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Algebra, STD2 A1 EO-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)

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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 shown 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 shown 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 
 

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\(B\)

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)

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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)

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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. 

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 (b) 50%.
♦ 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)

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

b.    \(86.9\)%

Show Worked Solution

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

\(\ce{CaC2O4.H2O(s) + 2NH4^+(aq)}\)
 

♦♦♦ 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)

Show Answers Only
  • 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.

Determine if this statement is correct or incorrect, giving reasons for your answer.   (2 marks)

Show Answers Only

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.
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 below.
 

  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.   Ground state to \(n=4\) state:

  • 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.   Ground state to \(n=4\) state:

  • 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 1.
 

     

Figure 2 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 3, 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.   Constructing the new graph:

  • 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 the diagram. 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 the diagram below.
 

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 the diagram. 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 the diagram?   (1 mark)

  2. What is the magnitude of the flux through the loop when it is in the position shown in the diagram? 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 the diagram.
  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.    Forces acting on satellites:

  • 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.

b.    \(4.25 \times 10^4\ \text{s}\)

Show Worked Solution

a.    Forces acting on satellites:

  • 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

The diagram below 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 the diagram 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

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.

\(\Rightarrow B\)
 

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

The diagram 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.   Wavelengths of \(\ce{He}\) spectrum:

  • Electrons in the helium atom exist within electron shells.
  • Electrons in each shell have a fixed amount of energy. 
  • The 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.   Wavelengths of \(\ce{He}\) spectrum:

  • Electrons in the helium atom exist within electron shells.
  • Electrons in each shell have a fixed amount of energy. 
  • The 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 below. 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 Diagram A.

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 Diagram B.
  2. On Diagram B, 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 Diagram B?   (1 mark)

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

Show Answers Only

a.  The photocurrent will also be tripled.

b.    The stopping voltage.

c.    Zero photocurrent at \(\text{A}\):

  • 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.    Zero photocurrent at \(\text{A}\):

  • 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.

PHYSICS, M7 2020 VCE 14

The diagram below shows a representation of an electromagnetic wave.

Correctly label the diagram below using the following symbols.   (3 marks)

 

Show Answers Only

Show Worked Solution

  •  The wavelength is from the first peak on the electric wave to the second peak on the electric wave (not from an electric wave peak to magnetic wave peak).

PHYSICS, M5 2020 VCE 8

The diagram below shows a small ball of mass 1.8 kg travelling in a horizontal circular path at a constant speed while suspended from the ceiling by a 0.75 m long string.
 


 

  1. Use labelled arrows on the diagram above to indicate the two physical forces acting on the ball.   (2 marks)
  1. Calculate the speed of the ball. Show your working.   (4 marks)
Show Answers Only

a.  
       

b.    \(1.2\ \text{ms}^{-1}\)

Show Worked Solution

a.    Only two forces are acting on the ball: \(F_w\) and \(F_T\):
 

   

Mean mark 58%.
COMMENT: Many students incorrectly identified centripetal force (this is a result of the tension force).

b.   
     

\(\text{Using}\ \ \tan\theta=\dfrac{F_{\text{net}}}{mg}:\)

\( F_{\text{net}}\) \(=mg\,\tan\theta\)  
\(\dfrac{mv^2}{r}\) \(=mg\,\tan\theta\)  
\(\dfrac{v^2}{r}\) \(=g\,\tan\theta\)  
\(\therefore v\) \(=\sqrt{g\,r\,\tan\theta}\)  
  \(=\sqrt{9.8 \times 0.317 \times \tan 25}\ \ ,\ \ (r = 0.75 \times \sin25=0.317\ \text{m})\)  
  \(=1.2\ \text{ms}^{-1}\)  
♦ Mean mark (b) 50%.

PHYSICS, M6 2020 VCE 6

Two Physics students hold a coil of wire in a constant uniform magnetic field, as shown in Figure 5a. The ends of the wire are connected to a sensitive ammeter. The students then change the shape of the coil by pulling each side of the coil in the horizontal direction, as shown in Figure 5b. They notice a current register on the ammeter.
 

  1. Will the magnetic flux through the coil increase, decrease or stay the same as the students change the shape of the coil?   (1 mark)
  1. Explain, using physics principles, why the ammeter registered a current in the coil and determine the direction of the induced current.   (3 marks)
  1. The students then push each side of the coil together, as shown in Figure 6a, so that the coil returns to its original circular shape, as shown in Figure 6b, and then changes to the shape shown in Figure 6c. 
     

