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BIOLOGY, M3 2004 HSC 18a

A plant species found in the area immediately around Sydney has also been found in a small area in the Gibraltar Range in the far north of NSW. 
 

Predict what might happen to the TWO populations over the next 5 million years, in terms of Darwin/Wallace's theory of evolution.  (3 marks)

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→ According to Darwin/Wallace’s theory of evolution these two plant populations will undergo divergent evolution over the next 5 million years due to their seperate geographical location.

→ The Sydney population will likely experience significantly more rainfall than the Gibralter range population. As a result, this plant will evolve to require more water and have less resistance to drier conditions.

→ The Gibralter range species exist further inland and at a higher altitude and are likely to experience less rainfall and a greater range of temperatures. This species could evolve to be more drought resistant and hardier in extreme temperatures.

→ While the theory of evolution can provide a prediction, the actual outcome may be vastly different due to other abiotic and biotic factors, such as the presence of predators.

Show Worked Solution

→ According to Darwin/Wallace’s theory of evolution these two plant populations will undergo divergent evolution over the next 5 million years due to their seperate geographical location.

→ The Sydney population will likely experience significantly more rainfall than the Gibralter range population. As a result, this plant will evolve to require more water and have less resistance to drier conditions.

→ The Gibralter range species exist further inland and at a higher altitude and are likely to experience less rainfall and a greater range of temperatures. This species could evolve to be more drought resistant and hardier in extreme temperatures.

→ While the theory of evolution can provide a prediction, the actual outcome may be vastly different due to other abiotic and biotic factors, such as the presence of predators.

BIOLOGY, M3 2009 HSC 27

Most offspring resemble their parents in a number of characteristics, but there are often some characteristics in the offspring that are unexpected.  

Explain, using examples, how genetics and the environment can affect the phenotype of individuals.  (8 marks)

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→ Offspring inherit characteristics based off their parents. Their are various factors which can influence what characteristics, or phenotype, they actually exhibit.

→ When offspring exhibit characteristics that are unexpected, this is often due to the characteristic being a recessive trait. In this way, both parents had the allele for it but it was masked by the dominant trait. The offspring then inherited both of these recessive alleles, meaning it could no longer be masked.

→ One such example is if the parents are both type A but their son/daughter is type O as a result of inheriting the recessive allele for no surface proteins.

→ Traits can also be unexpected if they are sex-linked, meaning the allele lies on an X sex chromosome. In this way, males will only have to have one recessive X allele for it to be expressed, while females need them both. In this way, the ratios for sex-linked characteristics is different than if they were autosomal.

→ When genes are being copied, mistakes (mutations) are made in the process and new genes or combinations of genes can be generated in the process.

→ These can be passed on to offspring, giving them characteristics different from their parents. For example, the peppered moth. Dark coloured peppered moths appeared in the population due to a mutation.

→ The environment can affect the way in which genes are expressed so that an individuals phenotype is affect by environmental conditions. For example, malnutrition can lead to individuals being shorter in height compared to their genetic potential.

Show Worked Solution

→ Offspring inherit characteristics based off their parents. Their are various factors which can influence what characteristics, or phenotype, they actually exhibit.

→ When offspring exhibit characteristics that are unexpected, this is often due to the characteristic being a recessive trait. In this way, both parents had the allele for it but it was masked by the dominant trait. The offspring then inherited both of these recessive alleles, meaning it could no longer be masked.

→ One such example is if the parents are both type A but their son/daughter is type O as a result of inheriting the recessive allele for no surface proteins.

→ Traits can also be unexpected if they are sex-linked, meaning the allele lies on an X sex chromosome. In this way, males will only have to have one recessive X allele for it to be expressed, while females need them both. In this way, the ratios for sex-linked characteristics is different than if they were autosomal.

→ When genes are being copied, mistakes (mutations) are made in the process and new genes or combinations of genes can be generated in the process.

→ These can be passed on to offspring, giving them characteristics different from their parents. For example, the peppered moth. Dark coloured peppered moths appeared in the population due to a mutation.

→ The environment can affect the way in which genes are expressed so that an individuals phenotype is affect by environmental conditions. For example, malnutrition can lead to individuals being shorter in height compared to their genetic potential.

BIOLOGY, M3 2010 HSC 30

Geological and biological history of New Zealand

Use this information and other relevant knowledge to demonstrate how the practice of biology has led to the validation of current theories of evolution.  (7 marks)

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Show Worked Solution

→ In the New Zealand information, many practices of biology were undertaken to compile the data.

→ The observations of fauna in New Zealand represent an important practice in biological science. Making clear, dispassionate and unbiased observations, and recording such observations are intrinsic to science and provide an essential pathway to validating theories.

→ These observations validated the idea of convergent evolution; birds in New Zealand, without competition from mammals, developed similar characteristics and were able to live in the same environments as mammals do elsewhere.

→ Practices in palaeontology of ageing fossils by radiometric dating or stratigraphy, describing fossils and comparing fossils, allows a history of fauna of New Zealand to be compiled, and history of islands (eg. deducing from fossils found to be almost completely marine that New Zealand was covered by oceans).

→ The history of fauna and geologic events validates Charles Darwin’s theory of evolution in that populations of birds, which began as occasional visitors, gradually changed into permanent endemic species.

→ Divergent evolution is validated as new unique species develop from migratory species, as shown by the fossils of new, unique birds.

→ The data also validates the punctuated equilibrium theory in that the diversification of bird species happened very quickly in the space of only 2 million years.

BIOLOGY, M3 2013 SM-Bank 21

Explain how ONE secondary source has provided support for the theory of evolution.  (4 marks)

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→ The theory of evolution suggests that species occur by variation of a common ancestor.

→ Variation within an ancient species caused certain characteristics to be favoured in certain environments, causing divergent evolution and overtime the development of new species.

→ This theory can be validated by comparative anatomy, the practise of comparing body structures of different species.

→ When comparing forelimbs of vertebrate, such as canines, humans and whales, we are presented with similarities in their structures. The bone structure is especially similar, with the named animals all having the same pentadactyl (five-digit) limb plan, as they all possess some form of humerus, ulna, radius, carpels and metacarpals. This suggests common ancestry.

Show Worked Solution

→ The theory of evolution suggests that species occur by variation of a common ancestor.

→ Variation within an ancient species caused certain characteristics to be favoured in certain environments, causing divergent evolution and overtime the development of new species.

→ This theory can be validated by comparative anatomy, the practise of comparing body structures of different species.

→ When comparing forelimbs of vertebrate, such as canines, humans and whales, we are presented with similarities in their structures. The bone structure is especially similar, with the named animals all having the same pentadactyl (five-digit) limb plan, as they all possess some form of humerus, ulna, radius, carpels and metacarpals. This suggests common ancestry.

BIOLOGY, M3 2014 HSC 26

Explain how Darwin/Wallace's theory of evolution by natural selection and isolation accounts for convergent evolution. Use an example to support your answer.  (5 marks)

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→ Convergent evolution occurs when two species evolve to possess similar characteristics by natural selection in similar environments.

→ The organisms cannot interbreed to share any new DNA generated by mutation: they are genetically isolated from each other yet they develop similar characteristics.

→ When variation exists in a population, the theory of evolution by natural selection states that this variation will cause some individuals to be better suited to their environment. In this way, they are more likely to survive and hence reproduce, passing on their adaptive characteristics to further generations. After many generations, the traits of those variants are common in the population.

→ Dolphins and sharks demonstrate convergent evolution. The dolphin is a mammal and the shark is a fish. They inhabit the marine environment which imposes the same selection pressures on both types of organism. 

→ Despite being genetically isolated, they both exhibit a streamlined body shape and possess fins for propulsion and stability.

→ These features are adaptive for movement in a highly viscous environment.

Show Worked Solution

→ Convergent evolution occurs when two species evolve to possess similar characteristics by natural selection in similar environments.

→ The organisms cannot interbreed to share any new DNA generated by mutation: they are genetically isolated from each other yet they develop similar characteristics.

→ When variation exists in a population, the theory of evolution by natural selection states that this variation will cause some individuals to be better suited to their environment. In this way, they are more likely to survive and hence reproduce, passing on their adaptive characteristics to further generations. After many generations, the traits of those variants are common in the population.

→ Dolphins and sharks demonstrate convergent evolution. The dolphin is a mammal and the shark is a fish. They inhabit the marine environment which imposes the same selection pressures on both types of organism. 

→ Despite being genetically isolated, they both exhibit a streamlined body shape and possess fins for propulsion and stability.

→ These features are adaptive for movement in a highly viscous environment.

Mean mark 51%.

BIOLOGY, M2 2014 HSC 24

  1. Use labelled diagrams to distinguish between the structure of an artery and that of a capillary.   (2 marks)

  2. Relate one structure of a capillary to its function.   (2 marks)

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a.   
       

b.    Structure: One cell thick wall.

Related function: Allows diffusion of small molecules through the capillary wall to allow substances in and out of the bloodstream.

Show Worked Solution

a.   
       

b.    Structure: One cell thick wall.

Related function: Allows diffusion of small molecules through the capillary wall to allow substances in and out of the bloodstream.

♦ Mean mark 49%.

BIOLOGY, M2 2015 HSC 27

  1. Outline TWO differences between whole blood and plasma.  (2 marks)

The steps below show the preparation and use of blood products in the treatment of Ebola Virus Disease. This disease is characterised by significant blood loss.

  1. Explain why this protocol produces an effective treatment for Ebola Virus Disease.  (3 marks)

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a.    Whole blood and plasma differences:

→ Whole blood contains cells including WBC’s, RBC’s and platelets while plasma does not.

→ Plasma is a straw coloured liquid and whole blood is a thick red liquid.
 

b.   Answers could include:

→ The plasma will contain antibodies for this disease because it has been taken from someone who has survived the disease. This will help to immobilise the virus in recipients blood stream.

→ Screening blood prevents the spread of Ebola and other blood-borne diseases from donor to recipient.

→ The plasma is separated from the whole blood meaning no blood type match is needed as there are no cells in the plasma but it still contains the beneficial antibodies.

Show Worked Solution

a.    Whole blood and plasma differences:

→ Whole blood contains cells including WBC’s, RBC’s and platelets while plasma does not.

→ Plasma is a straw coloured liquid and whole blood is a thick red liquid.

♦ Mean mark (a) 49%.

b.   Answers could include:

→ The plasma will contain antibodies for this disease because it has been taken from someone who has survived the disease. This will help to immobilise the virus in recipients blood stream.

