SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

BIOLOGY, M3 EQ-Bank 3 MC

Consider the following adaptations in Australian organisms:

\(\text{I.}\)   The ability of koalas to detoxify eucalyptus leaves.  
\(\text{II.}\)   The migration of bogong moths to alpine areas in summer.  
\(\text{III.}\)   The spines on an echidna.  
\(\text{IV.}\)   The camouflage patterns on a leaf-tailed gecko.  

 
Which combination correctly identifies one of each type of adaptation (structural, physiological, and behavioural)?

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

\(D\)

Show Worked Solution

Consider each statement:

\(\text{I.}\) Physiological adaptation. The ability of koalas to detoxify eucalyptus leaves involves internal metabolic processes.

\(\text{II.}\) Behavioral adaptation. The migration of bogong moths is a seasonal behavior that helps them survive and reproduce.

\(\text{III.}\) Structural adaptation. The spines on an echidna are physical features that provide protection.

\(\text{IV.}\) Structural adaptation. The camouflage patterns on a leaf-tailed gecko are physical characteristics, though they can change slightly which could be considered physiological.

\(\Rightarrow D\)

CHEMISTRY, M3 EQ-Bank 8 MC

Which statement best explains how molecular orientation affects the rate of a chemical reaction according to collision theory?

  1. Reactant molecules must collide with enough energy, but the orientation of the molecules is irrelevant.
  2. Correct orientation only affects reactions that involve more than two reactant molecules.
  3. Incorrect molecular orientation increases the activation energy required for the reaction to proceed.
  4. Even with sufficient energy, reactants must collide in the correct orientation to allow bonds to break and form.
Show Answers Only

\(D\)

Show Worked Solution

→ For a reaction to occur, not only must reactants collide with sufficient energy, but they must also collide in the correct orientation to break existing bonds and form new ones.

→ Even if the reactants have enough energy, incorrect orientation can prevent a successful reaction.

\(\Rightarrow D\)

CHEMISTRY, M3 EQ-Bank 14

A student conducted an experiment to investigate how the concentration of hydrochloric acid affects the rate of its reaction with magnesium ribbon. They measured the volume of hydrogen gas produced at regular intervals in reactions using 1.0 M, 2.0 M, and 3.0 M hydrochloric acid, keeping other variables constant.

  1. Explain how the concentration of hydrochloric acid influences the rate of reaction with magnesium, using collision theory.   (2 marks)
  1. Describe how the student could modify the experiment to investigate the effect of temperature on the reaction rate.   (2 marks)
  1. Identify and explain one advantage of a student using digital technologies such as a gas pressure sensor to collect or analyse data from the experiment.   (2 marks)
Show Answers Only

a.   \(\ce{[HCl]} influence on the rate of reaction:

→ As the concentration increases, the number of acid particles in a given volume also increases.

→ This leads to more frequent collisions between hydrochloric acid molecules and the magnesium surface.

→ As a result, a higher concentration of hydrochloric acid increases the chance of successful collisions, which increases the reaction rate and causes hydrogen gas to be produced more quickly.
 

b.   Experiment modifications:

→ To investigate the effect of temperature on the reaction rate, the student could modify the experiment by conducting the reaction at different temperatures while keeping the concentration of hydrochloric acid constant.

→ They could use a water bath or hot plate to control the temperature for each trial. For example, the student could perform the reaction at 20°C, 30°C, and 40°C, then measure and compare the volume of hydrogen gas produced over time at each temperature.
 

c.   Advantage of digital technologies:

→ Using a gas pressure sensor in the experiment allows for continuous and precise data collection without human intervention.

→ This reduces the chances of human error when measuring the volume of gas produced manually at intervals, leading to more accurate and reliable results.

→ Additionally, the sensor can automatically record data over time, providing detailed information about the reaction rate that can be easily analysed using digital tools such as graphing software.

Show Worked Solution

a.   \(\ce{[HCl]} influence on the rate of reaction:

→ As the concentration increases, the number of acid particles in a given volume also increases.

→ This leads to more frequent collisions between hydrochloric acid molecules and the magnesium surface.