  1. Describe the direction of any induced currents in the coil during these changes. Give your reasoning.   (2 marks)

Show Answers Only

a.    Decrease 

b.    Ammeter registers a current:

  • By Faraday’s law of induction, a change in flux will result in an induced EMF through the coil and induced current. 
  • This induced current acts to oppose the original change in flux that created it. As the magnetic flux is decreasing, the induced current will act to strengthen the magnetic field.
  • Hence by using the right-hand grip rule, the current will run clockwise through the coil.

c.    There will be an induced current from 6a to 6b.

  • As the magnetic flux is increasing, the induced current will act to decrease the flux and oppose the magnetic field.
  • By the right-hand grip rule, the current will flow anti-clockwise around the coil.
  • There will also be an induced current from 6b to 6c.
  • The magnetic flux this time is decreasing, so the induced current will act to increase the flux and strengthen the magnetic field.
  • Using the right-hand grip rule, the current through the coil will flow clockwise.
Show Worked Solution

a.    Magnetic flux will decrease.

  • The magnetic flux is proportional to how many field lines are passing through a given area.
  • As the number of lines passing through the coil of wire decreases, the magnetic flux will also decrease.

b.    Ammeter registers a current:

  • By Faraday’s law of induction, a change in flux will result in an induced EMF through the coil and induced current. 
  • This induced current acts to oppose the original change in flux that created it. As the magnetic flux is decreasing, the induced current will act to strengthen the magnetic field.
  • Hence by using the right-hand grip rule, the current will run clockwise through the coil.
♦ Mean mark (b) 42%.

c.    There will be an induced current from 6a to 6b.

  • As the magnetic flux is increasing, the induced current will act to decrease the flux and oppose the magnetic field.
  • By the right-hand grip rule, the current will flow anti-clockwise around the coil.
  • There will also be an induced current from 6b to 6c.
  • The magnetic flux this time is decreasing, so the induced current will act to increase the flux and strengthen the magnetic field.
  • Using the right-hand grip rule, the current through the coil will flow clockwise.
♦♦♦ Mean mark (c) 28%.
COMMENT: Lenz’s law was poorly understood in this question.

Functions, MET1 EQ-Bank 2

Consider the functions \(f\) and \(g\), where

\begin{aligned}
& f: R \rightarrow R, f(x)=x^2-9 \\
& g:[0, \infty) \rightarrow R, g(x)=\sqrt{x}
\end{aligned}

  1. State the range of \(f\).  (1 mark)

  2. Determine the rule for the equation and state the domain of the function \(f \circ g\).  (2 marks)

  3. Let \(h\) be the function \(h: D \rightarrow R, h(x)=x^2-9\).
  4. Determine the maximal domain, \(D\), such that \(g \circ h\) exists.  (1 marks)

Show Answers Only

a.    \([-9, \infty)\)

b.    \(f\circ g(x)=x-9, \text{Domain}\ [0, \infty)\)

c.    \((-\infty, -3)\cap (3, \infty)\)

Show Worked Solution

a.    \(\text{Range}\ \rightarrow\  [-9, \infty)\)

b.     \(f\circ g(x)\) \(=(g(x))^2-9\)
    \(=(\sqrt{x})^2-9\)
    \(=x-9\)

\(g(x)=\sqrt{x} \ \rightarrow x\ \text{must be }\geq 0\)

\(\therefore\ \text{Domain}\ f\circ g(x) \text{ is }[0, \infty)\)

c.     \(g\circ h(x)\) \(=\sqrt{h(x)}\)
    \(=\sqrt{x^2-9}\)

\(\text{For }g\circ h(x)\ \text{to exist}\ h(x)\geq 0\)

\(x\text{-intercepts for }h(x)\ \text{are } x=-3, 3\)

\(\text{and }h(x)\ \text{is positive for } x\leq -3\ \text{and }x\geq 3\)

\(\therefore\ \text{Maximal domain} = (-\infty, -3)\cap (3, \infty)\)

PHYSICS, M5 2020 VCE 4*

The Ionospheric Connection Explorer (ICON) space weather satellite, constructed to study Earth's ionosphere, was launched in October 2019. ICON will study the link between space weather and Earth's weather at its orbital altitude of 600 km above Earth's surface. Assume that ICON's orbit is a circular orbit. 

