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SACE Stage 2 Chemistry study materials
SACE Stage 2 Chemistry study resources

SACE Stage 2 Chemistry practice questions

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Question 1Monitoring the environment

Which gas is released mainly by the combustion of sulfur-containing fossil fuels and is a major contributor to acid rain?

  1. Methane
  2. Carbon monoxide
  3. Dichlorodifluoromethane (CFC-12)
  4. Sulfur dioxide Correct answer.

Explanation

Sulfur impurities in coal and oil are oxidised during combustion to sulfur dioxide. In the atmosphere this gas is slowly oxidised further to sulfur trioxide, which dissolves in cloud water to form sulfuric acid, a major component of acid rain.

Question 2Monitoring the environment

Which gas makes the largest contribution to the enhanced greenhouse effect caused by human activity?

  1. Tropospheric ozone
  2. Methane
  3. Water vapour
  4. Carbon dioxide Correct answer.

Explanation

Water vapour is the most abundant greenhouse gas overall, but its concentration is not controlled directly by human emissions and it acts mainly as a feedback. Carbon dioxide released by fossil fuel combustion accounts for the largest share of the extra radiative forcing attributed to human activity.

Question 3Monitoring the environment

Which statement best describes the biochemical oxygen demand (BOD) of a water sample?

  1. The quantity of oxygen released into the water by photosynthesising plants over five days at 20 °C
  2. The total mass of dissolved mineral salts left behind when a sample is evaporated to dryness
  3. The oxygen used by microbes breaking down organic matter in a sample over a set period Correct answer.
  4. The maximum mass of oxygen that can dissolve in a given volume of water at a stated temperature

Explanation

BOD is measured by keeping a sealed sample in the dark at 20 °C for five days and comparing the dissolved oxygen at the start and at the end. The difference represents the oxygen consumed by microorganisms oxidising dissolved and suspended organic matter.

Question 4Monitoring the environment

Chlorofluorocarbons are present in the stratosphere at only a few parts per trillion, yet they cause significant ozone loss. Which explanation best accounts for this?

  1. CFCs only destroy ozone when they dissolve in liquid water droplets of polar stratospheric clouds, which concentrate them greatly.
  2. CFCs absorb ultraviolet radiation in the troposphere, so less of it is available to form ozone in the stratosphere.
  3. UV light breaks the C–Cl bond, and each chlorine atom then destroys many ozone molecules catalytically. Correct answer.
  4. Each CFC molecule permanently absorbs exactly one ozone molecule, so the damage is fixed by the CFC concentration present.

Explanation

In the stratosphere, short-wavelength UV radiation cleaves the carbon–chlorine bond of a CFC to release a chlorine radical. That chlorine atom reacts with ozone to give chlorine monoxide, which then regenerates the chlorine atom, so a single chlorine can destroy thousands of ozone molecules before it is removed.

Question 5Monitoring the environment

In atomic absorption spectroscopy, why is the sample solution aspirated into a flame or heated graphite furnace?

  1. To precipitate the metal as a finely divided solid that scatters light from the hollow-cathode lamp
  2. To convert the metal into molecular species that fluoresce strongly in the visible region
  3. To ionise the metal completely so that it emits light at a characteristic wavelength of its own
  4. To convert metal ions into ground-state atoms that absorb at a characteristic wavelength Correct answer.

Explanation

Atomic absorption spectroscopy measures light absorbed by free atoms in their lowest energy state. The flame or furnace evaporates the solvent and breaks chemical bonds so that ground-state atoms are produced; these absorb radiation at sharp, element-specific wavelengths, allowing the concentration of that element to be determined from a calibration curve.

Question 6Monitoring the environment

In a Winkler determination of dissolved oxygen, a 100.0 mL water sample required 8.20 mL of 0.0100 mol/L sodium thiosulfate. Given that 1 mol of O₂ is equivalent to 4 mol of S₂O₃²⁻, what is the dissolved oxygen concentration in mg/L?

  1. 1.64 mg/L
  2. 6.56 mg/L Correct answer.
  3. 3.28 mg/L
  4. 26.2 mg/L

Explanation

Amount of thiosulfate is 0.00820 L × 0.0100 mol/L = 8.20 × 10⁻⁵ mol, so n(O₂) = 8.20 × 10⁻⁵ ÷ 4 = 2.05 × 10⁻⁵ mol. This is 2.05 × 10⁻⁵ × 32.0 = 6.56 × 10⁻⁴ g, or 0.656 mg, in 0.100 L, giving 6.56 mg/L.

