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BIOLOGY, M2 EQ-Bank 7 MC

Consider the following statements about alveoli in mammals and the internal structure of plant leaves:

\(\text{I.}\)    Both structures have a thin, moist surface to facilitate gas exchange.  
\(\text{II.}\)    Alveoli exchange gases with blood, while leaves exchange gases with air spaces.  
\(\text{III.}\)    Alveoli are specialised for both oxygen uptake and carbon dioxide release,
while leaves are specialised only for oxygen release.		  
\(\text{IV.}\)    Both structures are kept moist to facilitate the diffusion of gases.  

 
Which combination of statements is correct?

  1. \(\text{I}\) and \(\text{II}\) only
  2. \(\text{I, II}\) and \(\text{IV}\) only
  3. \(\text{II, III}\), and \(\text{IV}\) only
  4. \(\text{I, II, III}\) and \(\text{IV}\) only
Show Answers Only

\(B\)

Show Worked Solution

Consider each statement.

\(\text{I:}\) Correct. Both alveoli and the internal structure of plant leaves have a thin, moist surface to facilitate gas exchange.

\(\text{II:}\) Correct. Alveoli exchange gases directly with blood in the surrounding capillaries, while leaves exchange gases with air spaces within their internal structure (intercellular spaces). 

\(\text{III:}\) Incorrect. Both alveoli and leaves are specialised for both oxygen uptake and carbon dioxide release.

\(\text{IV:}\) Correct. Both structures are kept moist to facilitate the diffusion of gases. 

\(\Rightarrow B\)

BIOLOGY, M2 EQ-Bank 5

Explain how the respiratory structures of a terrestrial mammal and a bony fish are adapted to their respective environments.

In your answer, discuss how the structure of each system maximises gas exchange.   (4 marks)

Show Answers Only

→ The main respiratory organ in terrestrial mammals is the lungs, while in bony fish it’s the gills.

→ Lungs have a large internal surface area created by millions of alveoli.

→ This highly efficient structure, combined with the diaphragm-driven breathing mechanism, allows for rapid oxygen uptake and carbon dioxide release. This ensures that all cells receive adequate gas exchange despite the lower oxygen content in air compared to water.

→ Gills consist of many thin filaments that spread out in the water to provide a large surface area for gas exchange.

→ Fish are able to take in water through their mouths and force the water over their gills. This creates a consistent one-way flow of oxygen-rich water for gas exchange.

→ Fish use a counter-current flow in their gills to maximise oxygen uptake, which is not necessary in lungs. 

Show Worked Solution

→ The main respiratory organ in terrestrial mammals is the lungs, while in bony fish it’s the gills.

→ Lungs have a large internal surface area created by millions of alveoli.

→ This highly efficient structure, combined with the diaphragm-driven breathing mechanism, allows for rapid oxygen uptake and carbon dioxide release. This ensures that all cells receive adequate gas exchange despite the lower oxygen content in air compared to water.

→ Gills consist of many thin filaments that spread out in the water to provide a large surface area for gas exchange.

→ Fish are able to take in water through their mouths and force the water over their gills. This creates a consistent one-way flow of oxygen-rich water for gas exchange.

→ Fish use a counter-current flow in their gills to maximise oxygen uptake, which is not necessary in lungs. 

BIOLOGY, M2 EQ-Bank 2

Discuss one similarity and one difference between using transmission electron microscopy (TEM) versus scanning electron microscopy (SEM) in the study of plant structures.   (2 marks)

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Similarities (include one of the following):

→ Both transmission electron microscopy (TEM) and scanning electron microscopy (SEM) use beams of electrons rather than light to create high-resolution images.

→ Extremely high magnification is achievable in both TEM and SEM.

→ Specimens for both technologies are very thin sections of plants made up of non-living cells.

→ Specimens for both technologies need to be in a low pressure environment or vacuum.
  

Difference (include one of the following):

→ TEMs produce two-dimensional images while SEMs produce three-dimensional images.

→ Magnification and resolution of images if greater in TEM vs SEM.

Show Worked Solution

Similarities (include one of the following):

→ Both transmission electron microscopy (TEM) and scanning electron microscopy (SEM) use beams of electrons rather than light to create high-resolution images.

→ Extremely high magnification is achievable in both TEM and SEM.

→ Specimens for both technologies are very thin sections of plants made up of non-living cells.

→ Specimens for both technologies need to be in a low pressure environment or vacuum.
  

Difference (include one of the following):

→ TEMs produce two-dimensional images while SEMs produce three-dimensional images.

→ Magnification and resolution of images if greater in TEM vs SEM.

BIOLOGY, M2 EQ-Bank 10

"The circulatory and excretory systems in humans are intricately linked, each depending on the other for optimal function."

  1. Describe two ways in which the circulatory system and the excretory system of humans exhibit interdependence.   (2 marks)
  2. Provide one example of a disorder that could affect both systems.   (2 marks)
Show Answers Only

a.   Examples of system interdependence:

→ The circulatory system delivers blood to the kidneys, allowing waste products and excess water to be filtered out, which is crucial for the excretory system’s function.

→ Conversely, the kidneys produce a hormone that stimulates red blood cell production in the bone marrow, directly influencing the circulatory system.
 

b.   Disorder that can affect both systems:

→ High blood pressure (hypertension) is a disorder that affects both the circulatory and excretory systems.

→ Persistent high blood pressure can damage the blood vessels in the kidneys, impairing their ability to filter blood effectively and clear waste.

→ This kidney damage can lead to a toxic environment that, in turn, adversely affects the function of the circulatory system.

Show Worked Solution

a.   Examples of system interdependence:

→ The circulatory system delivers blood to the kidneys, allowing waste products and excess water to be filtered out, which is crucial for the excretory system’s function.

→ Conversely, the kidneys produce a hormone that stimulates red blood cell production in the bone marrow, directly influencing the circulatory system.
 

b.   Disorder that can affect both systems:

→ High blood pressure (hypertension) is a disorder that affects both the circulatory and excretory systems.

→ Persistent high blood pressure can damage the blood vessels in the kidneys, impairing their ability to filter blood effectively and clear waste.

→ This kidney damage can lead to a toxic environment that, in turn, adversely affects the function of the circulatory system.

CHEMISTRY, M4 EQ-Bank 5 MC

Determine the enthalpy change  \((\Delta H_4)\)  for the following reaction

\(\ce{CH4(g) -> C(s) + 2H2(g)}\)

Using the given chemical reactions and their associated enthalpy changes

\(1.\)   \(\ce{2CH4(g) + 4O2(g) -> 2CO2(g) +4H2O(l)}\)    \(\Delta H_1=-1780\ \text{kJ mol}^{-1}\)
\(2.\)   \(\ce{C(s) + O2 (g) -> CO2(g)}\)    \(\Delta H_2=-393\ \text{kJ mol}^{-1}\)
\(3.\)   \(\ce{2H2(g) + O2(g) -> 2H2O(l)}\)     \(\Delta H_3=-572\ \text{kJ mol}^{-1}\)

 

  1. \(-75\ \text{kJ mol}^{-1}\)
  2. \(75\ \text{kJ mol}^{-1}\)
  3. \(-485\ \text{kJ mol}^{-1}\)
  4. \(485\ \text{kJ mol}^{-1}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\frac{1}{2} \times\ \text{Equation 1:}\)

\(\ce{CH4(g) + 2O2(g) -> CO2(g) +2H2O(l)}\)    \(\dfrac{1}{2}\Delta H_1=-890\ \text{kJ mol}^{-1}\)

\(\text{Reverse Equation 2 and Equation 3:}\)

\(\ce{CO2(g) -> C(s) + O2(g)}\)    \(-\Delta H_2=393\ \text{kJ mol}^{-1}\)

\(\ce{2H2O(l) -> 2H2(g) + O2(g)}\)    \(-\Delta H_3=572\ \text{kJ mol}^{-1}\)
 

\(\text{Add the equations and their}\ \Delta H\ \text{values and cancel any reactants and products:}\)

\( \ce{CH4(g)} + \cancel{ \ce{2O2(g)}}\) \( \rightarrow \cancel{\ce{CO2(g)}} + \cancel{ \ce{H2O(l)}}\)    \(\dfrac{1}{2}\Delta H_1=-890\ \text{kJmol}^{-1}\)
\(\ce{2C(s)} +\cancel{\ce{2O2(g)}}\) \(\rightarrow \cancel{ \ce{2CO2(g)}}\)     \(-\Delta H_2=393\ \text{kJ mol}^{-1}\)
\(\ce{H2(g)} + \cancel{\ce{\frac{1}{2}O2(g)}}\) \(\rightarrow \cancel{\ce{H2O(l)}}\)    \(-\Delta H_3=572\ \text{kJ mol}^{-1}\)
\(\ce{CH4(g)}\) \(\ce{\rightarrow C(s) + 2H2(g)}\)    \(\Delta H_4=75\ \text{kJ mol}^{-1}\)
 
\(\Rightarrow B\)

BIOLOGY, M2 EQ-Bank 9

  1. Identify two different types of tissues found in a leaf.   (1 mark)
  2. Discuss how the overall structure of the leaf as an organ enhances its ability to perform photosynthesis.   (2 marks)
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a.   Types of tissues can include (choose two):

→ Epidermis, xylem and ground tissue.
 

b.   Leaf structure and photosynthesis:

→ A leaf’s broad, flat shape maximises the surface area exposed to sunlight, allowing for efficient light capture.

→ The epidermis on the upper surface is often transparent, allowing light to penetrate to the photosynthetic tissues beneath, while also providing protection and controlling water loss.

→ The xylem, part of the vascular tissue network throughout the leaf, ensures efficient transport of water and minerals to photosynthetic cells, supporting the process of photosynthesis.

Show Worked Solution

a.   Types of tissues can include (choose two):

→ Epidermis, xylem and ground tissue.
 

b.   Leaf structure and photosynthesis:

→ A leaf’s broad, flat shape maximises the surface area exposed to sunlight, allowing for efficient light capture.

→ The epidermis on the upper surface is often transparent, allowing light to penetrate to the photosynthetic tissues beneath, while also providing protection and controlling water loss.

→ The xylem, part of the vascular tissue network throughout the leaf, ensures efficient transport of water and minerals to photosynthetic cells, supporting the process of photosynthesis.

BIOLOGY, M2 EQ-Bank 8

Epidermal tissues in plants can be compared to the epithelium in animals.

  1. Describe one structural similarity between these tissue types.   (1 mark)

  2. Explain one functional similarity they share.   (1 mark)

  3. Discuss how both tissues contribute to their respective organism's interaction with the environment.   (2 marks)

Show Answers Only

a.   Structural similarity could include one of the following:

→ Both plant epidermal tissue and animal epithelium form a single layer of tightly packed cells that cover the outer surfaces of the organism.

→ Both types of tissue are typically attached to a basement membrane that separates them from underlying tissues.
 

b.   Functional similarity could include one of the following:

→ Both tissue types serve as a protective barrier, shielding the underlying tissues from physical damage.

→ Both tissue types can provide protection from pathogens and excessive water loss.
 

c.   Interaction with the environment:

→ In plants, the epidermis regulates gas exchange through stomata and may produce structures like trichomes for protection or water conservation.

→ Similarly, in animals, the epithelium can be specialised for absorption, allowing the organism to respond to various environmental stimuli.

Show Worked Solution

a.   Structural similarity could include one of the following:

→ Both plant epidermal tissue and animal epithelium form a single layer of tightly packed cells that cover the outer surfaces of the organism.

→ Both types of tissue are typically attached to a basement membrane that separates them from underlying tissues.
 

b.   Functional similarity could include one of the following:

→ Both tissue types serve as a protective barrier, shielding the underlying tissues from physical damage.

→ Both tissue types can provide protection from pathogens and excessive water loss.
 

c.   Interaction with the environment:

→ In plants, the epidermis regulates gas exchange through stomata and may produce structures like trichomes for protection or water conservation.

→ Similarly, in animals, the epithelium can be specialised for absorption, allowing the organism to respond to various environmental stimuli.

BIOLOGY, M2 EQ-Bank 5

  1. Define cell differentiation.   (1 mark)

  2. Explain one factor that influences cell differentiation.   (1 mark)

  3. Discuss why cell differentiation is crucial for the functioning of complex organisms.   (2 marks)

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a.   Cell differentiation:

→ The process by which a less specialised cell becomes a more specialised cell type with a specific structure and function.
 

b.   Factors that influence cell differentiation (include one of the following):

→ The timing of gene activation during development, which can determine what type of cell a stem cell becomes

→ The presence of specific growth factors or signalling molecules in the cell’s environment, which can activate or repress certain genes.

→ A cell’s position within an embryo or tissue, which exposes it to different chemical signals from neighbouring cells.
 

c.    Reasons cell differentiation is crucial for the functioning of complex organisms:

→ It allows for the development of specialised tissues and organs, each performing specific functions more efficiently than generalised cells could.

→ This specialisation enables the division of labor within the organism, leading to more complex and sophisticated biological processes.

→ Cell differentiation also allows multicellular organisms to develop intricate body plans and respond more effectively to their environment, ultimately enhancing their survival and reproductive success.