→ Screening blood prevents the spread of Ebola and other blood-borne diseases from donor to recipient.

→ The plasma is separated from the whole blood meaning no blood type match is needed as there are no cells in the plasma but it still contains the beneficial antibodies.

BIOLOGY, M2 2015 HSC 24

Data can be provided by a pulse oximeter pegged to a person's finger, as shown in the diagram.

  1. What is the oxygen saturation for this person?  (1 mark)

  2. Outline TWO limitations of using only the information provided in the diagram to determine the 'health' of a person.  (2 marks)

  3. Explain TWO advantages in using a pulse oximeter to measure oxygen saturation compared to using another named technology in a specific setting.  (4 marks)

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a.   97%
 

b.   Answers could include:

→ Data requires context (eg. is the user exercising or resting).

→ There are many other parameters of physical health: disease status, blood levels of toxic environmental substances, presence of cancers etc…

→ There are other aspects of non-physical health (eg. mental health).

→ The pulse oximeter could read faulty i.e. if not properly calibrated, so comparing with results from another form of test such as a blood pressure monitor would give a better indication of cardiorespiratory health.
 

c.   Answers could include:

→ The pulse oximeter can give you immediate and continuous data. This makes it better than arterial blood gas determination from a blood sample that requires the sample to be analysed in a laboratory with a delayed and one-off reading.

→ The pulse oximeter is very portable being compact and battery operated.

→ The pulse oximeter is better for an ambulance where fast and continuous data delivery in an emergency setting is required.

Show Worked Solution

a.   97%
 

b.   Answers could include:

→ Data requires context (eg. is the user exercising or resting).

→ There are many other parameters of physical health: disease status, blood levels of toxic environmental substances, presence of cancers etc…

→ There are other aspects of non-physical health (eg. mental health).

→ The pulse oximeter could read faulty i.e. if not properly calibrated, so comparing with results from another form of test such as a blood pressure monitor would give a better indication of cardiorespiratory health.
 

c.   Answers could include:

→ The pulse oximeter can give you immediate and continuous data. This makes it better than arterial blood gas determination from a blood sample that requires the sample to be analysed in a laboratory with a delayed and one-off reading.

→ The pulse oximeter is very portable being compact and battery operated.

→ The pulse oximeter is better for an ambulance where fast and continuous data delivery in an emergency setting is required.

BIOLOGY, M2 2016 HSC 31

As altitude increases, the partial pressure of oxygen \( \text{(p} \ce{O_2)}\) in air decreases. 

Species A and B are closely related endotherms that live in different habitats in Asia. The minimum \( \text{p} \ce{O_2}\) required for 100% blood oxygen saturation differs in these species because of differences in their haemoglobin structure. Data related to these two species are shown below.

\begin{equation}
\begin{array}{|c|c|c|}
\hline \text { Endotherm species } & \text { Habitat altitude } & \text { Minimum } \mathrm{pO}_2 \text { for } 100 \%\ \mathrm{Hb} \text { saturation } \\
\hline \mathrm{A} & \mathrm{High} & 54 \\
\mathrm{~B} & \text { Low } & 80 \\
\hline
\end{array}
\end{equation}

Explain how the differences in these species could have arisen, using the Darwin/Wallace theory of evolution and your understanding of the adaptive advantage of haemoglobin.  (8 marks)

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→ Haemoglobin is a protein that provides a mechanism for transport of oxygen around the body. As it is a protein, it’s structure is dependant on the individual’s genotype.

→ Species A and Species B are able to reach 100% saturation at differing partial pressures of oxygen, meaning they have different DNA which codes for different haemoglobin structures.

→ Species A and B are likely to have diverged from a common ancestor because of differing environmental pressures resulting in two different species. Within the ancestral population there was variation which resulted from random mutations. One mutation would have resulted in haemoglobin that is able to reach 100% saturation at a lower partial pressure of oxygen.

→ When members of the ancestral species moved to a higher altitude the ability of their haemoglobin to reach saturation at a lower \( \text{p} \ce{O_2}\) gave them a survival advantage. These individuals were then more likely to reproduce and pass on their favourable genes.

→ For individuals living at lower altitudes, there is no survival advantage to being able to reach 100% saturation at lower \( \text{p} \ce{O_2}\) which means this trait was not selected for.

→ Over time, due to the isolation at a higher altitude a new species evolved.

Show Worked Solution

→ Haemoglobin is a protein that provides a mechanism for transport of oxygen around the body. As it is a protein, it’s structure is dependant on the individual’s genotype.

→ Species A and Species B are able to reach 100% saturation at differing partial pressures of oxygen, meaning they have different DNA which codes for different haemoglobin structures.

→ Species A and B are likely to have diverged from a common ancestor because of differing environmental pressures resulting in two different species. Within the ancestral population there was variation which resulted from random mutations. One mutation would have resulted in haemoglobin that is able to reach 100% saturation at a lower partial pressure of oxygen.

→ When members of the ancestral species moved to a higher altitude the ability of their haemoglobin to reach saturation at a lower \( \text{p} \ce{O_2}\) gave them a survival advantage. These individuals were then more likely to reproduce and pass on their favourable genes.

→ For individuals living at lower altitudes, there is no survival advantage to being able to reach 100% saturation at lower \( \text{p} \ce{O_2}\) which means this trait was not selected for.

→ Over time, due to the isolation at a higher altitude a new species evolved.

♦♦ Mean mark 41%.

BIOLOGY, M2 2016 HSC 27

The diagram shows a vascular bundle from a flowering plant.

  1. On the diagram, clearly label a xylem vessel.  (1 mark)
  2. A adaptation of this species causes the walls of the xylem vessels to be significantly reduced in thickness.
  3. Explain why the leaves of these adapted strains will wilt more easily.  (3 marks)

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a.   
       

b.   → Leaves wilt when they lose water.

→ As xylem tissue is responsible for water transport from roots to leaves, plants with the adaptation will wilt easier as the xylem vessels are more likely to collapse with thinner walls.

Show Worked Solution

a.   
       

b.   → Leaves wilt when they lose water.

→ As xylem tissue is responsible for water transport from roots to leaves, plants with the adaptation will wilt easier as the xylem vessels are more likely to collapse with thinner walls.

BIOLOGY, M2 2017 HSC 23

Complete the table with reference to the two types of blood vessel shown.   (5 marks)

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Left: Artery

→ Arteries have thick muscular walls in order to withstand the high pressure at which blood is pumped from the heart.
 

Right: Vein

→ Nutrients are absorbed from the blood at the capillaries, and hence lowering the blood pressure. The valves are used to prevent backflow of blood at these lower pressures.

Show Worked Solution

Left: Artery

→ Arteries have thick muscular walls in order to withstand the high pressure at which blood is pumped from the heart.
 

Right: Vein

→ Nutrients are absorbed from the blood at the capillaries, and hence lowering the blood pressure. The valves are used to prevent backflow of blood at these lower pressures.

BIOLOGY, M1 2008 HSC 18

Using a light microscope, a student looked at a prepared slide of human blood, and drew a scaled diagram.

The diagram shown is a representation of the student's scaled diagram.
 

  1. Assess the accuracy of the diagram.  (3 marks)

  2. Why is it safer to use prepared slides instead of fresh blood?  (1 mark)

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a.    The diagram has the following inaccuracies:

→ The scale is wrong as red and white blood cells are much smaller. A more accurate scale would use micrometres as apposed to millimetres.

→ Relative size of red and white blood cells is wrong, as WBC’s are on average 1.5 – 2 times bigger than RBC’s.
 

b.   A slide with fresh blood could contain pathogenic bacteria, leading to disease.

Show Worked Solution

a.    The diagram has the following inaccuracies:

→ The scale is wrong as red and white blood cells are much smaller. A more accurate scale would use micrometres as apposed to millimetres.

→ Relative size of red and white blood cells is wrong, as WBC’s are on average 1.5 – 2 times bigger than RBC’s.
 

b.   A slide with fresh blood could contain pathogenic bacteria, leading to disease.

BIOLOGY, M1 2012 HSC 24

You conducted first-hand investigations to test the effects of temperature, pH and substrate concentration on enzyme activity.

  1. Complete the following table by identifying the variables for ONE of your investigations.  (1 mark)

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Independent variable} \rule[-1ex]{0pt}{0pt} & \textit{Dependent variable} & \textit{Kept constant} \\
\hline
\rule{0pt}{2.5ex}   \rule[-1ex]{0pt}{0pt} & & \text{pH,} \\ & & \text{substrate concentration,} \\ \text{................................} & \text{................................} & \text{enzyme concentration} \\
\hline
\end{array}

  1. Outline how you measured the activity of an enzyme in your investigation. In your answer, name the enzyme.  (2 marks)

  2. Describe how a condition needed for optimal enzyme activity would be expected to vary between endotherms and ectotherms.   (2 marks)

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a.    Independent variable: Temperature

Dependent variable: Enzyme Activity
 

b.    Activity measure:

→ The activity of catalase is measured by the height of oxygen in a test tube when a fresh piece of raw potato is added to hydrogen peroxide.
 

c.    Optimal enzyme activity:

→ The body temperature of ectotherms changes to be similar to the environment, while endotherms maintain a relatively stable internal temperature.

→ Enzyme activity is dependant on temperature.

→ Therefore, it is expected that an ectotherm’s enzymes can function effectively over a much greater temperature range as apposed to endotherms.

Show Worked Solution

a.    Independent variable: Temperature

Dependent variable: Enzyme Activity
 

b.    Activity measure:

→ The activity of catalase is measured by the height of oxygen in a test tube when a fresh piece of raw potato is added to hydrogen peroxide.
 

c.    Optimal enzyme activity:

→ The body temperature of ectotherms changes to be similar to the environment, while endotherms maintain a relatively stable internal temperature.

→ Enzyme activity is dependant on temperature.

→ Therefore, it is expected that an ectotherm’s enzymes can function effectively over a much greater temperature range as apposed to endotherms.

BIOLOGY, M1 2013 HSC 29

  1. What chemicals are filtered out of the blood by the kidney?  (2 marks)

  2. Explain the steps involved in the formation of urine.   (4 marks)

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a.    Glucose and \(\ce{NaCl}\).

b.    Water (\(\ce{H2O}\)) and glucose.

c.    → Blood is filtered at the glomerulus through Bowman’s capsule.

→ Blood cells and large molecules remain in the blood and small molecules, salts and water enter the proximal tubule.

→ Glucose is actively taken back to the blood through the wall of the proximal tubule.

