This set mixes the five lessons in integrated chemical reasoning. All questions and data are original and invented for practice, and none is taken from an exam paper.
Try each question on paper before opening the answer. Use 24 dm³ per mole for gases, and Ar values of Mg = 24, Cu = 64, H = 1, C = 12, O = 16. Keep a note of which question types you miss in the mistake log and retest queue, and use the scientific investigation critic on the data-based questions.
Questions, from easier to harder
1. Substance A melts at 1,650 °C, is very hard and does not conduct electricity as a solid or as a liquid. Name its structure and explain the lack of conduction.
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A giant covalent structure. All its atoms are held by strong covalent bonds, which explains the very high melting point and hardness. There are no free ions or delocalised electrons to carry charge, so it does not conduct.
2. Substance B melts at 660 °C, conducts as a solid and can be hammered into shape. Name its structure and say what carries the current.
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A metallic structure. Delocalised electrons move freely through the lattice of positive ions and carry the current. Layers of ions slide over each other, which allows it to be shaped.
3. Substance C melts at −20 °C and does not conduct in any state. Explain the low melting point without saying that covalent bonds break.
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C is a simple molecular substance. The covalent bonds inside each molecule stay intact. Only the weak forces between molecules are overcome on melting, so little energy is needed.
4. 0.060 g of magnesium reacts with excess hydrochloric acid: Mg + 2HCl → MgCl₂ + H₂. What volume of hydrogen forms?
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Moles Mg = 0.060 ÷ 24 = 0.0025 mol. The ratio Mg : H₂ is 1 : 1, so 0.0025 mol H₂. Volume = 0.0025 × 24 = 0.060 dm³ = 60 cm³.
5. What volume of 0.50 mol/dm³ hydrochloric acid is needed to react exactly with the 0.060 g of magnesium in question 4?
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The ratio Mg : HCl is 1 : 2, so moles of HCl = 2 × 0.0025 = 0.0050 mol. Volume = moles ÷ concentration = 0.0050 ÷ 0.50 = 0.010 dm³ = 10 cm³.
6. A different reaction of magnesium with acid gave the gas volumes below.
| Time (s) | 0 | 10 | 20 | 30 | 40 | 50 | 60 |
|---|---|---|---|---|---|---|---|
| Volume (cm³) | 0 | 30 | 50 | 62 | 69 | 72 | 72 |
(a) Find the mean rate from 0 to 10 s and from 20 to 30 s. (b) Explain why the rate falls. (c) What mass of magnesium was used if acid was in excess?
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(a) 0 to 10 s: 30 ÷ 10 = 3.0 cm³/s. 20 to 30 s: (62 − 50) ÷ 10 = 1.2 cm³/s.
(b) The acid concentration falls and the magnesium surface is used up, so fewer collisions per second are successful.
(c) The plateau is 72 cm³ = 0.072 dm³. Moles H₂ = 0.072 ÷ 24 = 0.0030 mol. Ratio 1 : 1, so 0.0030 mol Mg. Mass = 0.0030 × 24 = 0.072 g.
7. Current passes through copper(II) sulfate solution using inert electrodes until 0.020 mol of copper has been deposited. How many moles of electrons were transferred, and what volume of oxygen formed at the anode? (The anode reaction releases 4 electrons per O₂.)
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Each Cu²⁺ gains 2 electrons, so 0.020 × 2 = 0.040 mol of electrons. Oxygen: 0.040 ÷ 4 = 0.010 mol. Volume = 0.010 × 24 = 0.24 dm³ = 240 cm³.
8. In another electrolysis, copper electrodes are used in copper(II) sulfate solution. The blue colour does not fade, the cathode gains mass and the anode loses mass. Explain using particles.
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At the anode, copper atoms lose electrons and enter the solution as Cu²⁺: Cu → Cu²⁺ + 2e⁻. At the cathode, Cu²⁺ ions gain electrons and form copper: Cu²⁺ + 2e⁻ → Cu. Copper ions are removed and added at the same rate, so the blue colour stays the same. The anode loses mass and the cathode gains it.
9. Dodecane, C₁₂H₂₆, is cracked into C₈H₁₈ and one other product. Give the formula of the other product, say whether it decolourises bromine water, and check by relative formula mass.
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Carbon: 12 − 8 = 4. Hydrogen: 26 − 18 = 8. The other product is C₄H₈, which fits CₙH₂ₙ, so it is an alkene and does decolourise bromine water. Mass check: Mr of C₁₂H₂₆ = 144 + 26 = 170. Mr of C₈H₁₈ = 114. Mr of C₄H₈ = 56. And 114 + 56 = 170.
10. Ethene reacts with steam to make ethanol: C₂H₄ + H₂O → C₂H₅OH. What mass of ethanol should 14 g of ethene give? If 17.25 g is collected, what is the percentage yield?
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Mr of C₂H₄ = 28 and Mr of C₂H₅OH = 24 + 5 + 16 + 1 = 46. Moles of ethene = 14 ÷ 28 = 0.50 mol. Ratio 1 : 1, so 0.50 mol ethanol, mass = 0.50 × 46 = 23 g. Percentage yield = 17.25 ÷ 23 × 100 = 75.0 percent.
11. An unlabelled solution gives an orange-red flame. With sodium hydroxide it forms a white precipitate that does not dissolve in excess. Which ion is most likely, and what did each test contribute?
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Calcium, Ca²⁺. The flame test suggests calcium and rules out magnesium, which gives no flame colour. The precipitate that stays in excess rules out aluminium, whose precipitate dissolves. The two tests together support the conclusion.
If you got these wrong
Match the error to the lesson and revisit it before retesting.
| Questions you missed | What to revise |
|---|---|
| 1, 2, 3 | Name the structure first: connecting bonding to a property |
| 4, 5, 6 | Ratio and rate steps: combining an equation ratio with a rate dataset |
| 7, 8 | Ions, electrons and electrodes: explaining an electrochemical observation |
| 9, 10 | Balancing and structure: linking organic structure with a transformation |
| 11 | Weighing two tests: comparing evidence from two methods |
After a retest, an honest gap in reasoning is worth talking through with someone. Our teachers in online one-to-one Co-ordinated Sciences tuition can go through the questions you missed and write new ones on the same weak point.