(r\)  is the radius of the sphere.

If the volume of a sphere is  \(385\ \text{cm}^3\), find the radius, to 1 decimal place.  (3 marks)

Show Answers Only

\(4.5\ \text{cm  (to 1 d.p.)}\)

Show Worked Solution
\(V\) \(=\dfrac{4}{3}\pi r^3\)
\(3V\) \(= 4\pi r^3\)
\(r^3\) \(=\dfrac{3V}{4\pi}\)

 

\(\text{When}\ \ V =385\)

\(r^3\) \(=\dfrac{3\times 385}{4\pi}\)
  \(=91.911\dots\)
\(\therefore\ r\) \(=\sqrt[3]{91.911\dots}\)
  \(=4.512\dots\ \ \text{(by calc)}\)
  \(=4.5\ \text{cm   (to 1 d.p.)}\)

EXAMCOPY Functions, MET2 2022 VCAA 4

Consider the function `f`, where `f:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, f(x)=\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right).`

Part of the graph of `y=f(x)` is shown below.
 

  1. State the range of `f(x)`.   (1 mark)

  2.  i. Find `f^{\prime}(0)`.   (2 marks)

  3. ii. State the maximal domain over which `f` is strictly increasing.   (1 mark)

  4. Show that `f(x)+f(-x)=0`.   (1 mark)

  5. Find the domain and the rule of `f^{-1}`, the inverse of `f`.   (3 marks)

  6. Let `h` be the function `h:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, h(x)=\frac{1}{k}\left(\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)\right)`, where `k \in R` and `k>0`.
  7. The inverse function of `h` is defined by `h^{-1}: R \rightarrow R, h^{-1}(x)=\frac{e^{k x}-1}{2\left(e^{k x}+1\right)}`.
  8. The area of the regions bound by the functions `h` and `h^{-1}` can be expressed as a function, `A(k)`.
  9. The graph below shows the relevant area shaded.
     

  1. You are not required to find or define `A(k)`.
  1. Determine the range of values of `k` such that `A(k)>0`.   (1 mark)

  2. Explain why the domain of `A(k)` does not include all values of `k`.   (1 mark)

Show Answers Only
a.     `R`
b.i `f^{\prime}(0)=4`
b.ii `\left(-\frac{1}{2}, \frac{1}{2}\right)`
c. `0`
d. `x \in \mathbb{R}`
e.i  ` k > 4`
e.ii No bounded area for `0<k \leq 4`
Show Worked Solution

a.   `R` is the range.

b.i    `f(x)`
 `= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)`  
  `f^{\prime}(x)` `= \frac{1}{x+\frac{1}{2}}+\frac{1}{\frac{1}{2}-x}`  
    `= \frac{2}{2 x+1}-\frac{2}{2 x-1}`  
  `f^{\prime}(0)` `= \frac{2}{2 xx 0+1}-\frac{2}{2 xx 0-1}`  
    `= 4`  

 
b.ii  `\left(-\frac{1}{2}, \frac{1}{2}\right)`

c.   `f(x)+f(-x)` `= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)+\log _e\left(-x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}+x\right)`  
  `= 0`  

 
d.   To find the inverse swap `x` and `y` in `y=f(x)`

`x` `= \log _e\left(y+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-y\right)`  
`x` `= \log _e\left(\frac{y+\frac{1}{2}}{\frac{1}{2}-y}\right)`  
`e^x` `=\frac{y+\frac{1}{2}}{\frac{1}{2}-y}`  
`y+\frac{1}{2}` `= e^x\left(-y+\frac{1}{2}\right)`  
`y+\frac{1}{2}` `= -e^x y+\frac{e^x}{2}`  
`y\left(e^x+1\right)` `= \frac{e^x-1}{2}`  
`:.\ f^(-1)(x)` `= \frac{e^x-1}{2(e^x + 1)}`  

 
  `:.`  Domain: `x \in \mathbb{R}`
  

e.i   The vertical dilation factor of  `f(x)` is  `1/k`

For `A(k)>=0` , `h^{\prime}(0)<1`

`\frac{1}{k}(4)<1`   [Using CAS]

`:.\  k > 4`


♦♦♦♦ Mean mark (e.i) 10%.
MARKER’S COMMENT: Incorrect responses included `k>0` and `4<k<33`.

e.ii  When `h \geq h^{-1}` for  `x>0` (or `h \leq h^{-1}` for  `x<0`) there is no bounded area.

`:.`  There will be no bounded area for `0<k \leq 4`.


♦♦♦♦ Mean mark (e.ii) 10%.

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 11 MC

The spectroscopy information for an organic molecule is given below.

number of peaks in \({}^{13}\text{C NMR}\) 2
number of sets of peaks in \({}^{1}\text{H NMR}\) 3
m/z of the last peak in the mass spectrum 60
infra-red (IR) spectrum an absorption peak appears at 3350 cm\(^{-1}\)

The organic molecule is 
 

Show Answers Only

\(B\)

Show Worked Solution

→ Only \(B\) and \(C\) have three different hydrogen environments (eliminate \(A\) and \(D\)). 

→ The molar mass of \(B\) is 60 g mol\(^{-1}\) which matches the last m/z peak in the mass spectrum.