Question 7Monitoring the environment

Which set of conditions most strongly favours the formation of photochemical smog over a city?

  1. Strong sunlight, warm air, high NOx and volatile organic compounds, and still conditions Correct answer.
  2. Weak sunlight, high humidity and strong winds that rapidly disperse the pollutants
  3. Heavy rainfall, low temperatures and high concentrations of carbon dioxide near the ground
  4. Cold, dark winter mornings with high sulfur dioxide and low concentrations of nitrogen oxides

Explanation

Photochemical smog forms when nitrogen oxides and volatile organic compounds are trapped near the ground and irradiated by strong sunlight. Warm, still, sunny conditions with a shallow mixing layer allow these pollutants to accumulate and react, producing ozone, aldehydes and peroxyacetyl nitrate.

Question 8Monitoring the environment

A groundwater sample lathers poorly with soap and gives a white precipitate when it is boiled. Which dissolved ions best explain both observations?

  1. Lead(II) and mercury(II) ions, which come from industrial discharge
  2. Sodium and chloride ions, which are the main ions in seawater
  3. Calcium and hydrogencarbonate ions, which are common in limestone areas Correct answer.
  4. Nitrate and phosphate ions, which come from fertiliser runoff

Explanation

Calcium ions react with soap to form an insoluble scum, which is why hard water lathers poorly. Hydrogencarbonate ions decompose on boiling to release carbonate, which precipitates with calcium as calcium carbonate; this is the temporary hardness that disappears after boiling.

Question 9Monitoring the environment

A rainwater sample contains 0.98 mg of sulfuric acid dissolved in 1.00 L of water. Assuming the acid is fully dissociated, what is the pH of the sample? (M(H₂SO₄) = 98.0 g/mol)

  1. 4.70 Correct answer.
  2. 9.70
  3. 2.00
  4. 5.00

Explanation

The amount of acid is 9.80 × 10⁻⁴ g ÷ 98.0 g/mol = 1.00 × 10⁻⁵ mol. Because sulfuric acid is diprotic and fully dissociated here, [H⁺] = 2.00 × 10⁻⁵ mol/L, so pH = −log(2.00 × 10⁻⁵) = 4.70, which is well below the pH 5.6 of unpolluted rain.

Question 10Monitoring the environment

Methane has a global warming potential of about 28 over 100 years. Which statement best explains what this number means?

  1. Methane makes up 28 times more of the atmosphere by volume than carbon dioxide does at present
  2. Methane absorbs 28 times more ultraviolet radiation than carbon dioxide does in the stratosphere
  3. Methane molecules remain in the atmosphere for 28 times as long as carbon dioxide molecules do
  4. Each kilogram of methane traps about 28 times as much heat as a kilogram of CO₂ over 100 years Correct answer.

Explanation

Global warming potential compares the heat trapped by a given mass of a gas with the heat trapped by the same mass of carbon dioxide over a chosen time horizon, here 100 years. It combines the gas's strength of infrared absorption with how long it persists, so a value of 28 applies to equal masses, not to equal volumes or to lifetimes.

Question 11Monitoring the environment

Which property of a gas molecule is required for it to act as a greenhouse gas?

  1. Its molecular vibrations absorb infrared radiation. Correct answer.
  2. It contains at least two atoms joined by covalent bonds.
  3. The molecule possesses a permanent electric dipole moment.
  4. It is released into the atmosphere by human activities.

Explanation

A greenhouse gas must be able to absorb infrared radiation emitted by Earth's surface. Absorption occurs when the radiation frequency matches a vibrational mode of the molecule, so the molecule gains vibrational energy which it can re-emit. Gases such as CO2 and CH4 satisfy this requirement even though they have no permanent dipole moment.

Question 12Monitoring the environment

A lake affected by acid deposition is surrounded by soil rich in limestone. Which process most directly limits the fall in the lake's pH?

  1. Sulfate ions precipitate out of the lake water as insoluble salts.
  2. Dissolved carbon dioxide is rapidly lost from the lake surface.
  3. Soil carbonate neutralises the added hydrogen ions. Correct answer.
  4. Nitrate ions are taken up by algae growing in the lake.