→ The filtrate then enters the loop of Henle where a countercurrent mechanism uses active transport (ascending limb) and osmosis causes a concentration gradient of ions down the loop of Henle (towards the medulla).

→ The filtrate then enters the distal tubule where its other ions are removed. The urine can be made more acidic here. It then enters the collecting tubule which is not permeable to water unless stimulated by ADH.

→ ADH causes it to be permeable to water as it descends past the loop of Henle in the medulla. This then leads to a concentration of the urine by osmosis.

Show Worked Solution

a.    Glucose and \(\ce{NaCl}\).

b.    Water (\(\ce{H2O}\)) and glucose.

c.    → Blood is filtered at the glomerulus through Bowman’s capsule.

→ Blood cells and large molecules remain in the blood and small molecules, salts and water enter the proximal tubule.

→ Glucose is actively taken back to the blood through the wall of the proximal tubule.

→ The filtrate then enters the loop of Henle where a countercurrent mechanism uses active transport (ascending limb) and osmosis causes a concentration gradient of ions down the loop of Henle (towards the medulla).

→ The filtrate then enters the distal tubule where its other ions are removed. The urine can be made more acidic here. It then enters the collecting tubule which is not permeable to water unless stimulated by ADH.

→ ADH causes it to be permeable to water as it descends past the loop of Henle in the medulla. This then leads to a concentration of the urine by osmosis.

Mean mark (c) 53%.

BIOLOGY, M1 2013 HSC 25

The graph below shows the results obtained from testing the activity of a bacterial enzyme.
 

  1. Name ONE variable, other than temperature, that would have been controlled in the experiment.  (1 mark)

  2. For what temperature range does the enzyme display the maximum rate of change in activity?   (1 mark)

  3. Account for the activity of the enzyme at the parts of the graph labelled A, B, C and D.  (4 marks)

  4. Based on the information in the graph, suggest the type of environment in which these bacteria might survive. (1 mark)

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a.    Answers could include:

→ pH

→ Enzyme concentration

→ Substrate concentration 

b.    89 – 93°C (D).

c.    Enzyme activity at each point:

A: A constant gradient indicates a constant increasing rate of enzyme activity as temperature increases.

B: Graph level at a maximum (10 μmol/s). Enzyme-substrate complex operating at maximum even though temperature is increasing.

C: Graph decreasing indicating a reduction in enzyme efficiency due to increasing temperature denaturing enzyme.

D: Graph rapidly goes to 0 indicating enzyme activity stops because the enzyme is denatured.

d.    Hotter environments, such as desserts or hot, geothermal springs.

Show Worked Solution

a.    Answers could include:

→ pH

→ Enzyme concentration

→ Substrate concentration 

b.    89 – 93°C (D).

c.    Enzyme activity at each point:

A: A constant gradient indicates a constant increasing rate of enzyme activity as temperature increases.

B: Graph level at a maximum (10 μmol/s). Enzyme-substrate complex operating at maximum even though temperature is increasing.

C: Graph decreasing indicating a reduction in enzyme efficiency due to increasing temperature denaturing enzyme.

D: Graph rapidly goes to 0 indicating enzyme activity stops because the enzyme is denatured.

♦♦♦ Mean mark (b) 22%.

d.    Hotter environments, such as desserts or hot, geothermal springs.

BIOLOGY, M1 2016 HSC 23

  1. Explain ONE reason why the concentration of water in cells should be maintained within a narrow range for optimal cell function.  (2 marks)

  2. A person has consumed large amounts of water. Complete the table to show the effect on each of the variables listed.  (3 marks)

 

Show Answers Only

a.    Correct answers could include:

→ Concentration of water in cells is maintained to regulate concentrations of solutes in cells.

→  Concentration of water in a narrow range provides appropriate substrate concentrations for metabolic function.

b.

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex}\ \ \ \ \ \ \ \ \textit{Variable}\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex}\textit{Effect}\rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex}\text{Urine Volume}\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex}\text{Increase}\rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex}\text{ADH Secretion}\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex}\text{Decrease}\rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex}\text{Salt concentration in blood}\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex}\text{Decrease}\rule[-1ex]{0pt}{0pt}\\
\hline
\end{array}

Show Worked Solution

a.    Correct answers could include:

→ Concentration of water in cells is maintained to regulate concentrations of solutes in cells.

→  Concentration of water in a narrow range provides appropriate substrate concentrations for metabolic function.

♦ Mean mark 43%.

b.

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex}\ \ \ \ \ \ \ \ \textit{Variable}\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex}\textit{Effect}\rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex}\text{Urine Volume}\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex}\text{Increase}\rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex}\text{ADH Secretion}\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex}\text{Decrease}\rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex}\text{Salt concentration in blood}\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex}\text{Decrease}\rule[-1ex]{0pt}{0pt}\\
\hline
\end{array}

BIOLOGY, M1 2017 HSC 26

A controlled experiment was performed to investigate the effect of substrate concentration on the rate of an enzyme-catalysed reaction. Data were collected and are presented in the graph.
 

  1. What is the independent variable in this experiment?  (1 mark)

  2. Explain the trends shown in the graph.  (3 marks)

Show Answers Only

a.    Substrate concentration.

b.    Graph trends:

→ As the substrate concentration increases, the rate of reaction will increase because more substrate is available to undergo the reaction.

→ At higher substrate concentrations the rate of reaction will plateau as all enzyme active sites are utilised, so no more substrate can undergo the reaction.

Show Worked Solution

a.    Substrate concentration.

b.    Graph trends:

→ As the substrate concentration increases, the rate of reaction will increase because more substrate is available to undergo the reaction.

→ At higher substrate concentrations the rate of reaction will plateau as all enzyme active sites are utilised, so no more substrate can undergo the reaction.

BIOLOGY, M1 2017 HSC 25

Explain the difference in the urine concentration of marine fish and freshwater fish.  (4 marks)

Show Answers Only

→ Marine fish live in an environment which has a higher salt concentration than the concentration inside their cells, causing them to lose too much water through osmosis.

→ To maintain their internal environment they release very concentrated but small amounts of urine, which allows them to conserve water.

→ Freshwater fish live in an environment which has a lower salt concentration than the concentration inside their cells, causing too much water to move into their cells through osmosis.

→ To maintain their internal environment they produce large quantities of dilute urine, which allows them to expel the ‘extra’ water.

Show Worked Solution

→ Marine fish live in an environment which has a higher salt concentration than the concentration inside their cells, causing them to lose too much water through osmosis.

→ To maintain their internal environment they release very concentrated but small amounts of urine, which allows them to conserve water.

→ Freshwater fish live in an environment which has a lower salt concentration than the concentration inside their cells, causing too much water to move into their cells through osmosis.

→ To maintain their internal environment they produce large quantities of dilute urine, which allows them to expel the ‘extra’ water.

BIOLOGY, M7 SM-Bank 23

Smallpox is widely believed to be the cause of the significant population decline in Indigenous populations during early European arrival in Australia. Recent research suggests that it may have been chickenpox that caused the change in population numbers rather than smallpox. It is thought that chickenpox did not exist in Aboriginal and Torres Strait Islander communities before European arrival. Chickenpox is fairly mild in young children and is easily transmitted. It can be a fatal disease in adults who were not infected as children.

The graph below shows population numbers during the period 1780 –1850:

    • the estimated Aboriginal Australian population
    • the colonist population
    • the estimated total Australian population
       

With reference to chickenpox, explain how infection may have caused such a large impact on the Aboriginal Australian population and yet did not affect the population numbers of the colonists significantly, and why the Aboriginal population increased from 1790 to 1810.  (3 marks)

Show Answers Only

→ Childhood immunity existed in colonists due to previous exposure prior to arrival in Australia.

→ No immunity existed for the Aboriginal Australian population due to no previous exposure.

→ Aboriginal children became mildly ill when exposed in the early days of colonialism and became immune.

→ Aboriginal adults caught chickenpox in the early days of colonialism and died in disproportionately high numbers.

→ The initial exposure caused a steep decline in overall Aboriginal population. However, once Aboriginal children were immune and susceptible adults had died, there was an increase of population numbers over the next 30 years, from 1790 – 1810.

Show Worked Solution

→ Childhood immunity existed in colonists due to previous exposure prior to arrival in Australia.

→ No immunity existed for the Aboriginal Australian population due to no previous exposure.

→ Aboriginal children became mildly ill when exposed in the early days of colonialism and became immune.

→ Aboriginal adults caught chickenpox in the early days of colonialism and died in disproportionately high numbers.

→ The initial exposure caused a steep decline in overall Aboriginal population. However, once Aboriginal children were immune and susceptible adults had died, there was an increase of population numbers over the next 30 years, from 1790 – 1810.

BIOLOGY, M7 SM-Bank 22

A summary of case-control studies conducted from 1997 to 2003 showed that a single dose of varicella (chickenpox) vaccine was 97% effective in the first year after vaccination and 86% effective in the second year. From the second to eighth year after vaccination, the vaccine's effectiveness remained stable at 81% to 86%. Most vaccinated children who developed chickenpox during the eight years after vaccination had a mild case of the disease.

Why do most vaccinated individuals, if they are infected, show only a mild case of the disease?  (2 marks)

Show Answers Only

Answers should include two of the following reasons:

→ Virus enters and is detected.

→ Memory cells present.

→ Rapid antibody production where antibodies are produced in large numbers

→ Virus is neutralised

Show Worked Solution

Answers should include two of the following reasons:

→ Virus enters and is detected.

→ Memory cells present.

→ Rapid antibody production where antibodies are produced in large numbers

→ Virus is neutralised

BIOLOGY, M7 SM-Bank 27

Influenza is an infectious respiratory disease. In humans, it can be caused by the influenza A or influenza B viruses.

Antigenic drift can result in small changes to the structure of the antigens on the surface of the influenza virus, as shown in the diagram below.
 

  1. What change would have occurred within the virus to bring about the change in the structure of the antigen?  (1 mark)

  2. Vaccines against influenza are available and it is recommended that people are vaccinated each year.
  3. Explain, in terms of antigenic drift, why vaccinations are recommended yearly for influenza rather than once every few years.  (2 marks)

Show Answers Only

a.   Mutation.

b.   Correct answers could include:

→ Vaccines contain new antigens resulting in the production of new antibodies OR memory cells specific to the antigen.

→ Memory cells allow for a faster OR larger immune response.

→ Any existing memory cells specific to the previous antigens would no longer be effective.

Show Worked Solution

a.   Mutation.

b.   Correct answers could include:

→ Vaccines contain new antigens resulting in the production of new antibodies OR memory cells specific to the antigen.