Show Worked Solution

a.   Cell differentiation:

→ The process by which a less specialised cell becomes a more specialised cell type with a specific structure and function.
 

b.   Factors that influence cell differentiation (include one of the following):

→ The timing of gene activation during development, which can determine what type of cell a stem cell becomes

→ The presence of specific growth factors or signalling molecules in the cell’s environment, which can activate or repress certain genes.

→ A cell’s position within an embryo or tissue, which exposes it to different chemical signals from neighbouring cells.
 

c.    Reasons cell differentiation is crucial for the functioning of complex organisms:

→ It allows for the development of specialised tissues and organs, each performing specific functions more efficiently than generalised cells could.

→ This specialisation enables the division of labor within the organism, leading to more complex and sophisticated biological processes.

→ Cell differentiation also allows multicellular organisms to develop intricate body plans and respond more effectively to their environment, ultimately enhancing their survival and reproductive success.

BIOLOGY, M2 EQ-Bank 5 MC

Consider the following statements about colonial organisms:

\(\text{I.}\)   They consist of genetically identical cells.  
\(\text{II.}\)   Each cell in the colony can perform all life functions independently.  
\(\text{III.}\)   They show a higher degree of cell specialization compared to multicellular organisms.  
\(\text{IV.}\)   They represent an evolutionary step between unicellular and multicellular organisms.  

 
Which combination of statements correctly describes colonial organisms?

  1. \(\text{I}\) and \(\text{II}\) only
  2. \(\text{I}\), \(\text{II}\) and \(\text{IV}\) only
  3. \(\text{II}\) and \(\text{III}\) only
  4. \(\text{I}\), \(\text{II}\) and \(\text{III}\) only
Show Answers Only

\(B\)

Show Worked Solution

Consider each statement:

\(\text{I.}\) Correct. Colonial organisms do consist of genetically identical cells, as they typically arise from the division of a single cell.

\(\text{II.}\) Correct. In most colonial organisms, each cell can perform all life functions independently, although they may benefit from living in close proximity to other cells.

\(\text{III.}\) Incorrect. Colonial organisms generally show less cell specialization than multicellular organisms, not more.

\(\text{IV.}\) Correct. Colonial organisms are often considered an evolutionary intermediate step between unicellular and multicellular organisms.

\(\Rightarrow B\)

CHEMISTRY, M4 EQ-Bank 2

  1. Identify Hess' Law.   (2 marks)

  2. Given these heats of formation, \(\Delta H_f\):

\begin{array} {|c|c|}
\hline \text{Chemical} & \Delta H_f \ \text{(kJ mol}^{-1}) \\
\hline  \ce{C2H6(g)} & -84.7 \\
\hline \ce{CO2(g)} & -393.5 \\
\hline \ce{H2O(l)} & -285.8 \\
\hline \ce{CO(g)} & -115 \\
\hline \end{array}

  1. Calculate \(\Delta H\) for the combustion of:

i.  1 mole of ethane (\(\ce{C2H6}\)) to produce carbon dioxide and water.   (2 marks)

ii. 1 mole of ethane (\(\ce{C2H6}\)) to produce carbon monoxide and water.   (2 marks)

Show Answers Only

a.    Hess’s Law:

→ The total enthalpy change for a reaction is the same, regardless of the route by which the chemical reaction occurs, provided the initial and final conditions are the same.

b.i.    \(\Delta H= -1560.8 \, \text{kJ/mol}\)

b.ii.   \(\Delta H = -1177.6 \, \text{kJ/mol}\)

Show Worked Solution

a.    Hess’s Law:

→ The total enthalpy change for a reaction is the same, regardless of the route by which the chemical reaction occurs, provided the initial and final conditions are the same.
 

b.i.   Enthalpy change:

→ complete combustion of 1 mole of ethane (\(\ce{C2H6}\)) to produce carbon dioxide and water.

\(\ce{C2H6(g) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)}\)

\(\Delta H = \Delta H_f \text{ products}-\Delta H_f \text{ reactants}\)

\(\Delta H = [2(-393.5) + 3(-285.8)]-[(-84.7)] = -1560.8 \, \text{kJ/mol}\)
 

b.ii.  Enthalpy change:

→ incomplete combustion of 1 mole of ethane (\(\ce{C2H6}\)) to produce carbon monoxide and water.

\(\ce{C2H6(g) + 2.5 O2(g) → 2 CO(g) + 3 H2O(l)}\)

\(\Delta H = \Delta H_f \text{ products}-\Delta H_f \text{ reactants}\)

\(\Delta H = [2(-110.5) + 3(-285.8)]-[(-84.7)] = -1177.6 \, \text{kJ/mol}\)

CHEMISTRY, M4 EQ-Bank 2 MC

Three equations and their \(\Delta H\) values are shown.

Equation 1:   \(\ce{C(s) + O2(g) → CO2(g)}\)     \(\Delta H_1 = -393.5 \, \text{kJ mol}^{-1}\)
Equation 2:   \(\ce{H2(g) + 1/2 O2(g) → H2O(l)}\)    \(\Delta H_2 = -285.8 \, \text{kJ mol}^{-1}\)
Equation 3:   \(\ce{CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)}\)    \(\Delta H_3 = -890.4 \, \text{kJ mol}^{-1}\)

 
Using this information, what is the \(\Delta H\) for the following reaction?

\(\ce{C(s) + 2 H2(g) → CH4(g)}\)

  1. \(+74.9 \, \text{kJ mol}^{-1}\)
  2. \(-74.9 \, \text{kJ mol}^{-1}\)
  3. \(+210.3 \, \text{kJ mol}^{-1}\)
  4. \(-210.3 \, \text{kJ mol}^{-1}\)
Show Answers Only

\(B\)

Show Worked Solution

Reverse Equation 3 to express \(\ce{CH4(g)}\) as a product:

Equation 3*: \(\ce{CO2(g) + 2 H2O(l) → CH4(g) + 2 O2(g)} \quad -\Delta H_1 = +890.4 \, \text{kJ mol}^{-1}\)

Double the coefficients of Equation 2 which also doubles \(\Delta H_2\).

Equation 2*: \(\ce{2H2(g) + O2(g) → 2H2O(l)} \qquad 2\Delta H_2 = -571.6 \, \text{kJ mol}^{-1}\)

Combine the corresponding enthalpies of Equation 1, 2* and 3* to find the total \(\Delta H\):

\(\Delta H = +890.4-393.5-571.6 = -74.9 \, \text{kJ mol}^{-1}\)

\(\Rightarrow B\)

CHEMISTRY, M4 EQ-Bank 1 MC

The combustion of methane \((\ce{CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)})\) occurs with the following bond energies:

    • \(\ce{C-H}:\ \text{412 kJ/mol}\)
    • \(\ce{O=O}:\ \text{498 kJ/mol}\)
    • \(\ce{C=O}:\ 799\ \text{kJ/mol}\)
    • \(\ce{O-H}:\ \text{463 kJ/mol}\)

Calculate the enthalpy change \((\Delta H)\) for this reaction:

  1. \(-802\ \text{kJ/mol}\)
  2. \(-482\ \text{kJ/mol}\)
  3. \(+482\ \text{kJ/mol}\)
  4. \(+802\ \text{kJ/mol}\)
Show Answers Only

\(A\)

Show Worked Solution

Energy required to break the reactant bonds:

Energy to break 4 \(\ce{(C-H)}\) bonds: \( 4 \times 412 = 1648 \, \text{kJ/mol}\)

Energy to break 2 \(\ce{(O=O)}\) bonds: \( 2 \times 498 = 996 \, \text{kJ/mol}\)

\(1648 + 996 = 2644 \, \text{kJ/mol} \)
 

Energy released in the formation of the product bonds:

Energy to form 2 \(\ce{(C=O)}\) bonds: \( 2 \times 799 = 1598 \, \text{kJ/mol}\)

Energy to form 4 \(\ce{(O-H)}\) bonds: \( 4 \times 463 = 1852 \, \text{kJ/mol}\)

\(1598 + 1852 = 3450 \, \text{kJ/mol}\)
 

\(\Delta H= \Sigma\,\text{bond energies broken}-\Sigma\,\text{bond energies formed}\)

\( \Delta H = 2644-3450 = -802 \, \text{kJ/mol} \)

\(\Rightarrow A\)

CHEMISTRY, M4 EQ-Bank 16

Given the following values for a reaction:

\(\Delta H = +150 \, \text{kJ/mol}\)  and  \(\Delta S = +250 \, \text{J/mol K}\)

  1. Calculate the Gibbs free energy at 298 K.   (2 mark)
  2. Is the reaction spontaneous at this temperature? Justify your answer.   (1 mark)

  3. Predict the temperature at which the reaction would become spontaneous.   (2 marks)
Show Answers Only

a.    \(\Delta G = +75.5 \, \text{kJ/mol}\)
 
b.    The reaction is non-spontaneous at 298 K.
 
c.    The reaction becomes spontaneous above 600 K.

Show Worked Solution

a.    \(\Delta G = \Delta H-T\Delta S\):

Convert \(\Delta S\) to \(\text{kJ/mol K}\):

\(\Delta S = 0.250 \, \text{kJ/mol K}\)

\(\Delta G = 150-(298 \times 0.250) = 150-74.5 = +75.5 \, \text{kJ/mol}\)
 

b.    \(\Delta G > 0\) at 298 K \(\Rightarrow\) reaction is non-spontaneous at 298 K.
 

c.    Find the temperature where the reaction becomes spontaneous:

→ Solve for \(T\) when \(\Delta G = 0\):

\(0= \Delta H-T\Delta S\)

\(T = \dfrac{\Delta H}{\Delta S} = \dfrac{150}{0.250} = 600 \, \text{K}\)

→ As the temperature rises, \(T\Delta S\) increases, therefore \(\Delta G\) decreases, increasing the spontaneity of the reaction.

→ The reaction becomes spontaneous above 600 K.

CHEMISTRY, M4 EQ-Bank 15

A wide range of chemical reactions take place in the combustion chamber of car engines. One such reaction occurs when carbon monoxide is converted to carbon dioxide:

\(\ce{2CO(g) + O2(g) \rightarrow 2CO2(g)}\)

The following values were obtained under standard conditions:

\(\Delta H = +283 \, \text{kJ/mol}, \quad \Delta S = +86.6 \, \text{J/mol K}\)

  1. Calculate the Gibbs free energy for the above reaction at 298 K.   (2 marks)
  1. Explain the effect of enthalpy and entropy on the spontaneity of this reaction.   (3 marks)
Show Answers Only

a.    \(\Delta G = +257.2 \, \text{kJ/mol}\)

b.    Effect of enthalpy and entropy on spontaneity:

→ Since the entropy value for this reaction is positive, the reaction is considered to be entropy-driven as this will drive towards a negative value for the Gibbs free energy and contribute to the reaction being spontaneous.

→ However, the enthalpy value of this reaction is also positive, which does not contribute towards a negative value for the Gibbs free energy and in turn contributes towards making this reaction non-spontaneous.

→ Since the magnitude of the enthalpy value is greater, the entropy drive is unable to overcome this and results in a non-spontaneous reaction overall.

Show Worked Solution

a.    Using the Gibbs free energy equation:

\(\Delta G = \Delta H-T \Delta S\)

Convert \(\Delta S\) to \(\text{kJ/mol K}\):

\(\Delta S = +0.0866 \, \text{kJ/mol K}\).

Now, substitute the values into the equation:

\(\Delta G = 283 \, \text{kJ/mol}-(298 \, \text{K} \times 0.0866 \, \text{kJ/mol K}) = 283-25.8 = +257.2 \, \text{kJ/mol}\)

\(\Rightarrow\) Since \(\Delta G > 0\), the reaction is non-spontaneous.

 

b.    Effect of enthalpy and entropy on spontaneity:

→ Since the entropy value for this reaction is positive, the reaction is considered to be entropy-driven as this will drive towards a negative value for the Gibbs free energy and contribute to the reaction being spontaneous.

→ However, the enthalpy value of this reaction is also positive, which does not contribute towards a negative value for the Gibbs free energy and in turn contributes towards making this reaction non-spontaneous.

→ Since the magnitude of the enthalpy value is greater, the entropy drive is unable to overcome this and results in a non-spontaneous reaction overall.

CHEMISTRY, M4 EQ-Bank 5 MC

The combustion of butane occurs via an exothermic reaction.

\(\ce{2C4H10(g) + 13 O2(g) \rightarrow 8 CO2(g) + 10 H2O(g)}\)

Which of the following correctly describes the spontaneity of this reaction?

  1. Spontaneous at all temperatures
  2. Spontaneous at low temperatures
  3. Spontaneous at high temperatures
  4. Non-spontaneous at all temperatures
Show Answers Only

\(A\)

Show Worked Solution

→ This reaction is exothermic (\(\Delta H < 0\)) and also exhibits an increase in entropy (\(\Delta S > 0\)) due to the production of gaseous products from smaller hydrocarbon molecules.

→ As both \(\Delta H\) and \(\Delta S\) are favourable, the reaction will be spontaneous at all temperatures according to the Gibbs free energy equation:

\(\Delta G = \Delta H-T \Delta S\)

→ Since \(\Delta H\) is negative and \(\Delta S\) is positive, \(\Delta G\) will always be negative, making the reaction spontaneous at all temperatures.