Explanation

Carbonate minerals such as calcite react with hydrogen ions according to CaCO3 + 2H+ → Ca2+ + H2O + CO2. This reaction consumes H+ and buffers the water, so the pH falls less than it otherwise would. Lakes surrounded by granite have little carbonate and acidify much more readily.

Question 13Monitoring the environment

In a freshwater lake, which nutrient is normally limiting and therefore causes algal blooms when supplied in excess?

  1. Sulfate ions
  2. Phosphate ions Correct answer.
  3. Sodium ions
  4. Chloride ions

Explanation

Phosphorus, usually present as phosphate, is the nutrient that most often limits algal growth in freshwater. Extra phosphate from fertilisers, detergents or sewage relieves this limitation and allows rapid algal growth, which is the first stage of eutrophication. Nitrogen is more often the limiting nutrient in marine systems.

Question 14Monitoring the environment

Why is the cathode of the hollow-cathode lamp in an atomic absorption spectrometer made from the element being determined?

  1. It supplies the oxidant needed to maintain the flame.
  2. It atomises the sample before it reaches the flame.
  3. It provides a continuous spectrum across all wavelengths.
  4. It emits sharp lines that the analyte atoms absorb. Correct answer.

Explanation

A hollow-cathode lamp with a cathode of the analyte element emits the same narrow atomic emission lines that the analyte atoms in the flame can absorb. This matching gives the technique high specificity, because other elements present in the sample do not absorb those particular wavelengths. A continuum source would be absorbed by many species and would give poorer sensitivity and linearity.

Question 15Monitoring the environment

A water sample from a bore has a conductivity of 1850 µS/cm, much higher than a nearby stream. Which conclusion is best supported by this measurement?

  1. It has a very low dissolved oxygen concentration.
  2. It contains a large amount of suspended clay.
  3. It contains many dissolved ions. Correct answer.
  4. It has a high biochemical oxygen demand.

Explanation

Electrical conductivity depends mainly on the concentration and mobility of dissolved ions, because ions are the species that carry current through the water. A value of 1850 µS/cm is far above that of clean stream water and indicates a substantial dissolved-ion load, such as saline groundwater or an industrial discharge. Conductivity does not measure oxygen demand, turbidity or dissolved oxygen.

Question 16Monitoring the environment

Which condition is most important in causing the very low springtime ozone concentrations observed over Antarctica?

  1. Polar stratospheric clouds release active chlorine at very low temperatures. Correct answer.
  2. Emissions of chlorofluorocarbons are much greater in the southern hemisphere than in the north.
  3. The long period of winter darkness prevents any ozone from being formed above the continent.
  4. Volcanic ash particles deposited in the stratosphere catalyse ozone destruction each spring.

Explanation

Antarctic stratospheric temperatures in winter fall low enough for polar stratospheric clouds to form. Reactions on the surfaces of these particles convert reservoir species such as HCl and ClONO2 into active chlorine, which is then photolysed when sunlight returns in spring. The resulting chlorine atoms catalytically destroy ozone, giving the severe springtime depletion observed over the continent.

Question 17Monitoring the environment

Which of the following is a secondary pollutant formed in photochemical smog?

  1. CO
  2. SO2
  3. NO
  4. O3 Correct answer.

Explanation

A secondary pollutant is not emitted directly from a source but is formed in the atmosphere by chemical reactions. In photochemical smog, ozone is produced when nitrogen oxides and volatile organic compounds react in the presence of sunlight. Nitrogen monoxide, carbon monoxide and sulfur dioxide are all emitted directly by combustion and are therefore primary pollutants.

Question 18Monitoring the environment

A rainwater sample has a hydrogen ion concentration of 2.5 × 10⁻⁵ mol/L. What is its pH?

  1. 5.60
  2. 9.40
  3. 4.40
  4. 4.60 Correct answer.

Explanation

pH is defined as the negative logarithm to base ten of the hydrogen ion concentration. Since log10(2.5) is approximately 0.40, pH = 5.00 − 0.40 = 4.60. This value is well below the pH of unpolluted rainwater and is consistent with acid deposition.

Question 19Monitoring the environment

The production of ozone from the photolysis of O2 is fastest near 40 km altitude, but the maximum ozone concentration occurs near 25 km. Which explanation best accounts for this?