→ Memory cells allow for a faster OR larger immune response.

→ Any existing memory cells specific to the previous antigens would no longer be effective.

BIOLOGY, M7 SM-Bank 7 MC

The human body does not normally produce an immune response against self cells.

Cells in the human immune system can distinguish self cells from non-self cells

  1. because non-self cells have a faster reproduction rate.
  2. due to the presence of different antigens on non-self cells.
  3. only when antibodies are attached to the surface of non-self cells.
  4. because non-self cells are always smaller in size than self cells.
Show Answers Only

`B`

Show Worked Solution

→ The human body is able to distinguish self cells from non-self cells via the presence of different surface proteins.

→ Self cells have an MHC complex while non-self cells have antigens.

`=>B`

BIOLOGY, M5 SM-Bank 25

In fruit flies, eye colour is a sex-linked trait inherited on the X chromosome. The red-eye allele R is dominant over the white-eye allele (r). A red-eyed male and white-eyed female have 50 offspring.

Use a Punnett square to predict the number of male and female offspring and their eye colour.   (3 marks)

Show Answers Only

\begin{array} {|c|c|c|}
\hline  & \text{X}^{\text{R}} & \text{Y} \\
\hline \text{X}^{\text{r}} & \text{X}^{\text{R}}\text{X}^{\text{r}} & \text{X}^{\text{R}}\text{Y} \\
\hline \text{X}^{\text{r}} & \text{X}^{\text{R}}\text{X}^{\text{r}} & \text{X}^{\text{R}}\text{Y} \\  
\hline \end{array}

→ Predicted offspring: 25 red-eyed females and 25 white-eyed males.

Show Worked Solution

\begin{array} {|c|c|c|}
\hline  & \text{X}^{\text{R}} & \text{Y} \\
\hline \text{X}^{\text{r}} & \text{X}^{\text{R}}\text{X}^{\text{r}} & \text{X}^{\text{R}}\text{Y} \\
\hline \text{X}^{\text{r}} & \text{X}^{\text{R}}\text{X}^{\text{r}} & \text{X}^{\text{R}}\text{Y} \\  
\hline \end{array}

→ Predicted offspring: 25 red-eyed females and 25 white-eyed males.

BIOLOGY, M5 SM-Bank 24

Explain how transcription factors control cell differentiation, using an example.  (3 marks)

Show Answers Only

→ Transcription factors are proteins that turn specific genes on or off by binding to nearby sections of DNA.

→ This allows different genes to be expressed in different cell types.

→ For example, the SRY gene on the Y chromosome promotes the development of testes by producing a protein that regulates the expression of genes (production of proteins) required by Sertoli cells.

Show Worked Solution

→ Transcription factors are proteins that turn specific genes on or off by binding to nearby sections of DNA.

→ This allows different genes to be expressed in different cell types.

→ For example, the SRY gene on the Y chromosome promotes the development of testes by producing a protein that regulates the expression of genes (production of proteins) required by Sertoli cells.

BIOLOGY, M5 SM-Bank 22

Describe the roles of messenger RNA and transfer RNA in protein synthesis.  (2 marks)

Show Answers Only

→ Messenger RNA (or mRNA) provides copies of the gene from the DNA template strand.

→ Transfer RNA (or tRNA) decodes mRNA sequences by matching amino acids to codons of the mRNA.

Show Worked Solution

→ Messenger RNA (or mRNA) provides copies of the gene from the DNA template strand.

→ Transfer RNA (or tRNA) decodes mRNA sequences by matching amino acids to codons of the mRNA.

BIOLOGY, M6 SM-Bank 24

In 1950 , the myxoma virus was released into Australian pest rabbit populations to reduce their numbers. The resulting disease, myxomatosis, initially wiped out 95% of the rabbit population; however, it quickly became less effective as a population control measure.

This graph shows the frequency of myxomatosis resistance in Australia's rabbit population from 1949 to 1956.
 

Use evidence from the graph and the principles of natural selection to explain how myxomatosis became ineffective as a population control measure.  (3 marks)

Show Answers Only

→ The introduction of the myxoma virus conferred a selective advantage on the myxomatosis-resistant phenotype.

→ Myxomatosis-resistant rabbits had an increased chance of surviving to reproduce and pass on the trait.

→ This led to an increase in the frequency of this phenotype in the population from approximately 9% in 1950 to approximately 96% in 1953 .

→ After 1953, most of the remaining rabbits possessed the myxomatosis-resistant phenotype, so the disease was no longer effective at reducing rabbit numbers.

Show Worked Solution

→ The introduction of the myxoma virus conferred a selective advantage on the myxomatosis-resistant phenotype.

→ Myxomatosis-resistant rabbits had an increased chance of surviving to reproduce and pass on the trait.

→ This led to an increase in the frequency of this phenotype in the population from approximately 9% in 1950 to approximately 96% in 1953 .

→ After 1953, most of the remaining rabbits possessed the myxomatosis-resistant phenotype, so the disease was no longer effective at reducing rabbit numbers.

BIOLOGY, M5 SM-Bank 23

Explain the purpose of gel electrophoresis in DNA profiling.  (2 marks)

Show Answers Only

→ Gel electrophoresis separates DNA segments based on size.

→ This process produces characteristic banding patterns.

Show Worked Solution

→ Gel electrophoresis separates DNA segments based on size.

→ This process produces characteristic banding patterns.

PHYSICS, M8 EQ-Bank 21

With reference to the earliest conditions present following the Big Bang, describe the processes that led to the transformation of radiation into matter.  (4 marks)

Show Answers Only

→ According to the Big Bang theory, the universe began as a hot, dense, and homogeneous state of pure energy.

→ As the universe expanded and cooled, this energy was converted into matter and antimatter particles, such as quarks, leptons, and their antiparticles.

→ These particles and antiparticles annihilated each other in pairs, releasing more energy in the form of radiation.

→ However, there was a slight asymmetry between matter and antimatter, which resulted in a small excess of matter over antimatter. This excess matter is what makes up the observable universe today.

→ The transformation of radiation into matter occurred in several stages (or epochs), depending on the temperature and energy of the universe.

Show Worked Solution

→ According to the Big Bang theory, the universe began as a hot, dense, and homogeneous state of pure energy.

→ As the universe expanded and cooled, this energy was converted into matter and antimatter particles, such as quarks, leptons, and their antiparticles.

→ These particles and antiparticles annihilated each other in pairs, releasing more energy in the form of radiation.

→ However, there was a slight asymmetry between matter and antimatter, which resulted in a small excess of matter over antimatter. This excess matter is what makes up the observable universe today.

→ The transformation of radiation into matter occurred in several stages (or epochs), depending on the temperature and energy of the universe.

PHYSICS, M6 EQ-Bank 29

Prior to 2019, the ampere was characterized as the unvarying current that, when sustained in two infinitely long, straight parallel conductors with negligible circular cross-section and spaced one meter apart in a vacuum, would result in a force of  `10 xx 10^{-7}` newton per metre of length between the conductors.

  1. How does Newton’s Third Law apply to this definition of the ampere?  (2 marks)
  1. Two parallel current carrying wires are 1.2 metres long each, both carrying 0.55 amperes at a distance of 25 mm apart. Calculate the force between the wires.  (2 marks)
Show Answers Only

a.   
       

→ Since the two wires in the definition carry current in the same direction, wire 1 will impart an attraction force on wire 2.

→ Similarly, wire 2 will impart an attractive force on wire 1. This force will be equal but in an opposite direction to the attractive force imparted on wire 2 (above).

→ Newton’s Third Law states that two objects interact with each other, they exert forces on each other that are equal in magnitude and opposite in direction, also known as action and reaction forces.

→ In this way, Newton’s Third Law directly describes the forces between conductors within the definition of the ampere.
 

b.  `2.9 xx 10^{-5}\ text{N}`

Show Worked Solution

a.   
         

→ Since the two wires in the definition carry current in the same direction, wire 1 will impart an attraction force on wire 2.

→ Similarly, wire 2 will impart an attractive force on wire 1. This force will be equal but in an opposite direction to the attractive force imparted on wire 2 (above).

→ Newton’s Third Law states that two objects interact with each other, they exert forces on each other that are equal in magnitude and opposite in direction, also known as action and reaction forces.

→ In this way, Newton’s Third Law directly describes the forces between conductors within the definition of the ampere.
 

b.    `F/l` `=mu_0/(2pi) xx (I_1I_2)/r`
  `F` `=(4pi xx 10^{-7})/(2pi) xx (0.55)^2/(2.5 xx 10^{-3}) xx 1.2`
    `=(2 xx 10^{-7})xx (0.55)^2/(2.5 xx 10^{-3}) xx 1.2`
    `=2.9 xx 10^{-5}\ text{N}`

CHEMISTRY, M5 EQ-Bank 29

The information in the table shows how the solubility of lead chloride is affected by temperature.  
 

Using a graph, calculate the solubility product \((K_{sp})\) of the dissolution of lead chloride at 50°C. Include a fully labelled graph and a relevant chemical equation in your answer   (6 marks)
 

Show Answers Only

\(\ce{K_{sp} = 6.4 \times 10^{-5}}\)

Show Worked Solution

 

\(\ce{PbCl2(s) \rightleftharpoons Pb^2+(aq) + 2Cl^-(aq)}\)

\(\ce{Using the graph:}\)

\(\ce{Solubility (50°) = 0.7 g/100 g water = 7 g/L}\)

\(\ce{Converting to mol L^{-1}:}\)

\[\ce{MM(PbCl2) = 207.2 + 2 \times 35.45 = 278.1}\]

\[\ce{n = \frac{m}{MM} = \frac{7}{278.1} = 0.0252 mol L^{-1}}\]

\(\ce{[Pb^2+(aq)] = 0.0252 mol L^{-1}}\)
 

\(\ce{Mole ratio \ Pb^2+ : Cl^- = 1:2}\)

\(\Rightarrow \ce{[Cl^-]  = 2 \times 0.0252 = 0.0504 mol L^{-1}}\)
 

\begin{aligned}
\ce{$K_{sp}$} & \ce{= [Pb^2+][Cl^-]^{2}} \\
 & \ce{=0.0252 \times (0.0504)^{2}}  \\
 & \ce{= 6.4 \times 10^{-5}}  \\
\end{aligned}

CHEMISTRY, M5 EQ-Bank 28

A 100 mL saturated solution of calcium hydroxide at 25°C contains 0.173 g of calcium hydroxide.