→ As a result, a higher concentration of hydrochloric acid increases the chance of successful collisions, which increases the reaction rate and causes hydrogen gas to be produced more quickly.
 

b.   Experiment modifications:

→ To investigate the effect of temperature on the reaction rate, the student could modify the experiment by conducting the reaction at different temperatures while keeping the concentration of hydrochloric acid constant.

→ They could use a water bath or hot plate to control the temperature for each trial. For example, the student could perform the reaction at 20°C, 30°C, and 40°C, then measure and compare the volume of hydrogen gas produced over time at each temperature.
 

c.   Advantage of digital technologies:

→ Using a gas pressure sensor in the experiment allows for continuous and precise data collection without human intervention.

→ This reduces the chances of human error when measuring the volume of gas produced manually at intervals, leading to more accurate and reliable results.

→ Additionally, the sensor can automatically record data over time, providing detailed information about the reaction rate that can be easily analysed using digital tools such as graphing software.

CHEMISTRY, M3 EQ-Bank 13

A certain chemical reaction begins with a high reaction rate but over time this rate of reaction slows until the reaction is complete.

Explain this statement using collision theory.   (3 marks)

Show Answers Only

→ At the start of the reaction, there is a high concentration of reactant molecules, which increases the frequency of collisions between them. According to collision theory, more frequent and effective collisions lead to a higher reaction rate.

→ As the reaction progresses, the concentration of reactants decreases, leading to fewer collisions. With fewer successful collisions occurring over time, the reaction rate slows down.

→ Eventually, when most or all of the reactants are used up, the reaction reaches completion, and the rate drops to zero.

Show Worked Solution

→ At the start of the reaction, there is a high concentration of reactant molecules, which increases the frequency of collisions between them. According to collision theory, more frequent and effective collisions lead to a higher reaction rate.

→ As the reaction progresses, the concentration of reactants decreases, leading to fewer collisions. With fewer successful collisions occurring over time, the reaction rate slows down.

→ Eventually, when most or all of the reactants are used up, the reaction reaches completion, and the rate drops to zero.

BIOLOGY, M3 EQ-Bank 6

"The prickly pear cactus, introduced to Australia in the 1800s, became a significant ecological problem before being brought under control."

Describe two significant changes to the ecosystem caused by the prickly pear invasion and the control method used to manage its population. In your answer, discuss the effectiveness of this control measure and its ecological implications.   (4 marks)

Show Answers Only

→ The prickly pear outcompeted native vegetation, forming dense thickets that reduced grazing land for livestock and native animals.

→ It also altered habitat structures, displacing native fauna and changing local biodiversity patterns. eg. it harboured pest species like rabbits.

→ To control the prickly pear population, scientists introduced the Cactoblastis moth, whose larvae feed exclusively on prickly pear species.

→ This biological control method proved remarkably effective, with the moths feeding almost exclusively on the prickly pear. As prickly pear numbers reduced, the moths also became rarer.

→ This control measure rapidly reduced prickly pear populations to manageable levels within a decade, restoring native ecosystems and millions of hectares of land for agricultural use.

→ This demonstrates the potential power and risks of biological control agents, as the moth’s effectiveness in Australia led to its introduction in other parts of the world, where it has sometimes become a pest itself.

Show Worked Solution

→ The prickly pear outcompeted native vegetation, forming dense thickets that reduced grazing land for livestock and native animals.

→ It also altered habitat structures, displacing native fauna and changing local biodiversity patterns. eg. it harboured pest species like rabbits.

→ To control the prickly pear population, scientists introduced the Cactoblastis moth, whose larvae feed exclusively on prickly pear species.

→ This biological control method proved remarkably effective, with the moths feeding almost exclusively on the prickly pear. As prickly pear numbers reduced, the moths also became rarer.

→ This control measure rapidly reduced prickly pear populations to manageable levels within a decade, restoring native ecosystems and millions of hectares of land for agricultural use.

→ This demonstrates the potential power and risks of biological control agents, as the moth’s effectiveness in Australia led to its introduction in other parts of the world, where it has sometimes become a pest itself.

BIOLOGY, M3 EQ-Bank 1

Describe two biotic factors that could act as selection pressures in a grassland ecosystem.

For each factor, briefly explain how it might influence the organisms living in that environment.   (4 marks)

Show Answers Only

Predation:

→ Predation by herbivores like grazing animals could favour plants with defensive structures or compounds.