  1. Calculate the orbital radius of the ICON satellite.   (1 mark)

  2. Calculate the orbital period of the ICON satellite correct to three significant figures. Show your working.   (4 marks)

  3. Explain how the ICON satellite maintains a stable circular orbit without the use of propulsion engines.   (2 marks)

Show Answers Only

a.    \(r_{orbit}= 6.971 \times 10^6\ \text{m}\)

b.    \(5780\ \text{s}\)

c.   Stable circular orbit:

  • The icon satellite is only subject to the gravitational force of the Earth.
  • This force is a centripetal force of constant magnitude and hence the satellite maintains a stable circular orbit. 
Show Worked Solution

a.    Radius of orbit is equal to the altitude plus the radius of the Earth.

\(r_{orbit}=600\ 000\ \text{m} + 6.371 \times 10^6\ \text{m} = 6.971 \times 10^6\ \text{m}\)

 

b.     \(\dfrac{r^3}{T^2}\) \(=\dfrac{GM}{4\pi^2}\)
  \(T\) \(=\sqrt{\dfrac{4\pi^2r^3}{GM}}\)
    \(=\sqrt{\dfrac{4\pi^2 \times (6.971 \times 10^6)^3}{6.67 \times 10^{-11} \times 6 \times 10^{24}}}\)
    \(=5780\ \text{s}\ \ \text{(3 sig.fig.)}\)

 

c.   Stable circular orbit:

  • The icon satellite is only subject to the gravitational force of the Earth.
  • This force is a centripetal force of constant magnitude and hence the satellite maintains a stable circular orbit. 
♦♦♦ Mean mark (c) 23%.

Calculus, MET2 EQ-Bank 2

Jac and Jill have built a ramp for their toy car. They will release the car at the top of the ramp and the car will jump off the end of the ramp.

The cross-section of the ramp is modelled by the function \(f\), where

\(f(x)= \begin{cases}\displaystyle \ 40 & 0 \leq x<5 \\ \dfrac{1}{800}\left(x^3-75 x^2+675 x+30\ 375\right) & 5 \leq x \leq 55\end{cases}\)

\(f(x)\) is both smooth and continuous at \(x=5\).

The graph of  \(y=f(x)\)  is shown below, where \(x\) is the horizontal distance from the start of the ramp and \(y\) is the height of the ramp. All lengths are in centimetres.

  1. Find \(f^{\prime}(x)\) for \(0<x<55\).   (2 marks)

  2.   i. Find the coordinates of the point of inflection of \(f\).   (1 mark)

  3.  ii. Find the interval of \(x\) for which the gradient function of the ramp is strictly increasing.   (1 mark)

  4. iii. Find the interval of \(x\) for which the gradient function of the ramp is strictly decreasing.  (1 mark)

Jac and Jill decide to use two trapezoidal supports, each of width \(10 cm\). The first support has its left edge placed at \(x=5\) and the second support has its left edge placed at \(x=15\). Their cross-sections are shown in the graph below.

  1. Determine the value of the ratio of the area of the trapezoidal cross-sections to the exact area contained between \(f(x)\) and the \(x\)-axis between \(x=5\) and \(x=25\). Give your answer as a percentage, correct to one decimal place.   (3 marks)

  2. Referring to the gradient of the curve, explain why a trapezium rule approximation would be greater than the actual cross-sectional area for any interval \(x \in[p, q]\), where \(p \geq 25\).   (1 mark)

  3. Jac and Jill roll the toy car down the ramp and the car jumps off the end of the ramp. The path of the car is modelled by the function \(P\), where

      1. \(P(x)=\begin{cases}f(x) & 0 \leq x \leq 55 \\ g(x) & 55<x \leq a\end{cases}\)
  4. \(P\) is continuous and differentiable at \(x=55, g(x)=-\frac{1}{16} x^2+b x+c\), and \(x=a\) is where the car lands on the ground after the jump, such that \(P(a)=0\).
    1. Find the values of \(b\) and \(c\).   (2 marks)

    2. Determine the horizontal distance from the end of the ramp to where the car lands. Give your answer in centimetres, correct to two decimal places.   (1 mark)

Show Answers Only

a.    \(f^{\prime}(x)= \begin{cases}\displaystyle \ 0 & 0 < x<5 \\ \dfrac{1}{800}\left(3x^2-150 x+675 \right) & 5 \leq x < 55\end{cases}\)

b.i   \((25, 20)\)

b.ii  \([25, 55]\)

b.iii \([5, 25]\)

c.    \(98.1\%\)

d.    \(\text{In the interval }x\in [25, q]\text{ the curve is concave up.}\)

\(\therefore\ \text{The sloped edges of the trapeziums would be above }f(x)\)

\(\text{making the trapezium rule approximation greater than the}\)