\(\Rightarrow A\)

Trigonometry, EXT1 T2 EQ-Bank 3

  1. Show that  \(\cos 30^{\circ} \cos 15^{\circ}=\dfrac{1}{2}\left[\cos 15^{\circ}+\dfrac{1}{\sqrt{2}}\right]\).  (2 marks)
  2. Hence, or otherwise, find the exact value of  \(\cos 15^{\circ}\).   (2 marks)

Show Answers Only

a.   \(\text{Show:}\ \cos 30^{\circ} \cos 15^{\circ}=\dfrac{1}{2}\left[\cos 15^{\circ}+\dfrac{1}{\sqrt{2}}\right]\)

\(\cos(30^{\circ}+15^{\circ})=\cos 30^{\circ} \cos 15^{\circ}-\sin 30^{\circ} \sin 15^{\circ}\ …\ (1)\)

\(\cos(30^{\circ}-15^{\circ})=\cos 30^{\circ} \cos 15^{\circ}+\sin 30^{\circ} \sin 15^{\circ}\ …\ (2)\)

  \(\text{Add (1) + (2):}\)

\(2\cos 30^{\circ} \cos 15^{\circ}\) \(=\cos 45^{\circ}+\cos 15^{\circ}\)  
\(\cos 30^{\circ} \cos 15^{\circ}\) \(=\dfrac{1}{2}\Big[\cos 15^{\circ}+\dfrac{1}{\sqrt2}\Big] \)  

 
b.  \(\cos 15^{\circ}=\dfrac{\sqrt6+\sqrt2}{4} \)

Show Worked Solution

a.   \(\text{Show:}\ \cos 30^{\circ} \cos 15^{\circ}=\dfrac{1}{2}\left[\cos 15^{\circ}+\dfrac{1}{\sqrt{2}}\right]\)

\(\cos(30^{\circ}+15^{\circ})=\cos 30^{\circ} \cos 15^{\circ}-\sin 30^{\circ} \sin 15^{\circ}\ …\ (1)\)

\(\cos(30^{\circ}-15^{\circ})=\cos 30^{\circ} \cos 15^{\circ}+\sin 30^{\circ} \sin 15^{\circ}\ …\ (2)\)

  \(\text{Add (1) + (2):}\)

\(2\cos 30^{\circ} \cos 15^{\circ}\) \(=\cos 45^{\circ}+\cos 15^{\circ}\)  
\(\cos 30^{\circ} \cos 15^{\circ}\) \(=\dfrac{1}{2}\Big[\cos 15^{\circ}+\dfrac{1}{\sqrt2}\Big] \)  

 

b.    \(2\cos 30^{\circ} \cos 15^{\circ}\) \(=\cos 15^{\circ}+\dfrac{1}{\sqrt2}\)
  \(\cos 15^{\circ}(2\cos 30^{\circ}-1)\) \(=\dfrac{1}{\sqrt2}\)
  \(\cos 15^{\circ}(\sqrt3-1)\) \(=\dfrac{1}{\sqrt2}\)
  \(\cos 15^{\circ}\) \(=\dfrac{1}{\sqrt2(\sqrt3-1)}\)
    \(=\dfrac{1}{\sqrt6-\sqrt2} \times \dfrac{\sqrt6+\sqrt2}{\sqrt6+\sqrt2}\)
    \(=\dfrac{\sqrt6+\sqrt2}{4} \)

CHEMISTRY, M4 EQ-Bank 3 MC

Given the relationship  \(\Delta G = \Delta H-T\Delta S\), which of the following statements is correct?

  1. A reaction is always spontaneous if \(\Delta S\) is negative and \(\Delta H\) is negative.
  2. A reaction is always spontaneous if \(\Delta S\) is negative and \(\Delta H\) is positive.
  3. A reaction is always spontaneous if \(\Delta S\) is positive and \(\Delta H\) is negative.
  4. A reaction is always spontaneous if \(\Delta S\) is positive and \(\Delta H\) is positive.
Show Answers Only

\(C\)

Show Worked Solution

→ A reaction is spontaneous when \(\Delta G < 0\).

→ For this to always happen, \(\Delta S\) must be positive (increasing disorder), and \(\Delta H\) must be negative (exothermic).

→ Therefore, \(\Delta G\) will be negative at all temperatures.

CHEMISTRY, M4 EQ-Bank 1 MC

Which of the following scenarios will result in a decrease in entropy?

  1. Water transitioning to steam
  2. Formation of silver chloride as a precipitate from a solution containing silver and chloride ions
  3. Ice melting into liquid water
  4. Ethanol dissolving in water
Show Answers Only

\(B\)

Show Worked Solution

→ The formation of silver chloride precipitate involves the arrangement of silver and chloride ions into a solid structure, decreasing the system’s disorder.

→ This decrease in disorder corresponds to a reduction in entropy.

\(\Rightarrow B\)

CHEMISTRY, M4 EQ-Bank 13

The reduction of iron (III) oxide takes place as follows:

\(\ce{Fe2O3(s) + 3H2(g) \rightarrow 2Fe(s) + 3H2O(l)}\)

\( \Delta H = 98.8\ \text{kJ mol}^{-1}\),  \(\Delta S = 141\ \text{J mol}^{-1}\text{K}^{-1}\)

Predict the temperature at which this reaction will become spontaneous. Show your working.   (2 marks)

Show Answers Only

\(700.7\ \text{K}\)

Show Worked Solution

Using the Gibbs free energy equation:

\(\Delta G = \Delta H-T \Delta S\)

For spontaneity, \(\Delta G < 0\) , so we solve for \(T\) when \(\Delta G = 0\)

\(0 = \Delta H-T \Delta S\ \Rightarrow \ T = \dfrac{\Delta H}{\Delta S}\)

\(\Rightarrow\) Convert \(\Delta S\) from \(\text{J mol}^{-1}\text{K}^{-1}\) to \(\text{kJ mol}^{-1}\text{K}^{-1}\):

\(T = \dfrac{98.8\ \text{kJ mol}^{-1}}{0.141\ \text{kJ mol}^{-1}\text{K}^{-1}} = 700.7\ \text{K}\)

\(\Rightarrow\) The reaction becomes spontaneous above approximately 700.7 \(\text{K}\)

CHEMISTRY, M1 EQ-Bank 37

  1. Draw Lewis dot diagrams for both water AND methane.   (2 marks)
  1. Methane has a boiling point of –161.5°C, while water has a melting point of 0°C and a boiling point of 100°C.
  2. Account for the differences in these physical properties for these TWO common substances in our atmosphere.   (2 marks)

Show Answers Only

a.     Methane \((\ce{CH4})\)       Water \((\ce{H2O})\)

 

→ All electrons must be shown as dots, not crosses, circles, etc.

→ No lines should be present to represent covalent bonds.
 

b.    Differences in the physical properties of methane and water:

→ Can be explained by the differences in their molecular structures and intermolecular forces.

→ Methane \(\ce{(CH4)}\) is a nonpolar molecule with weak dispersion forces between its molecules. These weak forces result in a very low boiling point of -161.5°C.

→ Water \(\ce{(H2O)}\), on the other hand, is a polar molecule that forms strong hydrogen bonds between its molecules. These strong intermolecular forces require more energy to be broken and result in much higher melting and boiling points (0°C and 100°C, respectively).

Show Worked Solution

a.     Methane \((\ce{CH4})\)       Water \((\ce{H2O})\)

 

→ All electrons must be shown as dots, not crosses, circles, etc.

→ No lines should be present to represent covalent bonds.
 

b.    Differences in the physical properties of methane and water:

→ Can be explained by the differences in their molecular structures and intermolecular forces.

→ Methane \(\ce{(CH4)}\) is a nonpolar molecule with weak dispersion forces between its molecules. These weak forces result in a very low boiling point of -161.5°C.

→ Water \(\ce{(H2O)}\), on the other hand, is a polar molecule that forms strong hydrogen bonds between its molecules. These strong intermolecular forces require more energy to be broken and result in much higher melting and boiling points (0°C and 100°C, respectively).

CHEMISTRY, M2 EQ-Bank 4

A student is required to dilute 150.00 mL solution of 3.00 mol L\(^{-1}\) hydrochloric acid to produce 250.00 mL of 0.54 mol L\(^{-1}\) hydrochloric acid. 

Explain how the student should perform this dilution in a school laboratory. Include relevant calculations in your answer AND explain how the student should prepare any equipment they would use.   (4 marks)

Show Answers Only

→ The student should wash a 250 mL volumetric flask with distilled water only, if any hydrochloric acid was in the flask it would increase the final concentration to an unknown value.

→ A 50 mL volumetric pipette should be rinsed with distilled water to remove impurities then rinsed with the 3.00 M HCl solution to be diluted, as if any water was left in the pipette it would dilute the acid by an unknown factor.

→ The initial and final concentrations and volumes are related by the dilution equation:

\(C_1 V_1 = C_2 V_2 \), where

\( C_1 = 3.00 \, \text{M} \) (initial concentration)

\( V_1 \) = volume to be pipetted

\( C_2 = 0.54 \, \text{M} \) (final concentration)

\( V_2 = 250.00 \, \text{mL} \) (final volume)
 

→ Rearrange to solve for \( V_1 \):

\( V_1 = \dfrac{C_2 V_2}{C_1}  = \dfrac{0.54 \times 250.00}{3.00}= \dfrac{125.00}{3.00} = 45 \, \text{mL} \)

→ Thus, the student would pipette 45 mL of the 3.00 M acid and deliver it to the 250 mL volumetric flask, making it up to the 250 mL line with distilled water. This would produce a 0.50 M dilution.

Show Worked Solution

→ The student should wash a 250 mL volumetric flask with distilled water only, if any hydrochloric acid was in the flask it would increase the final concentration to an unknown value.

→ A 50 mL volumetric pipette should be rinsed with distilled water to remove impurities then rinsed with the 3.00 M HCl solution to be diluted, as if any water was left in the pipette it would dilute the acid by an unknown factor.

→ The initial and final concentrations and volumes are related by the dilution equation:

\(C_1 V_1 = C_2 V_2 \), where

\( C_1 = 3.00 \, \text{M} \) (initial concentration)

\( V_1 \) = volume to be pipetted

\( C_2 = 0.54 \, \text{M} \) (final concentration)

\( V_2 = 250.00 \, \text{mL} \) (final volume)
 

→ Rearrange to solve for \( V_1 \):

\( V_1 = \dfrac{C_2 V_2}{C_1}  = \dfrac{0.54 \times 250.00}{3.00}= \dfrac{125.00}{3.00} = 45 \, \text{mL} \)

→ Thus, the student would pipette 45 mL of the 3.00 M acid and deliver it to the 250 mL volumetric flask, making it up to the 250 mL line with distilled water. This would produce a 0.50 M dilution.

CHEMISTRY, M2 EQ-Bank 3

A student is required to dilute 100.00 mL solution of 2.00 mol L\(^{-1}\) hydrochloric acid to produce 200.00 mL of 0.20 mol L\(^{-1}\) hydrochloric acid. 

Explain how the student should perform this dilution in a school laboratory. Include relevant calculations in your answer AND explain how the student should prepare any equipment they would use.    (4 marks)

Show Answers Only

→ The student should wash a 200 mL volumetric flask with distilled water only, as if any hydrochloric acid was in the flask it would increase the final concentration to an unknown value.

→ A 20 mL volumetric pipette should be rinsed with distilled water to remove impurities then rinsed with the 2.00 M HCl solution to be diluted, as if any water was left in the pipette it would dilute the acid by an unknown factor.

→ The initial and final concentrations and volumes are related by the dilution equation:

\( C_1 V_1 = C_2 V_2 \), where 

\( C_1 = 2.00 \, \text{M} \) (initial concentration)

\( V_1 \) = volume to be pipetted

\( C_2 = 0.20 \, \text{M} \) (final concentration)

\( V_2 = 200.00 \, \text{mL} \) (final volume)
 

→ Solve for \( V_1 \):

\( V_1 = \dfrac{C_2 V_2}{C_1} = \dfrac{0.20 \times 200.00}{2.00} = \dfrac{40.00}{2.00} = 20.00 \ \text{mL} \)

→ Thus, the student would pipette 20.00 mL of the 2.00 M acid and deliver it to the 200 mL volumetric flask, making it up to the 200 mL line with distilled water. This would produce a 0.200 M or \(1:10\) dilution.

Show Worked Solution

→ The student should wash a 200 mL volumetric flask with distilled water only, as if any hydrochloric acid was in the flask it would increase the final concentration to an unknown value.

→ A 20 mL volumetric pipette should be rinsed with distilled water to remove impurities then rinsed with the 2.00 M HCl solution to be diluted, as if any water was left in the pipette it would dilute the acid by an unknown factor.

→ The initial and final concentrations and volumes are related by the dilution equation:

\( C_1 V_1 = C_2 V_2 \), where 

\( C_1 = 2.00 \, \text{M} \) (initial concentration)

\( V_1 \) = volume to be pipetted

\( C_2 = 0.20 \, \text{M} \) (final concentration)

\( V_2 = 200.00 \, \text{mL} \) (final volume)
 

→ Solve for \( V_1 \):

\( V_1 = \dfrac{C_2 V_2}{C_1} = \dfrac{0.20 \times 200.00}{2.00} = \dfrac{40.00}{2.00} = 20.00 \ \text{mL} \)

→ Thus, the student would pipette 20.00 mL of the 2.00 M acid and deliver it to the 200 mL volumetric flask, making it up to the 200 mL line with distilled water. This would produce a 0.200 M or \(1:10\) dilution.