→ eg. the Acacia tree, common in Australian grasslands, has developed long, sharp thorns as a defensive structure against herbivores.
 

Competition:

→ Competition for resources such as light or nutrients might select for faster-growing or taller plant species. 

→ Gum trees can grow to impressive heights which gives them an advantage in capturing sunlight. 
 

Parasites:

→ Parasites causing fungal infections, could drive the evolution of resistance mechanisms in both plants and animals.

→ Ticks are a parasite that can attach to the skin of animals like kangaroos and cattle, feeding on their blood and transmitting diseases. Animals with tick resistance will have a survival advantage.

Show Worked Solution

Answers could include two of the following.

Predation:

→ Predation by herbivores like grazing animals could favour plants with defensive structures or compounds.

→ eg. the Acacia tree, common in Australian grasslands, has developed long, sharp thorns as a defensive structure against herbivores.
 

Competition:

→ Competition for resources such as light or nutrients might select for faster-growing or taller plant species. 

→ Gum trees can grow to impressive heights which gives them an advantage in capturing sunlight.
 

Parasites:

→ Parasites causing fungal infections, could drive the evolution of resistance mechanisms in both plants and animals.

→ Ticks are a parasite that can attach to the skin of animals like kangaroos and cattle, feeding on their blood and transmitting diseases. Animals with tick resistance will have a survival advantage.

CHEMISTRY, M3 EQ-Bank 3 MC

When molecules react, the final rate of a chemical reaction is determined by which of the following factors?

  1. The proportion of collisions where the number of molecules are keep the same.
  2. The proportion of collisions where molecules have the correct orientation
  3. The proportion of collisions where atoms are destroyed
  4. The proportion of collisions where molecules form larger structures
Show Answers Only

\(B\)

Show Worked Solution

→ For a reaction to occur, it is not enough for molecules to simply collide.

→ They must collide with the correct orientation for bonds to break and new ones to form. Even if the energy is sufficient, an incorrect orientation will not result in a successful reaction.

→ Therefore, the rate of a chemical reaction is largely determined by the proportion collisions that occur with the proper orientation between reacting molecules.

\(\Rightarrow B\)

CHEMISTRY, M3 EQ-Bank 10

A student conducted a series of  investigations where 8.50 g of sodium carbonate was reacted with excess nitric acid \(\ce{(HNO3)}\) at a temperature of 25°C and 100 kPa. The volume of carbon dioxide gas produced was measured at regular intervals during each investigation. In experiment A, sodium carbonate was provided as large crystals, and in experiment B, it was supplied in powdered form.

Both reactions produced 1.988 L of \(\ce{CO2(g)}\) however experiment B finished reacting before experiment A finished reacting.

  1. Explain why experiment B had a faster rate of reaction than experiment A.   (1 mark)
  1. Using the volume of \(\ce{CO2(g)}\) produced, calculate the maximum mass of carbon dioxide produced in the reaction.   (3 marks)
  1. Explain why both experiments produced in the same volume of carbon dioxide.   (1 mark)
Show Answers Only

a.   Experiment B faster than experiment A:

→ In experiment B, the surface area of the sodium carbonate was greater than in experiment A due to it being in a powdered form.

→ Thus, there are a greater number of collisions able to occur in experiment B, leading to a greater number of successful collisions.

→ Overall this allows the rate of reaction of experiment B to be greater than that of experiment A.

b.    \(3.52\ \text{g}\)

c.   Reasons why same volume \(\ce{CO2}\) produced:

→ Both experiments used the same amount of sodium carbonate reacting with excess hydrochloric acid.

→ The maximum amount of carbon dioxide produced is dependent on how much sodium carbonate reacted.

→ As the initial mass of sodium carbonate is the same in both reaction and both reactions went until completion, the volume of carbon dioxide will be the same despite the different rates of reaction between experiment A and B.

Show Worked Solution

a.   Experiment B faster than experiment A:

→ In experiment B, the surface area of the sodium carbonate was greater than in experiment A due to it being in a powdered form.

→ Thus, there are a greater number of collisions able to occur in experiment B, leading to a greater number of successful collisions.