  1. Ultraviolet intensity is greatest at about 25 km above the surface.
  2. Temperature increases with altitude throughout the stratosphere.
  3. Air density and collision frequency rise at lower altitudes. Correct answer.
  4. Chlorofluorocarbons destroy ozone most rapidly at about 25 km.

Explanation

Photolysis of O2 requires ultraviolet radiation of wavelength below about 240 nm, which is only available high in the stratosphere, so oxygen atom production peaks near 40 km. Ozone is then formed by the termolecular reaction O + O2 + M → O3 + M, whose rate depends on the concentration of air molecules. Because air density falls rapidly with altitude, this third-body reaction is much faster lower down, shifting the ozone maximum to about 25 km.

Question 20Monitoring the environment

The ozone depletion potential of CFC-12 is about 1.0, while that of HCFC-22 is about 0.055. Which statement best explains this difference?

  1. CFC-12 is present in the stratosphere at a much higher concentration than HCFC-22.
  2. CFC-12 absorbs ultraviolet radiation more strongly than HCFC-22 does.
  3. HCFC-22 is largely destroyed by OH radicals in the troposphere. Correct answer.
  4. HCFC-22 contains fewer chlorine atoms in each molecule than CFC-12 does.

Explanation

Ozone depletion potential compares the amount of ozone a gas can destroy with that destroyed by CFC-11. CFC-12 is chemically inert in the troposphere, so almost all of the gas released reaches the stratosphere, where ultraviolet light liberates chlorine atoms that catalytically destroy ozone. HCFC-22 contains hydrogen, so it reacts with hydroxyl radicals in the troposphere and only a small fraction reaches the stratosphere, giving it a much lower ozone depletion potential.

Question 21Monitoring the environment

Air-quality reports often give the concentration of PM2.5 rather than only total suspended particles. Why is the PM2.5 fraction of greatest concern to human health?

  1. These particles react with stratospheric ozone and thin the ozone layer.
  2. These particles settle out of the air very quickly onto soil.
  3. These particles are small enough to reach the alveoli deep in the lungs. Correct answer.
  4. These particles absorb ultraviolet light before it reaches ground level.

Explanation

PM2.5 refers to particles with an aerodynamic diameter of 2.5 micrometres or less. Because they are so small they largely bypass the nose and upper airways, deposit deep in the alveoli and can pass into the bloodstream, where they are linked to respiratory and cardiovascular disease. Coarser particles are trapped higher in the respiratory tract.

Question 22Monitoring the environment

Ozone present in the troposphere is best described as:

  1. a gas that screens out harmful ultraviolet radiation from the Sun
  2. a gas released directly from the exhaust of motor vehicles
  3. a gas that has no measurable effect on human health
  4. a respiratory irritant formed by reactions driven by sunlight Correct answer.

Explanation

Ground-level ozone is a secondary pollutant, produced when nitrogen oxides and volatile organic compounds react in the presence of sunlight. It is a strong oxidant that irritates the airways, reduces lung function and damages crops. The ozone that screens ultraviolet radiation sits in the stratosphere, roughly 15 to 35 km above the surface.

Question 23Monitoring the environment

Carbon monoxide is monitored continuously in urban air because of its acute health effects. Which statement best explains its toxicity?

  1. It reacts with oxygen inside the lungs and removes it from the air.
  2. It irritates the airways so that fluid accumulates in the lungs.
  3. It binds to haemoglobin far more strongly than oxygen does. Correct answer.
  4. It dissolves in the blood plasma and lowers the pH of the blood.

Explanation

Carbon monoxide forms carboxyhaemoglobin, binding to haemoglobin roughly 200 to 250 times more strongly than oxygen. This blocks the transport of oxygen to tissues and also shifts the oxygen-binding curve, so even quite low concentrations can cause dizziness, loss of consciousness and death.

Question 24Monitoring the environment

A sewage sample is diluted 20-fold with aerated dilution water. Over five days at 20 degrees Celsius the dissolved oxygen in the diluted sample falls from 8.60 mg/L to 3.40 mg/L. What is the 5-day BOD of the undiluted sewage?

  1. 5.2 mg/L
  2. 52 mg/L
  3. 0.26 mg/L
  4. 104 mg/L Correct answer.

Explanation

The diluted sample consumed 8.60 minus 3.40, which equals 5.20 mg/L of oxygen. Because the sewage was diluted 20-fold, the undiluted sample would consume 20 times 5.20, giving a BOD5 of 104 mg/L.