CHEMISTRY, M4 EQ-Bank 1

Ethanol \(\ce{(C2H5OH)}\)) is a common hydrocarbon fuel used in alcohol burners. It combusts completely according to the following equation:

\(\ce{C2H5OH (l) + 3O2 (g) -> 2CO2 (g) + 3H2O (g)}\)

A student conducts an experiment to determine the heat energy released by ethanol during combustion.

They burn 0.25 g of ethanol in excess oxygen, capturing the heat released in a calorimeter containing 75.0 g of water. The water's temperature increases from 21.0 °C to 34.5 °C. Calculate the experimental molar heat of combustion \((\Delta H^{\circ}_{\text{comb}})\) for ethanol based on these results.   (3 marks)

Show Answers Only

\(\Delta H^{\circ}_{\text{comb}}=-775.46\ \text{kJ/mol}\).

Show Worked Solution

Heat absorbed by the water \((q_{\text{water}})\):

\[q = m \cdot c \cdot \Delta T = 75.0\ \text{g} \cdot 4.18\ \text{J/g}^{\circ}\text{C} \cdot (34.5\ \text{°C}-21.0\ \text{°C}) = 4204.65\ \text{J} = 4.205\ \text{kJ}\]

Moles of ethanol burned:

\[\ce{MM(C2H5OH)} = 46.07\ \text{g/mol} \]

\[ \ce{n(C2H5OH)} = \frac{0.25\ \text{g}}{46.07\ \text{g/mol}} = 0.00542\ \text{mol} \]

Experimental molar heat of combustion:

\[ \Delta H^{\circ}_{\text{comb}} = \frac{4.205\ \text{kJ}}{0.00542\ \text{mol}} = -775.46\ \text{kJ/mol}\]

CHEMISTRY, M4 EQ-Bank 38v4

Methanol (\(\ce{CH3OH}\)) is a hydrocarbon fuel which combusts completely according to the following equation:

\[\ce{2CH3OH (l) + 3O2 (g) -> 2CO2 (g) + 4H2O (g)}\]

A chemist burns 0.40 g of methanol in excess oxygen during a calorimetry study, and the heat energy is used to heat 80.0 g of water. The initial temperature of the water in the calorimeter is 24.0 °C, which rises to a maximum of 52.0 °C after absorbing the energy from the combustion reaction. Based on these results, calculate the experimental molar heat of combustion (\(\Delta H^\circ_{\text{comb}}\)) for methanol.   (4 marks)

Show Answers Only

The experimental molar heat of combustion (\(\Delta H^\circ_{\text{comb}}\)) for methanol is \( -770 \, \text{kJ/mol} \). (2 sig.fig.)

Show Worked Solution

Calculate the energy absorbed by the water:

\[q = mc\Delta T = (80.0 \, \text{g})(4.18 \, \text{J/g}^\circ\text{C})(52.0^\circ\text{C} – 24.0^\circ\text{C}) = 9363.2 \, \text{J}\]

Calculate the number of moles of methanol burned:

\[n = \frac{m}{M} = \frac{0.40 \, \text{g}}{32.042 \, \text{g/mol}} = 0.0125 \, \text{mol}\]

Calculate the molar heat of combustion:

\[\Delta H^\circ_{\text{comb}} = -\frac{q}{n} = -\frac{9363.2 \, \text{J}}{0.0125 \, \text{mol}} = -773056 \, \text{J/mol} = -770 \, \text{kJ/mol} \]

→ The experimental molar heat of combustion (\(\Delta H^\circ_{\text{comb}}\)) for methanol is \( -770 \, \text{kJ/mol} \). (2 sig.fig.)

CHEMISTRY, M4 EQ-Bank 38v3

Propane (\(\ce{C3H8}\)) is a hydrocarbon fuel which combusts completely according to the following equation:

\[\ce{C3H8 (g) + 5O2 (g) -> 3CO2 (g) + 4H2O (g)}\]

A chemist burns 0.25 g of propane in excess oxygen during a calorimetry study, and the heat energy is used to heat 60.0 g of water. The initial temperature of the water in the calorimeter is 21.0 °C, which rises to a maximum of 42.0 °C after absorbing the energy from the combustion reaction. Based on these results, calculate the experimental molar heat of combustion (\(\Delta H^\circ_{\text{comb}}\)) for propane.   (4 marks)

Show Answers Only

The experimental molar heat of combustion (\(\Delta H^\circ_{\text{comb}}\)) for propane is \( -930 \, \text{kJ/mol} \). (2 sig.fig.)

Show Worked Solution

Calculate the energy absorbed by the water:

\[q = mc\Delta T = (60.0 \, \text{g})(4.18 \, \text{J/g}^\circ\text{C})(42.0^\circ\text{C} – 21.0^\circ\text{C}) = 5266.8 \, \text{J}\]

Calculate the number of moles of propane burned:

\[n = \frac{m}{M} = \frac{0.25 \, \text{g}}{44.094 \, \text{g/mol}} = 0.00567 \, \text{mol}\]

Calculate the molar heat of combustion:

\[\Delta H^\circ_{\text{comb}} = -\frac{q}{n} = -\frac{5266.8 \, \text{J}}{0.00567 \, \text{mol}} = -928888.9 \, \text{J/mol} = -929 \, \text{kJ/mol}\]

→ The experimental molar heat of combustion (\(\Delta H^\circ_{\text{comb}}\)) for propane is \( -930 \, \text{kJ/mol} \). (2 sig.fig.)

CHEMISTRY, M4 EQ-Bank 38v2

Methane (\(\ce{CH4}\)) is a hydrocarbon fuel which combusts completely according to the following equation:

\[\ce{CH4 (g) + 2O2 (g) -> CO2 (g) + 2H2O (g)}\]

A chemist burns 0.30 g of methane in excess oxygen during a calorimetry study, and the heat energy is used to heat 75.0 g of water. The initial temperature of the water in the calorimeter is 22.0 °C, which rises to a maximum of 36.5 °C after absorbing the energy from the combustion reaction. Based on these results, calculate the experimental molar heat of combustion (\(\Delta H^\circ_{\text{comb}}\)) for methane.    (4 marks)

Show Answers Only

The experimental molar heat of combustion (\(\Delta H^\circ_{\text{comb}}\)) for methane is \( -240 \, \text{kJ/mol} \). (2 sig.fig)

Show Worked Solution

Calculate the energy absorbed by the water:

\[q = mc\Delta T = (75.0 \, \text{g})(4.18 \, \text{J/g}^\circ\text{C})(36.5^\circ\text{C} – 22.0^\circ\text{C}) = 4545.75 \, \text{J}\]

Calculate the number of moles of methane burned:

\[n = \frac{m}{M} = \frac{0.30 \, \text{g}}{16.042 \, \text{g/mol}} = 0.0187 \, \text{mol}\]

Calculate the molar heat of combustion:

\[\Delta H^\circ_{\text{comb}} = -\frac{q}{n} = -\frac{4545.75 \, \text{J}}{0.0187 \, \text{mol}} = -243088.24 \, \text{J/mol} = -240 \, \text{kJ/mol}\]

→ The experimental molar heat of combustion (\(\Delta H^\circ_{\text{comb}}\)) for methane is \( -240 \, \text{kJ/mol} \). (2 sig.fig)

CHEMISTRY, M4 EQ-Bank 2

Ethanol \(\ce{(C2H5OH)}\) is a hydrocarbon fuel which combusts completely according to the following equation:

\[\ce{C2H5OH (l) + 3O2 (g) -> 2CO2 (g) + 3H2O (g)}\]

A chemist burns 0.50 g of ethanol in excess oxygen during a calorimetry study, and the heat energy is used to heat 100.0 g of water. The initial temperature of the water in the calorimeter is 20.0 °C, which rises to a maximum of 45.0 °C after absorbing the energy from the combustion reaction. Based on these results, calculate the experimental molar heat of combustion \((\Delta H^\circ_{\text{comb}})\) for ethanol.   (4 marks)

Show Answers Only

\( -960 \, \text{kJ/mol  (2 sig.fig.)}\)

Show Worked Solution

Energy absorbed by the water:

\[q = mc\Delta T = (100.0 \, \text{g})(4.18 \, \text{J/g}^\circ\text{C})(45.0^\circ\text{C}-20.0^\circ\text{C}) = 10\,450 \, \text{J}\]

Number of moles of ethanol burned:

\[n\ce{(C2H5OH)} = \frac{m}{MM} = \frac{0.50 \, \text{g}}{46.068 \, \text{g/mol}} = 0.01085 \, \text{mol}\]

Experimental molar heat of combustion:

\[
\Delta H^\circ_{\text{comb}} = -\frac{q}{n} = -\frac{10\,450 \, \text{J}}{0.01085 \, \text{mol}} = -963\,133 \, \text{J/mol} = -960 \, \text{kJ/mol  (2 sig.fig.)}
\]

CHEMISTRY, M4 EQ-Bank 11

Consider the reaction below at 298 K:

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

The standard enthalpies of formation and standard entropies are as follows:

\(\begin{aligned}
\ce{N2(g)}: & \quad \Delta H_f^\circ = 0 \, \text{kJmol}^{-1}, & S^\circ = 191.5 \, \text{J/mol·K} \\
\ce{H2(g)}: & \quad \Delta H_f^\circ = 0 \, \text{kJmol}^{-1}, & S^\circ = 130.7 \, \text{J/mol·K} \\
\ce{NH3(g)}: & \quad \Delta H_f^\circ = -45.9 \, \text{kJmol}^{-1}, & S^\circ = 192.3 \, \text{J/mol·K} \\
\end{aligned}\)

Calculate the Gibbs free energy (\( \Delta G \)) for the reaction and determine whether the reaction is spontaneous at 298 K. Show all relevant calculations.   (4 marks)

Show Answers Only

\(\Delta G =-32.7\ \text{kJ/mol} \), indicating the reaction is spontaneous at 298 K.

Show Worked Solution

To calculate the Gibbs free energy (\( \Delta G \)) for the reaction, we use the following equation:

   \( \Delta G = \Delta H-T \Delta S \)

First, calculate the change in enthalpy (\( \Delta H \)) for the reaction:

   \( \Delta H_{\text{reaction}}^\circ = \sum \Delta H_f^\circ (\text{products}) -\ \sum \Delta H_f^\circ (\text{reactants})\)

   \( \Delta H_{\text{reaction}}^\circ = [2 \times (-45.9)]-[1 \times (0) + 3 \times (0)] = -91.8 \, \text{kJ/mol}\)
 

Calculate the change in entropy (\( \Delta S \)) for the reaction:

\( \Delta S_{\text{reaction}}^\circ\) \(= \sum S^\circ (\text{products})-\sum S^\circ (\text{reactants}) \)  
  \(= (2 \times (192.3))-(1 \times (191.5) + 3 \times (130.7))\)  
  \(= 384.6-(191.5 + 392.1)\)  
  \(= -199.0\ \text{J/mol·K}\)  
  \(=-0.199\ \text{kJ/mol·K}\)  

 
Calculate \( \Delta G \):

\(\Delta G= \Delta H-T \Delta S= -91.8-(298 \times -0.199)= -32.5\ \text{kJ/mol}\)

\(\Rightarrow\) The reaction is spontaneous at 298 K.

CHEMISTRY, M4 EQ-Bank 12

When aqueous solutions of \(\ce{Na^+}\) and \(\ce{Cl^-}\) are mixed, the following exothermic reaction occurs:

\(\ce{Na+(aq) + Cl-(aq) -> NaCl(aq)}\)

If the temperature of the mixture is increased, explain how this will affect the Gibbs free energy and whether the reaction will become more or less spontaneous. In your response, refer to both the enthalpy and entropy changes involved in the reaction.   (3 marks)

Show Answers Only

→ Gibbs free energy is calculated using  \(\Delta G = \Delta H\ \ -\ \ T\Delta S\).

→ Since the reaction is exothermic, the enthalpy change \(\Delta H\) will be negative.

→ In this reaction, two ions combine to form a single ion, leading to a decrease in entropy \(\Delta S\), making it negative.

→ As the temperature \(T\) increases, the term \(-T \Delta S\) becomes more positive, which in turn causes \(\Delta G\) to increase.

→ As a result, the reaction becomes less spontaneous at higher temperatures.

Show Worked Solution

→ Gibbs free energy is calculated using  \(\Delta G = \Delta H\ \ -\ \ T\Delta S\).

→ Since the reaction is exothermic, the enthalpy change \(\Delta H\) will be negative.

→ In this reaction, two ions combine to form a single ion, leading to a decrease in entropy \(\Delta S\), making it negative.

→ As the temperature \(T\) increases, the term \(-T \Delta S\) becomes more positive, which in turn causes \(\Delta G\) to increase.

→ As a result, the reaction becomes less spontaneous at higher temperatures.