→ Overall this allows the rate of reaction of experiment B to be greater than that of experiment A.
 

b.   The maximum volume of carbon dioxide produced is 1.988 L.

→ At SLC (standard laboratory conditions), the 1 mole of gas takes up 24.79 L.

→ \(\ce{n(CO2(g))}= \dfrac{1.988}{24.79} = 0.08\ \text{mol}\)

→ \(\ce{m(CO2(g))} = n \times MM = 0.08 \times 44.01 =3.52\ \text{g}\)
 

c.   Reasons why same volume \(\ce{CO2}\) produced:

→ Both experiments used the same amount of sodium carbonate reacting with excess hydrochloric acid.

→ The maximum amount of carbon dioxide produced is dependent on how much sodium carbonate reacted.

→ As the initial mass of sodium carbonate is the same in both reaction and both reactions went until completion, the volume of carbon dioxide will be the same despite the different rates of reaction between experiment A and B.

Functions, EXT1 F1 EQ-Bank 4

Let  \(f(x)=x^2-4 x+3\)  for  \(x \leqslant 2\).

  1. State the domain of  \(y=f^{-1}(x)\).  (1 mark)

  2. What is the equation of  \(y=f^{-1}(x)\) ?  (3 marks)

  3. For the restricted domain, sketch the function and the inverse function on the same number plane below.
  4. Clearly identify all axis intercepts.  (2 marks)

Show Answers Only

a.   \(f(x)=(x-1)(x-3)\ \Rightarrow \ \text{Axis of symmetry at}\ \ x=2. \)

\(f(2)=-1\)

\(\text{Range}\ f(x): \  y \geqslant -1\ \Rightarrow \ \text{Domain}\ f^{-1}(x): \  x \geqslant -1 \)
 

b.   \(y=2-\sqrt{x+1}\)

c.   \(f(x)\ \text{intercepts:}\ (1,0), (0,3) \)

\(f^{-1}(x)\ \text{intercepts:}\ (3,0), (0,1) \)

Show Worked Solution

a.   \(f(x)=(x-1)(x-3)\ \Rightarrow \ \text{Axis of symmetry at}\ \ x=2. \)

\(f(2)=-1\)

\(\text{Range}\ f(x): \  y \geqslant -1\ \Rightarrow \ \text{Domain}\ f^{-1}(x): \  x \geqslant -1 \)
 

b.   \(\text{Inverse: swap}\ x ↔ y \)

\(x\) \(=y^2-4y+3\)  
\(x\) \(=(y-2)^2-1\)  
\(x+1\) \(=(y-2)^2\)  
\(y-2\) \(=\pm \sqrt{x+1} \)  
\(y\) \(=2 \pm \sqrt{x+1} \)  
\(y\) \(=2-\sqrt{x+1}\ \ \ \ (\text{Range}\ f^{-1}(x): \ y \leqslant 2) \)  

 
c.   \(f(x)\ \text{intercepts:}\ (1,0), (0,3) \)

\(f^{-1}(x)\ \text{intercepts:}\ (3,0), (0,1) \)

Calculus, EXT1 C1 EQ-Bank 5 MC

Simplify  \(\dfrac{\sqrt{1+\tan ^2 \theta} \sqrt{1-\sin ^2 \theta}}{\sqrt{\operatorname{cosec}^2 \theta-1}}, \operatorname{cosec}^2 \theta \neq 1\)

  1. \(\tan \theta\)
  2. \(\cot \theta\)
  3. \(\sec \theta\)
  4. \(1\)
Show Answers Only

\(A\)

Show Worked Solution
  \(\dfrac{\sqrt{1+\tan ^2 \theta} \sqrt{1-\sin ^2 \theta}}{\sqrt{\operatorname{cosec}^2 \theta-1}} \) \(=\dfrac{\sqrt{\sec ^2 \theta} \sqrt{\cos ^2 \theta}}{\sqrt{\cot ^2 \theta}} \)
    \( =\dfrac{\sec \theta\, \cos \theta}{\cot \theta}\)
    \( =\dfrac{1}{\cot \theta}\)
    \( =\tan \theta\)

 
\(\Rightarrow A\)

BIOLOGY, M2 EQ-Bank 6

Describe the pulmonary circuit and the changes in blood composition as it circulates through this system.   (4 marks)

Show Answers Only

→ The pulmonary circuit is the part of the circulatory system that carries deoxygenated blood from the heart to the lungs for oxygenation and then returns oxygenated blood back to the heart

→ Deoxygenated blood entering the lungs has low oxygen levels and high carbon dioxide levels. As it passes through the alveolar capillaries, oxygen diffuses into the blood while carbon dioxide diffuses out.