\(\text{actual area.}\)

e.i   \(b=8.75, c=-283.4375\)

e.ii  \(34.10\ \text{cm}\)

Show Worked Solution

a.    \(\text{Using CAS: Define f(x) then}\)

\(\text{OR}\quad f^{\prime}(x)= \begin{cases}\displaystyle \ 0 & 0 < x<5 \\ \dfrac{1}{800}\left(3x^2-150 x+675 \right) & 5 \leq x < 55\end{cases}\)

b.i   \(\text{Using CAS: Find }f^”(x)\ \text{then solve }=0\)

\(\therefore\ \text{Point of inflection at }(25, 20)\)
  

b.ii  \(\text{Gradient function strictly increasing for }x\in [25, 55]\)

b.iii \(\text{Gradient function strictly decreasing for }x\in [5, 25]\)

c.    \(\text{Using CAS:}\)

\(\text{Area}\) \(=\dfrac{h}{2}\Big(f(5)+2f(15)+f(25)\Big)\)
  \(=\dfrac{10}{2}\Big(40+2\times 33.75+20\Big)=637.5\)

  

\(\therefore\ \text{Estimate : Exact}\) \(=637.5:650\)
  \(=\dfrac{637.5}{650}\times 100\)
  \(=98.0769\%\approx 98.1\%\)

    
d.    \(\text{In the interval }x\in [25, q]\text{ the curve is concave up.}\)

\(\therefore\ \text{The sloped edges of the trapeziums would be above }f(x)\)

\(\text{making the trapezium rule approximation greater than the}\)

\(\text{actual area.}\)
 

e.i   \(f(55)=\dfrac{35}{4}\ \text{(Using CAS)}\)

\(\text{and }f(55)=g(55)\rightarrow g(55)=\dfrac{35}{4}\)

\(\therefore -\dfrac{1}{16}\times 55^2+55b+c\) \(=\dfrac{35}{4}\)
\(c\) \(=\dfrac{35}{4}+\dfrac{3025}{16}-55b\)
\(c\) \(=\dfrac{3165}{16}-55b\quad (1)\)

  
\(g'(x)=-\dfrac{x}{8}+b\)

\(\text{and }g'(55)=f'(55)\)

\(\rightarrow\ f'(55)=\dfrac{1}{800}(3\times 55^2-150\times 55+675)=\dfrac{15}{8}\)

\(\rightarrow\ -\dfrac{55}{8}+b=\dfrac{15}{8}\)

\(\therefore b=\dfrac{35}{4}\quad (2)\)

\(\text{Sub (2) into (1)}\)

\(c=\dfrac{3165}{16}-55\times\dfrac{35}{4}\)

\(c=-\dfrac{4535}{16}\)

\(\therefore\ b=\dfrac{35}{4}\ \text{or}\ 8.75 , c=-\dfrac{4535}{16}\ \text{or}\ -283.4375\)
 

e.ii  \(\text{Using CAS: Solve }g(x)=0|x>55\)

\(\text{Horizontal distance}=\sqrt{365}+70-55=34.104\dots\approx 34.10\ \text{cm (2 d.p.)}\)

PHYSICS, M6 2020 VCE 3

Electron microscopes use a high-precision electron velocity selector consisting of an electric field, \(E\), perpendicular to a magnetic field, \(B\).

Electrons travelling at the required velocity, \(v_0\), exit the aperture at point \(\text{Y}\), while electrons travelling slower or faster than the required velocity, \(v_0\), hit the aperture plate, as shown in Figure 2.
 

  1. Show that the velocity of an electron that travels straight through the aperture to point \(\text{Y}\) is given by  \( v_{0} \) = \( \dfrac{E}{B}\).   (1 mark)

  2. Calculate the magnitude of the velocity, \(v_0\), of an electron that travels straight through the aperture to point \(\text{Y}\) if  \(E\) = 500 kV m\(^{-1}\)  and  \(B\) = 0.25 T. Show your working.   (2 marks)

  3.  i. At which of the points – \(\text{X, Y}\), or \(\text{Z}\) – in Figure 2 could electrons travelling faster than \(v_0\) arrive?   (1 mark)

  4. ii. Explain your answer to part c.i.   (2 marks)

Show Answers Only

a.    Find \(v_0\) where forces due to magnetic and electric field are balanced

\(F_E\) \(=F_B\)  
\(qE\) \(=qv_0B\)  
\(v_0\) \(=\dfrac{E}{B}\)  

 
b.    \(v_0=2 \times 10^6\ \text{ms}^{-1}\)

c.i.    Point \(\text{Z.}\)

c.ii.  When electrons travel faster than \(v_0\):