L&E, 2ADV E1 EQ-Bank 2

Show  \(f(x)=\dfrac{1}{2}-\dfrac{1}{2^x+1}\)  is an odd function.  (3 marks)

Show Answers Only

\(\text{Odd}\ \ \Rightarrow \  \ f(-x)=-f(x)\)

\(\begin{aligned}
f(x) & =\dfrac{1}{2}-\dfrac{1}{2^x+1} \\
f(-x) & =\dfrac{1}{2}-\dfrac{1}{2^{-x}+1} \times \dfrac{2^x}{2^x} \\
& =\dfrac{1}{2}-\dfrac{2^x}{1+2^x} \\
& =\dfrac{1}{2}-\dfrac{2^x+1-1}{2^x+1} \\
& =\dfrac{1}{2}-1+\dfrac{1}{2^x+1} \\
& =-\dfrac{1}{2}+\dfrac{1}{2^x+1} \\
& =-f(x)
\end{aligned}\)

 
\(\therefore f(x) \text { is odd.}\)

Show Worked Solution

\(\text{Odd}\ \ \Rightarrow \  \ f(-x)=-f(x)\)

\(\begin{aligned}
f(x) & =\dfrac{1}{2}-\dfrac{1}{2^x+1} \\
f(-x) & =\dfrac{1}{2}-\dfrac{1}{2^{-x}+1} \times \dfrac{2^x}{2^x} \\
& =\dfrac{1}{2}-\dfrac{2^x}{1+2^x} \\
& =\dfrac{1}{2}-\dfrac{2^x+1-1}{2^x+1} \\
& =\dfrac{1}{2}-1+\dfrac{1}{2^x+1} \\
& =-\dfrac{1}{2}+\dfrac{1}{2^x+1} \\
& =-f(x)
\end{aligned}\)

 
\(\therefore f(x) \text { is odd.}\)

Trigonometry, 2ADV T1 EQ-Bank 4

The circle below has centre \(O\), radius \(r\) and arc length \(l\).

The ratio of the radius to the length of the arc is \(2: 5\).
 

Show that the area of the sector is  \(A=\dfrac{5 r^2}{4}\).   (2 marks)

Show Answers Only

\(\dfrac{r}{l} = \dfrac{2}{5}\ \ \Rightarrow\ \ \ l=\dfrac{5r}{2} \)

\(\text{Area}\ = \dfrac{\theta}{360} \times \pi r^2\ \ …\ (1) \)

\(\text{Arc length}\ (l): \)

\(\dfrac{\theta}{360} \times 2\pi r\) \(=l\)  
\(\dfrac{\theta}{360}\) \(=\dfrac{l}{2\pi r} = \dfrac{\frac{5r}{2}}{2\pi r} = \dfrac{5}{4 \pi}\)  

 
\(\text{Substitute}\ \ \dfrac{\theta}{360}=\dfrac{5}{4 \pi}\ \ \text{into (1):}\)

\(\text{Area}\ = \dfrac{5}{4\pi} \times \pi r^2 = \dfrac{5r^2}{4}\ \ …\ \text{as required}\)

Show Worked Solution

\(\dfrac{r}{l} = \dfrac{2}{5}\ \ \Rightarrow\ \ \ l=\dfrac{5r}{2} \)

\(\text{Area}\ = \dfrac{\theta}{360} \times \pi r^2\ \ …\ (1) \)

\(\text{Arc length}\ (l): \)

\(\dfrac{\theta}{360} \times 2\pi r\) \(=l\)  
\(\dfrac{\theta}{360}\) \(=\dfrac{l}{2\pi r} = \dfrac{\frac{5r}{2}}{2\pi r} = \dfrac{5}{4 \pi}\)  

 
\(\text{Substitute}\ \ \dfrac{\theta}{360}=\dfrac{5}{4 \pi}\ \ \text{into (1):}\)

\(\text{Area}\ = \dfrac{5}{4\pi} \times \pi r^2 = \dfrac{5r^2}{4}\ \ …\ \text{as required}\)

Trigonometry, 2ADV T1 EQ-Bank 3

A tower \(T C\) is \(h\) metres high.

At point \(A\), due south of the tower, the angle of elevation to the top of the tower, point \(T\), is 13°.

Point \(B\) is due east of the tower, with an angle of elevation to the top of 24°, as shown in the diagram.

Point \(A\) is 1.1 kilometres from Point \(B\) and both points are on the same ground level as the base of the tower, point \(C\).
 


  1. Show \(B C=h \times \tan 66^{\circ}\).   (1 mark)

  2. Find a similar expression for \(A C\).   (1 mark)

  3. Hence, or otherwise, determine the height, \(h\), of the tower. Give your answer correct to the nearest metre.   (3 marks)

Show Answers Only

a.   \(\text{See Worked Solutions}\)

b.   \(\text{See Worked Solutions}\)

c.   \(\text{225 metres}\)

Show Worked Solution

a.   \(\text{In}\ \Delta TBC\ \Rightarrow \angle CTB=90-24=66^{\circ}\)

\(\tan 66^{\circ}\) \(=\dfrac{BC}{h} \)  
\(BC\) \(=h \times \tan 66^{\circ}\)  

 

b.   \(\text{In}\ \Delta TAC\ \Rightarrow \angle CTA=90-13=77^{\circ}\)

\(\tan 77^{\circ}\) \(=\dfrac{AC}{h} \)  
\(AC\) \(=h \times \tan 77^{\circ}\)  

 
c.   \(\Delta ACB\ \text{is right-angled.}\)
  

\(\text{By Pythagoras:}\)

\(AC^{2}+BC^{2}\) \(=1100^{2}\)  
\(h^2 \times \tan^{2} 77^{\circ} + h^2 \times \tan^{2}66^{\circ}\) \(=1100^2\)  
\(h^2(\tan^{2} 77^{\circ}+\tan^{2} 66^{\circ})\) \(=1100^2\)  
\(h^2\) \(=\dfrac{1100^2}{(\tan^{2} 77^{\circ}+\tan^{2} 66^{\circ})}\)  
\(h\) \(=225.44…\)  
  \(=225\ \text{m (nearest m)}\)  

Trigonometry, 2ADV T2 EQ-Bank 4

Prove  \(\dfrac{\operatorname{cosec} \theta+\sec \theta}{1+\tan \theta}=\operatorname{cosec} \theta\).   (3 marks)

Show Answers Only

\(\text{Prove:}\ \ \dfrac{\operatorname{cosec} \theta+\sec \theta}{1+\tan \theta}=\operatorname{cosec} \theta\)

  \(\text{LHS}\) \(= \dfrac{\operatorname{cosec} \theta + \sec \theta}{1 + \tan \theta}\)
    \(=\dfrac{\dfrac{1}{\sin \theta} + \dfrac{1}{\cos \theta}}{1 + \dfrac{\sin \theta}{\cos \theta}} \times \dfrac{\cos \theta}{\cos \theta} \)
    \(=\dfrac{\dfrac{\cos \theta}{\sin \theta}+1}{\cos \theta+\sin \theta}\)
    \(=\dfrac{\dfrac{\cos \theta+\sin \theta}{\sin \theta}}{\cos \theta+\sin \theta}\)
    \(=\dfrac{1}{\sin \theta}\)
    \(=\operatorname{cosec} \theta \quad \text{… as required.}\)
Show Worked Solution

\(\text{Prove:}\ \ \dfrac{\operatorname{cosec} \theta+\sec \theta}{1+\tan \theta}=\operatorname{cosec} \theta\)

  \(\text{LHS}\) \(= \dfrac{\operatorname{cosec} \theta + \sec \theta}{1 + \tan \theta}\)
    \(=\dfrac{\dfrac{1}{\sin \theta} + \dfrac{1}{\cos \theta}}{1 + \dfrac{\sin \theta}{\cos \theta}} \times \dfrac{\cos \theta}{\cos \theta} \)
    \(=\dfrac{\dfrac{\cos \theta}{\sin \theta}+1}{\cos \theta+\sin \theta}\)
    \(=\dfrac{\dfrac{\cos \theta+\sin \theta}{\sin \theta}}{\cos \theta+\sin \theta}\)
    \(=\dfrac{1}{\sin \theta}\)
    \(=\operatorname{cosec} \theta \quad \text{… as required.}\)

Functions, 2ADV F1 EQ-Bank 17

The tangent to the parabola  \(y=x^2+2 x-4\)  is  \(y=px-5\)  where  \(p>0\).

Find the value of \(p\).   (2 marks)

Show Answers Only

\(p=4\)

Show Worked Solution

\(\text{Intersection occurs when:}\)

\(x^2+2x-4\) \(=px-5\)  
\(x^2+(2-p)x+1\) \(=0\)  

 
\(\text{Tangent touches once}\ \Rightarrow\ \text{Discriminant}\ \Delta=0\)

\((2-p)^2-4 \times 1 \times 1\) \(=0\)  
\(4-4p+p^2-4\) \(=0\)  
\(p(p-4)\) \(=0\)  
\(p\) \(=4\ \ \ (p\gt 0)\)  
COMMENT: Key is to recognise this is a discriminant question, not a calculus application.

L&E, 2ADV E1 EQ-Bank 3

Solve the equation  \(4^x-2^{x+2}=32\), showing all working.   (3 marks)

Show Answers Only

\(x=3\)

Show Worked Solution
  \(2^{2 x}-2^{x+2}\) \(=2^5\)
  \(2^{2 x}-2^2 \times 2^x-32\) \(=0\)

 
\(\text{Let} \ \ 2^x=X\)

  \(X^2-4X-32\) \(=0\)
  \((X-8)(X+4)\) \(=0\)

\(X\) \(=8\) \(\quad\text{or}\quad\) \(X\) \(=-4\)
\(2^x\) \(=8\)   \(2^x\) \(=-4\ \ \text{(no solution)}\)
\(x\) \(=3\)      

 
\(\therefore x=3\)

Calculus, 2ADV C1 EQ-Bank 3 MC

At which point on the curve  \(y=2x^{2}-11x+3\)  can a tangent be drawn such that it is inclined at 45° when it crosses the positive \(x\)-axis?

  1. \((-3,54)\)
  2. \((-2,33)\)
  3. \((2,-11)\)
  4. \((3,-12)\)
Show Answers Only

\(D\)

Show Worked Solution
\(y\) \(=2x^{2}-11x+3\)  
\(y^{′}\) \(=4x-11\)  

 
\(\text{If a tangent crosses (positive) x-axis at 45}^{\circ},\)

\(m_{\text{tang}}=1  \ \ (\tan 45^{\circ}=1) \)

\(\text{Find}\ x\ \text{when}\ \ y^{′}=1: \)

\(4x-11\) \(=1\)  
\(4x\) \(=12\)  
\(x\) \(=3\)  

 
\(\text{Tangent at point}\ (3,-12) \)

\(\Rightarrow D\)

Trigonometry, 2ADV T2 EQ-Bank 1 MC

Determine the number of values of \(\theta\) in the range  \(0^{\circ} \leqslant \theta \leqslant 360^{\circ}\)  that satisfy the equation

\((\tan \theta-\sqrt{3})(\cos^{2}\theta-1)=0 \)

  1. \(3\)
  2. \(4\)
  3. \(5\)
  4. \(6\)
Show Answers Only

\(C\)

Show Worked Solution

\(\tan \theta = \sqrt{3}\ \rightarrow \text{2 solutions} \)

\(\cos^{2}\theta\) \(=1\)  
\(\cos\theta\) \(= \pm 1\)  
\(\theta\) \(=0^{\circ}, 180^{\circ}, 360^{\circ}\ \rightarrow \text{3 solutions} \)  

 
\(\Rightarrow C\)

BIOLOGY, M2 EQ-Bank 1

"Multicellular organisms exhibit different levels of cell complexity, from simple cells to highly specialised ones."

Justify this statement, providing examples to support your answer.   (4 marks)

Show Answers Only

→ Multicellular organisms show different levels of cell complexity, starting with simple cells that perform basic functions, like skin cells, to highly specialised cells, such as neurons, which have unique structures for transmitting signals.

→ Cell groups within multicellular organisms cannot survive without other cell types performing their role.

→ A hierarchy arises, organising cells into tissues, organs and other systems, each with distinct functions that work together to support life.

→ For example, muscle cells form muscle tissue, which contracts to allow movement, while red blood cells transport oxygen through the circulatory system but cannot reproduce.

→ The specialisation of cells allows multicellular organisms to perform complex tasks more efficiently, contributing to the organism’s overall survival and adaptability.

→ This organisation enhances both the functionality and efficiency of biological processes

Show Worked Solution

→ Multicellular organisms show different levels of cell complexity, starting with simple cells that perform basic functions, like skin cells, to highly specialised cells, such as neurons, which have unique structures for transmitting signals.

→ Cell groups within multicellular organisms cannot survive without other cell types performing their role.

→ A hierarchy arises, organising cells into tissues, organs and other systems, each with distinct functions that work together to support life.

→ For example, muscle cells form muscle tissue, which contracts to allow movement, while red blood cells transport oxygen through the circulatory system but cannot reproduce.

→ The specialisation of cells allows multicellular organisms to perform complex tasks more efficiently, contributing to the organism’s overall survival and adaptability.

