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BIOLOGY, M3 EQ-Bank 7

Describe one type of biochemical evidence that provides strong support for Darwin and Wallace's Theory of Evolution by Natural Selection.   (2 marks)

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Answers could include one of the following:

Oxygen transporting proteins

→ Biochemical evidence supporting evolutionary relationships can be found in the similarities of oxygen-transporting proteins across species.

→ Haemoglobin, the primary oxygen-carrying protein in vertebrates, shows structural similarities across diverse animal groups, indicating a common ancestral origin.

→ This evidence supports Darwin and Wallace’s theory by demonstrating common descent with modification.
 

DNA

→ One type of biochemical evidence supporting evolutionary relationships is the similarity in DNA sequences and proteins across different species.

→ The more closely related species are, the more similar their DNA and protein sequences tend to be, indicating common ancestry.

→ In Australian marsupials, DNA analysis has revealed close genetic relationships between seemingly diverse species like the Tasmanian devil and the numbat.

→ By supporting their common evolutionary origin, DNA provides support for Darwin and Wallace’s Theory.

Show Worked Solution

Answers could include one of the following:

Oxygen transporting proteins

→ Biochemical evidence supporting evolutionary relationships can be found in the similarities of oxygen-transporting proteins across species.

→ Haemoglobin, the primary oxygen-carrying protein in vertebrates, shows structural similarities across diverse animal groups, indicating a common ancestral origin.

→ This evidence supports Darwin and Wallace’s theory by demonstrating common descent with modification.
 

DNA

→ One type of biochemical evidence supporting evolutionary relationships is the similarity in DNA sequences and proteins across different species.

→ The more closely related species are, the more similar their DNA and protein sequences tend to be, indicating common ancestry.

→ In Australian marsupials, DNA analysis has revealed close genetic relationships between seemingly diverse species like the Tasmanian devil and the numbat.

→ By supporting their common evolutionary origin, DNA provides support for Darwin and Wallace’s Theory.

CHEMISTRY, M3 EQ-Bank 17

How does the presence of a catalyst influence the rate of a chemical reaction?   (2 marks)

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→ Catalysts are substances that increase the rate of a chemical reaction without being consumed in the process.

→ They work by providing an alternative reaction pathway with a lower activation energy, which is the energy required for the reaction to occur. Because of this, more reactant particles can successfully collide with enough energy to undergo the reaction at a given temperature.

→ As a result, the reaction occurs faster when a catalyst is present.

Show Worked Solution

→ Catalysts are substances that increase the rate of a chemical reaction without being consumed in the process.

→ They work by providing an alternative reaction pathway with a lower activation energy, which is the energy required for the reaction to occur. Because of this, more reactant particles can successfully collide with enough energy to undergo the reaction at a given temperature.

→ As a result, the reaction occurs faster when a catalyst is present.

CHEMISTRY, M3 EQ-Bank 4 MC

The fruits of the cycad Macrozamia, are known to be highly toxic for consumption.

Which of the following methods conducted by Indigenous peoples is most effective in removing toxins from this cycad?

  1. Letting the seeds mature before consuming them.
  2. Removing the individual seeds from the fruit before cooking them in a fire.
  3. Cooking the seeds with fire before leaching them out in running water overnight.
  4. Removing the individual seeds from the fruit and baking them in the sun.
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\(C\)

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→ Cooking of the seeds in fire helps to break down the toxins within the seeds.

→ The toxins within the seeds are soluble compounds and so the leaching process helps eliminate any remaining harmful substances as the toxins are dissolved and diffused into the water and then washed away.

\(\Rightarrow C\)

CHEMISTRY, M3 EQ-Bank 1 MC

For centuries, Indigenous Australians have used chemical processes to detoxify certain foods. The seeds of the nardoo fern (Marsilea) were a traditional food source, but they contain harmful toxins. To safely consume the seeds, they were first ground into a paste and soaked in water for several days.

Soaking the nardoo seeds aids in detoxification by:

  1. Causing the water to neutralise the harmful substances.
  2. Altering the chemical structure of the toxins through grinding.
  3. Allowing the water to remove the toxins from the paste.
  4. Increasing the temperature to speed up the breakdown of toxins.
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\(C\)

Show Worked Solution

→ Water removes the toxins from nardoo fern seeds during the leaching process because the toxic substance is water-soluble, meaning they can dissolve in water.

→ When the seeds are soaked, the toxins dissolve and diffuse into the surrounding water. By changing or draining the water, the dissolved toxins are effectively removed from the food.

→ This process relies on the solubility of the toxins, and repeated soaking ensures that a significant amount of the harmful compounds is washed away, leaving the food safe for consumption.

\(\Rightarrow C\)

BIOLOGY, M3 EQ-Bank 3

"Natural selection is the architect of biodiversity."

Justify this statement. In your answer, describe how natural selection can lead to changes in a population and over time, increased biodiversity. Provide an example from the Australian ecosystems that demonstrates this relationship.   (5 marks)

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→ Natural selection is the architect of biodiversity as it drives the adaptation of populations to their environments, leading to the evolution of new species over time.

→ This process begins with genetic variations within a population, where individuals with traits better suited to their environment are more likely to survive and reproduce, passing these beneficial traits to their offspring.

→ Over generations, this can lead to significant changes in the population, potentially resulting in the emergence of new species adapted to specific environmental niches.

→ As populations adapt to different environments or ecological roles, the overall biodiversity increases.

→ Australian ecosystem example: the radiation of marsupials, such as the diverse kangaroo species.

→ From a common ancestor, natural selection has led to the evolution of various kangaroo species adapted to different habitats, from the large red kangaroos of the arid interior to the tree-dwelling tree-kangaroos of the rainforests.

→ Each fill a unique ecological niche and contribute to Australia’s biodiversity.

Show Worked Solution

→ Natural selection is the architect of biodiversity as it drives the adaptation of populations to their environments, leading to the evolution of new species over time.

→ This process begins with genetic variations within a population, where individuals with traits better suited to their environment are more likely to survive and reproduce, passing these beneficial traits to their offspring.

→ Over generations, this can lead to significant changes in the population, potentially resulting in the emergence of new species adapted to specific environmental niches.

→ As populations adapt to different environments or ecological roles, the overall biodiversity increases.

→ Australian ecosystem example: the radiation of marsupials, such as the diverse kangaroo species.

→ From a common ancestor, natural selection has led to the evolution of various kangaroo species adapted to different habitats, from the large red kangaroos of the arid interior to the tree-dwelling tree-kangaroos of the rainforests.

→ Each fill a unique ecological niche and contribute to Australia’s biodiversity.

BIOLOGY, M3 EQ-Bank 2

  1. Define convergent evolution within the context of Darwin and Wallace's Theory of Evolution by Natural Selection.   (1 mark)

  2. Describe two key principles of natural selection that lead to convergent evolution.   (2 marks)

  3. Provide two examples of convergent evolution, one involving Australian fauna and one non-Australian, explaining how each demonstrates the process of natural selection.   (2 marks)

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a.   Convergent evolution definition:

→ The process by which unrelated organisms develop similar traits or features as a result of adapting to similar environmental pressures or ecological niches.
 

b.   Key principles that lead to convergent evolution:

→ Natural selection acts on existing variations within populations, favouring traits that enhance survival and reproduction in specific environments.

→ When unrelated organisms face similar environmental challenges, natural selection can lead to the evolution of similar adaptations, even in distantly related species.

→ This process occurs independently in each lineage, resulting in analogous structures or behaviours that serve similar functions but have different evolutionary origins.
 

c.   Australian fauna (example):

→ The similarity between the marsupial Tasmanian tiger (thylacine) and the placental grey wolf is an example of convergent evolution.

→ Despite their distant relationship, both evolved similar body shapes, jaw structures, and striped patterns due to adapting to similar predatory lifestyles. 
 

Non-Australian (example):

→ The similar body shapes of sharks and dolphins is another example of convergent evolution.