\(\Rightarrow B\)

Statistics, STD2 S1 2016 VCE-G 2*

A weather station records daily maximum temperatures

The five-number summary for the distribution of maximum temperatures for the month of February is displayed in the table below. 

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt} & \textbf{Temperature (°C)} \\
\hline
\rule{0pt}{2.5ex} \text{Minimum} \rule[-1ex]{0pt}{0pt} & 16 \\
\hline
\rule{0pt}{2.5ex} Q_1 \rule[-1ex]{0pt}{0pt} & 21 \\
\hline
\rule{0pt}{2.5ex} \text{Median} \rule[-1ex]{0pt}{0pt} & 25 \\
\hline
\rule{0pt}{2.5ex} Q_3 \rule[-1ex]{0pt}{0pt} & 31 \\
\hline
\rule{0pt}{2.5ex} \text{Maximum} \rule[-1ex]{0pt}{0pt} & 39 \\
\hline
\end{array}

  1. Use the five-number summary above to construct a boxplot on the grid below.   (1 mark)
      


  1. Show, using calculations, that there are no outliers in the dataset.   (2 marks)

Show Answers Only
Show Worked Solution
a.   

 

b.   `IQR = Q_3-Q_1=31-21=10`

`text(Lower fence)` `= Q_1-1.5 xx IQR`
  `= 21-1.5 xx 10`
  `= 6`

 

`text(Upper fence)` `= Q_3 + 1.5 xx IQR`
  `= 31 + 1.5 xx 10`
  `= 46`

 

`text{Since 6 < 16 (minimum) and 46 > 39 (maximum)}`

`=>\ \text{There are no outliers.}`

Statistics, STD2 S1 2019 VCE-G 2*

The five-number summary below was determined from the sleep time, in hours, of a sample of 59 types of mammals.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \ \ \ \textbf{Statistic} \rule[-1ex]{0pt}{0pt} & \textbf{Sleep time (hours)} \\
\hline
\rule{0pt}{2.5ex} \text{minimum} \rule[-1ex]{0pt}{0pt} & \text{2.5} \\
\hline
\rule{0pt}{2.5ex} \text{first quartile} \rule[-1ex]{0pt}{0pt} & \text{8.0} \\
\hline
\rule{0pt}{2.5ex} \text{median} \rule[-1ex]{0pt}{0pt} & \text{10.5} \\
\hline
\rule{0pt}{2.5ex} \text{third quartile} \rule[-1ex]{0pt}{0pt} & \text{13.5} \\
\hline
\rule{0pt}{2.5ex} \text{maximum} \rule[-1ex]{0pt}{0pt} & \text{20.0} \\
\hline
\end{array}

  1. Show with calculations, that a boxplot constructed from this five-number summary will not include outliers.   (2 marks)

  2. Construct the boxplot below.   (1 mark)

     

Show Answers Only
  1. `text(Proof (See Worked Solution))`
  2.  
Show Worked Solution

a.    `IQR = Q_3-Q_1 = 13.5-8.0 = 5.5`

`text(Lower fence)` `= Q_1-1.5 xx IQR`
  `= 8-1.5 xx 5.5`
  `= -0.25`

 

`text(Upper fence)` `= Q_3 + 1.5 xx IQR`
  `= 13.5 + 1.5 xx 5.5`
  `= 21.75`

 
`text(S) text(ince) \ -0.25 < 2.5 \ text{(minimum value) and} \ 21.75 > 20.0 \ text{(maximum value)}`

`=> \ text(no outliers)`
 

b. 

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 2013 VCE 2

The strength of the eggshell of birds is determined by the calcium carbonate, \(\ce{CaCO3}\), content of the eggshell.

The percentage of calcium carbonate in the eggshell can be determined by gravimetric analysis.

0.412 g of clean, dry eggshell was completely dissolved in a minimum volume of dilute hydrochloric acid.

\(\ce{CaCO3(s) + 2H+(aq)\rightarrow Ca^2+(aq) + CO2(g) + H2O(l)}\)

An excess of a basic solution of ammonium oxalate, \(\ce{(NH4)2C2O4}\), was then added to form crystals of calcium oxalate monohydrate, \(\ce{CaC2O4.H2O}\).

The suspension was filtered and the crystals were then dried to constant mass.

0.523 g of \(\ce{CaC2O4.H2O}\) was collected.

  1. Write a balanced equation for the formation of the calcium oxalate monohydrate precipitate.   (1 mark)

  2. Determine the percentage, by mass, of calcium carbonate in the eggshell.   (3 marks)

Show Answers Only

a.    \(\ce{(NH4)2C2O4(aq) + Ca^2+(aq) + H2O(l) \rightarrow CaC2O4.H2O(s) + 2NH4^+(aq)}\)

b.    \(86.9\)%

Show Worked Solution

a.    \(\ce{(NH4)2C2O4(aq) + Ca^2+(aq) + H2O(l) \rightarrow CaC2O4.H2O(s) + 2NH4^+(aq)}\)
 

♦♦♦ Mean mark (a) 30%.

b.    \(\ce{M(CaC2O4.H2O)= 40.08 + 2(12.01) + 4(16) + 2(1.008) + 16 = 146.116\ \text{g mol}^{-1}}\)

\(\ce{n(CaC2O4.H2O)=\dfrac{0.523}{146.116}=0.003579\ \text{mol}}\)

\(\ce{n(CaC2O4.H2O) = n(Ca^2+)= n(CaCO3) = 0.003579\ \text{mol}}\)

\(\ce{m(CaCO3)=0.003579 \times (40.08 +12.01 + 3(16))= 0.358\ \text{g}}\)

\(\text{% Mass}\ =\dfrac{0.358}{0.412} \times 100 = 86.9\%\)

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, M7 2018 VCE 1a

Organic compounds are numerous and diverse due to the nature of the carbon atom. There are international conventions for the naming and representation of organic compounds.