→ This process results in blood leaving the lungs with higher oxygen content and lower carbon dioxide levels.

→ Additionally, the blood slightly cools as it passes through the lungs due to its proximity to the air in the alveoli and the evaporation of water vapour during exhalation.

→ The pH of the blood also slightly increases as carbon dioxide (which forms carbonic acid in blood) is removed.

Show Worked Solution

→ The pulmonary circuit is the part of the circulatory system that carries deoxygenated blood from the heart to the lungs for oxygenation and then returns oxygenated blood back to the heart

→ Deoxygenated blood entering the lungs has low oxygen levels and high carbon dioxide levels. As it passes through the alveolar capillaries, oxygen diffuses into the blood while carbon dioxide diffuses out.

→ This process results in blood leaving the lungs with higher oxygen content and lower carbon dioxide levels.

→ Additionally, the blood slightly cools as it passes through the lungs due to its proximity to the air in the alveoli and the evaporation of water vapour during exhalation.

→ The pH of the blood also slightly increases as carbon dioxide (which forms carbonic acid in blood) is removed.

BIOLOGY, M2 EQ-Bank 5

Compare and contrast the structure of red blood cells (erythrocytes) and platelets (thrombocytes), explaining how their unique features contribute to their efficiency in performing their respective functions in the circulatory system.   (4 marks)

Show Answers Only

→ Red blood cells are biconcave discs lacking a nucleus, which maximises their capacity to carry haemoglobin and increases their flexibility to squeeze through narrow capillaries, enhancing oxygen transport efficiency.

→ Platelets are flat, irregularly-shaped cell fragments that lack both organelles and nuclei. When blood vessel damage occurs, they change shape and aggregate to form a blood clot and stop bleeding.

→ The absence of a nucleus in red blood cells allows for more haemoglobin but limits their lifespan to about 120 days, while platelets typically survive for 8-10 days.

→ Red blood cells’ structure optimises oxygen carrying capacity and delivery, while platelets’ structure facilitates rapid response to vascular damage. 

→ Both red blood cells and platelets have high surface area to volume ratios which allows them to efficiently carry out their primary functions.

Show Worked Solution

→ Red blood cells are biconcave discs lacking a nucleus, which maximises their capacity to carry haemoglobin and increases their flexibility to squeeze through narrow capillaries, enhancing oxygen transport efficiency.

→ Platelets are flat, irregularly-shaped cell fragments that lack both organelles and nuclei. When blood vessel damage occurs, they change shape and aggregate to form a blood clot and stop bleeding.

→ The absence of a nucleus in red blood cells allows for more haemoglobin but limits their lifespan to about 120 days, while platelets typically survive for 8-10 days.

→ Red blood cells’ structure optimises oxygen carrying capacity and delivery, while platelets’ structure facilitates rapid response to vascular damage. 

→ Both red blood cells and platelets have high surface area to volume ratios which allows them to efficiently carry out their primary functions.

BIOLOGY, M2 EQ-Bank 3

The diagram below shows features that are observed in cross-sections of three types of blood vessel.
 

   

  1. Complete the diagram by identifying the type of blood vessel identified by \(\text{A}\) and \(\text{C}\).   (2 marks)
  2. Describe how the structural features of \(\text{B}\) contribute to its role in the circulatory system.   (2 marks)
Show Answers Only

a.   

 
b.   \(\text{B}\) is an artery.

→ Arteries have thick, elastic walls with multiple layers of smooth muscle, allowing them to withstand and maintain the high pressure of blood pumped directly from the heart.

→ This elasticity also enables arteries to expand and recoil with each heartbeat, helping to propel blood forward in a phenomenon known as the “pulse.”

→ Additionally, the smooth inner lining of arteries reduces friction, facilitating efficient blood flow and preventing clot formation.

Show Worked Solution

a.   

 
b.   \(\text{B}\) is an artery.