→ This organisation enhances both the functionality and efficiency of biological processes

BIOLOGY, M1 EQ-Bank 9

Products of an enzyme-controlled reaction can sometimes inhibit the enzyme that produced them.

Discuss how this process can be advantageous for cellular metabolism.   (3 marks)

Show Answers Only

→ When products of an enzyme-controlled reaction inhibit the enzyme that produced them, it regulates the cell’s metabolism (in a process known as feedback).

→ This process prevents the accumulation of excess products, ensuring that resources like ATP and raw materials are used efficiently.

→ By controlling the enzyme activity, the cell can balance production according to its needs, conserving energy and maintaining metabolic stability.

Show Worked Solution

→ When products of an enzyme-controlled reaction inhibit the enzyme that produced them, it regulates the cell’s metabolism (in a process known as feedback).

→ This process prevents the accumulation of excess products, ensuring that resources like ATP and raw materials are used efficiently.

→ By controlling the enzyme activity, the cell can balance production according to its needs, conserving energy and maintaining metabolic stability.

BIOLOGY, M1 EQ-Bank 7

How do temperature and pH affect enzyme activity? In your answer, briefly explain how extreme conditions of each factor influence enzyme function.   (4 marks)

Show Answers Only

→ Temperature and pH significantly impact enzyme activity.

→ At high temperatures, enzymes can denature, losing their shape and functionality. At extreme heat, the active site can be destroyed or made inaccessible.

→ At low temperatures, enzyme activity slows down due to reduced molecular movement.

→ Certain enzymes have evolved to be most effective in specific pH environments (such as the acidity of the stomach) and are much less effective in more alkaline settings.

→ Extreme pH levels can alter the enzyme’s structure, affecting the active site’s ability to bind with substrates.

→ Optimal enzyme activity occurs within specific temperature and pH ranges for each enzyme.

Show Worked Solution

→ Temperature and pH significantly impact enzyme activity.

→ At high temperatures, enzymes can denature, losing their shape and functionality. At extreme heat, the active site can be destroyed or made inaccessible.

→ At low temperatures, enzyme activity slows down due to reduced molecular movement.

→ Certain enzymes have evolved to be most effective in specific pH environments (such as the acidity of the stomach) and are much less effective in more alkaline settings.

→ Extreme pH levels can alter the enzyme’s structure, affecting the active site’s ability to bind with substrates.

→ Optimal enzyme activity occurs within specific temperature and pH ranges for each enzyme.

BIOLOGY, M1 EQ-Bank 4

Compare the movement of a lipid-soluble substance and water across the cell membrane.

In your answer, explain how the structure of the membrane affects the transport of these molecules.   (2 marks)

Show Answers Only

→ Lipid-soluble substances can pass directly through the phospholipid bilayer by simple diffusion because the cell membrane contains lipids.

→ In contrast, water is unable to penetrate the cell through the phospholipid bilayer.

→ Instead, water molecules must pass through special protein channels (pores) in the cell membrane.

Show Worked Solution

→ Lipid-soluble substances can pass directly through the phospholipid bilayer by simple diffusion because the cell membrane contains lipids.

→ In contrast, water is unable to penetrate the cell through the phospholipid bilayer.

→ Instead, water molecules must pass through special protein channels (pores) in the cell membrane.

BIOLOGY, M1 EQ-Bank 4 MC

In active transport, how does the cell maintain a steep concentration gradient across the membrane?

  1. By using ion channels to allow the passive diffusion of ions along the gradient.
  2. By using carrier proteins and ATP to pump ions from an area of low concentration to an area of high concentration.
  3. By allowing water molecules to diffuse freely through the membrane, balancing the concentration.
  4. By increasing the surface area of the membrane to allow more diffusion of molecules.
Show Answers Only

\(B\)

Show Worked Solution

→ Active transport requires energy (ATP) to move ions or molecules against their concentration gradient, often with the help of carrier proteins.

→ This process helps cells maintain steep concentration gradients essential for functions like nerve impulse transmission and muscle contraction.

\(\Rightarrow B\)

BIOLOGY, M1 EQ-Bank 8

Explain how lysosomes contribute to the maintenance of normally functioning cells, and describe what happens when the function of lysosomes is impaired.   (2 marks)

Show Answers Only

→ Lysosomes maintain cellular homeostasis by breaking down waste materials, damaged organelles, and cellular debris using enzymes.

→ This prevents the accumulation of harmful substances.

→ When lysosomal function is impaired, waste products and damaged components build up, leading to cellular dysfunction and potentially contributing to diseases that impair cellular efficiency and overall health.

Show Worked Solution

→ Lysosomes maintain cellular homeostasis by breaking down waste materials, damaged organelles, and cellular debris using enzymes.

→ This prevents the accumulation of harmful substances.

→ When lysosomal function is impaired, waste products and damaged components build up, leading to cellular dysfunction and potentially contributing to diseases that impair cellular efficiency and overall health.

BIOLOGY, M1 EQ-Bank 9 MC

Which of the following correctly describes the roles of lysosomes and the Golgi apparatus?

  1. Lysosomes break down waste using enzymes, and the Golgi apparatus transports substances across the membrane.
  2. Lysosomes synthesize lipids, and the Golgi apparatus modifies and packages proteins for transport.
  3. Lysosomes break down waste using enzymes, and the Golgi apparatus modifies and packages proteins for transport.
  4. Lysosomes produce energy for the cell, and the Golgi apparatus breaks down cellular waste.
Show Answers Only

\(C\)

Show Worked Solution

→ Lysosomes contain digestive enzymes that break down cellular waste and debris.

→ Golgi apparatus is responsible for modifying, sorting, and packaging proteins for secretion or use within the cell.

\(\Rightarrow C\)

BIOLOGY, M1 EQ-Bank 5

According to the fluid mosaic model of the cell membrane

  1. describe the structure of the phospholipid bilayer.   (2 marks)

  2. explain the role of proteins.   (1 mark)

Show Answers Only

a.    The phospholipid bilayer:

→ Consists of two layers of phospholipids.

→ One layer involves hydrophilic (water-attracting) heads facing outward toward the water-based environment while the opposite layer has hydrophobic (water-repelling) tails facing inward, away from water.
 

b.     Proteins:

→ are embedded in the double phospholid layer and play a role in the cell structure and communication.

Show Worked Solution

a.    The phospholipid bilayer:

→ Consists of two layers of phospholipids.

→ One layer involves hydrophilic (water-attracting) heads facing outward toward the water-based environment while the opposite layer has hydrophobic (water-repelling) tails facing inward, away from water.
 

b.     Proteins:

→ are embedded in the double phospholid layer and play a role in the cell structure and communication.

BIOLOGY, M5 2020 VCE 11a

A student wanted to investigate the effect of two different endonucleases (restriction enzymes) on a linear DNA fragment.

The student used three tubes containing a buffered solution of linear DNA fragments, each fragment being 9500 base pairs in length.

Two different endonucleases were available: BamHI and HindIII.

The student followed the steps below.

After 45 minutes the student obtained the results shown below

    Tube 1 Rube 2 Tube 3
DNA   BamHI HindIII BamHI
ladder       HindIII

Analyse the results of the experiment performed by the student.  (5 marks)

Show Answers Only

Answers could include 5 of the following:

→ Smaller DNA fragments tend to migrate further during gel electrophoresis
    compared to larger fragments.

→ A standard or marker is used as a point of comparison and reference to
    determine the size of unknown DNA fragments.

→ The restriction enzyme BamH1 cuts the DNA molecule at a single location,
    resulting in the formation of two DNA fragments.

→ The restriction enzyme HindIII cuts the DNA molecule at two distinct
    locations, producing three DNA fragments.

→ When the DNA is cut with the HindIII enzyme, it results in the largest and
    smallest DNA fragments compared to other restriction enzyme digestions.

→ The DNA fragments generated by the BamH1 enzyme are 4000 bp and
    5500 bp in length, while the fragments produced by HindIII are 8000 bp,
    1000 bp, and 500 bp.

→ When the DNA is digested with both BamH1 and HindIII enzymes, a total
    of four DNA fragments are produced, with sizes of 5500 bp, 2500 bp,
    1000 bp, and 500 bp.

Show Worked Solution

Answers could include 5 of the following:

→ Smaller DNA fragments tend to migrate further during gel electrophoresis
    compared to larger fragments.

→ A standard or marker is used as a point of comparison and reference to
    determine the size of unknown DNA fragments.

→ The restriction enzyme BamH1 cuts the DNA molecule at a single location,
    resulting in the formation of two DNA fragments.

→ The restriction enzyme HindIII cuts the DNA molecule at two distinct
    locations, producing three DNA fragments.

→ When the DNA is cut with the HindIII enzyme, it results in the largest and
    smallest DNA fragments compared to other restriction enzyme digestions.

→ The DNA fragments generated by the BamH1 enzyme are 4000 bp and
    5500 bp in length, while the fragments produced by HindIII are 8000 bp,
    1000 bp, and 500 bp.

→ When the DNA is digested with both BamH1 and HindIII enzymes, a total
    of four DNA fragments are produced, with sizes of 5500 bp, 2500 bp,
    1000 bp, and 500 bp.


♦♦ Mean mark 50%.
COMMENT: 39% of responses received 0 or 1 mark

BIOLOGY, M5 2020 VCE 7 MC

The codon table below can be used to determine amino acids coded for by a nucleotide sequence.
 

\( \textbf{lst position} \)

 \( \textbf{(5}^{′}\ \textbf{end)} \)

\( \boldsymbol{\downarrow} \)

 \( \textbf{2nd position} \)

\( \textbf{3rd position} \)
\( \ \ \textbf{(3}^{′}\ \textbf{end)} \)

\( \boldsymbol{\downarrow} \)
\( \textbf{U} \) \( \textbf{C} \) \( \textbf{A} \) \( \textbf{G} \)
\( \textbf{U} \) \( \text{Phe}\) \( \text{Ser}\) \( \text{Tyr}\) \( \text{Cys}\) \( \textbf{U} \)
\( \text{Phe}\) \( \text{Ser}\) \( \text{Tyr}\)  \( \text{Cys}\) \( \textbf{C} \)
\( \text{Leu}\) \( \text{Ser}\) \( \text{STOP}\)  \( \text{STOP}\) \( \textbf{A} \)
\( \text{Leu}\) \( \text{Ser}\) \( \text{STOP}\) \( \text{Trp}\) \( \textbf{G} \)
\( \textbf{C} \) \( \text{Leu}\) \( \text{Pro}\) \( \text{His}\) \( \text{Arg}\) \( \textbf{U} \)
\( \text{Leu}\) \( \text{Pro}\) \( \text{His}\) \( \text{Arg}\) \( \textbf{C} \)
\( \text{Leu}\) \( \text{Pro}\) \( \text{Gln}\) \( \text{Arg}\) \( \textbf{A} \)
\( \text{Leu}\) \( \text{Pro}\) \( \text{Gln}\) \( \text{Arg}\) \( \textbf{G} \)
\( \textbf{A} \) \( \text{Ile}\) \( \text{Thr}\) \( \text{Asn}\) \( \text{Ser}\) \( \textbf{U} \)
\( \text{Ile}\) \( \text{Thr}\) \( \text{Asn}\) \( \text{Ser}\) \( \textbf{C} \)
\( \text{Ile}\) \( \text{Thr}\) \( \text{Lys}\) \( \text{Arg}\) \( \textbf{A} \)
\( \text{Met}\) \( \text{Thr}\) \( \text{Lys}\) \( \text{Arg}\) \( \textbf{G} \)
\( \textbf{G} \) \( \text{Val}\) \( \text{Ala}\) \( \text{Asp}\) \( \text{Gly}\) \( \textbf{U} \)
\( \text{Val}\) \( \text{Ala}\) \( \text{Asp}\)  \( \text{Gly}\) \( \textbf{C} \)
\( \text{Val}\) \( \text{Ala}\) \( \text{Glu}\)  \( \text{Gly}\) \( \textbf{A} \)
\( \text{Val}\) \( \text{Ala}\) \( \text{Glu}\) \( \text{Gly}\) \( \textbf{G} \)
 

It is correct to state

  1. identical amino acid sequences are found in all organisms.
  2. the genetic code is degenerate with respect to Met.
  3. the codon GGU adds Trp to a polypeptide chain.
  4. the DNA template sequence GAA codes for Leu.
Show Answers Only

\(D\)

Show Worked Solution

Consider each option:

Option A: The codon table shows that some codons code for different amino acids in
certain alternative translation systems → Incorrect

Option B: The genetic code is degenerate, meaning multiple codons can code for the
same amino acid, but this is not specific to Met → Incorrect

Option C: According to the codon table, the codon GGU codes for the amino acid
glycine (Gly), not tryptophan (Trp). The codon for Trp is UGG → Incorrect

Option D: Looking at the first position G, second position A, and third position A,
the codon GAA codes for the amino acid leucine (Leu) → Correct

\(\Rightarrow D\)


♦♦ Mean mark 41%. 
NOTE: 55% of students incorrectly chose A or B.

v1 Algebra, STD2 A4 2010 HSC 24b

Damo hires paddle boats in summertime as part of his water sports business. To calculate the cost,  \(C\), in dollars, of hiring  \(x\) paddle boats, he uses the equation  \(C=40+25x\).