  1. Calculate the solubility product \(\ce{($K_{sp}$)}\) of this salt at 25°C.  (3 marks)

  2. Explain why the undissolved solid is not included in the expression for the solubility product constant.  (1 marks)

Show Answers Only

a.    \(\ce{$K_{sp}$ = 5.06 \times 10^{-5}}\)

b.    → \(\ce{[CaOH2(s)]}\) is constant throughout the reaction.

→ Since it does not change, it is not included in the equilibrium expression.

Show Worked Solution

a.   \(\ce{CaOH2(s) \rightleftharpoons Ca^2+(aq) + 2OH^-(aq)}\)

\(\ce{Solubility = 0.173 g/100 mL = 1.73 g/L}\)

\[\ce{n = \frac{m}{MM} = \frac{1.73}{74.093} = 0.00233 mol}\]

\(\ce{[Ca^{2+}] = 0.0233 mol L^{-1}}\)
 

\(\ce{Mole ratio  \ Ca^2+ : OH^- = 1:2}\)

\(\ce{$K_{sp}$   =[Ca^2+][OH^-]^{2}}\)

 

\(\ce{[OH^-]  = 2 \times 0.0233 = 0.0466 mol L^{-1}}\)
 

\begin{aligned}
\ce{$K_{sp}$} & \ce{=[Ca^2+][OH^-]^{2}}  \\
 & \ce{= 0.0233 \times (0.0466)^{2}}  \\
 & \ce{= 5.06 \times 10^{-5}}  \\
\end{aligned}

 

b.    → \(\ce{[CaOH2(s)]}\) is constant throughout the reaction.

→ Since it does not change, it is not included in the equilibrium expression.

CHEMISTRY, M5 EQ-Bank 26

The diagrams represent equipment used in an investigation to determine the chloride ion concentration in a water sample.
 

Describe how the chloride ion concentration in a water sample can be determined using the equipment in the diagrams. Include a relevant chemical equation.   (3 marks)

Show Answers Only

→ When the silver nitrate solution is added, chloride ions present in the water sample will precipitate with the added silver ions described by the following equation:

 \(\ce{Ag+(aq) +Cl^-(aq) \rightleftharpoons AgCl(s)}\)

→ Adding excess silver nitrate ensures all chloride ions precipitate out.

→ The filtering apparatus is used to filter solid silver chloride. This solid is then dried to constant mass and weighed.

→ Using the molar mass of silver chloride, the number of moles of solid silver chloride produced is calculated. Silver chloride contains silver ions and chloride ions in a 1:1 molar ratio and using this ratio, the moles of chloride present is calculated.

→ The result represents the same number of moles of chloride in the original water sample. Using the volume of the water sample, its chloride ion concentration is calculated.

Show Worked Solution

→ When the silver nitrate solution is added, chloride ions present in the water sample will precipitate with the added silver ions described by the following equation:

 \(\ce{Ag+(aq) +Cl^-(aq) \rightleftharpoons AgCl(s)}\)

→ Adding excess silver nitrate ensures all chloride ions precipitate out.

→ The filtering apparatus is used to filter solid silver chloride. This solid is then dried to constant mass and weighed.

→ Using the molar mass of silver chloride, the number of moles of solid silver chloride produced is calculated. Silver chloride contains silver ions and chloride ions in a 1:1 molar ratio and using this ratio, the moles of chloride present is calculated.

→ The result represents the same number of moles of chloride in the original water sample. Using the volume of the water sample, its chloride ion concentration is calculated.

CHEMISTRY, M5 EQ-Bank 24

When a sample of solid silver chloride is added to a `1.00 xx10^(-2)` mol L−1 sodium chloride solution, only some of the silver chloride dissolves.

Calculate the equilibrium concentration of silver ions in the resulting solution, given that the `K_(sp)` of silver chloride is `1.8 xx10^(-10)`.   (3 marks)

Show Answers Only

\(\ce{[Ag^+] = 1.80 \times 10^{-8} mol L^{-1}}\)

Show Worked Solution

 \(\ce{AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)}\)

\begin{array} {|l|c|c|c|}
\hline  & \ce{[AgCl(s)]} & \ce{[Ag^+(aq)]} & \ce{[Cl^-(aq)]} \\
\hline \text{Initial} &  & 0 & 1.00 \times 10^{-2} \\
\hline \text{Change} &  & +x & +x \\
\hline \text{Equilibrium} &  & x & 1.00 \times 10^{-2} +x \\
\hline \end{array}

\(\ce{Let \ $x$ = [Ag^+]}\)

\[\ce{$K_{sp}$ = [Ag^+][Cl^-]}\]

\(\ce{$K_{sp} = x$(1.00 \times 10^{-2} + $x$) = 1.80 \times 10^{-10}}\)

\(\ce{Since $x$\ is small, 1.00 \times 10^{-2} + $x$ ≈ 1.00 \times 10^{-2}}\)

\begin{aligned}
\ce{$x$(1.00 \times 10^{-2})} & \ce{= 1.80 \times 10^{-10}}  \\
\ce{$x$} & \ce{= 1.80 \times 10^{-8}}  \\
\ce{\therefore [Ag^+]} & \ce{= 1.80 \times 10^{-8} mol L^{-1}}  \\
\end{aligned}

CHEMISTRY, M5 EQ-Bank 23

Butanoic acid is a natural product and a component of human sweat.

Calculate the value of \(\ce{$K_{a}$}\) for butanoic acid if a 0.10 mol L–1 solution has a pH of 2.9 at 298 K.   (3 marks)

Show Answers Only

\(\ce{$K_{a}$ = 1.6 \times 10^{-5}}\)

Show Worked Solution

\(\ce{C3H7COOH(aq) \rightleftharpoons C3H7COO^-(aq) + H^+(aq)}\)  \[\ce{$K_{a}$ = \frac{[C3H7COO^-][H^+]}{[C3H7COOH]}}\]

\(\ce{[H^+]}\) \(\ce{= 10^{-pH}}\)  
  \(\ce{= 10^{-2.9}}\)  
  \(\ce{= 1.26 \times 10^{-3} mol L^{-1}}\)  

\begin{array} {|l|c|c|c|}
\hline  & \ce{[C3H7COOH]} & \ce{[C3H7COO^-]} & \ce{[H^+]} \\
\hline \text{Initial} & 0.10 & 0 & 0 \\
\hline \text{Change} & -1.26 \times 10^{-3} & +1.26 \times 10^{-3} & +1.26 \times 10^{-3} \\
\hline \text{Equilibrium} & 0.0987 & 1.26 \times 10^{-3} & 1.26 \times 10^{-3} \\
\hline \end{array}

`K_(a)` `=(1.26 xx10^-3)^2/(0.0987)`  
  `=1.6 xx10^(-5)`  

CHEMISTRY, M6 EQ-Bank 13 MC

The pKa of trichloroacetic acid is 0.70 and the pKa of acetic acid is 4.8.

Which of the following identifies the acid with the higher pH and explain why?

  1. Acetic acid as it is less likely to lose a hydrogen ion
  2. Acetic acid as it is more likely to lose a hydrogen ion
  3. Trichloroacetic acid as it is less likely to lose a hydrogen ion
  4. Trichloroacetic acid as it is more likely to lose a hydrogen ion
Show Answers Only

`A`

Show Worked Solution

→ Higher pH corresponds to a weaker acid.

→ Higher pKa corresponds to a weaker acid (i.e. a higher pH).

→ Acetic acid the weaker acid (higher pKa), meaning it is less likely to dissociate in solution and lose a hydrogen ion.

`=> A`

CHEMISTRY, M6 EQ-Bank 28

The flowchart shown outlines the sequence of steps used to determine the concentration of an unknown hydrochloric acid solution.
 

Describe steps A, B and C including correct techniques, equipment and appropriate calculations. Determine the concentration of the hydrochloric acid.  (8 marks)

Show Answers Only

Step A

→ Prepare \(\ce{Na2CO3}\) by drying. Protect solid \(\ce{Na2CO3}\) from moisture in the air by storing in a desiccator.

→ Calculate mass of dried \(\ce{Na2CO3}\) required and weigh accurately.

\(\ce{m(Na2CO3) = 0.1 \times 0.5 \times 105.99 = 5.30 g}\)

→ Clean and rinse a 500 mL volumetric flask with distilled water.

→ Add 5.30 grams of \(\ce{Na2CO3}\) to the volumetric flask using a funnel and wash funnel using distilled water. Add distilled water to the flask to the bottom of the meniscus.

Step B

→ Clean and rinse a 50 mL burette. Fill burette with the unknown acid and place on a retort stand.

→ Clean and rinse a 250 mL conical flask with distilled water.

→ Clean a 25 mL pipette and rinse with 0.1 M \(\ce{Na2CO3}\) solution. Fill pipette with \(\ce{Na2CO3}\) solution to bottom of meniscus.

→ Transfer all pipette solution into conical flask and add an appropriate indicator. A white background (tile) should be placed under the flask to highlight any colour changes in the solution.

→ Slowly add acid solution from the burette into the conical flask and record the volume used when the indicator changes colour.

Step C

→ The initial titration represents a test run to establish an indicative volume. Three subsequent titrations should be performed with the average titration forming the basis of \(\ce{HCl}\) concentration calculations.

→ Calculate the concentration of \(\ce{HCl}\)

\(\ce{2H^+(aq) + CO3^{2-} (aq) \rightarrow H2CO3 (aq) \rightarrow H2O (l) + CO2 (g)}\)

\(\ce{M(Na2CO3) = 0.1 \times 0.025 = 2.5 \times 10^{-3}}\)

\(\ce{M(HCl) = 2 \times M(Na2CO3) = 5\times 10^{-3}}\)

\[\ce{[HCl] = \frac{M(HCl)}{Vol HCl} = \frac{5 \times 10^{-3}}{21.4 \times 10^{-3}} = 0.234 mol L^{-1}}\]

Show Worked Solution

Step A

→ Prepare \(\ce{Na2CO3}\) by drying. Protect solid \(\ce{Na2CO3}\) from moisture in the air by storing in a desiccator.

→ Calculate mass of dried \(\ce{Na2CO3}\) required and weigh accurately.

\(\ce{m(Na2CO3) = 0.1 \times 0.5 \times 105.99 = 5.30 g}\)

→ Clean and rinse a 500 mL volumetric flask with distilled water.

→ Add 5.30 grams of \(\ce{Na2CO3}\) to the volumetric flask using a funnel and wash funnel using distilled water. Add distilled water to the flask to the bottom of the meniscus.