  • The force due to the magnetic field will increase but the force due to the electric field will remain unchanged. 
  • Therefore, there will be a net force acting on the electron due to the magnetic force being greater than the electric force.
  • Using the right-hand rule, the force on the electron due to the magnetic field is down the page, hence the electron will arrive at point \(\text{Z.}\)
Show Worked Solution

a.    Find \(v_0\) where forces due to magnetic and electric field are balanced

\(F_E\) \(=F_B\)  
\(qE\) \(=qv_0B\)  
\(v_0\) \(=\dfrac{E}{B}\)  
♦ Mean mark (a) 46%.

b.    \(v_0=\dfrac{E}{B}=\dfrac{500\ 000}{0.25}=2 \times 10^6\ \text{ms}^{-1}\)
 

c.i.    Point \(\text{Z.}\)
 

c.ii.  When electrons travel faster than \(v_0\):

  • The force due to the magnetic field will increase but the force due to the electric field will remain unchanged. 
  • Therefore, there will be a net force acting on the electron due to the magnetic force being greater than the electric force.
  • Using the right-hand rule, the force on the electron due to the magnetic field is down the page, hence the electron will arrive at point \(\text{Z.}\)
♦ Mean mark (c.i.) 40%.
♦♦♦ Mean mark (c.ii.) 15%.
COMMENT: Students needed to include what would happen to the electric force.

PHYSICS, M8 2020 VCE 18 MC

Quantised energy levels within atoms can best be explained by

  1. electrons behaving as individual particles with different energies.
  2. electrons behaving as waves, with each energy level representing a diffraction pattern.
  3. protons behaving as waves, with only standing waves at particular wavelengths allowed.
  4. electrons behaving as waves, with only standing waves at particular wavelengths allowed.
Show Answers Only

\(D\)

Show Worked Solution
  • This was explained by de Broglie’s work in determining electrons (all matter) can exhibit wave properties.
  • Quantised energy levels in the atom correspond to where electrons form standing waves due to there being an integer number of electron wavelengths.

\(\Rightarrow D\)

PHYSICS, M7 2020 VCE 15 MC

\(\text{I}\) The energy of a light wave increases with increasing amplitude.
\(\text{II}\) The energy of a light wave increases with increasing frequency.
\(\text{III}\) The energy of a light wave increases with decreasing wavelength.

Which of the statements above about the energy of light waves is correct?

  1. \(\text{III}\) only
  2. \(\text{I}\) and \(\text{II}\) only
  3. \(\text{I}\) and \(\text{III}\) only
  4. all of the statements are correct
Show Answers Only

\(D\)

Show Worked Solution
  • Statement \(\text{I}\) is correct when considering the wave-model of light. The higher the amplitude, the further away the displacement of the wave from the origin which corresponds to higher energy.
  • Statements \(\text{II}\) and \(\text{III}\) are correct when considering the particle model of light.
  • The energy of a light photon, \(E=hf=\dfrac{hc}{\lambda}\). It follows that increasing the frequency or decreasing the wavelength both lead to an increase in energy.

\(\Rightarrow D\)

PHYSICS, M8 2020 VCE 13 MC

Matter is converted to energy by nuclear fusion in stars.

If the star Alpha Centauri converts mass to energy at the rate of 6.6 × 10\(^9\) kg s\(^{-1}\), then the power generated is closest to

  1. \(2.0 \times 10^{18}\ \text{W}\)
  2. \(2.0 \times 10^{18}\ \text{J}\)
  3. \(6.0 \times 10^{26}\ \text{W}\)
  4. \(6.0 \times 10^{26}\ \text{J}\)
Show Answers Only

\(C\)

Show Worked Solution
  • The energy produced by the star each second is the power generated by the star.
  • \(E=mc^2=6.6 \times 10^9 \times (3 \times 10^8)^2=6 \times 10^{26}\ \text{Js}^{-1}=6 \times 10^{26}\ \text{W}\)

\(\Rightarrow C\)

Mean mark 56%.

Statistics, SPEC2 2022 VCAA 6

A company produces soft drinks in aluminium cans.

The company sources empty cans from an external supplier, who claims that the mass of aluminium in each can is normally distributed with a mean of 15 grams and a standard deviation of 0.25 grams.

A random sample of 64 empty cans was taken and the mean mass of the sample was found to be 14.94 grams.