He hires the paddle boats for $35 per hour and determines his income,  \(I\), in dollars, using the equation  \(I=35x\).
 

Use the graph to solve the two equations simultaneously for \(x\) and explain the significance of this solution for Damo's business.   (2 marks)

Show Answers Only

\(x=4\ \ \text{See worked solution}\)

Show Worked Solution

\(\text{From the graph, intersection occurs at}\ x=4\)

\(\rightarrow\ \text{Break-even point occurs at}\ x=4\)

\(\text{i.e. when 4 hours of paddle board hire occurs}\)

\(\text{Income}\) \(=35\times 4=$140\ \ \text{is equal to}\)
\(\text{Costs}\) \(=40+(25\times 4)=$140\)

\(\text{If}\ <4\ \text{hours of board hire}\ \rightarrow\ \text{LOSS for business}\)

\(\text{If}\ >4\ \text{hours of board hire}\ \rightarrow\ \text{PROFIT}\)


♦ Mean mark 36%.
MARKER’S COMMENT: The intersection on the graph is the same point at which the two simultaneous equations are solved for the given value of \(x\).

BIOLOGY, M6 2021 VCE 9

The following table provides information on three commonly grown genetically modified (GM) crops in Australia.

\begin{array} {|l|l|l|}
\hline
\rule{0pt}{2.5ex} \quad \ \  \textbf{Crop} \rule[-1ex]{0pt}{0pt} & \quad \quad \textbf{Genetic modification} &  \quad  \ \textbf{Characteristic given by} \\
& & \quad \quad \quad \quad \textbf{modification} \rule[-1ex]{0pt}{0pt} \\
\hline
\rule{0pt}{2.5ex} \text{GM cotton} \rule[-1ex]{0pt}{0pt} & \text{several bacterial genes inserted} & \text{insect resistance and herbicide} \\
& & \text{tolerance} \rule[-1ex]{0pt}{0pt} \\
\hline
\rule{0pt}{2.5ex} \text{GM canola} \rule[-1ex]{0pt}{0pt} & \text{two genes from two different} & \text{tolerance to several herbicides} \\
& \text{bacterial species inserted} & \rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex} \text{GM safflower} \rule[-1ex]{0pt}{0pt} & \text{a selection of genes silenced within} & \text{elevated levels of oleic acid in its} \\
& \text{the safflower genome} & \text{seeds} \rule[-1ex]{0pt}{0pt} \\
\hline
\end{array}

  1. Select one of the GM crops in the table above and justify whether or not this crop could be described as transgenic.   (1 mark)

  2. One issue with GM canola is the accidental release, during transport, of seeds along roadsides. Usually, unwanted plants that grow on the side of the road are killed using the herbicide glyphosate. However, GM canola is resistant to glyphosate.
  3. Suggest one practical solution for treating GM canola that is found growing along roadsides.  (1 mark)

  4. A new GM canola crop has been approved for use in Australia. It contains increased levels of omega-3 fatty acids, which are important in humans for building healthy cell membranes and for general growth and development, and also protect against a wide variety of diseases.
  5. Omega-3 has traditionally been sourced from fish. Due to the growing demand for sources of omega-3 , bioengineers have been encouraged to continue developing GM canola crops as a sustainable alternative.
  6. Discuss one social implication and one biological implication of using GM canola with increased levels of omega-3. Use a different implication in each response.   (4 marks)

Show Answers Only

a.   Answers should include one of the following:

→ GM cotton contains genes from other species or bacteria and is therefore transgenic.

→ GM canola contains genes from other species or bacteria and is therefore transgenic.

→ GM safflower does not contain genes from another species and is therefore not transgenic.
 

b.    Possible answers include:

→ use alternative herbicides that the GM canola is not resistant to

→ digging out and removing roadside GM canola by hand

→ mowing the roadsides could help manage GM canola

→ burn the GM canola using controlled methods.
 

c.    Social Implication – possible answers could include:

→ Depending on market demand non-GM canola farmers may see increased sales and improved quality of life, or decreased sales and lower quality of life.

→ GM canola farmers may benefit from higher yields and lower production costs may lead to increased profits and a better quality of life.

→ Consumers may benefit from improved nutrition that may lead to reduced strain on the healthcare system.

→ If GM canola becomes more affordable and accessible than alternative sources of nutrition, such as fish, a wider range of consumers will enjoy the health benefits.

→ Decreased consumption of fish due to the availability of cheaper, more accessible GM canola could lead to reduced sales and lower incomes for fish farmers, potentially decreasing their quality of life.

→ Some consumers may be hesitant to purchase or consume GM food products. This could lead to decreased demand for GM canola and lower incomes for farmers growing it.
 

Biological Implication – possible answers could include:

→ Crossbreeding between GM canola and non-GM canola crops could lead to changes in the genome of the crops which may reduce the genetic variation within the GM canola crop

→ Consumer safety concerns regarding consumption of GM products could negatively impact the demand for GM canola and other GM crops.

→ If GM canola leads to a reduction in fish consumption, it could have a positive impact on fish populations by reducing over-fishing. This could help to restore and maintain healthy fish populations in the long run.

→ Improved nutrition for consumers through the consumption of GM crops could lead to better health outcomes and improved well-being for the broader population.

Show Worked Solution

a.   Answers should include one of the following:

→ GM cotton contains genes from other species or bacteria and is therefore transgenic.

→ GM canola contains genes from other species or bacteria and is therefore transgenic.

→ GM safflower does not contain genes from another species and is therefore not transgenic.
 

b.    Possible answers include:

→ use alternative herbicides that the GM canola is not resistant to

→ digging out and removing roadside GM canola by hand

→ mowing the roadsides could help manage GM canola

→ burn the GM canola using controlled methods.
 

c.    Social Implication – possible answers could include:

→ Depending on market demand non-GM canola farmers may see increased sales and improved quality of life, or decreased sales and lower quality of life.

→ GM canola farmers may benefit from higher yields and lower production costs may lead to increased profits and a better quality of life.

→ Consumers may benefit from improved nutrition that may lead to reduced strain on the healthcare system.

→ If GM canola becomes more affordable and accessible than alternative sources of nutrition, such as fish, a wider range of consumers will enjoy the health benefits.

→ Decreased consumption of fish due to the availability of cheaper, more accessible GM canola could lead to reduced sales and lower incomes for fish farmers, potentially decreasing their quality of life.

→ Some consumers may be hesitant to purchase or consume GM food products. This could lead to decreased demand for GM canola and lower incomes for farmers growing it.
 

Biological Implication – possible answers could include:

→ Crossbreeding between GM canola and non-GM canola crops could lead to changes in the genome of the crops which may reduce the genetic variation within the GM canola crop

→ Consumer safety concerns regarding consumption of GM products could negatively impact the demand for GM canola and other GM crops.

→ If GM canola leads to a reduction in fish consumption, it could have a positive impact on fish populations by reducing over-fishing. This could help to restore and maintain healthy fish populations in the long run.

→ Improved nutrition for consumers through the consumption of GM crops could lead to better health outcomes and improved well-being for the broader population.

♦ Mean mark (c) 47%.

BIOLOGY, M6 2021 VCE 30 MC

The short-beaked echidna can be found in Australia and New Guinea. 

A university Biology student set out to compare DNA from four different short-beaked echidnas living in four different locations. The student aimed to determine the country of origin of each echidna. This was achieved using primers that amplified a 430 base pair (bp) sequence from the echidna's mitochondrial genome.

A flow chart of the steps taken by the student is shown below.
 

Which one of the following would be a requirement of either Step 1 or Step 2?

  1. Step 2 would require the use of a thermocycling machine to maintain the temperature at 37 °C.
  2. Step 1 would require the quill samples to have been stored at a very high temperature.
  3. Step 1 would require the extraction of DNA from the nuclei of quill epithelial cells.
  4. Step 2 would usually require a reaction mixture containing two primers.
Show Answers Only

\(D\)

Show Worked Solution

→ Polymerase chain reaction (PCR), step 2,  needs two primers for the amplification of DNA.

→ Step 1 is the isolation of mtDNA, which is not present in the nucleus.
 

\(\Rightarrow D\)


♦ Mean mark 46%.
COMMENT: 31% of students chose option C.

CHEMISTRY, M8 2012 VCE 6

The iron content in multivitamin tablets was determined using atomic absorption spectroscopy.

The absorbances of four standards were measured.

Three multivitamin tablets were selected. Each tablet was dissolved in 100.0 mL of water. The absorbance of each of the three solutions was then measured.

The following absorbances were obtained.

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \ \textbf{Solution} \rule[1ex]{0pt}{0pt} & \textbf{Concentration} & \textbf{Absorbance} \\
& \textbf{mg/L} & \\
\hline
\rule{0pt}{2.5ex} \text{Standard 1} \quad \quad & 0.00 & 0.06 \\
\hline
\rule{0pt}{2.5ex} \text{Standard 2} & 100.0 & 0.16 \\
\hline
\rule{0pt}{2.5ex} \text{Standard 3} & 200.0 & 0.25 \\
\hline
\rule{0pt}{2.5ex} \text{Standard 4} & 300.0 & 0.36 \\
\hline
\rule{0pt}{2.5ex} \text{Standard 5} & 400.0 & 0.46 \\
\hline
\rule{0pt}{2.5ex} \text{Tablet 1} & - & 0.39 \\
\hline
\rule{0pt}{2.5ex} \text{Tablet 2} & - & 0.42 \\
\hline
\rule{0pt}{2.5ex} \text{Tablet 3} & - & 0.45 \\
\hline
\end{array}

  1.  i.  Use the grid below to construct a calibration graph of the absorbances of the standard solutions.  (2 marks)

  2. ii.  Determine the average iron content, in milligrams, of the multivitamin tablets.  (2 marks)

Spectroscopic techniques work on the principle that, under certain conditions, atoms, molecules or ions will interact with electromagnetic radiation. The type of interaction depends on the wavelength of the electromagnetic radiation.

  1. Name one spectroscopic technique that you have studied this year.
    1. Which part of the electromagnetic spectrum does this technique use?  (1 mark)

    2. How does this part of the electromagnetic spectrum interact with matter? What information does this spectroscopic technique provide?  (2 marks)

Show Answers Only

a.i.  

a.ii.  \(35.5\ \text{mg}\)
 

b.i. Answers could include:

→ AAS (visible light)

→ UV-Vis (UV or visible light)

→ IR (Infrared radiation)

→ NMR (radio waves)
 

b.ii. Spectroscopic technique: AAS (one of many possible – see b.i.)

→ During AAS energy of a certain frequency is transferred to electrons within atoms to move them into higher energy levels.

→ The absorption of the light indicates the concentration of the targeted element within the sample.

Show Worked Solution

a.i.  

a.ii. Average absorbance (tablets) \(=\dfrac{0.39+0.42+0.45}{3}=0.42\)

Using the graph: absorbance value of \(0.42 → 355\ \text{mg L}^{-1}\)

\(\ce{m(Fe) (100\ \text{ml}) = 355 \times 0.1 =35.5\ \text{mg}}\)
 

b.i.  Answers could include:

→ AAS (visible light)

→ UV-Vis (UV or visible light)

→ IR (Infrared radiation)

→ NMR (radiowaves)
 

b.ii. Spectroscopic technique: AAS (one of many possible – see b.i.)

→ During AAS energy of a certain frequency is transferred to electrons within atoms to move them into higher energy levels.

→ The absorption of the light indicates the concentration of the targeted element within the sample.

♦ Mean mark (b.ii) 42%.

v1 Algebra, STD2 A4 SM-Bank 4

Bec is a baker and makes cookies to sell every week.

The cost of making \(n\) cookies, $\(C\),  can be calculated using the equation

\(C=400+2.5n\)

Bec sells the cookies for $4.50 each, and her income is calculated using the equation

\(I=450n\)

  1. On the grid above, draw the graphs of  \(C\) and \(I\).  (2 marks)

  2. On the graph, label the breakeven point and the loss zone.  (2 marks)

Show Answers Only

(i) and (ii)

Show Worked Solution
i.   

 

ii.  \(\text{Loss zone occurs when}\ C > I,\ \text{which is shaded}\)

\(\text{in the diagram above.}\)

v1 Algebra, STD2 A4 2005 HSC 28b

Jake and Preston are planning a fund-raising event at the local swim centre. They can have access to the giant pool float for $550 and the party room hire for $250. A sausage sizzle and drinks will cost them $9 per person.

  1. Write a formula for the cost ($C) of running the event for \(x\) people. (1 mark)

The graph shows planned income and costs when the ticket price is $15. 
  

  1. Estimate the minimum number of people needed at the fund raising event to cover the costs.  (1 mark)

  2. How much profit will be made if 200 people attend the fund raiser? (1 mark)

Jake and Preston have 300 tickets to sell. They want to make a profit of $1510.