Question 25Monitoring the environment

Lead in a river-water sample is measured by atomic absorption spectroscopy. Compared with simple aqueous lead standards, the sample gives a much smaller absorbance because of its matrix. Which procedure is most appropriate?

  1. Subtract the absorbance of a distilled-water blank from the sample absorbance.
  2. Add known amounts of lead standard to several portions of the sample and extrapolate. Correct answer.
  3. Dilute the sample with distilled water until the matrix effect disappears.
  4. Reduce the slit width of the monochromator to increase the signal.

Explanation

In the method of standard additions, small known volumes of a lead standard are added to separate aliquots of the sample, so every calibration point experiences exactly the same matrix as the unknown. The resulting line is extrapolated back to zero absorbance to give the concentration in the original sample, which corrects for multiplicative interference.

Question 26Monitoring the environment

Measurements at a remote background station in the Northern Hemisphere show atmospheric carbon dioxide rising from year to year, but falling by several parts per million between May and September. Which explanation is correct?

  1. Sunlight destroys atmospheric carbon dioxide at a faster rate in summer.
  2. Warm summer oceans absorb more carbon dioxide than cold winter oceans do.
  3. Growing vegetation in the Northern Hemisphere takes up carbon dioxide by photosynthesis. Correct answer.
  4. Summer winds carry the station's local emissions away from the site.

Explanation

Most of the world's land and most of its forests lie in the Northern Hemisphere. During spring and summer, photosynthesis exceeds respiration and decay, drawing carbon dioxide down by several parts per million; in autumn and winter the balance reverses and the concentration climbs again, producing the sawtooth pattern of the Keeling curve.

Question 27Monitoring the environment

A power station discharges warm cooling water into a river. What change in dissolved oxygen is expected downstream of the discharge, and why?

  1. Dissolved oxygen is unchanged, because no organic matter is added to the river.
  2. Dissolved oxygen falls, because warm water carries extra dissolved salts.
  3. Dissolved oxygen falls, because gas solubility decreases as water warms. Correct answer.
  4. Dissolved oxygen rises, because gases are more soluble in warmer water.

Explanation

The solubility of oxygen in water decreases as temperature rises, so warmer water holds less dissolved oxygen. At the same time the metabolic oxygen demand of fish and micro-organisms increases with temperature, so the two effects combine to cause oxygen stress downstream.

Question 28Monitoring the environment

In a region, sulfur dioxide emissions have fallen sharply but rainfall has become more acidic as traffic has increased. Which statement best explains this?

  1. Ozone from vehicle exhausts dissolves in rain and lowers the pH.
  2. Carbon dioxide from vehicle exhausts dissolves in rain to form carbonic acid.
  3. Nitrogen oxides from exhausts are oxidised to nitric acid in the atmosphere. Correct answer.
  4. Unburned fuel from vehicles forms hydrochloric acid in cloud droplets.

Explanation

High-temperature combustion in engines produces nitrogen monoxide, which is oxidised to nitrogen dioxide and then to nitric acid and nitrate aerosol in the atmosphere. Nitric acid dissociates in cloud water to release hydronium ions and nitrate ions, so it acidifies rain. This is why acid deposition can persist even after sulfur controls are introduced.

Question 29Monitoring the environment

A 50.00 mL water sample is titrated with 0.0100 mol/L EDTA solution buffered at pH 10. The endpoint is reached after 12.40 mL of EDTA has been added, and each EDTA molecule binds one metal ion. What is the hardness of the water expressed as mg of calcium carbonate per litre?

  1. 248 mg/L Correct answer.
  2. 24.8 mg/L
  3. 496 mg/L
  4. 124 mg/L

Explanation

The amount of EDTA used is 0.01240 L multiplied by 0.0100 mol/L, which equals 1.24 x 10^-4 mol, and this equals the moles of metal ions in the 50.00 mL aliquot. Scaling to one litre gives 2.48 x 10^-3 mol/L, and multiplying by the molar mass of calcium carbonate, 100.1 g/mol, gives 0.248 g/L, or 248 mg/L.

Question 30Monitoring the environment

A river receives a continuous discharge of treated sewage rich in organic matter. Dissolved oxygen falls sharply just downstream, reaches a minimum, and then gradually returns to its upstream value. Which explanation is correct?