Uncertain about the supplier's claim, the company will conduct a one-tailed test at the 5% level of significance. Assume that the standard deviation for the test is 0.25 grams.

  1. Write down suitable hypotheses \(H_0\) and \(H_1\) for this test.   (1 mark)

  2. Find the \(p\) value for the test, correct to three decimal places.   (1 mark)

  3. Does the mean mass of the random sample of 64 empty cans support the supplier's claim at the 5% level of significance for a one-tailed test? Justify your answer.   (1 mark)

  4. What is the smallest value of the mean mass of the sample of 64 empty cans for \(H_0\) not to be rejected? Give your answer correct to two decimal places.   (1 mark)

The equipment used to package the soft drink weighs each can after the can is filled. It is known from past experience that the masses of cans filled with the soft drink produced by the company are normally distributed with a mean of 406 grams and a standard deviation of 5 grams.

  1. What is the probability that the masses of two randomly selected cans of soft drink differ by no more than 3 grams? Give your answer correct to three decimal places.   (2 marks)

Show Answers Only

a.   \(H_0: \mu=15, \quad H_1: \mu<15\)

b.   \(p=0.027\)

c.    \(\text{Since \(\ p<0.05\), claim is not supported.}\)

d.   \(a=14.95\)

e.  \(\text{Pr}(-3<D<3)=0.329\)

Show Worked Solution

a.    \(H_0: \mu=15, \quad H_1: \mu<15\)
 

b.    \(\mu=15, \ \ \sigma=0.25\)

\(\bar{x}=14.94, \ \sigma_{\bar{x}}=\dfrac{0.25}{\sqrt{64}}=0.03125\)

\(\text{By CAS:}\)

\(p=\text{Pr}\left(\bar{X}<14.94 \mid \mu=15\right)=0.027\ \text {(3 d.p.)}\)
 

c.    \(\text{Since \(\ p<0.05\), claim is not supported.}\)

\(\text{Evidence is against \(H_0\)  at the \(5 \%\) level.}\)
 

d.    \(\text{Pr}\left(\bar{X}<a \mid \mu=15\right)>0.05\)

\(\text{Pr}\left(Z<\dfrac{a-15}{0.03125}\right)>0.05\)

\(\text{By CAS:}\ \ a=14.95\ \text{(2 d.p.)}\)
 

e.    \(\text{Let}\ \ M=\ \text{mass of one can}\)

\(M \sim N\left(406,5^2\right)\)

\(E\left(M_1\right)=E\left(M_2\right)=\mu=406\)

\(\text {Let}\ \ D=M_1-M_2\)

\(E(D)=406-406=0\)

\(\text{Var}(D)=1^2 \times \text{Var}\left(M_1\right)+(-1)^2 \times  \text{Var}\left(M_2\right)=50\)

\(\sigma_D=\sqrt{50}\)

\(D \sim N\left(0,(\sqrt{50})^2\right)\)

\(\text{By CAS: Pr\((-3<D<3)=0.329\) (3 d.p.) }\)

♦ Mean mark (e) 46%.

Vectors, SPEC2 2022 VCAA 4

A student is playing minigolf on a day when there is a very strong wind, which affects the path of the ball. The student hits the ball so that at time  \(t=0\) seconds it passes through a fixed origin \(O\). The student aims to hit the ball into a hole that is 7 m from \(O\). When the ball passes through \(O\), its path makes an angle of \(\theta\) degrees to the forward direction, as shown in the diagram below.
 

The path of the ball \(t\) seconds after passing through \(O\) is given by

\(\underset{\sim}{\text{r}}(t)=\dfrac{1}{2} \sin \left(\dfrac{\pi t}{4}\right) \underset{\sim}{\text{i}}+2 t \underset{\sim}{\text{j}}\)  for  \(t \in[0,5]\)

where \(\underset{\sim}{i}\) is a unit vector to the right, perpendicular to the forward direction, \(\underset{\sim}{j}\) is a unit vector in the forward direction and displacement components are measured in metres.