→ Though one is a fish and the other a mammal, both have evolved streamlined bodies, dorsal fins, and tail flukes as adaptations for efficient swimming in marine environments.

→ This demonstrates how natural selection can produce similar outcomes in response to comparable environmental pressures.

Show Worked Solution

a.   Convergent evolution definition:

→ The process by which unrelated organisms develop similar traits or features as a result of adapting to similar environmental pressures or ecological niches.
 

b.   Key principles that lead to convergent evolution:

→ Natural selection acts on existing variations within populations, favouring traits that enhance survival and reproduction in specific environments.

→ When unrelated organisms face similar environmental challenges, natural selection can lead to the evolution of similar adaptations, even in distantly related species.

→ This process occurs independently in each lineage, resulting in analogous structures or behaviours that serve similar functions but have different evolutionary origins.
 

c.   Australian fauna (example):

→ The similarity between the marsupial Tasmanian tiger (thylacine) and the placental grey wolf is an example of convergent evolution.

→ Despite their distant relationship, both evolved similar body shapes, jaw structures, and striped patterns due to adapting to similar predatory lifestyles. 
 

Non-Australian (example):

→ The similar body shapes of sharks and dolphins is another example of convergent evolution.

→ Though one is a fish and the other a mammal, both have evolved streamlined bodies, dorsal fins, and tail flukes as adaptations for efficient swimming in marine environments.

→ This demonstrates how natural selection can produce similar outcomes in response to comparable environmental pressures.

BIOLOGY, M3 EQ-Bank 3 MC

Which of the following groups of organisms is the best example of divergent evolution?

  1. Sharks, dolphins, and penguins
  2. Darwin's finches in the Galapagos Islands
  3. Cacti in America and euphorbia (a succulent species) in Africa
  4. Marsupial and placental mammals
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\(B\)

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→ Darwin’s finches in the Galapagos Islands are a classic example of divergent evolution because they all descended from a common ancestor but evolved different beak shapes and sizes to exploit differing food sources.

→ This adaptation to different ecological niches led to the formation of distinct species from a single ancestral population.

\(\Rightarrow B\)

BIOLOGY, M3 EQ-Bank 2 MC

Which of the following pairs of organisms best demonstrates convergent evolution?

  1. Tasmanian tiger and grey wolf
  2. Kangaroo and wallaby
  3. Eucalyptus and acacia trees
  4. Platypus and echidna
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\(A\)

Show Worked Solution

→ The Tasmanian tiger (a marsupial) and grey wolf (a placental mammal) demonstrate convergent evolution because they developed similar body shapes and predatory features despite being unrelated and evolving in different parts of the world.

→ This similarity in form is due to adapting to similar ecological niches and hunting strategies.

\(\Rightarrow A\)

BIOLOGY, M3 EQ-Bank 1

  1. Explain the difference between micro-evolution and macro-evolution, highlighting their key characteristics and relationship.   (2 marks)
  2. Provide an example of how microevolutionary changes can accumulate to drive larger evolutionary changes and potentially lead to speciation.   (2 marks)

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a.   Difference between microevolution and macroevolution:

→ Microevolution refers to small-scale changes in gene frequencies within a population over time, such as changes in colour or size.

→ Macroevolution, on the other hand, involves large-scale changes that lead to the formation of new species or higher taxonomic groups.

→ While both rely on the same mechanics of change, microevolution occurs relatively quickly and can be observed within human timescales while macroevolution typically occurs over much longer periods and is inferred from fossil records.
 

b.   Answers could include one of the following examples.

Evolution of the platypus:

→ Over millions of years, the evolution of the platypus saw small genetic changes accumulate.

→ This process resulted in a combination of reptilian and mammalian features, such as egg-laying and milk production.

→ These gradual changes eventually led to the emergence of a distinct species that occupies a unique ecological niche.
 

Evolution of the horse:

→ Fossil evidence shows the evolution of the horse from a small, multi-toed ancestor to larger, single-toed modern horses.

→ Each small change, such as increases in size or reductions in toe number, represented micro-evolutionary steps.

→ Over millions of years, these accumulated changes resulted in the diverse horse species we see today, adapted to various environments.

Show Worked Solution

a.   Difference between microevolution and macroevolution:

→ Microevolution refers to small-scale changes in gene frequencies within a population over time, such as changes in colour or size.

→ Macroevolution, on the other hand, involves large-scale changes that lead to the formation of new species or higher taxonomic groups.

→ While both rely on the same mechanics of change, microevolution occurs relatively quickly and can be observed within human timescales while macroevolution typically occurs over much longer periods and is inferred from fossil records.
 

b.   Answers could include one of the following examples.

Evolution of the platypus:

→ Over millions of years, the evolution of the platypus saw small genetic changes accumulate.

→ This process resulted in a combination of reptilian and mammalian features, such as egg-laying and milk production.

→ These gradual changes eventually led to the emergence of a distinct species that occupies a unique ecological niche.
 

Evolution of the horse:

→ Fossil evidence shows the evolution of the horse from a small, multi-toed ancestor to larger, single-toed modern horses.

→ Each small change, such as increases in size or reductions in toe number, represented micro-evolutionary steps.

→ Over millions of years, these accumulated changes resulted in the diverse horse species we see today, adapted to various environments.

BIOLOGY, M3 EQ-Bank 4 MC

Which of the following scenarios best demonstrates a physiological adaptation rather than a structural or behavioural adaptation in an Australian organism?

  1. A kangaroo's ability to hop at high speeds over long distances.
  2. The drought deciduousness of mulga trees in arid regions.
  3. The mallee fowl's behaviour of building and maintaining large incubation mounds.
  4. The production of cyanide compounds by Swainsona plants when grazed upon.
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\(D\)

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→ Physiological adaptations refer to functions of an organism’s structural features that have evolved to help them survive in their specific environments.

\(\Rightarrow D\)

BIOLOGY, M3 EQ-Bank 6

Many plants in hot climates exhibit leaf droop, where their leaves hang downwards during the hottest parts of the day.

  1. Describe this type of adaptation.   (1 mark)

  2. Provide two reasons why this adaptation might be beneficial for plants in these environments.   (2 marks)

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a.   Behavioural adaptation.

→ Leaf droop is classified as a behavioural adaptation because it involves a dynamic response to environmental conditions rather than a fixed structural or physiological change.
 

b.   This adaption is beneficial because:

→ it reduces the surface area of the leaf directly exposed to intense sunlight, minimising water loss and protecting the leaf from heat damage.

→ when drooping leaves are arranged vertically, they direct water to the base of the plant where the roots can more efficiently absorb it. This is particularly beneficial in arid environments.

Show Worked Solution

a.   Behavioural adaptation.

→ Leaf droop is classified as a behavioural adaptation because it involves a dynamic response to environmental conditions rather than a fixed structural or physiological change.
 

b.   This adaption is beneficial because:

→ it reduces the surface area of the leaf directly exposed to intense sunlight, minimising water loss and protecting the leaf from heat damage.

→ when drooping leaves are arranged vertically, they direct water to the base of the plant where the roots can more efficiently absorb it. This is particularly beneficial in arid environments.

BIOLOGY, M3 EQ-Bank 5

Summarise the key principles of Darwin's theory of evolution by natural selection.   (3 marks)

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→ Darwin’s theory of evolution by natural selection proposes that species change over time through a process of natural selection acting on inherited variations within populations.

→ He observed that individuals in a population vary in their traits, some of which are heritable, and that more offspring are produced than can survive given limited resources.

→ Darwin concluded that individuals with advantageous traits are more likely to survive and reproduce, passing these beneficial traits to their offspring.

→ Over many generations, this process leads to changes in the frequency of traits within a population, potentially resulting in the evolution of new species adapted to their environments.

Show Worked Solution

→ Darwin’s theory of evolution by natural selection proposes that species change over time through a process of natural selection acting on inherited variations within populations.

→ He observed that individuals in a population vary in their traits, some of which are heritable, and that more offspring are produced than can survive given limited resources.