  1. Oxygen is consumed by the sewage itself and can only return once the discharge stops.
  2. Bacteria consume oxygen while breaking down the waste, and reaeration from the air restores it. Correct answer.
  3. The sewage raises the temperature, and oxygen returns when the river cools at night.
  4. Photosynthesis by algae raises oxygen, which then reacts with the sewage.

Explanation

Downstream of the outfall, micro-organisms oxidise the organic load and use dissolved oxygen faster than it can be replaced, producing an oxygen sag. Further downstream the organic matter is used up and reaeration from the atmosphere, together with photosynthesis, slowly restores oxygen towards saturation. The depth of the sag and the distance needed for recovery reflect the BOD load and the river's reaeration capacity.

Question 31Monitoring the environment

Which pair of gases are both direct greenhouse gases in the atmosphere?

  1. Carbon dioxide and methane Correct answer.
  2. Carbon dioxide and nitrogen gas
  3. Argon and helium
  4. Nitrogen and oxygen

Explanation

A greenhouse gas must absorb outgoing infrared radiation, which requires molecular vibrations that change the dipole moment. Carbon dioxide and methane both have such vibrations and absorb strongly in the infrared; nitrogen, oxygen, argon and helium do not absorb appreciably.

Question 32Monitoring the environment

In atomic absorption spectroscopy, what is the main purpose of the monochromator placed between the flame and the detector?

  1. To convert the sharp hollow-cathode lamp output into a continuous spectrum of light
  2. To isolate the analyte resonance line from other radiation reaching the detector Correct answer.
  3. To amplify the small electrical signal produced by the photomultiplier tube
  4. To break the sample into free atoms by heating it in the flame

Explanation

The hollow-cathode lamp emits a line spectrum containing the analyte resonance line, but the flame also emits background radiation and impurity lines may be present. The monochromator transmits only the chosen resonance wavelength, which improves selectivity and reduces stray light reaching the detector.

Question 33Monitoring the environment

At 25 °C, a water sample has a hydroxide ion concentration of 1.0 × 10⁻⁶ mol/L. What is the pH of the sample?

  1. 7.00
  2. 8.00 Correct answer.
  3. 6.00
  4. 10.00

Explanation

At 25 °C the ionic product of water gives [H⁺][OH⁻] = 1.0 × 10⁻¹⁴. Substituting [OH⁻] = 1.0 × 10⁻⁶ mol/L gives [H⁺] = 1.0 × 10⁻⁸ mol/L, so pH = −log(1.0 × 10⁻⁸) = 8.00.

Question 34Monitoring the environment

The five-day biochemical oxygen demand (BOD₅) of a water sample is determined by:

  1. comparing dissolved oxygen before and after five days of dark incubation at 20 °C Correct answer.
  2. measuring the mass of organic matter evaporated from a filtered sample
  3. measuring the light absorbed by the sample at 254 nm before and after filtration
  4. titrating the sample with EDTA to find the concentration of metal ions

Explanation

BOD₅ measures the oxygen consumed by microorganisms as they oxidise biodegradable organic matter in the sample. Incubation is carried out in the dark at 20 °C for five days so that photosynthesis cannot add oxygen and the conditions are standardised.

Question 35Monitoring the environment

Two lakes receive similar acid deposition, but only the lake in a granite catchment becomes acidic. Which statement best explains this difference?

  1. Granite catchments receive much less rainfall than limestone catchments
  2. Granite dissolves rapidly in rain and releases large amounts of dissolved carbonate into the lake water
  3. Granite has little acid-neutralising capacity, so hydrogen ions remain in the water Correct answer.
  4. Limestone catchments have no contact between soil water and the underlying bedrock

Explanation

Acid neutralisation depends on the supply of base cations and carbonate from the catchment. Limestone, mainly calcium carbonate, reacts with hydrogen ions and buffers the water, whereas granite is a slowly weathering silicate rock that provides very little alkalinity.

Question 36Monitoring the environment

A single chlorine atom released from a CFC can destroy thousands of ozone molecules because chlorine:

  1. is converted permanently into chlorine nitrate and therefore stays active in the stratosphere
  2. reacts directly with ozone to form dichlorine and oxygen gas
  3. is regenerated in a catalytic cycle after each ozone molecule is destroyed Correct answer.
  4. absorbs ultraviolet light and re-emits it as ozone-forming radiation

Explanation

The chlorine atom reacts with ozone to give chlorine monoxide and oxygen, and chlorine monoxide then reacts with another ozone molecule or a radical to release chlorine again. Because the chlorine atom is reformed, one atom destroys many ozone molecules before it is finally removed from the stratosphere.