  1. Draw the structural formula of 2-methyl-propan-2-ol.   (1 mark)

  2. Give the molecular formula of but-2-yne.   (1 mark)

  1. Give the IUPAC name of the compound that has the structural formula shown above.   (1 mark)

Show Answers Only

i.    
         

ii.    \(\ce{C4H6}\)

→ Structural or semi-structural formulas are not appropriate.
 

iii.  → Longest carbon chain is 6, hence hexane

→ Number carbons to give functional group lowest possible numbers: 2,3-dibromo, 4-methyl

→ Compound name: 2,3-dibromo-4-methylhexane

Show Worked Solution

i.    
         

ii.    \(\ce{C4H6}\)

→ Structural or semi-structural formulas are not appropriate.
 

iii.  → Longest carbon chain is 6, hence hexane

→ Number carbons to give functional group lowest possible numbers: 2,3-dibromo, 4-methyl

→ Compound name: 2,3-dibromo-4-methylhexane

♦♦♦ Mean mark (c) 27%.

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 6

Brass is an alloy of copper and zinc.

To determine the percentage of copper in a particular sample of brass, an analyst prepared a number of standard solutions of copper\(\text{(II)}\) ions and measured their absorbance using an atomic absorption spectrometer (AAS).

The calibration curve obtained is shown below.
 

  1. A 0.198 g sample of the brass was dissolved in acid and the solution was made up to 100.00 mL in a volumetric flask. The absorbance of this test solution was found to be 0.13
  2. Calculate the percentage by mass of copper in the brass sample.   (3 marks)

  3. If the analyst had made up the solution of the brass sample to 20.00 mL instead of 100.00 mL, would the result of the analysis have been equally reliable? Why?   (2 marks)

  4. Name another analytical technique that could be used to verify the result from part a.   (1 mark)

Show Answers Only

a.    55.6%

b.    → No, this would increase the concentration of the copper solution by a factor of 5.

→ The concentration of the solution and absorbance would be too high and outside the range of the calibration curve (it can’t be assumed that the calibration curve remains linear beyond the range of the known data).

c.    Answers could include:

→ UV-vis spectroscopy, colorimetry, volumetric analysis, gravimetric analysis.

Show Worked Solution

a.    Absorbance of 0.13 → \(\ce{Cu^2+}\) concentration of 1.1 gL\(^{-1}\)  (see graph)

\(\text{In 100 mL:}\)

\(\ce{m(Cu^2+)}=1.1 \times 0.1=0.11\ \text{g}\)

\(\Rightarrow \ce{\% Cu^2+}=\dfrac{0.11}{0.198} \times 100=55.6\%\)
 

b.    → No, this would increase the concentration of the copper solution by a factor of 5.

→ The concentration of the solution and absorbance would be too high and outside the range of the calibration curve.

→ It can’t be assumed that the calibration curve remains linear beyond the range of the known data.
 

♦♦♦ Mean mark (b) 25%.

c.    Answers could include:

→ UV-vis spectroscopy, colorimetry, volumetric analysis, gravimetric analysis.

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, M4 EQ-Bank 7 MC

The molar heat of combustion of pentan-1-ol \(\ce{(C5H12O)}\) is 2800 kJ mol\(^{-1}\). A quantity of pentan-1-ol was combusted, generating 108 kJ of heat.

What mass of pentan-1ol was combusted?

  1. 2.29 grams
  2. 2.86 grams
  3. 3.32 grams
  4. 3.40 grams
Show Answers Only

\(D\)

Show Worked Solution

\(\ce{MM(C5H12O = 5 \times 12.0 + 12 \times 1.008 + 16.0 = 88.096\ \text{g mol}^{-1}\)

\(\ce{m(C5H12O\ \text{required}}\ = \dfrac{108}{2800} \times 88.096 = 3.397\ \text{g}\)

\(\Rightarrow D\)

CHEMISTRY, M4 2012 VCE 4*

Enthalpy changes for the melting of iodine, \(\ce{I2}\), and for the sublimation of iodine are provided below.