  1. Find \(\theta\) correct to one decimal place.   (2 marks)

  2.  i. Find the speed of the ball as it passes through \(O\). Give your answer in metres per second, correct to two decimal places.   (2 marks)

  3. ii. Find the minimum speed of the ball, in metres per second, and the time, in seconds, at which this minimum speed occurs.   (2 marks)

  4. Find the minimum distance from the ball to the hole. Give your answer in metres, correct to three decimal places.   (3 marks)

  5. How far does the ball travel during the first four seconds after passing through \(O\) ? Give your answer in metres, correct to three decimal places.   (2 marks)

Show Answers Only

a.   \(\theta =11.1^{\circ}\)

b.i.  \(2.04\ \text{ms}^{-1}\)

b.ii.  \(\text{Minimum speed} =2 \text{ ms} ^{-1}\)

c.   \(\text {Minimum }\abs{d}=0.188\ \text{m}\)

d.   \(8.077\ \text{m}\)

Show Worked Solution

a.    \(r(t)=\dfrac{1}{2}\,\sin \left(\dfrac{\pi t}{4}\right) \underset{\sim}{i}+2 t \underset{\sim}{j}\)

\(\dot{r}(t)=\dfrac{\pi}{8}\, \cos \left(\dfrac{\pi t}{4}\right) \underset{\sim}{i}+2 \underset{\sim}{j}\)

\(\dot{r}(0)=\dfrac{\pi}{8}\,\underset{\sim}{i}+2\underset{\sim}{j}\)

\begin{aligned}
\tan (90-\theta) & =\dfrac{2}{\frac{\pi}{8}} \\
90-\theta & =\tan ^{-1}\left(\dfrac{16}{\pi}\right)=78.9^{\circ} \\
\theta & =11.1^{\circ}
\end{aligned}

♦ Mean mark (a) 46%.
b.i.     \(\text{Speed}\) \(=\abs{\dot{r}(0)}\)
    \(=\sqrt{\left(\dfrac{\pi}{8}\right)^2+2^2}\)
    \(=2.04\ \text{ms}^{-1}\ \text{(2 d.p.)}\)

 

b.ii.  \(\abs{\dot{r}}=\sqrt{\left(\dfrac{\pi}{8}\right)^2 \times \cos ^2\left(\dfrac{\pi t}{4}\right)+4}\)

 \(\abs{\dot{r}}\text { is a minimum when}\ \ \cos ^2\left(\dfrac{\pi t}{4}\right)=0\)

\(\Rightarrow t=2\)

\(\therefore\ \text{Minimum speed} =\sqrt{4}=2 \text{ ms} ^{-1}\)
 

c.    \(\text{Ball position}\ \Rightarrow \ \underset{\sim}{r}=\dfrac{1}{2}\,\sin \left(\dfrac{\pi t}{4}\right)\underset{\sim}{i}+2 \, t\underset{\sim}{j}\)

\(\text{Hole position}\ \Rightarrow \ \underset{\sim}{h}=0 \underset{\sim}{i}+7\underset{\sim}{j}\)

\(\underset{\sim}{d}=\underset{\sim}{r}-\underset{\sim}{h}=\dfrac{1}{2}\, \sin \left(\dfrac{\pi t}{4}\right) \underset{\sim}{i}+(2 t-7)\underset{\sim}{j}\)

\(\abs{\underset{\sim}{d}}=\sqrt{\left(\dfrac{1}{2}\right)^2 \sin ^2\left(\dfrac{\pi t}{4}\right)+(2 t-7)^2}\)

\(\text {Minimum }\abs{d}=0.188\text{ m (3 d.p.)}\)

♦ Mean mark (c) 45%.
d.     \(\text{Distance}\) \(=\displaystyle \int_0^4\left(\frac{\pi}{8}\, \cos \left(\frac{\pi t}{4}\right)\right)^2+4\, dt\)
    \(=8.077\ \text{m  (3 d.p.)}\)
♦ Mean mark (d) 48%.

CHEMISTRY, M3 EQ-Bank 16

Write balanced chemical equations for each of the following reactions (states of matter are not required).

  1. The reaction between lithium hydroxide and hydrochloric acid.   (1 mark)

  2. The reaction between potassium carbonate and hydrochloric acid.   (1 mark)

Show Answers Only

a.    \(\ce{LiOH + HCl -> LiCl + H2O}\)

b.    \(\ce{K2CO3 + 2HCl -> 2KCl + CO2 + H2O}\)

Show Worked Solution

a.    Acid-Base

\[\ce{LiOH + HCl -> LiCl + H2O}\]

b.    Acid-carbonate

\[\ce{K2CO3 + 2HCl -> 2KCl + CO2 + H2O}\]

CHEMISTRY, M3 EQ-Bank 15

Write balanced chemical equations for each of the following reactions (states of matter are not required).