→ Arteries have thick, elastic walls with multiple layers of smooth muscle, allowing them to withstand and maintain the high pressure of blood pumped directly from the heart.

→ This elasticity also enables arteries to expand and recoil with each heartbeat, helping to propel blood forward in a phenomenon known as the “pulse.”

→ Additionally, the smooth inner lining of arteries reduces friction, facilitating efficient blood flow and preventing clot formation.

CHEMISTRY, M3 EQ-Bank 28v4

A student stirs 2.50 g of silver (I) nitrate powder into 100.0 mL of 1.50 mol L\(^{-1}\) sodium chloride solution until it is fully dissolved. A reaction occurs and a precipitate appears.

  1. Write a balanced chemical equation for the reaction.   (1 mark)
  2. Calculate the theoretical mass of precipitate that will be formed.   (4 marks)

    The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).
  3. Identify one scientific reason why the precipitate mass was too high and suggest an improvement to the experimental method which would eliminate this error.   (2 marks)
Show Answers Only

a.    \(\ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)}\)

b.    \(2.11 \text{ g}\)

c.    One possible reason for the higher mass could be the presence of excess solution in the filter paper or incomplete drying. An improvement could be to ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities and dried completely in the incubator before weighing.

Show Worked Solution

a.    \(\ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)}\)

b.    \(\ce{n(AgNO3) = \frac{m}{M} = \frac{2.50}{107.9 + 14.01 + 16.00 \times 3} = \frac{2.50}{169.91} = 0.01471 \text{ mol}}\)

\(\ce{n(NaCl) = c \times V = 1.50 \times 0.100 = 0.150 \text{ mol}}\)

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

\(\ce{n(AgCl) = n(AgNO3) = 0.01471 \text{ mol}}\)

\(\ce{m(AgCl) = n \times M = 0.01471 \times (107.9 + 35.45) = 0.01471 \times 143.35 = 2.11 \text{ g}}\)

c.    One possible reason for the higher mass could be the presence of excess solution in the filter paper or incomplete drying. An improvement could be to ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities and dried completely in the incubator before weighing.

CHEMISTRY, M3 EQ-Bank 28v2

A student stirs 3.50 g of copper (II) nitrate powder into 150.0 mL of 1.50 mol L\(^{-1}\) sodium chloride solution until it is fully dissolved. A reaction occurs and a precipitate appears.

  1. Write a balanced chemical equation for the reaction.   (1 mark)
  2. Calculate the theoretical mass of precipitate that will be formed.   (4 marks)

    The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).
  3. Identify one scientific reason why the precipitate mass was too high and suggest an improvement to the experimental method which would eliminate this error.   (2 marks)
Show Answers Only

a.    \(\ce{Cu(NO3)2(aq) + 2NaCl(aq) -> CuCl2(s) + 2NaNO3(aq)}\)

b.    \(2.51 \text{ g}\)

c.    One possible reason for the higher mass could be the presence of excess solution in the filter paper or incomplete drying. An improvement could be to ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities and dried completely in the incubator before weighing.

Show Worked Solution

a.    \(\ce{Cu(NO3)2(aq) + 2NaCl(aq) -> CuCl2(s) + 2NaNO3(aq)}\)

b.    \(\ce{n(Cu(NO3)2) = \frac{m}{M} = \frac{3.50}{63.55 + 2 \times (14.01 + 16.00 \times 3)} = \frac{3.50}{187.57} = 0.01866 \text{ mol}}\)

\(\ce{n(NaCl) = c \times V = 1.50 \times 0.150 = 0.225 \text{ mol}}\)

\(\ce{Cu(NO3)2} \text{ is the limiting reagent}\)

\(\ce{n(CuCl2) = n(Cu(NO3)2) = 0.01866 \text{ mol}}\)

\(\ce{m(CuCl2) = n \times M = 0.01866 \times (63.55 + 2 \times 35.45) = 0.01866 \times 134.45 = 2.51 \text{ g}}\)

c.    One possible reason for the higher mass could be the presence of excess solution in the filter paper or incomplete drying. An improvement could be to ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities and dried completely in the incubator before weighing.