\(\ce{I2(s) \rightarrow I2(l) \quad \quad \Delta H = +16 kJ mol^{-1}}\)

\(\ce{I2(s) \rightarrow I2(g) \quad \quad \Delta H = +62 kJ mol^{-1}}\)

Determine the enthalpy change for the vaporisation of iodine that is represented by the equation  \(\ce{I2(l) \rightarrow I2(g)}\).   (2 marks)

Show Answers Only

\(+46\ \text{kJ mol}^{-1}\)

Show Worked Solution

\(\ce{I2(l) \rightarrow I2(s)}, \quad \Delta H = -16\ \text{kJ mol}^{-1}\)

\(\ce{I2(s) \rightarrow I2(g)}, \quad \Delta H = +62\ \text{kJ mol}^{-1}\)

\(\ce{I2(l) \rightarrow I2(g)}, \quad \Delta H = -16 +62 = +46\ \text{kJ mol}^{-1}\)

CHEMISTRY, M3 EQ-Bank 9

Consider the statement:

The effectiveness of a metal catalyst is not dependent upon its surface area.

Explain if this statement correct, giving reasons for your answer.   (2 marks)

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The statement is incorrect.

→ Metal catalysts are solid and reactions occur on the catalyst surface.

→ It follows that the larger the surface, the more effective the catalyst (i.e. its “effectiveness” is dependent on the surface area).

Show Worked Solution

The statement is incorrect.

→ Metal catalysts are solid and reactions occur on the catalyst surface.

→ It follows that the larger the surface, the more effective the catalyst (i.e. its “effectiveness” is dependent on the surface area).

CHEMISTRY, M5 EQ-Bank 30

A catalyst increases the value of the equilibrium constant, thus favouring the extent of the forward reaction, resulting in a greater yield of product.

Evaluate this statement, giving reasons why it is correct or incorrect.   (2 marks)

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→ A catalyst does not affect relative amounts of reactants and products at equilibrium.

→ Only a change in the position of equilibrium will change the value of the equilibrium constant (i.e. the yield of product).

→ A catalyst increases the rates of forward and reverse reactions equally, getting the system to equilibrium faster.

Show Worked Solution

The statement is incorrect.

→ A catalyst does not affect relative amounts of reactants and products at equilibrium.

→ Only a change in the position of equilibrium will change the value of the equilibrium constant (i.e. the yield of product).

→ A catalyst increases the rates of forward and reverse reactions equally, getting the system to equilibrium faster.

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 2013 VCE 23*

Determine the enthalpy change when 40 g of \(\ce{NaOH}\) is dissolved in one litre of water, given the temperature of the solution is measure to have increased by 10.6 °C. Give your answer to the nearest kilojoule.   (2 marks)

Show Answers Only

\(\Delta H = -44\ \text{kJ}\)

Show Worked Solution

\(\ce{m(1 litre H2O) = 1000\ \text{g}}\)

\(\Delta T = +10.6^{\circ}\text{C}\)

\(\text{Energy absorbed (water)}\) \(=4.18 \times 1000 \times 10.6\)  
  \(=4.43 \times 10^4\ \text{J}\)  
  \(=44\ \text{kJ (nearest kJ)}\)  

 
\(\therefore \Delta H = -44\ \text{kJ}\)

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 2015 VCE 18*

Consider the following equations.

\(\ce{\frac{1}{2}N2(g) + O2(g) → NO2(g)}\) \(\quad\Delta H = +30 \text{kJ mol}^{-1}\)
\(\ce{N2(g) + 2O2(g) → N2O4(g)}\) \(\quad \Delta H = +10 \text{kJ mol}^{-1}\)

 Calculate the enthalpy change for the reaction  \(\ce{N2O4(g)\rightarrow 2NO2(g)}\)   (2 marks)

Show Answers Only

\(\Delta H = + 50\ \text{kJ mol}^{-1}\)

Show Worked Solution

→ Double 1st equation:

\(\ce{N2(g) + 2O2(g) \rightarrow 2NO2(g)\ \ \ \ \Delta H = +60\ \text{kJ mol}^{-1}}\)

→ Reverse 2nd equation:

\(\ce{N2O4(g) \rightarrow N2(g) + 2O2(g)\ \ \ \ \Delta H = -10\ \text{kJ mol}^{-1}}\)

→ Combining both equations:

\(\ce{N2O4(g)\rightarrow 2NO2(g)\ \ \ \ \Delta H = +50\ \text{kJ mol}^{-1}}\)

CHEMISTRY, M4 2015 VCE 16 MC

Consider the following energy profile for a particular chemical reaction, where \(\text{I, II}\) and \(\text{III}\) represent enthalpy changes during the reaction.
 

Which one of the following statements is correct?

  1. The activation energy for the reverse reaction is \(\text{(III–II)}\).
  2. The net energy released for the forward reaction is represented by \(\text{II}\).
  3. The energy required to break the reactant bonds is represented by \(\text{II}\).
  4. The energy released by the formation of new bonds is represented by \(\text{I}\).
Show Answers Only

\(B\)

Show Worked Solution

→ \(\text{I:}\) energy needed to break reactant bonds (activation energy)

→ \(\text{II:}\) energy released by forward reaction

→ \(\text{III:}\) energy released when product bonds form (activation energy of reverse reaction)

\(\Rightarrow B\)

PHYSICS, M8 2019 VCE 18

The energy level diagram for a hydrogen atom is shown in Figure 19.
 

  1. A hydrogen atom in the ground state is excited to the \(n=4\) state.
  2. Explain how the hydrogen atom could be excited to the \(n=4\) state in one step.   (2 marks)
  1. List the possible photon energies that could be emitted as the atom goes from the \(n=4\) state to the \(n=2\) state.   (3 marks) 

Show Answers Only

a.    → The hydrogen atom would need to absorb exactly 12.8 eV of energy.