Question 37Monitoring the environment

Atmospheric CO₂ at a Northern Hemisphere background station falls each year between May and September. What causes this seasonal fall?

  1. A seasonal increase in the burning of fossil fuels for heating
  2. A seasonal increase in ocean outgassing caused by warmer surface water
  3. A seasonal increase in photosynthesis by Northern Hemisphere land plants Correct answer.
  4. A seasonal decrease in the rate of soil respiration across the tropical rainforests

Explanation

Most of the world's land area and its seasonal deciduous and boreal vegetation lie in the Northern Hemisphere. In spring and summer these plants take up CO₂ by photosynthesis faster than respiration releases it, drawing the atmospheric concentration down.

Question 38Monitoring the environment

An air sample contains carbon monoxide at 0.45 ppm by volume. What is this concentration in mg/m³ at 25 °C and 1 atm? (M(CO) = 28.0 g/mol; molar volume = 24.5 L/mol)

  1. 0.45 mg/m³
  2. 12.6 mg/m³
  3. 0.51 mg/m³ Correct answer.
  4. 1.15 mg/m³

Explanation

A volume fraction of 0.45 ppm means 4.5 × 10⁻⁴ L of CO in each cubic metre of air. Moles of CO = 4.5 × 10⁻⁴/24.5 = 1.84 × 10⁻⁵ mol, so mass = 1.84 × 10⁻⁵ × 28.0 = 5.15 × 10⁻⁴ g, which is 0.51 mg/m³.

Question 39Monitoring the environment

Dissolved oxygen in a 200.0 mL water sample is fixed and the liberated iodine is titrated with 5.20 mL of 0.0125 mol/L sodium thiosulfate. Given that 1 mol O₂ produces 2 mol I₂ and 2 mol I₂ reacts with 4 mol thiosulfate, what is the dissolved oxygen concentration in mg/L? (M(O₂) = 32.0 g/mol)

  1. 1.30 mg/L
  2. 5.20 mg/L
  3. 2.60 mg/L Correct answer.
  4. 10.4 mg/L

Explanation

Moles of thiosulfate = 0.00520 L × 0.0125 mol/L = 6.50 × 10⁻⁵ mol. The stoichiometry gives 1 mol O₂ per 4 mol thiosulfate, so moles of O₂ = 1.625 × 10⁻⁵ mol; dividing by 0.200 L and multiplying by 32.0 g/mol gives 2.60 mg/L.

Question 40Monitoring the environment

Which substance is a primary air pollutant?

  1. Nitric acid
  2. Ozone
  3. Sulfur dioxide Correct answer.
  4. Peroxyacetyl nitrate

Explanation

Primary pollutants are emitted directly from sources. Sulfur dioxide is released when sulfur-containing fossil fuels are burned; ozone, nitric acid and peroxyacetyl nitrate are mainly formed by atmospheric reactions.

Question 41Monitoring the environment

Which measurement is most directly used to indicate the amount of suspended material in a water sample?

  1. Turbidity Correct answer.
  2. Alkalinity
  3. Total hardness
  4. Conductivity

Explanation

Turbidity measures the scattering of light by suspended particles, so it gives a direct indication of material that makes water cloudy. Conductivity reflects dissolved ions, while hardness and alkalinity reflect chemical composition rather than suspended solids.

Question 42Monitoring the environment

Which water-quality parameter is used as a simple estimate of the total concentration of dissolved salts?

  1. Turbidity
  2. Colour
  3. Conductivity Correct answer.
  4. Dissolved oxygen

Explanation

Dissolved salts dissociate into ions, which carry electric current. Conductivity therefore increases with the concentration of dissolved ions and is used as a rapid estimate of salinity.

Question 43Monitoring the environment

Which statement best distinguishes biochemical oxygen demand (BOD) from chemical oxygen demand (COD)?

  1. BOD uses a strong chemical oxidant; COD uses bacteria.
  2. BOD measures biodegradable matter; COD measures all oxidisable matter. Correct answer.
  3. BOD is always greater than COD for the same water sample.
  4. BOD measures all oxidisable matter; COD measures only biodegradable matter.