CHEMISTRY, M4 EQ-Bank 5

The chemical equation for the complete combustion of ethane  \(\ce{(C2H6)}\)  is given below:

\(\ce{2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l)}\)

The structural formula for ethane and standard bond energies for provided for you.

\begin{array} {|c|c|}
\hline Bond & Enthalpy \text{ (kJ mol}^{-1})  \\     
\hline \ce{C-C} & 348 \\
\hline \ce{O-O} & 146 \\
\hline \ce{O=O} & 495 \\
\hline \ce{C-O} & 358 \\
\hline \ce{C=O} & 799 \\
\hline \ce{C-H} & 413 \\
\hline \ce{H-O} & 463 \\
\hline \end{array}

Using the bond energies provided, calculate the enthalpy for the complete combustion of one mole of ethane.   (3 marks)

Show Answers Only

\(-1415.5\ \text{kJ mol}^{-1}\)

Show Worked Solution
\(\Delta H\) \(=\Sigma\,{\text{bonds broken}}-\Sigma\,{\text{bonds formed}}\)  
  \(=((12 \times 413) + (2 \times 348) + (7 \times 495))-((8 \times 799) + (12 \times 463))\)  
  \(=9117-11948\)  
  \(=-2831\)  (for two moles of ethane, as per the equation)  

 
→ \(\Delta H\) for the combustion of one mole of ethane is \(-1415.5 \text{ kJ mol}^{-1}\)

BIOLOGY, M2 EQ-Bank 2

Describe the nature and significance of plasma in the circulatory system. In your answer, include two crucial functions that plasma performs in the body and how a deficiency or imbalance in plasma components might affect overall health.   (4 marks)

Show Answers Only

→ Plasma is the liquid component of blood, making up about 55% of its volume, and consists primarily of water with dissolved proteins, glucose, clotting factors, hormones, and electrolytes.

→ One crucial function of plasma is transport, carrying nutrients, hormones, and waste products throughout the body, fuelling cells and removing metabolic waste.

→ Another vital role of plasma is in maintaining blood pressure and volume, with plasma proteins keeping fluid within blood vessels.

→ A deficiency or imbalance in plasma components can reduce the ability to form blood clots and lead to excessive bleeding.

→ A decrease in plasma proteins can result in edema where fluid leaks from blood vessels into surrounding tissues, causing swelling.

Show Worked Solution

→ Plasma is the liquid component of blood, making up about 55% of its volume, and consists primarily of water with dissolved proteins, glucose, clotting factors, hormones, and electrolytes.

→ One crucial function of plasma is transport, carrying nutrients, hormones, and waste products throughout the body, fuelling cells and removing metabolic waste.

→ Another vital role of plasma is in maintaining blood pressure and volume, with plasma proteins keeping fluid within blood vessels.

→ A deficiency or imbalance in plasma components can reduce the ability to form blood clots and lead to excessive bleeding.

→ A decrease in plasma proteins can result in edema where fluid leaks from blood vessels into surrounding tissues, causing swelling.

BIOLOGY, M2 EQ-Bank 1 MC

Which of the following is a characteristic of an open circulatory system?

  1. Blood is always contained within blood vessels.
  2. It is typically found in larger, more complex animals.
  3. Hemolymph bathes organs directly in body cavities.
  4. It allows for rapid, high-pressure blood flow.
Show Answers Only

\(C\)

Show Worked Solution

→ In an open circulatory system, the circulatory fluid (hemolymph) is not always confined within blood vessels but instead bathes organs directly in body cavities. 

\(\Rightarrow C\)

BIOLOGY, M2 EQ-Bank 10 MC

Which of the following statements about phloem transport in plants is correct?

  1. Sap in the phloem always moves from areas of high sugar concentration to areas of low sugar concentration.
  2. The direction of sap movement in the phloem is solely determined by gravity.
  3. Moving sap through the phloem requires energy input and can occur both up and down the plant.
  4. The rate of sap movement in the phloem is constant throughout the day and night.
Show Answers Only

\(C\)

Show Worked Solution

Consider each option.

Option A: Incorrect. While sugar concentration plays a role, phloem transport is more complex and can move against concentration gradients.

Option B: Incorrect. Gravity is not the determining factor as phloem transport can occur both up and down the plant.