→ The energy source could be either an incident photon or electron.
 

b.    From \(n=4\) to \(n=3\ \ \Rightarrow \ 0.7\ \text{eV}\)

From \(n=3\) to \(n=2\ \ \Rightarrow \ 1.9\ \text{eV}\)

From \(n=4\) to \(n=2\ \ \Rightarrow \ 2.6\ \text{eV}\)

Show Worked Solution

a.    → The hydrogen atom would need to absorb exactly 12.8 eV of energy.

→ The energy source could be either an incident photon or electron.
 

b.    From \(n=4\) to \(n=3\ \ \Rightarrow \ 0.7\ \text{eV}\)

From \(n=3\) to \(n=2\ \ \Rightarrow \ 1.9\ \text{eV}\)

From \(n=4\) to \(n=2\ \ \Rightarrow \ 2.6\ \text{eV}\)

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, M5 2019 VCE 10

A projectile is launched from the ground at an angle of 39° and at a speed of 25 m s\(^{-1}\), as shown in Figure 10. The maximum height that the projectile reaches above the ground is labelled \(h\).
 

  1. Ignoring air resistance, show that the projectile's time of flight from the launch to the highest point is equal to 1.6 seconds. Give your answer to two significant figures. Show your working and indicate your reasoning.  (2 marks)

  2. Calculate the range, \(R\), of the projectile. Show your working.  (2 marks)

Show Answers Only

a.    \(t=1.6\ \text{s}\)

b.    \(62\ \text{m}\)

Show Worked Solution

a.    
       

\(v_v=25\,\sin39^{\circ}=15.733\ \text{ms}^{-1}\)

\(\text{At max height:}\ v_v=0\)

\(v\) \(=u+at_1\)  
\(0\) \(=15.733 -9.8t_1\)  
\(9.8t_1\) \(=15.733\)  
\(t_1\) \(=1.6\ \text{s  (2 sig.fig)}\)  

 

b.    \(\text{Since path is symmetrical:}\)

\(\text{Time of flight}\ (t_2) =1.6 \times 2=3.2\ \text{s}\).

\(v_h=25\,\cos39^{\circ}=19.43\ \text{ms}^{-1}\ \ \text{(see part (a) diagram)}\) 

\(\therefore R=v_ht_2=19.43 \times 3.2= 62\ \text{m}\)

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 3

Figure 3 shows a schematic diagram of a DC motor. The motor has a coil, \(JKLM\), consisting of 100 turns. The permanent magnets provide a uniform magnetic field of 0.45 T.

The commutator connectors, \(X\) and \(Y\), provide a constant DC current, \(I\), to the coil. The length of the side \(JK\) is 5.0 cm.

The current \(I\) flows in the direction shown in the diagram.
 

  1. Which terminal of the commutator is connected to the positive terminal of the current supply?   (1 mark)

  2. Draw an arrow on Figure 3 to indicate the direction of the magnetic force acting on the side \(JK\).   (1 mark)

  3. Explain the role of the commutator in the operation of the DC motor.   (2 marks)

  4. A current of 6.0 A flows through the 100 turns of the coil \(JKLM\).
  5. The side \(JK\) is 5.0 cm in length.
  6. Calculate the size of the magnetic force on the side \(JK\) in the orientation shown in Figure 3. Show your working.   (2 marks)

Show Answers Only

a.    → Current runs from the positive terminal to the negative terminal.

→ \(X\) is connected to the positive terminal of the current supply.
 

b.    Using the right-hand rule: force on \(JK\) is down the page.

c.    Role of the split-ring commutator:

→ Changes the direction of the current through the arms of the coil every 180\(^{\circ}\).

→ This ensures that the torque through the motor will always be in the same direction so the motor will rotate in the same direction.
 

d.    \(F=nlIB=100 \times 0.05 \times 6 \times 0.45 = 13.5\ \text{N}\)

Show Worked Solution

a.    → Current runs from the positive terminal to the negative terminal.

→ \(X\) is connected to the positive terminal of the current supply.
 

b.    Using the right-hand rule: force on \(JK\) is down the page.

c.    Role of the split-ring commutator:

→ Changes the direction of the current through the arms of the coil every 180\(^{\circ}\).

→ This ensures that the torque through the motor will always be in the same direction so the motor will rotate in the same direction.
 

d.    \(F=nlIB=100 \times 0.05 \times 6 \times 0.45 = 13.5\ \text{N}\)

PHYSICS, M7 2019 VCE 17 MC

Which one of the following is true when incandescent light is compared to laser light?

  1. Laser light has a very wide spectrum; incandescent light has a very narrow spectrum.
  2. Both laser light and incandescent light have a very narrow spectrum.
  3. Laser light is incoherent; incandescent light is coherent.
  4. Laser light is coherent; incandescent light is incoherent.
Show Answers Only

\(D\)

Show Worked Solution

→ Incandescent light has a very wide spectrum (eliminate \(A\) and \(B\)).

→ Laser light is coherent while incandescent light is incoherent.

