These twelve questions mix the five skills from water, air and environmental chemistry: separation steps, solutions and suspensions, data, carbon processes and evaluating claims. They run from easier to harder. All data is invented for practice.
Attempt each one on paper first. Open the answer only after you have written a full attempt with units.
Relative atomic masses: H = 1, C = 12, O = 16, Ca = 40. Use 24 dm³ as the volume of one mole of gas at room temperature and pressure.
Questions
1. Put these treatment stages in a sensible order and say which one is a separation step and which one kills microbes: chlorination, sedimentation, filtration.
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Order: sedimentation, filtration, chlorination. Sedimentation and filtration are separation steps because they remove solids. Chlorination kills microbes, so it is sterilising and not a separation.
2. A mixture contains sand, dissolved salt and water. The mixture is filtered. Name what is in the residue and what is in the filtrate.
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The residue is sand, because it is insoluble and too large to pass through the paper. The filtrate is salt dissolved in water, because dissolved particles are small enough to pass through.
3. Three samples are tested. P is cloudy and settles on standing. Q is clear, leaves nothing in the filter paper and leaves a white solid when evaporated. R is clear, leaves nothing in the paper and leaves nothing when evaporated. Classify each.
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P is a suspension, as it is cloudy and settles. Q is a solution, as a solute remains after evaporation. R is pure water (or contains nothing non-volatile), as nothing is left after evaporation.
4. Explain why a clear liquid is not necessarily pure.
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Dissolved substances have particles too small to scatter light or be seen, so the liquid stays clear. Pure means only one substance is present, and a clear solution still contains a solute. Evaporation or another test is needed to check.
5. Solid Q has a solubility of 25 g per 100 g of water at 20 °C. A student adds 60 g of Q to 200 g of water at 20 °C and stirs well. How much does not dissolve?
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200 g of water can dissolve 25 × 2 = 50 g. So 50 g dissolves and 60 − 50 = 10 g stays undissolved.
6. Nitrate readings at a stream site were 4.2, 4.6 and 4.4 mg/dm³. Calculate the mean.
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4.2 + 4.6 + 4.4 = 13.2. Then 13.2 ÷ 3 = 4.4 mg/dm³.
7. Phosphate was 0.5 mg/dm³ upstream and 2.0 mg/dm³ downstream of a discharge pipe. Calculate the percentage increase and say what the data suggests.
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Increase = 2.0 − 0.5 = 1.5. Then 1.5 ÷ 0.5 = 3, so the increase is 300%. The data suggests the discharge may add phosphate, but it does not prove it, because one pair of readings cannot rule out other sources.
8. Methane burns completely: CH₄ + 2O₂ → CO₂ + 2H₂O. Calculate the mass of carbon dioxide formed from 3.2 g of methane.
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Mr of CH₄ = 16, so moles = 3.2 ÷ 16 = 0.20 mol. The ratio is 1:1, so 0.20 mol CO₂. Mr of CO₂ = 44. Mass = 0.20 × 44 = 8.8 g.
9. Balance: CO + NO → CO₂ + N₂. Name the type of device that does this in a vehicle exhaust.
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2CO + 2NO → 2CO₂ + N₂. Check: carbon 2 and 2, oxygen 4 and 4, nitrogen 2 and 2. The device is a catalytic converter.
10. In photosynthesis, 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂. How many moles of oxygen form when 3.0 mol of carbon dioxide reacts?
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The ratio CO₂ to O₂ is 6:6, which is 1:1. So 3.0 mol CO₂ gives 3.0 mol O₂.
11. 5.0 g of calcium carbonate reacts fully with dilute hydrochloric acid: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Calculate the mass and the volume at room temperature and pressure of carbon dioxide formed.
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Mr of CaCO₃ = 40 + 12 + 48 = 100, so moles = 5.0 ÷ 100 = 0.050 mol. The ratio is 1:1, so 0.050 mol CO₂. Mass = 0.050 × 44 = 2.2 g. Volume = 0.050 × 24 = 1.2 dm³.
12. A report says a new filter “cuts lead in water by 90%”. In one test on one sample, lead fell from 10.0 µg/dm³ to 1.5 µg/dm³ (invented). Evaluate the claim.
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Decrease = 10.0 − 1.5 = 8.5. Percentage = 8.5 ÷ 10.0 × 100 = 85%. This is lower than the 90% claimed, so the claim is not fully supported. The test used one sample and no repeats, so a firmer conclusion needs more tests.
If you got these wrong
- Questions 1 to 4: reread explaining a separation step in water treatment and dissolved substances and suspensions. Focus on the particle-size reason.
- Question 5: scaling the solubility to the mass of water is covered in the dissolved substance lesson.
- Questions 6 and 7: means and percentage change are in interpreting pollutant evidence.
- Questions 8 to 11: balancing and mole ratios are in tracing a carbon-containing process. Use the mole and equation-ratio tutor and the equation balance reasoning trainer.
- Question 12 and any “suggests or proves” slip: see comparing a claim with its evidence.
Use the mistake log and retest queue to record the cause of each error, and return to the module overview if a whole skill needs rebuilding.
If you keep losing marks at the same step even after rereading, a teacher watching your working can find it faster than you can alone. That is what we do in online one-to-one Chemistry tuition.