  1. What should be the price of a ticket, assuming all 300 tickets will be sold?  (3 marks)

Show Answers Only
  1. \(C=800+9x\)
  2. \(\text{Approximately }135\)
  3. \($400\)
  4. \($16.70\)
Show Worked Solution
i.    \($C\) \(=550+250+(9\times x)\)
    \(=800+9x\)

 

ii.  \(\text{Using the graph intersection}\)

\(\text{Approximately 135 people are needed}\)

\(\text{to cover the costs.}\)

 

iii.  \(\text{If 200 people attend}\)

\(\text{Income}\) \(=200\times $15\)
  \(=$3000\)
\(\text{Costs}\) \(=800+(9\times 200)\)
  \(=$2600\)

 

\(\therefore\ \text{Profit}\) \(=3000-2600\)
  \(=$400\)

 

iv.  \(\text{Costs when}\ x=300:\)

\(C\) \(=800+(9\times 300)\)
  \(=$3500\)

 

\(\text{Income required to make }$1510\ \text{profit}\)

\(=3500+1510\)

\(=$5010\)
 

\(\therefore\ \text{Price per ticket}\) \(=\dfrac{5010}{300}\)
  \(=$16.70\)

v1 Algebra, STD2 A4 2020 HSC 24

There are two tanks at an industrial plant, Tank A and Tank B. Initially, Tank A holds 2520 litres of liquid fertiliser and Tank B is empty.

  1. Tank A begins to empty liquid fertiliser into a transport vehicle at a constant rate of 40 litres per minute.

     

    The volume of liquid fertiliser in Tank A is modelled by  \(V=1400-40t\)  where \(V\) is the volume in litres and  \(t\) is the time in minutes from when the tank begins to drain the fertiliser.

     

    On the grid below, draw the graph of this model and label it as Tank A.   (1 mark)
     

     

  2. Tank B remains empty until  \(t=10\)  when liquid fertiliser is added to it at a constant rate of 60 litres per minute.
     By drawing a line on the grid (above), or otherwise, find the value of  \(t\)  when the two tanks contain the same volume of liquid fertiliser.  (2 marks)

  3. Using the graphs drawn, or otherwise, find the value of  \(t\)  (where  \(t > 0\)) when the total volume of liquid fertiliser in the two tanks is 1400 litres.  (1 mark)

Show Answers Only
  1.  \(\text{Tank} \ A \ \text{will pass through (0, 1400) and (35, 0)}\)
      
  2. \(20 \ \text{minutes}\)
  3. \(30 \ \text{minutes}\)
Show Worked Solution

a.     \(\text{Tank} \ A \ \text{will pass through (0, 1400) and (35, 0)}\)
 

 

b.   \(\text{Tank} \ B \ \text{will pass through (10, 0) and (30, 1200)}\)  
 

 

\(\text{By inspection, the two graphs intersect at} \ \ t = 20 \ \text{minutes}\)

c.   \(\text{Strategy 1}\)

\(\text{By inspection of the graph, consider} \ \ t = 30\)

\(\text{Tank A} = 200 \ \text{L} , \ \text{Tank B} =1200 \ \text{L}\)

\(\therefore\ \text{Total volume = 1400 L when  t = 30}\)
  

\(\text{Strategy 2}\)

\(\text{Total Volume}\) \(=\text{Tank A} + \text{Tank B}\)
\(1400\) \(=1400-40t+(t-10)\times 60\)
\(1400\) \(=1400-40t+60t-600\)
\(20t\) \(= 600\)
\(t\) \(= 30 \ \text{minutes}\)

♦♦ Mean mark part (c) 22%.

v1 Algebra, STD2 A4 2019 HSC 36

A small business makes and dog kennels.

Technology was used to draw straight-line graphs to represent the cost of making the dog kennels \((C)\) and the revenue from selling dog kennels \((R)\). The \(x\)-axis displays the number of dog kennels and the \(y\)-axis displays the cost/revenue in dollars.
 


 

  1. How many dog kennels need to sold to break even?  (1 mark)

  2. By first forming equations for cost `(C)` and revenue `(R)`, determine how many dog kennels need to be sold to earn a profit of $2500.  (3 marks)

Show Answers Only
  1. \(20\)
  2. \(145\)
Show Worked Solution

a.  \(20\ \ (x\text{-value at intersection})\)

  

b.   \(\text{Find equations of both lines}:\)

\((0, 400)\ \text{and}\ (20, 600)\ \text{lie on}\ \ C\)

\(\text{gradient}_C = \dfrac{600-400}{20-0}=10\)

\(\rightarrow\ C=400+10x\)
   

\((0,0)\ \text{and}\ (20, 600)\ \text{lie on}\ \ R\)

\(\text{gradient}_R =\dfrac{600-0}{20-0}=30\)

\(\rightarrow\ R=30x\)
 

\(\text{Profit} = R-C\)

\(\text{Find}\ \ x\ \text{when Profit }= $2500:\)

\(2500\) \(=30x-(400+10x)\)
\(20x\) \(=2900\)
\(x\) \(=145\)

  
\(\therefore\ 145\ \text{dog kennels need to be sold to earn }$2500\ \text{profit}\)


♦♦ Mean mark 28%.

v1 Algebra, STD2 A4 2004 HSC 16 MC

Uri drew a correct diagram that gave the solution to the simultaneous equations

\(y=2x+3\)  and  \(y=x+4\).

Which diagram did he draw?
  

Show Answers Only

\(D\)

Show Worked Solution

\(\text{By elimination:}\)

\(y=2x+3\ \text{cuts the }y \text{-axis at}\ 3\)

\(\rightarrow\ \text{Eliminate be A and B}\)

 

\(y=x+4\ \text{cuts the }y\text{-axis at}\ 4\)

\(\text{AND has a positive gradient}\)

\(\rightarrow\ \text{Eliminate C}\)

\(\Rightarrow D\)

v1 Algebra, STD2 A4 2011 HSC 20 MC

A function centre hosts events for up to 500 people. The cost \(C\), in dollars, for the centre
to host an event, where \(x\) people attend, is given by:

\(C=20\ 000+40x\)

The centre charges $120 per person. Its income \(I\), in dollars, is given by:

\(I=120x\)
 

How much greater is the income of the function centre when 500 people attend an event, than its income at the breakeven point?

  1. \($10\ 000\)
  2. \($20\ 000\)
  3. \($30\ 000\)
  4. \($40\ 000\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{When}\ x=500,\ I=120\times 500=$60\ 000\)

\(\text{Breakeven when}\ \ x=250\ \ \text{(from graph)}\)

\(\text{When}\ \ x=250,\ I=120\times 250=$30\ 000\)

\(\text{Difference}\) \(=60\ 000-30\ 000\)
  \(=$30\ 000\)

 
\(\Rightarrow C\)


♦ Mean mark 50%
COMMENT: Students can read the income levels directly off the graph to save time and then check with the equations given.

v1 Algebra, STD2 A2 2009 HSC 13 MC

The volume of water in a tank changes over six months, as shown in the graph.
 

Consider the overall decrease in the volume of water.

What is the average percentage decrease in the volume of water per month over this time, to the nearest percent?

  1. 6%
  2. 12%
  3. 35%
  4. 64%
Show Answers Only

\(B\)

Show Worked Solution
\(\text{Initial Volume}\) \(=50\ 000\ \text{L}\)
\(\text{Final volume}\) \(=15\ 000\ \text{L}\)
\(\text{Decrease}\) \(=50\ 000-15\ 000\)
  \(=35\ 000\ \text{L   (over 6 months)}\)

 

\(\text{Loss per month}\) \(=\dfrac{35\ 000}{6}\)
  \(=5833.33\dots\ \text{L per month}\)
\(\text{% loss per month}\) \(=\dfrac{5833.33\dots}{50\ 000}\times 100\%\)
  \(=11.666\dots \%\)

 
\(\Rightarrow B\)


♦ Mean mark 48%.
 COMMENT: Remember that % decrease requires the decrease in volume to be divided by the original volume (50,000L)

v1 Algebra, STD2 A2 2009 HSC 24d

A factory makes both cloth and leather lounges. In any week

• the total number of cloth lounges and leather lounges that are made is 400
• the maximum number of leather lounges made is 270
• the maximum number of cloth lounges made is 325.

The factory manager has drawn a graph to show the numbers of leather lounges (\(x\)) and cloth lounges (\(y\)) that can be made.
 

 

  1. Find the equation of the line \(AD\).   (1 mark)

  2. Explain why this line is only relevant between \(B\) and \(C\) for this factory.     (1 mark)

  3. The profit per week, \($P\), can be found by using the equation  \(P = 2520x + 1570y\).

     

    Compare the profits at \(B\) and \(C\).     (2 marks)

Show Answers Only
  1. \(x+y=400\)
  2. \(\text{Since the max amount of leather lounges}=270\)

     

    \(\rightarrow\ x\ \text{cannot be}\ >270\)

     

    \(\text{Since the max amount of cloth lounges}=325\)

     

    \(\rightarrow\ y\ \text{cannot be}\ >325\)

     

    \(\therefore\ \text{The line}\ AD\ \text{is only possible between}\ B\ \text{and}\ C.\)

  3. \(\text{The profits at}\ C\ \text{are }$185\ 250\ \text{more than at}\ B.\)
Show Worked Solution

i.   \(\text{We are told the number of leather lounges}\ (x),\)

\(\text{and cloth lounges}\  (y),\ \text{made in any week} = 400\)

\(\rightarrow\ \text{Equation of}\ AD\ \text{is}\ x+y=400\)


♦♦♦ Mean mark part (i) 14%.
Using \(y=mx+c\) is a less efficient but equally valid method, using  \(m=–1\)  and  \(b=400\) (\(y\)-intercept).

ii.   \(\text{Since the max amount of leather lounges}=270\)

\(\rightarrow\ x\ \text{cannot}\ >270\)

\(\text{Since the max amount of cloth lounges}=325\)

\(\rightarrow\ y\ \text{cannot}\ >325\)

\(\therefore\ \text{The line}\ AD\ \text{is only possible between}\ B\ \text{and}\ C.\)


♦ Mean mark part (ii) 49%.

iii.  \(\text{At}\ B,\ x=75,\ y=325\)

\(\rightarrow\ $P  (\text{at}\ B)\) \(=2520\times 75+1570\times 325\)
  \(=189\ 000+510\ 250\)
  \(=$699\ 250\)

  
\(\text{At}\ C,\ x=270,\ y=130\)

\(\rightarrow\ $P  (\text{at}\ C)\) \(=2520\times 270+1570\times 130\)
  \(=680\ 400+204\ 100\)
  \(=$884\ 500\)

  
\(\text{Difference in profits}=$884\ 500-$699\ 250=$185\ 250\)

\(\text{The profits at}\ C\ \text{are } $185\ 250\ \text{more than at}\ B.\)


♦ Mean mark (iii)40%.

v1 Algebra, STD2 A2 2010 HSC 27c

The graph shows tax payable against taxable income, in thousands of dollars.
  

  1. Use the graph to find the tax payable on a taxable income of \($18\ 000\).  (1 mark)

  2. Use suitable points from the graph to show that the gradient of the section of the graph marked  \(A\)  is  \(\dfrac{7}{15}\).    (1 mark)

  3. How much of each dollar earned between  \($18\ 000\)  and  \($33\ 000\) is payable in tax? Give your answer correct to the nearest whole number.   (1 mark)

  4. Write an equation that could be used to calculate the tax payable, \(T\), in terms of the taxable income, \(I\), for taxable incomes between  \($18\ 000\)  and  \($33\ 000\).   (2 marks)

Show Answers Only
  1. \($3000\ \ \text{(from graph)}\)
  2. \(\text{See worked solution}\)
  3. \(46\frac{2}{3}\approx  47\ \text{cents per dollar earned}\)
  4. \(\text{Tax payable →}\ T=\dfrac{7}{15}I-5400\)
Show Worked Solution
i.   

\(\text{Income on}\ $18\ 000=$3000\ \ \text{(from graph)}\)

  

ii.  \(\text{Using the points}\ (18, 3)\ \text{and}\ (33, 10)\)

\(\text{Gradient at}\ A\) \(=\dfrac{y_2-y_1}{x_2-x_1}\)
  \(=\dfrac{10\ 000-3000}{33\ 000-18\ 000}\)
  \(=\dfrac{7000}{15\ 000}\)
  \(=\dfrac{7}{15}\ \ \ \ \text{… as required}\)

♦♦ Mean mark (ii) 25%.

iii.  \(\text{The gradient represents the tax applicable on each dollar}\)

\(\text{Tax}\) \(=\dfrac{7}{15}\ \text{of each dollar earned}\)
  \(=46\frac{2}{3}\approx 47\ \text{cents per dollar earned (nearest whole number)}\)

♦♦♦ Mean mark (iii) 12%!
MARKER’S COMMENT: Interpreting gradients is an examiner favourite, so make sure you are confident in this area.

iv.  \(\text{Tax payable up to }$18\ 000 = $3000\)

\(\text{Tax payable on income between }$18\ 000\ \text{and }$33\ 000\)

\(=\dfrac{7}{15}(I-18\ 000)\)

\(\therefore\ \text{Tax payable →}\ \ T\) \(=3000+\dfrac{7}{15}(I-18\ 000)\)
  \(=3000+\dfrac{7}{15} I-8400\)
  \(=\dfrac{7}{15}I-5400\)

♦♦♦ Mean mark (iv) 15%.
STRATEGY: The earlier parts of this question direct students to the most efficient way to solve this question. Make sure earlier parts of a question are front and centre of your mind when devising strategy.