\(\Rightarrow D\)

PHYSICS, M7 2019 VCE 13 MC

Joanna is an observer in Spaceship \(\text{A}\), watching Spaceship \(\text{B}\) fly past at a relative speed of 0.943\(c\). She measures the length of Spaceship \(\text{B}\) from her frame of reference to be 150 m.
 

Which one of the following is closest to the proper length of Spaceship \(\text{B}\)?

  1. 50 m
  2. 150 m
  3. 450 m
  4. 900 m
Show Answers Only

\(C\)

Show Worked Solution
\(l\) \(=l_0\sqrt{1-\frac{v^2}{c^2}}\)  
\(l_0\) \(=\dfrac{l}{\sqrt{1-\frac{v^2}{c^2}}}\)  
  \(=\dfrac{150}{\sqrt{1-0.943^2}}\)  
  \(=450\ \text{m}\)  

 

\(\Rightarrow C\)

PHYSICS, M5 2019 VCE 12 MC

A small ball is rolling at constant speed along a horizontal table. It rolls off the edge of the table and follows the parabolic path shown in the diagram below. Ignore air resistance.
 

Which one of the following statements about the motion of the ball as it falls is correct?

  1. The ball's speed increases at a constant rate.
  2. The momentum of the ball is conserved.
  3. The acceleration of the ball is constant.
  4. The ball travels at constant speed.
Show Answers Only

\(C\)

Show Worked Solution

→ During projectile motion the ball is only subject to the acceleration due to gravity.

→ Hence the acceleration of the ball is constant.

\(\Rightarrow C\)

PHYSICS, M6 2019 VCE 8 MC

An electrical generator is shown in the diagram below. The generator is turning clockwise.
 

   

The voltage between \(\text{P}\) and \(\text{Q}\) and the magnetic flux through the loop are both graphed as a function of time, with voltage versus time shown as a solid line and magnetic flux versus time shown as a dashed line. Which one of the following graphs best shows the relationships for this electrical generator?
 

Show Answers Only

\(A\)

Show Worked Solution

→ The magnetic flux initially is a maximum as the plane of the coil is perpendicular to the magnetic field lines.

→ The voltage can be determined by taking the negative gradient of the magnetic flux line. This is because  \(\varepsilon=-\dfrac{\Delta \Phi}{\Delta t}\).

\(\Rightarrow A\)

PHYSICS, M6 2019 VCE 7 MC

The coil of an AC generator completes 50 revolutions per second. A graph of output voltage versus time for this generator is shown below.
 

Which one of the following graphs best represents the output voltage if the rate of rotation is changed to 25 revolutions per second?
 

Show Answers Only

\(D\)

Show Worked Solution

→ \(f \propto \dfrac{1}{T}\), so if the frequency is halved to 25 revolutions per second, the period of the wave will be doubled.

→ The induced voltage is proportional to the rate of change of flux through the coil. As the rate of change of flux is halved, the induced voltage will also be halved.

\(\Rightarrow D\)

PHYSICS, M6 2019 VCE 5-6* MC

A 40 V AC generator and an ideal transformer are used to supply power. The diagram below shows the generator and the transformer supplying 240 V to a resistor with a resistance of 1200 \( \Omega \).
 

Question 5

Which of the following correctly identifies the parts labelled \(\text{X}\) and \(\text{Y}\), and the function of the transformer?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\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}{|l|l|l|}
\hline
\rule{0pt}{2.5ex}\quad \ \ \ \text{Part X}\quad \rule[-1ex]{0pt}{0pt}&\ \ \  \quad \text{Part Y} \quad& \text{Function of transformer} \\
\hline
\rule{0pt}{2.5ex}\text{primary coil}\rule[-1ex]{0pt}{0pt}&\text{secondary coil} & \text{step-down}\\
\hline
\rule{0pt}{2.5ex}\text{primary coil}\rule[-1ex]{0pt}{0pt}& \text{secondary coil}&\text{step-up}\\
\hline
\rule{0pt}{2.5ex}\text{secondary coil}\rule[-1ex]{0pt}{0pt}& \text{primary coil} &\text{step-down}\\
\hline
\rule{0pt}{2.5ex}\text{secondary coil}\rule[-1ex]{0pt}{0pt}& \text{primary coil} &\text{step-up}\\
\hline
\end{array}
\end{align*}

Question 6

Which one of the following is closest to the current in the primary circuit?

  1. \(0.04\ \text{A}\)
  2. \(0.20\ \text{A}\)
  3. \(1.20\ \text{A}\)
  4. \(1.50\ \text{A}\)
Show Answers Only

\(\text{Question 5:}\ B\)

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

Show Worked Solution

\(\text{Question 5}\)

→ The primary coil is connected to the power supply (i.e. Part \(\text{X}\)).

→ For a step-up transformer, there are more coils in the secondary coil then there are in the primary coil.

\(\Rightarrow B\)
 

\(\text{Question 6}\)

\(I_s=\dfrac{V_s}{r_s}=\dfrac{240}{1200}=0.2\ \text{A}\)

\(\dfrac{I_s}{I_p}\) \(=\dfrac{N_p}{N_s}\)  
\(I_p\) \(=\dfrac{N_s \times I_s}{N_p}\)  
  \(=\dfrac{6000 \times 0.2}{1000}\)  
  \(=1.2\ \text{A}\)  

 

\(\Rightarrow C\)

Mean mark Q6 57%.