→ Darwin concluded that individuals with advantageous traits are more likely to survive and reproduce, passing these beneficial traits to their offspring.

→ Over many generations, this process leads to changes in the frequency of traits within a population, potentially resulting in the evolution of new species adapted to their environments.

BIOLOGY, M3 EQ-Bank 4

The illustrations below depict three species of finches from the Galapagos Islands, as observed by Charles Darwin during his voyage.

Describe the reasons why these finch species inhabit distinct ecological niches.   (3 marks)

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→ The distinct beak shapes of these finch species are adaptations that allow them to exploit different food sources, leading to occupation of distinct ecological niches.

→ The two finches with wider, shorter beaks are adapted for eating seeds, with the variation in width likely corresponding to different seed sizes or hardness they can crack.

→ The finch with the thin, long beak is specialised for probing and catching insects, allowing it to access food sources that the seed-eating finches cannot.

→ These adaptations reduce direct competition for food between the species, enabling them to coexist in the same general area by utilising different resources, a concept known as niche partitioning.

Show Worked Solution

→ The distinct beak shapes of these finch species are adaptations that allow them to exploit different food sources, leading to occupation of distinct ecological niches.

→ The two finches with wider, shorter beaks are adapted for eating seeds, with the variation in width likely corresponding to different seed sizes or hardness they can crack.

→ The finch with the thin, long beak is specialised for probing and catching insects, allowing it to access food sources that the seed-eating finches cannot.

→ These adaptations reduce direct competition for food between the species, enabling them to coexist in the same general area by utilising different resources, a concept known as niche partitioning.

CHEMISTRY, M3 EQ-Bank 15

Using collision theory, explain how increasing the temperature of a reaction mixture affects the rate of the reaction.   (3 marks)

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→ According to collision theory, increasing the temperature of a reaction mixture raises the kinetic energy of the reactant molecules.

→ As the molecules move faster, the frequency of collisions increases. In addition, a higher proportion of the molecules have enough energy to overcome the activation energy barrier.

→ As a result, there are more successful collisions per unit of time, leading to a faster reaction rate.

Show Worked Solution

→ According to collision theory, increasing the temperature of a reaction mixture raises the kinetic energy of the reactant molecules.

→ As the molecules move faster, the frequency of collisions increases. In addition, a higher proportion of the molecules have enough energy to overcome the activation energy barrier.

→ As a result, there are more successful collisions per unit of time, leading to a faster reaction rate.

CHEMISTRY, M3 EQ-Bank 7 MC

Which of the following best describes the role of activation energy in a chemical reaction?

  1. It is the total energy released when bonds are broken in a reaction.
  2. It is the minimum energy required for reactants to collide and form products.
  3. It is the energy needed to cool the reactants and stop the reaction.
  4. It increases the number of collisions between molecules.
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\(B\)

Show Worked Solution

→ Activation energy is the minimum energy that reactants must have during a collision for a reaction to occur and form products.

→ If the reactants do not have sufficient energy, they will collide but fail to react.

\(\Rightarrow B\)

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)
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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 6 MC

How does increasing the surface area of a reactant, such as using a powdered solid instead of a solid block, affect the rate of a reaction?

  1. It allows more particles to be exposed to the reactant, increasing collision frequency.
  2. It decreases the energy required for successful collisions between particles.
  3. It decreases the temperature of the system, making the reaction faster.
  4. It increases the number of particles that remain unreacted at the end.
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\(A\)

Show Worked Solution

→ Increasing the surface area of a reactant (e.g., using powdered solids) exposes more of the particles to the other reactants, resulting in a greater number of collisions per unit of time.

→ Since collision frequency is a key factor in the rate of reaction, having more available particles in contact increases the likelihood of successful collisions, which speeds up the reaction.

\(\Rightarrow A\)

BIOLOGY, M3 EQ-Bank 3

Analyse the role of behavioural adaptations in enhancing species survival. In your response, describe a specific behavioural adaptation in an Australian plant species and a specific Australian native animal.   (3 marks)

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→ Behavioural adaptations play a crucial role in enhancing species survival by allowing organisms to respond to environmental challenges through learned or instinctive actions.

→ In Australian plants, trigger plants display a distinctive behavioural adaptation in which their column rapidly snaps forward to deposit pollen onto insect pollinators, thereby enhancing their reproductive success.

→ Among Australian native animals, the frill-neck lizard erects a large frill around its neck and opens its brightly-coloured mouth when threatened. This makes itself appear larger and more intimidating to potential predators, increasing its chances of survival.

Show Worked Solution

→ Behavioural adaptations play a crucial role in enhancing species survival by allowing organisms to respond to environmental challenges through learned or instinctive actions.

→ In Australian plants, trigger plants display a distinctive behavioural adaptation in which their column rapidly snaps forward to deposit pollen onto insect pollinators, thereby enhancing their reproductive success.

→ Among Australian native animals, the frill-neck lizard erects a large frill around its neck and opens its brightly-coloured mouth when threatened. This makes itself appear larger and more intimidating to potential predators, increasing its chances of survival.

BIOLOGY, M3 EQ-Bank 2

Explain the concept of physiological adaptation in living organisms. In your answer, describe one example of a physiological adaptation in a plant species and one example in an animal species.   (3 marks)

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→ Physiological adaptations refer to functions of an organism’s structural features that have evolved to help them survive in their specific environments.

→ Plant example: Photosynthesis used by many succulent plants in arid regions, such as the Australian pigface, allows them to conserve water by opening their stomata at night to collect \(\ce{CO2}\).

→ Animal example: The salt glands of marine birds like the Australian pelican are a physiological adaptation that allows them to excrete excess salt from their bodies. This enables them to maintain proper osmotic balance despite drinking seawater.

Show Worked Solution

→ Physiological adaptations refer to functions of an organism’s structural features that have evolved to help them survive in their specific environments.

→ Plant example: Photosynthesis used by many succulent plants in arid regions, such as the Australian pigface, allows them to conserve water by opening their stomata at night to collect \(\ce{CO2}\).

→ Animal example: The salt glands of marine birds like the Australian pelican are a physiological adaptation that allows them to excrete excess salt from their bodies. This enables them to maintain proper osmotic balance despite drinking seawater.

CHEMISTRY, M4 EQ-Bank 12

The energy profile diagram below shows the energy changes that occur during a specific chemical reaction in the absence of a catalyst.
 

 

  1. Draw the energy profile for this reaction with a catalyst on the diagram above.   (1 mark)

  2. Describe the impact of a catalyst on this reaction, using the diagram as a reference.   (2 marks)
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a.    
     

b.   Catalyst impact:

→ A catalyst offers an alternative pathway for the reaction to occur that requires less activation energy.

→ On the energy profile diagram, this would appear as a lower peak, indicating the reduction in the energy needed for the reaction to proceed.

Show Worked Solution

a.   
     
 

b.   Catalyst impact:

→ A catalyst offers an alternative pathway for the reaction to occur that requires less activation energy.

→ On the energy profile diagram, this would appear as a lower peak, indicating the reduction in the energy needed for the reaction to proceed.

BIOLOGY, M3 EQ-Bank 7

"While biological control methods can be effective, they are not without risks and limitations."

Describe three potential disadvantages of biological controls used to manage introduced species populations.   (3 marks)

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→ Disadvantage 1: The potential for introducing new diseases that may affect non-target species. For example, a pathogen introduced to control an invasive plant might mutate and infect native plant species, causing unintended ecological damage.

→ Disadvantage 2: Another drawback is the long-term viability of the control agent, as the target species may develop resistance over time, rendering the control ineffective. This occurred with the myxoma virus introduced to control rabbits in Australia.

→ Disadvantage 3: The risk of the control agent itself becoming invasive. The cane toad in Australia, originally introduced to control beetle pests in sugar cane crops, became a significant pest itself.

Show Worked Solution

→ Disadvantage 1: The potential for introducing new diseases that may affect non-target species. For example, a pathogen introduced to control an invasive plant might mutate and infect native plant species, causing unintended ecological damage.