Explanation

BOD relies on microorganisms consuming dissolved oxygen while they oxidise biodegradable organic material. COD uses a strong chemical oxidant and therefore includes both biodegradable and non-biodegradable oxidisable substances.

Question 44Monitoring the environment

In the Winkler method for dissolved oxygen, what is the main purpose of adding manganese(II) sulfate and alkaline iodide?

  1. To fix oxygen as a precipitate Correct answer.
  2. To adjust the sample to pH 7
  3. To precipitate calcium carbonate
  4. To remove chlorine from the sample

Explanation

Manganese(II) is oxidised by dissolved oxygen to a brown manganese(III) precipitate under alkaline conditions. This fixes the oxygen so that it can later be determined by iodometric titration.

Question 45Monitoring the environment

In gas chromatography used to monitor volatile organic compounds, what does the retention time of a peak indicate?

  1. The molar mass of the compound
  2. The identity of the compound Correct answer.
  3. The concentration of the compound in the sample
  4. The toxicity of the compound

Explanation

Under fixed column, temperature and carrier-gas conditions, each compound has a characteristic retention time. Comparing the retention time with standards allows qualitative identification, while peak area gives concentration.

Question 46Monitoring the environment

Which set of conditions most favours the formation of photochemical smog?

  1. Cold, dark conditions with high sulfur dioxide
  2. Sunny, warm conditions with NOx and VOCs Correct answer.
  3. Cool, humid conditions with high carbon dioxide
  4. Windy, dry conditions with low nitrogen oxides

Explanation

Photochemical smog forms when sunlight drives reactions between nitrogen oxides and volatile organic compounds. Warm, sunny and still conditions allow these secondary pollutants, including ozone, to accumulate.

Question 47Monitoring the environment

A colorimetric nitrate method gives absorbances of 0.20, 0.40 and 0.80 for standards of 1.0, 2.0 and 4.0 mg/L. A sample gives an absorbance of 0.60. What is the nitrate concentration?

  1. 2.5 mg/L
  2. 3.0 mg/L Correct answer.
  3. 2.8 mg/L
  4. 3.5 mg/L

Explanation

The standards show a linear response: absorbance increases by 0.20 for each 1.0 mg/L. An absorbance of 0.60 lies halfway between 0.40 and 0.80, so the concentration is 3.0 mg/L.

Question 48Monitoring the environment

Which property of a pollutant best explains why it biomagnifies through an aquatic food chain?

  1. High water solubility and rapid breakdown
  2. High lipid solubility and persistence Correct answer.
  3. Low molecular mass and high volatility
  4. Strong adsorption to mineral particles

Explanation

Biomagnification occurs when a substance is absorbed and stored in fatty tissue faster than it is excreted or metabolised. A lipid-soluble, persistent pollutant therefore increases in concentration at each higher trophic level.

Question 49Monitoring the environment

In a UV-absorption ozone monitor, ambient air is passed through an ozone scrubber before entering one of the two cells. Why is this done?

  1. To increase the ozone concentration in the reference cell
  2. To remove water vapour from the sample
  3. To convert nitrogen dioxide into ozone
  4. To provide an ozone-free reference Correct answer.

Explanation

Ozone absorbs UV light strongly at 254 nm. Comparing air that has passed through the scrubber with untreated air gives the absorbance due to ozone alone, allowing its concentration to be calculated using the Beer–Lambert law.

Question 50Monitoring the environment

Water hardness is often monitored by titrating a sample with EDTA. Why is the sample first buffered to about pH 10?

  1. To oxidise dissolved organic matter that would interfere with the endpoint
  2. To convert dissolved calcium ions into calcium carbonate precipitate before titration
  3. To allow both calcium and magnesium ions to be complexed completely by EDTA Correct answer.
  4. To remove chloride ions that compete with EDTA for metal ions

Explanation

EDTA forms stable 1:1 complexes with Ca²⁺ and Mg²⁺, but the fraction of EDTA present as the fully deprotonated Y⁴⁻ ion depends strongly on pH. At about pH 10 there is enough Y⁴⁻ for both ions to be titrated, while the pH is not so high that metal hydroxides precipitate. The indicator used for the endpoint also requires this alkaline range.