PHYSICS, M5 2019 VCE 4 MC

The magnitude of the acceleration due to gravity at Earth's surface is \(g\).

Planet \(\text{Y}\) has twice the mass and half the radius of Earth. Both planets are modelled as uniform spheres.

Which one of the following best gives the magnitude of the acceleration due to gravity on the surface of Planet \(\text{Y}\)?

  1. \(\dfrac{1}{2} g\)
  2. \(1g\)
  3. \(4g\)
  4. \(8g\)
Show Answers Only

\(D\)

Show Worked Solution

\(g=\dfrac{GM}{r^2}\)

\(\text{Planet Y:}\ \dfrac{G \times 2M}{(\frac{r}{2})^2}=\dfrac{2GM}{\frac{r^2}{4}}=8 \times \dfrac{GM}{r^2}=8g\)

\(\Rightarrow D\)

Mean mark 58%.

Probability, MET1 2023 VCAA SM-Bank 7

The duration of telemarketing calls to mobile phone users is a continuous random variable \(T\) minutes, with probability density function

\(f(t)= \begin{cases} \dfrac{2}{5} e^{-\frac{2}{5} t} & t \geq 0 \\ \ 0 & \text {elsewhere }\end{cases}\)

Find the value of \(k\) such that 90% of telemarketing calls last less than \(k\) minutes. Express your answer in the form  \(\dfrac{a}{b} \,\log _e(c)\),  where \(a, b\) and \(c\) are positive integers.   (3 marks)

Show Answers Only

\(k=\dfrac{5}{2}\log_{e}{10}\)

Show Worked Solution

\(\text{Pr}(T<k)=\displaystyle\int_{0}^{k}\dfrac{2}{5} e^{-\frac{2}{5} t}\,dt=0.9\)

\(\dfrac{2}{5}\Bigg[-\dfrac{5}{2}e^{-\frac{2t}{5}}\displaystyle\Bigg]_{0}^{k}\) \(=0.9\)
\(-e^{-\frac{2k}{5}}+1\) \(=0.9\)
\(e^{-\frac{2k}{5}}\) \(=0.1=\dfrac{1}{10}\)
\(e^{\frac{2k}{5}}\) \(=10\)
\(\dfrac{2k}{5}\) \(=\log_{3}{10}\)
\(\therefore\ k\) \(=\dfrac{5}{2}\log_{3}{10}\)

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

Functions, MET1 EQ-Bank 4

Consider the simultaneous equations below, where \(a\) and \(b\) are real constants.

\begin{aligned}
& (a+3) x+9 y=3 b\\
& 2 x+a y=5
\end{aligned}

Find the values of \(a\) and \(b\) for which the simultaneous equations have no solutions.   (4 marks)

Show Answers Only

\(a=-6\ \text{and}\ b\neq -\dfrac{5}{2}\ \ \text{OR}\ \ a=3\ \text{and}\ b\neq 5\)

Show Worked Solution

(\(a+3)x+9y\) \(=3b\)  
\(y\) \(=-\dfrac{a+3}{9}x+\dfrac{b}{3}\) \(\dots (1)\)
\(\rightarrow \ m_1\) \(=-\dfrac{a+3}{9}, \ c_1=\dfrac{b}{3}\)  

 

\(2x+ay\) \(=5\)  
\(y\) \(=-\dfrac{2}{a}x+\dfrac{5}{a}\) \(\dots (2)\)
\(\rightarrow \ m_2\) \(=-\dfrac{2}{a}, \ c_2=\dfrac{5}{a}\)  

  
\(\text{No solution if  }m_1=m_2,\ \text{and}\ c_1\neq c_2.\)

\(\rightarrow\ -\dfrac{a+3}{9}=-\dfrac{2}{a}\ \text{and }\dfrac{b}{3}\neq \dfrac{5}{a}\rightarrow\ b\neq \dfrac{15}{a}\)

   

\(-\dfrac{a+3}{9}=-\dfrac{2}{a}\)

\(a^2+3a-18\) \(=0\)  
\((a+6)(a-3)\) \(=0\)  
\(\therefore a=-6\ \) \(\ \text{or}\ \) \(\ a=3\)

  

\(\text{When }\ a=-6,\  b\neq \dfrac{15}{-6}\rightarrow\ b\neq -\dfrac{5}{2}\)

\(\text{OR}\)

\(\text{When }\ a=3,\  b\neq \dfrac{15}{3}\rightarrow\ b\neq 5\)

CHEMISTRY, M6 2008 HSC 14 MC

20 mL 0.08 mol L\(^{-1}\ \ce{HCl}\) is mixed with 30 mL of 0.05 mol L\(^{-1}\ \ce{NaOH}\).

What is the pH of the resultant solution?

  1. 1.1
  2. 2.7
  3. 4.0
  4. 7.0
Show Answers Only

\(B\)

Show Worked Solution

\(\ce{n(HCl)_{excess} = 0.0016-0.0015=0.0001 mol}\)

\[\ce{[HCl] = \frac{0.0001}{0.020+0.030} = 0.002 mol L^{-1}}\]

\(\ce{[HCl] = [H+]}\)

\(\ce{pH = -log(0.002) = 2.7}\)

\(\Rightarrow B\)

PHYSICS, M8 2020 VCE 17

Figure 17 shows the emission spectrum for helium gas.
 