→ Disadvantage 2: Another drawback is the long-term viability of the control agent, as the target species may develop resistance over time, rendering the control ineffective. This occurred with the myxoma virus introduced to control rabbits in Australia.

→ Disadvantage 3: The risk of the control agent itself becoming invasive. The cane toad in Australia, originally introduced to control beetle pests in sugar cane crops, became a significant pest itself.

CHEMISTRY, M3 EQ-Bank 4 MC

There could be a risk of explosion in a coal mine due to the presence of fine coal dust particles, so strict safety precautions are followed in these environments.
Which of the following statements best explains this risk?

  1. The activation energy for the reaction of fine particles is much lower than for large particles.
  2. The surface area of the coal dust particles is very large, which makes combustion reactions occur more rapidly.
  3. It is very easy to heat the fine coal dust particles to their ignition temperature.
  4. When coal is ground into fine particles, the energy profile diagram for the combustion reaction is altered.
Show Answers Only

\(B\)

Show Worked Solution

→ Fine coal dust particles have a much greater surface area compared to larger pieces of coal.

→ This increased surface area allows for more particles to be exposed to oxygen, leading to a faster and more intense combustion reaction. When ignited, the reaction can happen very quickly, which increases the risk of an explosion.

→ This is why environments with fine particles, like coal mines, must adhere to strict safety precautions to prevent such dangerous reactions.

\(\Rightarrow B\)

CHEMISTRY, M3 EQ-Bank 11

Brian measured the reaction rate of 0.5 g of sodium metal in 2.0 mol/L nitric acid. The volume of hydrogen gas produced per minute was recorded both without a catalyst and with copper as a catalyst.

\begin{array} {|c|c|c|}
\hline \text{Time (minutes)} & \text{Volume of hydrogen gas (mL)} & \text{Volume of hydrogen gas (mL)} \\ & \text{no catalyst} & \text{catalyst}  \\
\hline \text{0}  & 0 & 0  \\
\hline \text{1} & 16 & 7 \\
\hline \text{2} & 30 &  13\\
\hline \text{3} & 43 & 19 \\
\hline \text{4}  & 44 & 26 \\
\hline \text{5} & 44 &  32 \\
\hline \text{6} & 44 & 38 \\
\hline \text{7} & 44 &  44 \\
\hline \end{array}

  1. On the grid below, plot a line graph for the data in the table.   (3 marks)
     
  2. Write a conclusion for the experiment.   (2 marks)

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a.    
       

b.   Experiment conclusion:

→ The rate of reaction was significantly faster when copper was used as a catalyst compared to when no catalyst was present.

→ This is because the copper catalyst provided an alternative reaction pathway with a lower activation energy, allowing the reactant particles to collide more frequently and effectively.

Show Worked Solution

a.   
     

b.   Experiment conclusion:

→ The rate of reaction was significantly faster when copper was used as a catalyst compared to when no catalyst was present.

→ This is because the copper catalyst provided an alternative reaction pathway with a lower activation energy, allowing the reactant particles to collide more frequently and effectively.

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)

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→ 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 5

Describe the potential changes in native species populations as an introduced species becomes established and its population grows. In your answer, consider both direct and indirect effects on the ecosystem.   (4 marks)

Show Answers Only

Cane toads:

→ As cane toad populations grew, many native predators such as quolls, goannas, and certain snake species experienced rapid declines due to poisoning when they attempted to eat the toxic toads.

→ This direct effect led to a decrease in these predator populations, which in turn caused indirect effects on their usual prey species.

→ For example, with fewer predators, some native frog and small mammal populations initially increased.

→ However, the cane toads also compete with native species for food and habitat, potentially leading to declines in some native insect and small animal populations over time.
 

Rabbits:

→ As rabbit populations exploded, they caused significant damage to vegetation through overgrazing, reducing food and habitat for native herbivores such as kangaroos and wallabies.

→ This direct competition led to declines in these native species.

→ The changes in vegetation also indirectly affected other species; for example, small ground-dwelling animals lost protective cover, making them more vulnerable to predation.

→ In some areas, vegetation loss and rabbit burrowing has led to increased soil erosion, further altering habitats.

→ Over time, some native predators like dingoes and wedge-tailed eagles may have benefited from the abundant rabbit population as a new food source, potentially leading to complex changes in predator-prey dynamics.

Show Worked Solution

Answers could include one of the following.

Cane toads:

→ As cane toad populations grew, many native predators such as quolls, goannas, and certain snake species experienced rapid declines due to poisoning when they attempted to eat the toxic toads.

→ This direct effect led to a decrease in these predator populations, which in turn caused indirect effects on their usual prey species.

→ For example, with fewer predators, some native frog and small mammal populations initially increased.

→ However, the cane toads also compete with native species for food and habitat, potentially leading to declines in some native insect and small animal populations over time.
 

Rabbits:

→ As rabbit populations exploded, they caused significant damage to vegetation through overgrazing, reducing food and habitat for native herbivores such as kangaroos and wallabies.

→ This direct competition led to declines in these native species.

→ The changes in vegetation also indirectly affected other species; for example, small ground-dwelling animals lost protective cover, making them more vulnerable to predation.

→ In some areas, vegetation loss and rabbit burrowing has led to increased soil erosion, further altering habitats.

→ Over time, some native predators like dingoes and wedge-tailed eagles may have benefited from the abundant rabbit population as a new food source, potentially leading to complex changes in predator-prey dynamics.

BIOLOGY, M3 EQ-Bank 3

"In the post industrial age, human activities have become increasingly influential on nature, reshaping the very foundations of our planet's ecosystems."

Describe two abiotic selection pressures caused by human activity and their potential impact on organisms in the ecosystem.   (4 marks)

Show Answers Only

Answers could include two of the following.

Climate warming:

→ One major abiotic pressure is the increase in global temperatures due to climate change. This increase has affected both the atmosphere and the oceans.

→ Ocean warming has led to heat-tolerant coral species in the Great Barrier Reef having a survival advantage over less tolerant species.
 

Pollution (plastics):

→ Another significant selection pressure is from plastic waste, particularly in rivers and oceans.

→ Sea turtles frequently mistake plastic bags for jellyfish, leading to blockages in their digestive systems and sometimes fatal outcomes.
 

Light pollution:

→ Light pollution is another abiotic selection pressure created by humans.

→ In urban areas, light pollution has been shown to affect the breeding behaviour of birds, favouring those that can adapt to nesting and foraging under artificial light conditions.

Show Worked Solution

Answers could include two of the following.

Climate warming:

→ One major abiotic pressure is the increase in global temperatures due to climate change. This increase has affected both the atmosphere and the oceans.

→ Ocean warming has led to heat-tolerant coral species in the Great Barrier Reef having a survival advantage over less tolerant species.
 

Pollution (plastics):

→ Another significant selection pressure is from plastic waste, particularly in rivers and oceans.

→ Sea turtles frequently mistake plastic bags for jellyfish, leading to blockages in their digestive systems and sometimes fatal outcomes.
 

Light pollution:

→ Light pollution is another abiotic selection pressure created by humans.

→ In urban areas, light pollution has been shown to affect the breeding behaviour of birds, favouring those that can adapt to nesting and foraging under artificial light conditions.

BIOLOGY, M3 EQ-Bank 2

Describe, providing an example, an abiotic factor that could act as a selection pressure in

  1. a desert ecosystem.   (2 marks)
  2. an aquatic ecosystem.   (2 marks)
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a.   Desert ecosystem:

→ An abiotic factor acting as a selection pressure is extremely low rainfall/moisture.

→ For example, the thorny devil lizard has evolved a unique skin structure that allows it to collect and channel water to its mouth from any part of its body, enabling it to survive in extremely arid conditions.
 

b.   Aquatic ecosystem:

→ In an aquatic ecosystem, water temperature can act as a significant abiotic selection pressure.

→ The Great Barrier Reef, for example, experiences coral bleaching events when water temperatures rise. These events stress the coral and if prolonged, can potentially kill the coral. 