Step B

→ Clean and rinse a 50 mL burette. Fill burette with the unknown acid and place on a retort stand.

→ Clean and rinse a 250 mL conical flask with distilled water.

→ Clean a 25 mL pipette and rinse with 0.1 M \(\ce{Na2CO3}\) solution. Fill pipette with \(\ce{Na2CO3}\) solution to bottom of meniscus.

→ Transfer all pipette solution into conical flask and add an appropriate indicator. A white background (tile) should be placed under the flask to highlight any colour changes in the solution.

→ Slowly add acid solution from the burette into the conical flask and record the volume used when the indicator changes colour.

Step C

→ The initial titration represents a test run to establish an indicative volume. Three subsequent titrations should be performed with the average titration forming the basis of \(\ce{HCl}\) concentration calculations.

→ Calculate the concentration of \(\ce{HCl}\)

\(\ce{2H^+(aq) + CO3^{2-} (aq) \rightarrow H2CO3 (aq) \rightarrow H2O (l) + CO2 (g)}\)

\(\ce{M(Na2CO3) = 0.1 \times 0.025 = 2.5 \times 10^{-3}}\)

\(\ce{M(HCl) = 2 \times M(Na2CO3) = 5\times 10^{-3}}\)

\[\ce{[HCl] = \frac{M(HCl)}{Vol HCl} = \frac{5 \times 10^{-3}}{21.4 \times 10^{-3}} = 0.234 mol L^{-1}}\]

CHEMISTRY, M7 EQ-Bank 21

With the use of a labelled diagram, describe how grease can be removed from fabric using soap.  (3 marks)

Show Answers Only

→ Soap molecules, as shown in the diagram, have a negatively charged (hydrophilic) “head” and a hydrophobic ‘tail’ consisting made up of fatty acids.

→ Hydrophobic tails are attracted to the grease particles and attach themselves. 

→ The hydrophilic heads, at the same time, are attracted to water molecules. This chain of attractions will lift grease off the fabric and suspend it in the water.

Show Worked Solution

→ Soap molecules, as shown in the diagram, have a negatively charged (hydrophilic) “head” and a hydrophobic ‘tail’ consisting made up of fatty acids.

→ Hydrophobic tails are attracted to the grease particles and attach themselves. 

→ The hydrophilic heads, at the same time, are attracted to water molecules. This chain of attractions will lift grease off the fabric and suspend it in the water.

CHEMISTRY, M6 EQ-Bank 27

What determines the pH of a buffer solution?   (2 marks)

Show Answers Only

→ The strength of the acid. The greater the pKa of (weaker) the acid, the greater the buffer pH.

→ Relative concentrations of acid and base. 

Show Worked Solution

→ The strength of the acid. The greater the pKa of (weaker) the acid, the greater the buffer pH.

→ Relative concentrations of acid and base. 

CHEMISTRY, M6 EQ-Bank 23

Propanoic acid dissociation in water can is represented in the following equation:

\(\ce{CH3CH2COOH($aq$) + H2O($l$) \rightleftharpoons CH3CH2COO^-($aq$) + H3O^{+}($aq$)}\)

Explain how the pH of the propanoic acid solution would change if it was diluted.   (3 marks)

Show Answers Only

→ Propanoic acid only partially ionises in solution and is defined as a weak acid.

→ Any dilution of the acid will result in a decrease of the concentration of all species (including the hydronium ion).

→ According to Le Chatelier, the decreasing concentrations of dissolved species will cause the equilibrium to shift to the right.

→ While this effect causes an increase in ionisation, it is not sufficient to counter the decrease in hydronium ion concentration caused by the original dilution.

→ Since the net effect causes the hydronium ion concentration to decrease, the solution will become less acidic and the pH will increase.

Show Worked Solution

→ Propanoic acid only partially ionises in solution and is defined as a weak acid.

→ Any dilution of the acid will result in a decrease of the concentration of all species (including the hydronium ion).

→ According to Le Chatelier, the decreasing concentrations of dissolved species will cause the equilibrium to shift to the right.

→ While this effect causes an increase in ionisation, it is not sufficient to counter the decrease in hydronium ion concentration caused by the original dilution.

→ Since the net effect causes the hydronium ion concentration to decrease, the solution will become less acidic and the pH will increase.

CHEMISTRY, M6 EQ-Bank 8 MC

Which of the following is NOT a Bronsted-Lowry reaction?

  1. \(\ce{NH4^+ + NH2^- \rightleftharpoons 2NH3}\)
  2. \(\ce{CO2 + OH^- \rightleftharpoons HCO3^-}\)
  3. \(\ce{HClO4 + CH3COOH \rightleftharpoons CH3COOH2^+ + ClO4^-}\)
  4. \(\ce{CH3CH2O^- + CH3NH3^+ \rightleftharpoons CH3CH2OH + CH3NH2}\)
Show Answers Only

`B`

Show Worked Solution

→ A Bronsted-Lowry reaction occurs when one species (acid) transfers a proton to another species (base).

→ Although it is an acid-base reaction, no proton transfer occurs in  \(\ce{CO2 + OH^- \rightleftharpoons HCO3^-}\)

`=> B`

CHEMISTRY, M7 EQ-Bank 27

Contrast ONE addition polymer and ONE condensation polymer in terms of their structures, properties and uses. Include structural formulae in your answers.   (7 marks)

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Addition polymer – polyethylene. Condensation polymer – nylon.

Structures

→ Polyethylene is made by the addition of ethylene monomers with the following structural formula:

→ Nylon is a condensation polymer made from 1,6-diaminohexane and adipic acid, producing a by-product of water molecules. It has the following structural formula.

Properties

→ Polyethylene is inexpensive, weatherproof and relatively resistant to chemicals.

→ There are two main types of polyethylene which have different properties. Low-density polyethylene (LDPE) is semi-rigid while high density polyethylene (HDPE) is fluid.

→ Nylon is strong and relatively resistant to moisture absorptivity. It is longer lasting than polyethylene, resistant to chemicals and is used to make nylon fibre.

Applications

→ LDPE is used to produce products such as plastic soft drink bottles, flexible water pipes and cling wrap. HDPE’s fluidity make it an appropriate material for producing shopping bags, plastic crates and drums for storage.

→ Nylon is very versatile in its manufacturing uses. It is a common input for clothing, can be used for injection-moulded parts for vehicles and also as reinforcement for rubber tyres. 

→ Nylon is also used as the main material input for nylon thread which has a myriad of uses, including stitching for clothes and the production of rope.

Show Worked Solution

Addition polymer – polyethylene. Condensation polymer – nylon.

Structures

→ Polyethylene is made by the addition of ethylene monomers with the following structural formula:

→ Nylon is a condensation polymer made from 1,6-diaminohexane and adipic acid, producing a by-product of water molecules. It has the following structural formula.

Properties

→ Polyethylene is inexpensive, weatherproof and relatively resistant to chemicals.

→ There are two main types of polyethylene which have different properties. Low-density polyethylene (LDPE) is semi-rigid while high density polyethylene (HDPE) is fluid.

→ Nylon is strong and relatively resistant to moisture absorptivity. It is longer lasting than polyethylene, resistant to chemicals and is used to make nylon fibre.

Applications

→ LDPE is used to produce products such as plastic soft drink bottles, flexible water pipes and cling wrap. HDPE’s fluidity make it an appropriate material for producing shopping bags, plastic crates and drums for storage.

→ Nylon is very versatile in its manufacturing uses. It is a common input for clothing, can be used for injection-moulded parts for vehicles and also as reinforcement for rubber tyres. 

→ Nylon is also used as the main material input for nylon thread which has a myriad of uses, including stitching for clothes and the production of rope.

CHEMISTRY, M8 EQ-Bank 22

A bottle of solution is missing its label. It is either \(\ce{Pb(NO_3)_2, Ba(NO_3)_2 or Fe(NO_3)_2}\).

Using only \(\ce{HCl, NaOH}\) and \(\ce{H_2SO_4}\) solutions, outline a sequence of steps that could be followed to confirm the identity of the solution in the bottle. Include observed results and ionic equations in your answer.   (4 marks)

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Step 1: Prepare samples of the original solution and \(\ce{HCl, NaOH}\)  and  \(\ce{H_2SO_4}\)  solutions and then follow the flow chart below.

Show Worked Solution

Step 1: Prepare samples of the original solution and \(\ce{HCl, NaOH}\)  and  \(\ce{H_2SO_4}\) solutions and then follow the flow chart below. 

CHEMISTRY, M5 EQ-Bank 10 MC

At a certain temperature, the \(\ce{$K_{eq}$}\) for the following reaction is 75.

\( \ce{2O3(g) \rightleftharpoons 3O2(g)}\)

0.4 mol of \(\ce{O3}\) and 1.2 mol of \(\ce{O2}\) were introduced to a 5 L reaction vessel.

Which row of the table correctly identifies the direction of the equilibrium shift and the reason for the shift?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Direction favoured}\rule[-1ex]{0pt}{0pt}& \quad \quad \textit{Reason} \quad \quad \\
\hline
\rule{0pt}{2.5ex}\text{Left}\rule[-1ex]{0pt}{0pt}&Q>K_{e q}\\
\hline
\rule{0pt}{2.5ex}\text{Left}\rule[-1ex]{0pt}{0pt}& Q<K_{e q}\\
\hline
\rule{0pt}{2.5ex}\text{Right}\rule[-1ex]{0pt}{0pt}&Q>K_{e q} \\
\hline
\rule{0pt}{2.5ex}\text{Right}\rule[-1ex]{0pt}{0pt}& Q<K_{e q}\\
\hline
\end{array}
\end{align*}

Show Answers Only

\(D\)

Show Worked Solution

\[\ce{O2: c = \frac{n}{V} = \frac{1.2}{5} = 0.24 mol L^{-1}} \]

\[\ce{O3: c = \frac{n}{V} = \frac{0.4}{5} = 0.08 mol L^{-1}} \]

\[\ce{$K_{eq}$ = \frac{[O2]^{3}}{[O2]^{2}} = 75 (given)}\]

\[\ce{$Q$ = \frac{0.24^{3}}{0.08^{2}} = 2.16} \]

\(\ce{Since $Q<K_{eq}$, the reaction will shift right to favour the products until  $Q = K_{eq}$} \)

\(\Rightarrow D\)

Algebra, STD1 A1 2020 HSC 18

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

`v^2 = u^2-100 d`

What distance does a car travel while slowing down from 70 km/h to 40 km/h?   (2 marks)

Show Answers Only

`33 \ text{metres}`

Show Worked Solution

`u = 70 \ , \ v = 40`

`v^2` `= u^2-100d`
`40^2` `= 70^2-100d`
`100d` `= 70^2-40^2`
`:. d` `= frac{70^2-40^2}{100}`
  `= 33 \ text{metres}`

Algebra, STD1 A1 2019 HSC 34

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

Show Answers Only

`4.76`

Show Worked Solution

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

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

Measurement, STD1 M1 2019 HSC 25

The diagram shows a sector with an angle of 120° cut from a circle with radius 10 m.