   

  1. Which spectral line indicates the photon with the lowest energy?  (1 mark)
  1. Calculate the frequency of the photon emitted at the 588 nm line. Show your working.  (2 marks)
  1. Explain why only certain wavelengths and, therefore, certain energies are present in the helium spectrum.  (2 marks)

Show Answers Only

a.    \(668\ \text{nm}\)

b.    \(5.1 \times 10^{14}\ \text{Hz}\)

c.    → Electrons in the helium atom exist within electron shells.

→ Electrons in each shell have a fixed amount of energy. 

→ he electrons can absorb certain amounts of energy and move up into the next energy shell which is equal to the difference in energy between the shells.

→ When electrons fall back to their original energy state, they emit a photon with energy equal to the difference in energy between the shells.

→ As these are always fixed energy levels, only certain wavelengths and energies are emitted and therefore present in the helium spectrum.

Show Worked Solution

a.    \(E=\dfrac{hc}{\lambda}\),  therefore \(E \propto \dfrac{1}{\lambda}\)

→ The photon with the lowest energy will have the highest wavelength.

→ Lowest energy spectral line = 668 nm
 

b.    Convert: 588 nm = 588 × 10\(^{-9}\) m

\(f=\dfrac{c}{\lambda}=\dfrac{3 \times 10^8}{588 \times 10^{-9}}=5.1 \times 10^{14}\ \text{Hz}\)
 

c.    → Electrons in the helium atom exist within electron shells.

→ Electrons in each shell have a fixed amount of energy. 

→ he electrons can absorb certain amounts of energy and move up into the next energy shell which is equal to the difference in energy between the shells.

→ When electrons fall back to their original energy state, they emit a photon with energy equal to the difference in energy between the shells.

→ As these are always fixed energy levels, only certain wavelengths and energies are emitted and therefore present in the helium spectrum.

♦♦ Mean mark (c) 34%.

PHYSICS, M8 2020 VCE 16

A beam of electrons travelling at 1.72 × 10\(^5\) m s\(^{-1}\) illuminates a crystal, producing a diffraction pattern as shown in Figure 16. Ignore relativistic effects.
 

  1. Calculate the kinetic energy of one of the electrons. Show your working.  (2 marks)

  2. The electron beam is now replaced by an X-ray beam. The resulting diffraction pattern has the same spacing as that produced by the electron beam.

    Calculate the energy of one X-ray photon. Show your working.  (3 marks)

Show Answers Only

a.    \(0.084\ \text{eV}\)

b.    \(293\ \text{eV}\)

Show Worked Solution

a.     \(KE\) \(=\dfrac{1}{2}mv^2\)
    \(=\dfrac{1}{2} \times 9.109 \times 10^{-31} \times (1.72 \times 10^5)^2\)
    \(=1.3474\times 10^{-20}\ \text{J}\)
    \(=\dfrac{1.3474 \times 10^{-20}}{1.602 \times 10^{-19}}\)
    \(=0.084\ \text{eV}\)

 

b.     \(\lambda_e\) \(=\dfrac{h}{mv}\)
    \(=\dfrac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 1.72 \times 10^5}\)
    \(=4.23 \times 10^{-9}\ \text{m}\)
    \(=\lambda_{\text{x-ray}}\)

 

  \(E_{\text{x-ray}}\) \(=\dfrac{hc}{\lambda}\)
    \(=\dfrac{6.626 \times 10^{-34} \times 3 \times 10^8}{4.23 \times 10^{-9}}\)
    \(=4.7 \times 10^{-17}\ \text{J}\)
    \(=\dfrac{4.7 \times 10^{-17}}{1.602 \times 10^{-19}}\)
    \(=293\ \text{eV}\)
♦ Mean mark (b) 41%.

PHYSICS, M7 2020 VCE 15

The metal surface in a photoelectric cell is exposed to light of a single frequency and intensity in the apparatus shown in Figure 14.

The voltage of the battery can be varied in value and reversed in direction.
  

  1. A graph of photocurrent versus voltage for one particular experiment is shown in Figure 15.
  2. On Figure 15, draw the trace that would result for another experiment using light of the same frequency but with triple the intensity.  (2 marks)
     

  1. What is a name given to the point labelled \(\text{A}\) on Figure 15?  (1 mark)

  2. Why does the photocurrent fall to zero at the point labelled \(\text{A}\) on Figure 15 ?  (1 mark)

Show Answers Only

a.   
         

b.    The stopping voltage.

c.    → The energy of the stopping voltage is equal to the most energetic photoelectrons. 

→ Therefore, the stopping voltage will have enough energy to turn back/stop all of the photoelectrons.

Show Worked Solution

a.    The photocurrent will also be tripled.
 

b.    The stopping voltage.
 

c.    → The energy of the stopping voltage is equal to the most energetic photoelectrons. 

→ Therefore, the stopping voltage will have enough energy to turn back/stop all of the photoelectrons.

♦♦ Mean mark (c) 32%.
COMMENT: The work function of the metal had no relevance to this question.