Option C: Correct. Phloem transport is an active process requiring energy, and it can move substances bidirectionally based on the plant’s needs.

Option D: Incorrect. The rate of phloem transport varies depending on factors such as photosynthetic activity and the plant’s growth needs.

\(\Rightarrow C\)

BIOLOGY, M2 EQ-Bank 8

The transpiration-cohesion-tension theory provides an explanation for water movement in tall trees.

Describe this theory, including the role of transpiration in this process.   (4 marks)

Show Answers Only

→ Transpiration creates a negative water potential at the leaves (tension), driving water movement upwards through the xylem.

→ This process acts like a ‘pull’ force, drawing water from the roots to replace water lost through the leaves.

→ The theory states that this movement also relies critically on cohesion, which refers to the tendency of water molecules to stick together due to hydrogen bonding.

→ This property allows water to form a continuous column in the xylem, enabling it to be pulled upwards without breaking. 

→ This tension, combined with the cohesive properties of water, allows trees to transport water against gravity to great heights. 

Show Worked Solution

→ Transpiration creates a negative water potential at the leaves (tension), driving water movement upwards through the xylem.

→ This process acts like a ‘pull’ force, drawing water from the roots to replace water lost through the leaves.

→ The theory states that this movement also relies critically on cohesion, which refers to the tendency of water molecules to stick together due to hydrogen bonding.

→ This property allows water to form a continuous column in the xylem, enabling it to be pulled upwards without breaking. 

→ This tension, combined with the cohesive properties of water, allows trees to transport water against gravity to great heights. 

BIOLOGY, M2 EQ-Bank 6

Compare and contrast the microscopic structures involved in gas exchange in mammals and plants.

In your answer, describe one structural similarity between these structures that aids in gas exchange and explain one key difference in how these structures function.   (4 marks)

Show Answers Only

→ The primary microscopic structure for gas exchange in mammals is the alveolus, while in plants it is the leaf.
 

Structural similarities that aid in gas exchange (include one):

→ Both have a large surface area to volume ratio to maximise gas exchange.

→ Both have thin, moist surfaces to facilitate the diffusion of gases.
 

Two key differences in how these structures function in gas exchange are:

→ Alveoli primarily exchange oxygen and carbon dioxide with blood, while leaves exchange these gases with air in intercellular spaces.

→ Gas exchange in alveoli occurs continuously for respiration, while in leaves it varies with light availability due to its role in photosynthesis.

Show Worked Solution

→ The primary microscopic structure for gas exchange in mammals is the alveolus, while in plants it is the leaf.
 

Structural similarities that aid in gas exchange (include one):

→ Both have a large surface area to volume ratio to maximise gas exchange.

→ Both have thin, moist surfaces to facilitate the diffusion of gases.
 

Two key differences in how these structures function in gas exchange are:

→ Alveoli primarily exchange oxygen and carbon dioxide with blood, while leaves exchange these gases with air in intercellular spaces.

→ Gas exchange in alveoli occurs continuously for respiration, while in leaves it varies with light availability due to its role in photosynthesis.

CHEMISTRY, M4 EQ-Bank 7 MC

The chemical equation for the decomposition of calcium carbonate is shown below

\(\ce{CaCO3(s) -> CO2(g) + CaO(s)}\)

Which of the following best describes the type of reaction given that the bond energy of the reactant is greater than the bond energy of the products.

  1.  An exothermic reaction because energy is absorbed
  2. An exothermic reaction because energy is released
  3. An endothermic reaction because energy is absorbed
  4. An endothermic reaction because energy is released
Show Answers Only

\(C\)

Show Worked Solution

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

→ If the bond energy of the reactant is greater than the bond energy of the products, then more energy is required to break apart the reactant bonds then being produced in the formation of the product bonds.

→ Hence \(\Delta H\) will be positive as energy is being absorbed.

\(\Rightarrow C\)

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)

Show Answers Only

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)
Show Answers Only

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)

Show Answers Only

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.    (insert image here)

 

 b.    The 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 which require more energy to be broken resulting in much higher melting and boiling points (0°C and 100°C, respectively).

Show Worked Solution

a.    (insert image here)

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

→ No lines should be present to represent covalent bonds.

 

b.    The 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 which require more energy to be broken resulting 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