Show Worked Solution

a.   Desert ecosystem:

→ An abiotic factor acting as a selection pressure is extremely low rainfall/moisture.

→ For example, the thorny devil lizard has evolved a unique skin structure that allows it to collect and channel water to its mouth from any part of its body, enabling it to survive in extremely arid conditions.
 

b.   Aquatic ecosystem:

→ In an aquatic ecosystem, water temperature can act as a significant abiotic selection pressure.

→ The Great Barrier Reef, for example, experiences coral bleaching events when water temperatures rise. These events stress the coral and if prolonged, can potentially kill the coral. 

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)

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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.

BIOLOGY, M3 EQ-Bank 2 MC

A new predatory bird species is introduced to an island ecosystem. Which of the following best describes the likely impact on the small mammal populations on the island?

  1. All small mammal species will be equally affected.
  2. Nocturnal species will have an immediate advantage.
  3. Species with better camouflage may have higher survival rates.
  4. The mammal population size will remain stable.
Show Answers Only

\(C\)

Show Worked Solution

→ The introduction of a new predator creates a selection pressure that favours individuals with traits that help them avoid predation.

→ Small mammals with better camouflage are less likely to be spotted by the predatory birds, giving them a survival advantage and potentially higher reproductive success.

\(\Rightarrow C\)

BIOLOGY, M3 EQ-Bank 1 MC

In a grassland ecosystem, a prolonged drought occurs. Which of the following is most likely to be an immediate effect on the plant population?

  1. Increased genetic diversity.
  2. Rapid evolutionary changes.
  3. Selection for drought-tolerant individuals.
  4. Uniform growth across all plant species.
Show Answers Only

\(C\)

Show Worked Solution

→ In a prolonged drought, plants that are better adapted to conserve water or tolerate dry conditions are more likely to survive and reproduce.

→ This selection pressure favours drought-tolerant individuals in the short term, although evolutionary changes would take many generations to occur.

\(\Rightarrow C\)

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 2

A particle is moving along the \(x\)-axis where its displacement, in metres from the origin, after \(t\) seconds, is given by:

\(x=t^3-5t^2+8t-6\).

  1. Determine the times when the particle is at rest.   (2 marks)

  2. Find and describe the velocity of the particle at the time where there is no accelerating force acting on it.  (2 marks)

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i.    \(t=\dfrac{4}{3}\ \text{s  or}\  \ t=2\ \text{s}\)

ii.  \(\dot{x}=-\dfrac{1}{3}\ \text{m/s} \ \text{(towards the left)}\)

Show Worked Solution

i.    \(x=t^3-5t^2+8t-6\)

\(\dot{x}=3 t^2-10 t+8\)

\(\ddot{x}=6 t-10\)
 

\(\text {Particle at rest}\ \Rightarrow \ \dot{x}=0:\)

\(3 t^2-10 t+8\) \(=0\)  
\((3t-4)(t-2)\) \(=0\)  

 
\(t=\dfrac{4}{3}\ \text{s  or}\  \ t=2\ \text{s}\)
 

ii.  \(\text {No acceleration}\ \Rightarrow\ \ddot{x}=0\)

\(6 t-10=0\ \Rightarrow\ \ t=\dfrac{5}{3}\ \text{s}\)

\(\text{Find}\ \dot{x}\ \text{when}\ \ t=\dfrac{5}{3}: \)

  \(\dot{x}\) \( =3\left(\dfrac{5}{3}\right)^2-10\left(\dfrac{5}{3}\right)+8\)
    \(=\dfrac{25}{3}-\dfrac{50}{3}+\dfrac{24}{3} \)
    \( =-\dfrac{1}{3}\ \text{m/s}\)

 
\(\therefore\ \text{No accelerating force when}\ \ \dot{x}=-\dfrac{1}{3}\ \text{m/s} \ \text{(towards the left)}\)

Functions, EXT1 F1 EQ-Bank 4

  1. Given  \(f(x)=3 x-1\), determine the domain and range for  \(y=\sqrt{f(x)}\).   (2 marks)

  2. Sketch the graph  \(y=\sqrt{f(x)}\), clearly identifying any axis intercepts.   (1 mark)

Show Answers Only

i.     \(f(x)\) \(=3x-1\)
  \(y\) \(=\sqrt{3x-1}\)

\(\text{Domain:}\ \ 3 x-1 \ \geqslant 0 \ \Rightarrow \ x \geqslant \dfrac{1}{3}\)

\(\text {Range:}\ \ y \geqslant 0\)
 

ii.

Show Worked Solution

i.     \(f(x)\) \(=3x-1\)
  \(y\) \(=\sqrt{3x-1}\)

\(\text{Domain:}\ \ 3 x-1 \ \geqslant 0 \ \Rightarrow \ x \geqslant \dfrac{1}{3}\)

\(\text {Range:}\ \ y \geqslant 0\)
 

ii.

Functions, EXT1 F2 EQ-Bank 2

In the equation  \(x^3-8 x^2+11x+20=0\)  one of the roots is equal to the sum of the other two roots.

Find the value of the three roots.   (3 marks)

Show Answers Only

\(\text{Roots}\ = -1, 4, 5\)

Show Worked Solution

\(x^3-8 x^2+11x+20=0\)

\(\text{Sum of roots}\ =-\dfrac{b}{a}:\)

\(\alpha+\beta+(\alpha + \beta)\) \(=8\)  
\(2\alpha + 2\beta\) \(=8\)  
\(\beta\) \(=4-\alpha\ \ …\ (1)\)  

 

\(\text{Product of roots}\ =-\dfrac{d}{a}:\)

\(\alpha \beta(\alpha + \beta)=-20\ \ …\ (2) \)

\(\text{Substitute (1) into (2):}\)

\(\alpha(4-\alpha)(4)\) \(=-20\)  
\(16\alpha-4\alpha^{2}\) \(=-20\)  
\(\alpha^{2}-4\alpha-5\) \(=0\)  
\((\alpha-5)(\alpha+1)\) \(=0\)  

\(\alpha = 5\ \ \text{or}\ -1\)

\(\beta = 5\ \ \text{or}\ -1\ \ \text{(using (1))}\)

\(\alpha+ \beta = 5-1=4\)

\(\therefore \text{Roots}\ = -1, 4, 5\)

Trigonometry, EXT1 T1 EQ-Bank 5

  1. State the domain and range for  \(f(x)=4 \sin ^{-1}(2 x-3)\).   (2 marks)

  2. Sketch the function  \(f(x)=4 \sin ^{-1}(2 x-3)\).  (1 mark)

Show Answers Only

a.    \(\text{Domain:}\ \ -1 \leqslant\) \(2 x-3 \leqslant 1 \)
  \( 2 \leqslant\) \(2 x \leqslant 4\)
  \( 1 \leqslant\)  \(x \leqslant 2\)

 

        \(\text{Range:}\ \ -\dfrac{\pi}{2} \leqslant\) \(y \leqslant \dfrac{\pi}{2}\)
  \( -2 \pi \leqslant \) \(4 y \leqslant 2 \pi\)

 
b.

Show Worked Solution

a.    \(\text{Domain:}\ \ -1 \leqslant\) \(2 x-3 \leqslant 1 \)
  \( 2 \leqslant\) \(2 x \leqslant 4\)
  \( 1 \leqslant\)  \(x \leqslant 2\)

 

        \(\text{Range:}\ \ -\dfrac{\pi}{2} \leqslant\) \(y \leqslant \dfrac{\pi}{2}\)
  \( -2 \pi \leqslant \) \(4 y \leqslant 2 \pi\)

 
b.

Functions, EXT1 F1 EQ-Bank 3

Using inequalities, define the shaded region in the diagram below.   (2 marks)
 

Show Answers Only

\(y \geqslant \abs{x+1}\  \cap \  y \leqslant 2 \)

Show Worked Solution

\(\text{Shaded area is defined by the inequalities:}\)

\(y \geqslant \abs{x+1}\  \cap \  y \leqslant 2 \)

Trigonometry, EXT1 T1 EQ-Bank 5 MC

Which equation describes the graph shown?
 