What is the perimeter of the sector? Write your answer correct to 1 decimal place.  (3 marks)

Show Answers Only

`40.9\ \ (text(1 d. p.))`

Show Worked Solution
`text(Arc length)` `= 120/360 xx 2 xx pi xx 10`
  `= 20.94`

 

`:.\ text(Perimeter)` `= 20.94 + 2 xx 10`
  `= 40.94`
  `= 40.9\ \ (text(1 d. p.))`

Measurement, STD1 M2 2020 HSC 15

The time in Melbourne is 11 hours ahead of Coordinated Universal Time (UTC). The time in Honolulu is 10 hours behind UTC. A plane departs from Melbourne at 7 pm on Tuesday and lands in Honolulu 9 hours later.

What is the time and day in Honolulu when the plane lands?   (2 marks)

Show Answers Only

`7\ text{am (Tuesday)}`

Show Worked Solution

`text{Melbourne} \ to \ text{UTC}\ + 11`

`text{Honolulu} \ to \ text{UTC}\ – 10`

`=>\ text{Honolulu is 21 hours behind Melbourne}`
 

`therefore \ text{Plane landing time}`

STRATEGY: 21 hours behind = 1 day behind + 3 hours.

`= 7\ text{pm} + 9 \ text{hours}`

`= 4\ text{am (Wednesday – Melb)}`

`= 7\ text{am (Tuesday – Honolulu)}`

Financial Maths, STD1 F3 2021 HSC 30

Blake opens a new credit card account on 1 May. He uses it, for the first time, on 4 May to buy concert tickets for $850.

He makes no further purchases or repayments during the month of May.

A statement for the credit card is issued on the last day of each month.

The statement for May shows that interest is charged at 19.75% per annum, compounding daily, from 20 May (included) until 31 May (included).

  1. What is the compound interest shown on the statement issued on 31 May?  (3 marks)

  2. The minimum payment is calculated as 3% of the closing balance on 31 May. Calculate the minimum payment.  (1 mark)

Show Answers Only
  1. `$5.54`
  2. `$25.67`
Show Worked Solution

a.   `text(Daily interest rate) = 19.75/365 = 0.05411text(%) = 0.0005411`

`text(Days incurring interest = 12)`

`text{Card balance (31 May)}` `= PV(1+r)^n`
  `= 850 (1 + 0.0005411)^12`
  `= $855.54`

 

`:.\ text(Interest)` `= 855.54-850`
  `= $5.54`

 

b.    `text(Minimum payment)` `= 855.54 xx 0.03`
    `= $25.67`

CHEMISTRY, M7 EQ-Bank 26

This flow chart shows reactions involving six different organic compounds (A to F).
 

 

Draw the structures of compounds A to F, justifying your diagrams with reference to the information provided.   (7 marks)

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A:  
       

B:  
       

C:  
           

D:  
       

E:  
       

F:  
       

Show Worked Solution

→ A combines with dilute  \(\ce{H2SO4}\)  to produce two alcohols.

→ C doesn’t react with strong oxidant (i.e. it is a tertiary alcohol).

C:  
           

→ B must therefore be 2-methylbutan-1-ol or 3-methylbutan-2-ol.

→ Since B gives two products when heated with concentrated \(\ce{H2SO4}\), it must be 3-methylbutan-2-ol, as 2-methylbutan-1-ol will only produce one product.

B:  
       

→ D is a ketone, produced by B reacting with a strong oxidant.

D:  
       

→ A can be dehydrated using concentrated  \(\ce{H2SO4}\)  on either B or C.

A:  
       

→ B dehydrates to A or E. Given A‘s structure above,

E:  
       

→ C dehydrates to A or F. Again, given A’s structure above,

F:  
       

CHEMISTRY, M7 EQ-Bank 23

  1. Design a procedure that can be used to produce the ester, ethyl ethanoate, in a school laboratory.   (4 marks)

  2. Describe a safety precaution in the production of an ester in a school laboratory.   (2 marks)

Show Answers Only

a.   Methodology:

STEP 1: Reflux equipment using a hot plate heating source.

STEP 2: Pour 10 mL of ethanol into a round bottom flask. Add 20 mL of glacial (water-free) acetic acid.

STEP 3: Add 1 mL of concentrated \(\ce{H2SO4}\) (sulfuric acid) to the mixture. Boiling stones/chips can be added to smooth the boiling process before heating under reflux for 45 minutes.

STEP 4: Allow the solution to cool. The contents should then be transferred to a separating funnel.

STEP 5: Distilled water should be used to wash the contents which will remove excess alcohol and acetic acid. The ester layer has a density less than 1 g/mL and will therefore form a layer above the aqueous solution.

STEP 5: Remove the aqueous solution using the separating funnel.

STEP 6: By adding a solution of 1 mol L ¯1  \(\ce{Na2CO3}\) (sodium carbonate), all remaining acid will be removed. Similarly to Step 4, the ester can be isolated by removing the lower layer of aqueous solution using the separating funnel.
 

b.    → Ethanol presents a danger due to its combustibility.

→ A safety precaution that reduces the fire threat is through the use of a hot plate for heating rather than a naked flame.

Show Worked Solution

a.   Methodology:

STEP 1: Reflux equipment using a hot plate heating source.

STEP 2: Pour 10 mL of ethanol into a round bottom flask. Add 20 mL of glacial (water-free) acetic acid.

STEP 3: Add 1 mL of concentrated \(\ce{H2SO4}\) (sulfuric acid) to the mixture. Boiling stones/chips can be added to smooth the boiling process before heating under reflux for 45 minutes.

STEP 4: Allow the solution to cool. The contents should then be transferred to a separating funnel.

STEP 5: Distilled water should be used to wash the contents which will remove excess alcohol and acetic acid. The ester layer has a density less than 1 g/mL and will therefore form a layer above the aqueous solution.

STEP 5: Remove the aqueous solution using the separating funnel.

STEP 6: By adding a solution of 1 mol L ¯1  \(\ce{Na2CO3}\) (sodium carbonate), all remaining acid will be removed. Similarly to Step 4, the ester can be isolated by removing the lower layer of aqueous solution using the separating funnel.
 

b.    → Ethanol presents a danger due to its combustibility.

→ A safety precaution that reduces the fire threat is through the use of a hot plate for heating rather than a naked flame.

CHEMISTRY, M7 EQ-Bank 22

Calculate the mass of methanol that must be burnt to increase the temperature of 325 g of water by 65°C, if exactly half of the heat released by this combustion is lost to the surroundings.

The heat of combustion of methanol is 726 kJ mol ¯1.  (3 marks)

Show Answers Only

7.79 g

Show Worked Solution

\(\ce{Using $q = mc\Delta T$}\):

\(\ce{$q$ = 325 \times 10^{-3} \times 4.18 \times 10^{3} \times 65 = 88 302.5 J = 88.3025 kJ}\)

\(\ce{Methanol heat combustion = 726 kJ mol^{-1} (given)}\)

\[\ce{n(CH3(OH)) = \frac{88.3025}{726} }\]

\(\ce{MM(CH3(OH)) = 12.01 + 3 \times 1.008 + 16 + 1.008 = 32.042 }\)

\[\ce{m(CH3(OH)) = \frac{88.3025}{726} \times 32.042 }\]

\(\text{Since 50% of heat is lost → twice as much methanol is needed}\)

\[\ce{m(CH3(OH)) initial = 2 \times \frac{88.3025}{726} \times 32.042 = 7.79 g} \]

CHEMISTRY, M5 EQ-Bank 5 MC

The conversion of calcium carbonate to calcium oxide and carbon dioxide is a reversible reaction and will reach equilibrium under certain conditions.

In which diagram is the system most likely to have reached equilibrium?
 

 

Show Answers Only

`D`

Show Worked Solution

→ In order to reach equilibrium, the reaction must occur in a closed system. In this system, a lid must be on the container.

→ When the reaction  \(\ce{CaCO3(s) \rightleftharpoons CaO(s) + CO2(g)} \)  has reached equilibrium, both the forwards and reverse reactions will be occurring.

→ All three species will be present.

`=>D`

CHEMISTRY, M7 EQ-Bank 24

Primary, unbranched alcohols and alkanes of the same carbon length have quite different boiling points.

Explain the difference in boiling point of these organic compounds, showing all intermolecular forces. Support your answer with diagrams.   (4 marks)

Show Answers Only

→ Alkanes are saturated hydrocarbons – i.e. they are made up of carbon and hydrogen atoms only and all atoms are joined together by single covalent bonds which are non-polar.

→ Weak intermolecular forces (Van der Waals) are therefore the only forces holding alkane molecules together and as a result, alkanes have low boiling points.

→ In contrast, alcohols have an \(\ce{OH}\) functional group. The \(\ce{OH}\) bond is polar with an oxygen “pole” that is slightly negatively charged and hydrogen “pole” that is slightly positively charged.

→ The hydrogen atom on one molecule will form an electrostatic bond with the oxygen atom on another atom creating a hydrogen bond.

→ Since hydrogen bonds are much stronger intermolecular forces than dispersion forces, the boiling points of alcohols are significantly higher than those of alkanes with the same carbon lengths.

Show Worked Solution

→ Alkanes are saturated hydrocarbons – i.e. they are made up of carbon and hydrogen atoms only and all atoms are joined together by single covalent bonds which are non-polar.

→ Weak intermolecular forces (Van der Waals) are therefore the only forces holding alkane molecules together and as a result, alkanes have low boiling points.

→ In contrast, alcohols have an \(\ce{OH}\) functional group. The \(\ce{OH}\) bond is polar with an oxygen “pole” that is slightly negatively charged and hydrogen “pole” that is slightly positively charged.

→ The hydrogen atom on one molecule will form an electrostatic bond with the oxygen atom on another atom creating a hydrogen bond.

→ Since hydrogen bonds are much stronger intermolecular forces than dispersion forces, the boiling points of alcohols are significantly higher than those of alkanes with the same carbon lengths.