  1. The reaction between zinc and sulfuric acid.   (1 mark)

  2. The decomposition of calcium carbonate.   (1 mark)

  3. The incomplete combustion of ethane \(\ce{(C2H6)}\) with a limited amount of oxygen.   (1 mark)

Show Answers Only

a.    \(\ce{Zn + H2SO4 -> ZnSO4 + H2}\)

b.    \(\ce{CaCO3 -> CaO + CO2}\)

c.    \(\ce{2C2H6 + 5O2 -> 4CO + 6H2O}\)

Show Worked Solution

a.    Active metal and acid

\[\ce{Zn + H2SO4 -> ZnSO4 + H2}\]

b.    Decomposition

\[\ce{CaCO3 -> CaO + CO2}\]

c.    Combustion

\[\ce{2C2H6 + 5O2 -> 4CO + 6H2O}\]

CHEMISTRY, M3 EQ-Bank 14

Write balanced chemical equations for each of the following reactions (states of matter are not required).

  1. The reaction between potassium hydroxide and sulfuric acid.   (1 mark)
  2. The reaction between sodium bicarbonate \(\ce{(NaHCO3)}\) and acetic acid \(\ce{(CH3COOH)}\).   (1 mark)
Show Answers Only

a.    \(\ce{2KOH + H2SO4 -> K2SO4 + 2H2O}\)

b.    \(\ce{NaHCO3 + CH3COOH -> CH3COONa + CO2 + H2O}\)

Show Worked Solution

a.   Acid-base 

\(\ce{2KOH + H2SO4 -> K2SO4 + 2H2O}\)

b.   Acid-carbonate 

\(\ce{NaHCO3 + CH3COOH -> CH3COONa + CO2 + H2O}\)

CHEMISTRY, M3 EQ-Bank 13

Write balanced chemical equations for each of the following reactions (states of matter are not required).

  1. The reaction between magnesium and hydrochloric acid.   (1 mark)

  2. The decomposition of ammonia.   (1 mark)

  3. The incomplete combustion of butane (\(\ce{C4H10}\)) with a limited amount of oxygen.   (1 mark)

Show Answers Only

a.    \(\ce{Mg + 2HCl -> MgCl2 + H2}\)

b.    \(\ce{2NH3 -> N2 + 3H2}\)

c.    \(\ce{2C4H10 + 9O2 -> 8CO + 10H2O}\)

Show Worked Solution

a.   Reaction between active metal and acid

\(\ce{Mg + 2HCl -> MgCl2 + H2}\)

b.   Decomposition

\(\ce{2NH3 -> N2 + 3H2}\)

c.   Combustion

\(\ce{2C4H10 + 9O2 -> 8CO + 10H2O}\)

CHEMISTRY, M3 EQ-Bank 12

A student stirs 2.80 g of silver \(\text{(I)}\) nitrate powder into 250.0 mL of 1.00 mol L\(^{-1}\) sodium hydroxide solution until it is fully dissolved. A reaction occurs and a precipitate appears.

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

  2. Calculate the theoretical mass of precipitate that will be formed.   (3 marks)

The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).

  1. Identify one scientific reason why the precipitate mass was too high and suggest an improvement to the experimental method which would eliminate this error.   (2 marks)

Show Answers Only

a.    \(\ce{AgNO3(aq) + NaOH(aq) -> AgOH(s) + NaNO3(aq)}\)

b.    \(2.06 \text{ g}\)

c.    Possible reasons for the higher mass:

  • Presence of excess solution in the filter paper or incomplete drying.
  • Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.

Experimental adjustment to eliminate error:

  • Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.
  • Dry precipitate completely in the incubator before weighing.
Show Worked Solution

a.    \(\ce{AgNO3(aq) + NaOH(aq) -> AgOH(s) + NaNO3(aq)}\)
 

b.   \(\ce{MM(AgNO3) = 107.9 + 14.01 + 16.00 \times 3 = 169.91}\)

\(\ce{n(AgNO3) = \dfrac{\text{m}}{\text{MM}} = \dfrac{2.80}{169.91} = 0.01648 \text{ mol}}\)

\(\ce{n(NaOH) = c \times V = 1.00 \times 0.250 = 0.250 \text{ mol}}\)

\(\Rightarrow \ce{AgNO3}\ \text{is the limiting reagent}\)

\(\ce{n(AgOH) = n(AgNO3) = 0.01648 \text{ mol}}\)

\(\ce{m(AgOH) = n \times MM = 0.01648 \times (107.9 + 16.00 + 1.008) = 2.06 \text{ g}}\)
 

c.    Possible reasons for the higher mass:

  • Presence of excess solution in the filter paper or incomplete drying.
  • Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.

Experimental adjustment to eliminate error:

  • Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.
  • Dry precipitate completely in the incubator before weighing.