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

\(C\)

Show Worked Solution

\(\text{Graph shape (general)}\ \ → \cos^{-1}(x) \)

\(\text{Graph translated 1 unit to the right}\ → \cos^{-1}(x-1) \)

\(\text{Range:}\ 0 \leq \cos^{-1}(x-1) \leq \pi\ \ →\ 0 \leq 2\cos^{-1}(x-1) \leq 2\pi \)

\(\Rightarrow C\)

BIOLOGY, M2 EQ-Bank 7

Select two human organs and explain any alterations in blood composition as it passes through each organ.   (4 marks)

Show Answers Only

Answers could include two of the following.

Lungs:

→ As blood passes through the lungs, deoxygenated blood becomes oxygenated.

→ Carbon dioxide is also removed from the blood and exhaled.
 

Kidneys:

→ In the kidneys, blood is filtered to remove waste products like urea, excess salts, and water, which are excreted as urine.

→ The blood leaving the kidneys has a reduced concentration of these waste products, and its composition is more balanced in terms of water and electrolyte levels.
 

Liver:

→ The liver modifies the blood by metabolising nutrients, detoxifying harmful substances, and storing or releasing glucose as needed.

→ Blood exiting the liver has a regulated concentration of glucose and may have reduced levels of toxins, as the liver converts harmful compounds into safer by-products for excretion.

Show Worked Solution

Answers could include two of the following.

Lungs:

→ As blood passes through the lungs, deoxygenated blood becomes oxygenated.

→ Carbon dioxide is also removed from the blood and exhaled.
 

Kidneys:

→ In the kidneys, blood is filtered to remove waste products like urea, excess salts, and water, which are excreted as urine.

→ The blood leaving the kidneys has a reduced concentration of these waste products, and its composition is more balanced in terms of water and electrolyte levels.
 

Liver:

→ The liver modifies the blood by metabolising nutrients, detoxifying harmful substances, and storing or releasing glucose as needed.

→ Blood exiting the liver has a regulated concentration of glucose and may have reduced levels of toxins, as the liver converts harmful compounds into safer by-products for excretion.

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 2 MC

Blood was analysed after passing through a particular organ. It was observed that oxygen levels increased, while carbon dioxide and urea levels decreased. Which tissue is most likely responsible for these changes in blood composition?

  1. Brain
  2. Heart
  3. Kidneys
  4. Lungs
Show Answers Only

\(D\)

Show Worked Solution

→ In the lungs, oxygen diffuses into the blood while carbon dioxide diffuses out, resulting in an increase in oxygen levels and a decrease in carbon dioxide levels in the blood leaving the lungs.

→ The decrease in urea is not typical of lung function, but could be explained by small amounts of urea being excreted in exhaled air.

\(\Rightarrow D\)

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, M2 EQ-Bank 7v3

Consider a sample of gas confined in a cylinder equipped with a movable piston. The initial pressure of the gas is 1.2 atmospheres, and the initial volume is 6.0 litres. The piston is then adjusted to change the volume of the gas.

Assuming the temperature of the gas remains constant throughout the process, calculate the new pressure of the gas when the volume is adjusted to:

  1. 3.0 litres  (2 marks)
  2. 12.0 litres  (2 marks)
Show Answers Only

a.    \(2.4\ \text{atm}\)

b.    \(0.6\ \text{atm}\)

Show Worked Solution

Boyle’s Law is mathematically expressed as \( P_1V_1 = P_2V_2 \).

a.    \( P_1 = 1.2 \, \text{atm} \)

\( V_1 = 6.0 \, \text{L} \)

\( V_2 = 3.0 \, \text{L} \)

\( P_2 = \dfrac{P_1 \times V_1}{V_2} = \dfrac{1.2 \, \text{atm} \times 6.0 \, \text{L}}{3.0 \, \text{L}} = 2.4 \, \text{atm}\)
 

b.    \( V_2 = 12.0 \, \text{L} \)

\( P_2 = \dfrac{P_1 \times V_1}{V_2} = \dfrac{1.2 \, \text{atm} \times 6.0 \, \text{L}}{12.0 \, \text{L}} = 0.6 \, \text{atm}\)

CHEMISTRY, M2 EQ-Bank 7v4

Consider a sample of gas confined in a cylinder equipped with a movable piston. The initial pressure of the gas is 2.5 atmospheres, and the initial volume is 2.0 litres. The piston is then adjusted to change the volume of the gas.

Assuming the temperature of the gas remains constant throughout the process, calculate the new pressure of the gas when the volume is adjusted to:

  1. 1.0 litre  (2 marks)
  2. 4.0 litres  (2 marks)
Show Answers Only

a.    \(5.0\ \text{atm}\)

b.    \(1.25\ \text{atm}\)

Show Worked Solution

a.   Boyle’s Law is mathematically expressed as \( P_1V_1 = P_2V_2 \).

\( P_1 = 2.5 \, \text{atm} \)

\( V_1 = 2.0 \, \text{L} \)

\( V_2 = 1.0 \, \text{L} \)

\( P_2 = \dfrac{P_1 \times V_1}{V_2} = \dfrac{2.5 \, \text{atm} \times 2.0 \, \text{L}}{1.0 \, \text{L}} = 5.0 \, \text{atm}\)
 

b.    \( V_2 = 4.0 \, \text{L} \)

\( P_2 = \dfrac{P_1 \times V_1}{V_2} = \dfrac{2.5 \, \text{atm} \times 2.0 \, \text{L}}{4.0 \, \text{L}} = 1.25 \, \text{atm}\)

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, M2 EQ-Bank 7v3

Consider a sample of gas confined in a cylinder equipped with a movable piston. The initial pressure of the gas is 1.8 atmospheres, and the initial volume is 5.0 litres. The piston is then adjusted to change the volume of the gas.

Calculate the new pressure of the gas when the volume is adjusted to:

  1. 2.5 litres  (2 marks)
  2. 10.0 litres  (2 marks)

Assume the temperature of the gas remains constant throughout the process.

Show Answers Only

a.    \(3.6\ \text{atm}\)

b.    \(0.9\ \text{atm}\)

Show Worked Solution

a.   Boyle’s Law is mathematically expressed as \( P_1V_1 = P_2V_2 \).

\( P_1 = 1.8 \, \text{atm} \)

\( V_1 = 5.0 \, \text{L} \)

\( V_2 = 2.5 \, \text{L} \)

\( P_2 = \dfrac{P_1 \times V_1}{V_2} = \dfrac{1.8 \, \text{atm} \times 5.0 \, \text{L}}{2.5 \, \text{L}} = 3.6 \, \text{atm}\)
 

b.    \( V_2 = 10.0 \, \text{L} \)

\( P_2 = \dfrac{P_1 \times V_1}{V_2} = \dfrac{1.8 \, \text{atm} \times 5.0 \, \text{L}}{10.0 \, \text{L}} = 0.9 \, \text{atm}\)

CHEMISTRY, M3 EQ-Bank 29v4

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

  1. The reaction between aluminium and nitric acid.   (1 mark)
  2. The decomposition of mercury (II) oxide.    (1 mark)
  3. The complete combustion of pentane (\(\ce{C5H12}\)) with excess oxygen.   (1 mark)
  4. The reaction between magnesium hydroxide and acetic acid (\(\ce{CH3COOH}\)).   (1 mark)
  5. The reaction between calcium carbonate and hydrochloric acid.   (1 mark)
Show Answers Only

a.    \(\ce{2Al + 6HNO3 -> 2Al(NO3)3 + 3H2}\)

b.    \(\ce{2HgO -> 2Hg + O2}\)

c.    \(\ce{C5H12 + 8O2 -> 5CO2 + 6H2O}\)

d.    \(\ce{Mg(OH)2 + 2CH3COOH -> Mg(CH3COO)2 + 2H2O}\)

e.    \(\ce{CaCO3 + 2HCl -> CaCl2 + CO2 + H2O}\)