CHEMISTRY, M8 EQ-Bank 22

  1. The diagram is a schematic representation of a mass spectrometer.
     

  1. Name and outline the function of the part labelled `A` in the diagram.   (2 marks)

  2. Outline the advantages of using mass spectrometry for analysis of a compound.   (3 marks)

Show Answers Only

a.    → \(\ce{A}\) is a magnet.

→ It bends the charged particles that are accelerated through an electric field and directed through it.

→ The amount of bending can then be used to distinguish between and identify the particles/ions.
 

b.   Advantages of mass spectrometry

→ The mass spectrometer can identify the mass/charge ratio of ions that pass through it. This data can then be used to identify components of a mixture or components in a compound.

→ Different isotopes of elements in a compound can be identified by mass spectrometers. This allows the compound to be matched to a sample.

→ Mass spectrometry can provide both qualitative and quantitative data on a compound.

→ Finally, mass spectrometers only require a small quantity of a compound to perform detailed analysis, as well as being fast and sensitive.

Show Worked Solution

a.    → \(\ce{A}\) is a magnet.

→ It bends the charged particles that are accelerated through an electric field and directed through it.

→ The amount of bending can then be used to distinguish between and identify the particles/ions.
 

b.   Advantages of mass spectrometry

→ The mass spectrometer can identify the mass/charge ratio of ions that pass through it. This data can then be used to identify components of a mixture or components in a compound.

→ Different isotopes of elements in a compound can be identified by mass spectrometers. This allows the compound to be matched to a sample.

→ Mass spectrometry can provide both qualitative and quantitative data on a compound.

→ Mass spectrometers only require a small quantity of a compound to perform detailed analysis, as well as being fast and sensitive.

CHEMISTRY, M8 EQ-Bank 25

The diagram shows the infrared spectrum of a compound.

 

The molecular weight of the compound analysed is approximately 60 g mol ¯1. Suggest TWO possible compounds that could fit this spectrum and justify your selection.   (4 marks)

Show Answers Only

→ The absorbance shows an initial trough in the range 3100 – 3600 cm¯1. This absorbance maximises at around 3350 cm¯1. This data indicates the likely presence of a hydroxyl group within the compound.

→ The next trough occurs in the range 3100 – 3600 cm¯1. Its intensity trough is not broad. This data is consistent with the presence of a  \(\ce{C-H}\)  bond.

→ There is little absorbance between 2500 – 1500 cm¯1. This data indicates the absence of any  \(\ce{C=O}\)  group, discounting the possibility of the compound being carboxylic acid or a ketone.

→ An absorption peak between 1250 – 1050 cm¯1 indicates the presence of a  \(\ce{C-O}\)  bond.

→ Propanol is a compound whose structure is consistent with the data above:

       

→ The molecular weight of propanol is 60.01 g mol ¯1 which is also supported by the data.

→ Since the spectrum of isomers is similar, the two possible compounds that fit this are propan-1-ol or propan-2-ol.

Show Worked Solution

→ The absorbance shows an initial trough in the range 3100 – 3600 cm¯1. This absorbance maximises at around 3350 cm¯1. This data indicates the likely presence of a hydroxyl group within the compound.

→ The next trough occurs in the range 3100 – 3600 cm¯1. Its intensity trough is not broad. This data is consistent with the presence of a  \(\ce{C-H}\)  bond.

→ There is little absorbance between 2500 – 1500 cm¯1. This data indicates the absence of any  \(\ce{C=O}\)  group, discounting the possibility of the compound being carboxylic acid or a ketone.

→ An absorption peak between 1250 – 1050 cm¯1 indicates the presence of a  \(\ce{C-O}\)  bond.

→ Propanol is a compound whose structure is consistent with the data above:

       

→ The molecular weight of propanol is 60.01 g mol ¯1 which is also supported by the data.

→ Since the spectrum of isomers is similar, the two possible compounds that fit this are propan-1-ol or propan-2-ol.

CHEMISTRY, M8 EQ-Bank 23

\(\ce{Fe^2^+}\) and \(\ce{X} \) react to form an ionic compound according to the general equation

\(\ce{aFe^2^+ + $b$(X)\rightleftharpoons [Fe_a(X)_b]^2^a^+}\)

where \(\ce{$a$}\) and \(\ce{$b$}\) are numbers representing the ratio in which \(\ce{Fe^2^+}\) and \(\ce{X} \) combine.

Spectrophotometry was used to determine the stoichiometric ratio between \(\ce{Fe^2^+}\) and \(\ce{X} \). To do this, eight 10 mL samples were prepared by reacting solutions of \(\ce{Fe^2^+}\) with solutions of \(\ce{X} \) in varying ratios. All \(\ce{Fe^2^+}\) and \(\ce{X} \) solutions had the same concentration. The absorbance of the samples is tabulated below.
 

  1. On the grid, construct a graph of absorbance against volume of \(\ce{Fe^2^+}\) solution from 0.00 mL to 6.00 mL, and draw TWO lines of best fit.  (3 marks)
     

     

  1. The reaction proceeds according to the general equation
  2.       \(\ce{aFe^2^+ + $b$X \rightleftharpoons [Fe_a(X)_b]^2^a^+}\).
  3. Find the values of \(\ce{$a$}\) and \(\ce{$b$}\) . Justify your answer with reference to the data given and the graph in part (a).   (3 marks)

Show Answers Only

a.  
           

 

b.   Find values of \(\ce{$a$}\) and \(\ce{$b$}\):

→ The graph shows that as the volume of compound \(\ce{X} \) decreases between 10mL and 7.5mL, there is an increase in \(\ce{Fe^{2+}}\) ions. This indicates an excess of \(\ce{Fe^{2+}}\) ions limiting the products.

→ The absorbance reaches a maximum when the 2.5 mL of compound \(\ce{Fe^{2+}}\) is added to 7.5 mL of compound \(\ce{X}\). Since the concentrations of the initial solutions are equal (given), equal volumes contain equal moles.

→ The ratio of volumes at peak absorbance = 2.5 : 7.5 = 1 : 3 (i.e. the correct stoichiometric ratio).

→ Hence \(\ce{$a$ = 1}\) and \(\ce{$b$ = 3}\).

→ The curve then turns down sharply, indicating a reduction in \(\ce{Fe^{2+}}\).

Show Worked Solution

a.  
           

 

b.   Find values of \(\ce{$a$}\) and \(\ce{$b$}\):

→ The graph shows that as the volume of compound \(\ce{X} \) decreases between 10mL and 7.5mL, there is an increase in \(\ce{Fe^{2+}}\) ions. This indicates an excess of \(\ce{Fe^{2+}}\) ions limiting the products.

→ The absorbance reaches a maximum when the 2.5 mL of compound \(\ce{Fe^{2+}}\) is added to 7.5 mL of compound \(\ce{X}\). Since the concentrations of the initial solutions are equal (given), equal volumes contain equal moles.

→ The ratio of volumes at peak absorbance = 2.5 : 7.5 = 1 : 3 (i.e. the correct stoichiometric ratio).

→ Hence \(\ce{$a$ = 1}\) and \(\ce{$b$ = 3}\).

→ The curve then turns down sharply, indicating a reduction in \(\ce{Fe^{2+}}\).

CHEMISTRY, M8 EQ-Bank 12

A colorimeter was used to calculate the percentage of iron in a 0.200 gram tablet. The tablet was dissolved and oxidised, then reacted with thiosulfate according to the equation

\(\ce{Fe^3^+($aq$) + SCN^-($aq$)\rightarrow [FeSCN]^2^+($aq$)}\)

The resulting solution was made up to 200 mL with distilled water. The absorbance of the final solution was measured to be 0.6105.

The calibration curve shows the absorbance of various concentrations of \(\ce{Fe^3^+}\).
 

Calculate the percentage of iron in the tablet.   (3 marks)

Show Answers Only

0.307%

Show Worked Solution

\(\ce{Fe^{3+}($aq$) + SCN^-($aq$)\rightarrow [FeSCN]^2^+($aq$)}\)

\(\ce{\text{Using the graph:}}\)

\(\ce{Absorbance = 0.615 }\)

\(\ce{\rightarrow Initial [Fe^{3+}] in 200 mL = 55 \times 10^{-6} = 5.5 \times 10^{-5} mol L^{-1}}\)

\(\ce{n(Fe^{3+}) = c \times V = 5.5 \times 10^{-5} \times 0.2 = 1.1 \times 10^{-5} mol} \)

\(\ce{MM(Fe) = 55.85 g mol^{-1}} \)

\(\ce{m(Fe) = 1.1 \times 10^{-5} \times 55.85 = 6.14 \times 10^{-4} g} \)

\[ \ce{\therefore \text{% Fe} = \frac{6.14 \times 10^{-4}}{0.200} \times 100 = 0.307\text{%} } \]

CHEMISTRY, M8 EQ-Bank 24

A common antacid tablet contains aluminium hydroxide to neutralise stomach acid. In order for the antacid to be effective, each 500 mg tablet must contain a minimum of 200 mg of aluminium hydroxide.

Two antacid tablets were crushed and reacted with 70 mL of 0.60 mol L¯ 1 hydrochloric acid. After the antacid had reacted with the acid, the remaining hydrochloric acid was titrated against 0.60 mol L¯ 1 sodium hydroxide. The average volume of sodium hydroxide used was 35 mL.

Calculate the amount of aluminium hydroxide present in each tablet and justify whether the tablets will be effective as an antacid.   (4 marks)

Show Answers Only

\(\text{Tablets will be effective (See Worked Solutions)}\)

Show Worked Solution

\(\ce{3HCl(aq) + Al(OH)3(aq) \rightarrow AlCl3(s) + 3H2O(l)} \)

\(\ce{n(HCl^-) = c \times V = 0.6 \times 0.070 = 0.042 mol} \)

\(\ce{n(Na(OH)) = c \times V = 0.6 \times 0.035 = 0.021 mol} \)

\(\ce{Moles reacted with antacid = 0.042 – 0.021 = 0.021 mol} \)
 

\(\ce{Mole ratio = 3:1}\)

\[\ce{Moles of Al(OH)3 in 2 tablets = \frac{0.021}{3} = 0.007 mol}\]

\(\ce{Formula weight of Al(OH)3 = 27 + 3 \times (1 + 16) = 78 g}\)

\(\ce{m(Al(OH)3) in 2 tablets = 0.007 \times 78 = 0.546 g}\)

\[\ce{m(Al(OH)3) in 1 tablet = \frac{0.546}{2} = 273 g}\]

\(\text{Since 273 > 200, tablets will be effective.}\)