Show Worked Solution

a.    Active metal and acid

\(\ce{2Al + 6HNO3 -> 2Al(NO3)3 + 3H2}\)

b.    Decomposition 

\(\ce{2HgO -> 2Hg + O2}\)

c.    Combustion

\(\ce{C5H12 + 8O2 -> 5CO2 + 6H2O}\)

d.    Acid-base

\(\ce{Mg(OH)2 + 2CH3COOH -> Mg(CH3COO)2 + 2H2O}\)

e.    Acid-carbonate

\(\ce{CaCO3 + 2HCl -> CaCl2 + CO2 + H2O}\)

CHEMISTRY, M3 EQ-Bank 29v3

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

  1. The reaction between iron and sulfuric acid.   (1 mark)
  2. The decomposition of sodium carbonate.   (1 mark)
  3. The incomplete combustion of propene (\(\ce{C3H6}\)) with a limited amount of oxygen.   (1 mark)
  4. The reaction between calcium hydroxide and phosphoric acid.   (1 mark)
  5. The reaction between sodium carbonate and nitric acid.   (1 mark) 
Show Answers Only

a.    \(\ce{Fe + H2SO4 -> FeSO4 + H2}\)

b.    \(\ce{Na2CO3 -> Na2O + CO2}\)

c.    \(\ce{2C3H6 + 9O2 -> 6CO + 6H2O}\)

d.    \(\ce{3Ca(OH)2 + 2H3PO4 -> Ca3(PO4)2 + 6H2O}\)

e.    \(\ce{Na2CO3 + 2HNO3 -> 2NaNO3 + CO2 + H2O}\)

Show Worked Solution

a.    Active metal and acid

\[\ce{Fe + H2SO4 -> FeSO4 + H2}\]

b.    Decomposition

\[\ce{Na2CO3 -> Na2O + CO2}\]

c.   Combustion

\[\ce{2C3H6 + 9O2 -> 6CO + 6H2O}\]

d.    Acid-Base

\[\ce{3Ca(OH)2 + 2H3PO4 -> Ca3(PO4)2 + 6H2O}\]

e.    Acid-Carbonate

\[\ce{Na2CO3 + 2HNO3 -> 2NaNO3 + CO2 + H2O}\]

CHEMISTRY, M2 EQ-Bank 5

Consider a sample of gas confined in a cylinder equipped with a movable piston. The initial pressure of the gas is 2.0 atmospheres, and the initial volume is 3.0 litres. The piston is then adjusted to change the volume of the gas.

Assuming the temperature of the gas remains constant throughout the process, calculate the new pressure of the gas when the volume is adjusted to:

  1. 1.5 litres   (2 marks)
  2. 6.0 litres   (2 marks)
Show Answers Only

a.    \(4.0\ \text{atm}\)

b.    \(1.0\ \text{atm}\)

Show Worked Solution

a.   Boyle’s Law is mathematically expressed as \( P_1V_1 = P_2V_2 \).

\( P_1 = 2.0 \, \text{atm} \)

\( V_1 = 3.0 \, \text{L} \)

\( V_2 = 1.5 \, \text{L} \)

\( P_2 = \dfrac{P_1 \times V_1}{V_2} = \dfrac{2.0 \, \text{atm} \times 3.0 \, \text{L}}{1.5 \, \text{L}} = 4.0 \, \text{atm}\)
 

b.    \( V_2 = 6.0 \, \text{L} \)

\( P_2 = \dfrac{P_1 \times V_1}{V_2} = \dfrac{2.0 \, \text{atm} \times 3.0 \, \text{L}}{6.0 \, \text{L}} = 1.0 \, \text{atm}\)

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, M2 EQ-Bank 8v3

  1. Consider the compounds ethanol (\(\ce{C2H6O}\)), formaldehyde (\(\ce{CH2O}\)), and acetic acid (\(\ce{C2H4O2}\)).
    Identify which TWO of these compounds have the same empirical formula and justify your choice.    (2 marks)

  2. The empirical formula of a compound is \(\ce{C4H5N2}\) and its molar mass is determined to be 243.3 g mol\(^{-1}\).
    Calculate the molecular formula of this compound.    (3 marks)
Show Answers Only

a.    Formaldehyde (\(\ce{CH2O}\)) and acetic acid (\(\ce{C2H4O2}\)) have the same empirical formula of \(\ce{CH2O}\).

b.    The molecular formula of the compound is \(\ce{C12H15N6}\).

Show Worked Solution

a.    Determine the empirical formula of each compound:

Ethanol (\(\ce{C2H6O}\)):
\[\text{Empirical formula} = \ce{C2H6O}\]

Formaldehyde (\(\ce{CH2O}\)):
\[\text{Empirical formula} = \ce{CH2O}\]

Acetic acid (\(\ce{C2H4O2}\)):
\[\text{Empirical formula} = \ce{CH2O}\]

Formaldehyde and acetic acid have the same empirical formula of \(\ce{CH2O}\).

b.     Calculate the molar mass of the empirical formula \(\ce{C4H5N2}\):

\[\text{Molar mass of} \ \ce{C4H5N2} = 4 \times 12.01 + 5 \times 1.01 + 2 \times 14.01 = 81.1 \ \text{g/mol}\]

Determine the ratio of the molar mass of the compound to the molar mass of the empirical formula:

\[\text{Ratio} = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{243.3 \ \text{g/mol}}{81.1 \ \text{g/mol}} = 3\]

Multiply the subscripts in the empirical formula by 3 to get the molecular formula:

\[\text{Molecular formula} = \ce{(C4H5N2)} \times 3 = \ce{C12H15N6}\]

Thus, the molecular formula of the compound is \(\ce{C12H15N6}\).

CHEMISTRY, M2 EQ-Bank 8v2

  1. Consider the compounds glyceraldehyde (\(\ce{C3H6O3}\)), glycolic acid (\(\ce{C2H4O3}\)), and ribose (\(\ce{C5H10O5}\)).
    Identify which TWO of these compounds have the same empirical formula and justify your choice.    (2 marks)

  2. The empirical formula of a compound is \(\ce{C3H5O}\) and its molar mass is determined to be 114.14 g mol\(^{-1}\).
    Calculate the molecular formula of this compound.    (3 marks)
Show Answers Only

a.    glyceraldehyde (\(\ce{C3H6O3}\)) and ribose (\(\ce{C5H10O5}\)) have the same empirical formula of \(\ce{CH2O}\).

b.    The molecular formula of the compound is \(\ce{C6H10O2}\).

Show Worked Solution

a.    Determine the empirical formula of each compound:

Glyceraldehyde (\(\ce{C3H6O3}\)):
\[\text{Empirical formula} = \ce{CH2O}\]

Glycolic acid (\(\ce{C2H4O3}\)):
\[\text{Empirical formula} = \ce{C2H4O3}\]

Ribose (\(\ce{C5H10O5}\)):
\[\text{Empirical formula} = \ce{CH2O}\]

Acetone and ribose have the same empirical formula of \(\ce{CH2O}\).

b.    Calculate the molar mass of the empirical formula \(\ce{C3H5O}\):

\[\text{Molar mass of} \ \ce{C3H5O} = 3 \times 12.01 + 5 \times 1.008 + 1 \times 16.00 = 57.07 \ \text{g/mol}\]

Determine the ratio of the molar mass of the compound to the molar mass of the empirical formula:

\[\text{Ratio} = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{114.14 \ \text{g/mol}}{57.07 \ \text{g/mol}} = 2\]

Multiply the subscripts in the empirical formula by 2 to get the molecular formula:

\[\text{Molecular formula} = \ce{(C3H5O)} \times 2 = \ce{C6H10O2}\]

Thus, the molecular formula of the compound is \(\ce{C6H10O2}\).

CHEMISTRY, M4 EQ-Bank 10 MC

Using Hess’s Law, if the enthalpy changes for the following reactions are known:

\(1.\ \ce{C(s) + O2(g) -> CO2(g)}\ \ \ \ \ \ \ \ \ \ \ \Delta H_1 = -393\ \text{kJ mol}^{-1}\)

\(2.\ \ce{CO(g) + 1/2 O2(g) -> CO2(g)}\ \ \ \ \Delta H_2 = -283\ \text{kJ mol}^{-1}\)

Determine the enthalpy change for the reaction:

\(\ce{C(s) + 1/2 O2(g) -> CO(g)}\)
 

  1. \(+110\ \text{kJ mol}^{-1}\)
  2. \(-110\ \text{kJ mol}^{-1}\)
  3. \(-676\ \text{kJ mol}^{-1}\)
  4. \(+676\ \text{kJ mol}^{-1}\)
Show Answer Only

\(B\)

Show Worked Solution

→ Reverse equation 2, making  \(\ce{Co(g)}\)  a product and it changes the sign for \(\Delta H_2\).

Equation 2*: \(\ce{CO2(g) -> C2(g) + 1/2 O2(g)} \qquad -\Delta H_2 = +283\ \text{kJ mol}^{-1}\)

→ Combine equation 2* with equation 1 which cancels out \(\ce{CO2(g)}\) and adds the enthalpy values together.

→ Thus, the enthalpy for the final reaction \(=-393 + 283= -110\ \text{kJ mol}^{-1}\)

\(\Rightarrow B\)

CHEMISTRY, M4 EQ-Bank 9 MC

Which statement best describes Hess's Law?

  1. The total enthalpy change for a reaction depends on the pathway taken to reach the final products.
  2. The total enthalpy change for a reaction is zero if the reaction is reversible.
  3. The total enthalpy change for a reaction is independent of the pathway taken, as long as the initial and final states are the same.
  4. The enthalpy change for a reaction is the same regardless of whether it occurs in one step or multiple steps, provided heat is lost to the surroundings.
Show Answer Only

\(C\)

Show Worked Solution

→ Hess’s Law states that the total enthalpy change for a reaction is independent of the pathway taken, as long as the initial and final states are the same.

→ This means that whether a reaction happens in one step or multiple steps, the overall enthalpy change will be the same if the reactants and products are the same.

\(\Rightarrow C\)

CHEMISTRY, M4 EQ-Bank 7

Bond energies can be used to estimate the enthalpy change of a reaction. The equation for the combustion of methane \(\ce{CH4}\) is shown below:

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

Given the following bond energies:

  • \(\ce{C-H}\) bond: 412 kJ mol\(^{-1}\)
  • \(\ce{O=O}\) bond: 498 kJ mol\(^{-1}\)
  • \(\ce{C=O}\) bond: 805 kJ mol\(^{-1}\)
  • \(\ce{O-H}\) bond: 463 kJ mol\(^{-1}\)

Calculate the total bond energy of the products.   (2 marks)

Show Answer Only

\(3462\ \text{kJ mol}^{-1}\)

Show Worked Solution

→ Total bond energy of 1 mole of \(\ce{CO2} = 2 \times 805 = 1610\ \text{kJ mol}^{-1}\)

→ Total bond energy of 2 mole of \(\ce{H2O} = 4 \times 463 = 1852\ \text{kJ mol}^{-1}\)

→ Thus the total bond energy of the products is \(3462\ \text{kJ mol}^{-1}\)

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 4

Describe two structural differences between arteries and veins and explain how these structural differences relate to their respective functions in blood circulation.   (3 marks)

Show Answers Only

→ Arteries have thicker walls with more elastic tissue and smooth muscle compared to veins, which have thinner walls and less muscle.

→ This structural difference allows arteries to withstand and maintain the high pressure of blood pumped directly from the heart, while also enabling them to expand and contract to help propel blood forward.

→ Veins, on the other hand, have valves along their length, which are absent in arteries.

→ These valves prevent the backflow of blood as it moves towards the heart against gravity, especially from the lower extremities, a feature not needed in arteries where blood flows under high pressure away from the heart.

Show Worked Solution

→ Arteries have thicker walls with more elastic tissue and smooth muscle compared to veins, which have thinner walls and less muscle.

→ This structural difference allows arteries to withstand and maintain the high pressure of blood pumped directly from the heart, while also enabling them to expand and contract to help propel blood forward.

→ Veins, on the other hand, have valves along their length, which are absent in arteries.

→ These valves prevent the backflow of blood as it moves towards the heart against gravity, especially from the lower extremities, a feature not needed in arteries where blood flows under high pressure away from the heart.

BIOLOGY, M2 EQ-Bank 1

Compare and contrast open and closed circulatory systems. In your answer, describe one advantage of each system, providing an example of an organism with each type of circulatory system.   (4 marks)

Show Answers Only

→ An open circulatory system has blood flowing freely in body cavities, while a closed circulatory system contains blood within vessels throughout its journey.

→ An advantage of an open system is its simplicity and low energy requirement, while a closed system allows for more efficient and rapid distribution of materials.

→ Closed systems are necessary in larger animals because they can maintain higher blood pressures, allowing for more efficient oxygen delivery to tissues far from the heart.

→ Insects (eg. grasshoppers) have an open circulatory system, while mammals (eg. humans) have a closed circulatory system.

Show Worked Solution

→ An open circulatory system has blood flowing freely in body cavities, while a closed circulatory system contains blood within vessels throughout its journey.

→ An advantage of an open system is its simplicity and low energy requirement, while a closed system allows for more efficient and rapid distribution of materials.

→ Closed systems are necessary in larger animals because they can maintain higher blood pressures, allowing for more efficient oxygen delivery to tissues far from the heart.

→ Insects (eg. grasshoppers) have an open circulatory system, while mammals (eg. humans) have a closed circulatory system.

BIOLOGY, M2 EQ-Bank 16

Compare and contrast the methods by which autotrophs and heterotrophs obtain essential minerals, giving an example of each.   (3 marks)

Show Answers Only

→ Autotrophs primarily obtain essential minerals directly from their environment, such as soil or water.

→ Heterotrophs mainly acquire minerals by consuming other organisms or their products, obtaining these nutrients that have already been accumulated and processed by their food sources.

→ This means autotrophs must invest energy in converting inorganic minerals into usable organic forms, whereas heterotrophs receive minerals that have already undergone this conversion.

→ For example, a plant (autotroph) absorbs nitrogen in the form of nitrates from the soil through its root system, while a lion (heterotroph) obtains nitrogen by consuming proteins in the meat of its prey. 

Show Worked Solution

→ Autotrophs primarily obtain essential minerals directly from their environment, such as soil or water.

→ Heterotrophs mainly acquire minerals by consuming other organisms or their products, obtaining these nutrients that have already been accumulated and processed by their food sources.

→ This means autotrophs must invest energy in converting inorganic minerals into usable organic forms, whereas heterotrophs receive minerals that have already undergone this conversion.

→ For example, a plant (autotroph) absorbs nitrogen in the form of nitrates from the soil through its root system, while a lion (heterotroph) obtains nitrogen by consuming proteins in the meat of its prey. 

BIOLOGY, M2 EQ-Bank 15

"Autotrophs and heterotrophs have distinct nutritional strategies."

Describe one similarity and one difference in the oxygen requirements between these two groups.   (2 marks)

Show Answers Only

→ Both autotrophs (eg. green plants) and heterotrophs (eg. mammals) require oxygen for cellular respiration.

→ Autotrophs also produce oxygen as a byproduct of photosynthesis.

→ This means that during daylight hours, many autotrophs may produce more oxygen than they consume, while heterotrophs are always net consumers of oxygen.

Show Worked Solution

→ Both autotrophs (eg. green plants) and heterotrophs (eg. mammals) require oxygen for cellular respiration.

→ Autotrophs also produce oxygen as a byproduct of photosynthesis.

→ This means that during daylight hours, many autotrophs may produce more oxygen than they consume, while heterotrophs are always net consumers of oxygen.