25.0 cm³ of aqueous sodium hydroxide is neutralised by 26.7 cm³ of 0.100 mol/dm³ hydrochloric acid. What is the concentration of the sodium hydroxide?
- A 0.000107 mol/dm³
- B 0.0936 mol/dm³
- C 0.107 mol/dm³
- D 2.67 mol/dm³
Show answer
Answer: C
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25.0 cm³ of aqueous sodium hydroxide is neutralised by 26.7 cm³ of 0.100 mol/dm³ hydrochloric acid. What is the concentration of the sodium hydroxide?
Answer: C
Flour dust in mills can explode.
Explain why fine powders react much faster than lumps.
Suggest one way of reducing the risk of explosion.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | large surface area, more collisions a fine powder has a very large surface area B1 so there are many more collisions per second B1 | 2 |
| (b) | ventilation; no sparks good ventilation / remove dust / avoid naked flames and sparks B1 | 1 |
Magnesium ribbon reacts with excess dilute hydrochloric acid. The volume of hydrogen is measured every minute.
| time / min | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| volume of hydrogen / cm³ | 0 | 22 | 36 | 44 | 50 | 54 | 56 |
Calculate the average rate of reaction during the first two minutes, in cm³/min.
Describe and explain how the rate changes during the experiment. Use ideas about particles.
The experiment is repeated at a higher temperature with all other conditions the same. State the effect on the initial rate and on the total volume of hydrogen. Explain both answers.
Suggest why the total volume of hydrogen does not increase after a certain time.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 18 cm³/min 36 ÷ 2 M1 18 A1 | 2 |
| (b) | Rate decreases: the concentration of acid falls and the surface area of magnesium decreases, so there are fewer collisions per second rate decreases (as time goes on) B1 concentration of acid / amount of magnesium decreases B1 fewer (successful) collisions per second B1 | 3 |
| (c) | Initial rate is higher because particles move faster and have more energy, so more frequent and more successful collisions; total volume is the same because the same mass of magnesium (limiting) is used rate increases B1 particles have more (kinetic) energy, more frequent / more energetic collisions B1 total volume unchanged B1 same amount of the limiting reactant (magnesium) B1 | 4 |
| (d) | All the magnesium has reacted all the magnesium has been used up B1 | 1 |
This question is about oxidation and reduction.
Define oxidation in terms of electrons and in terms of oxidation number.
Iron(III) oxide reacts with aluminium.
Fe₂O₃ + 2Al → Al₂O₃ + 2Fe
Give the oxidation number of iron in Fe₂O₃ and in Fe.
Identify the reducing agent. Explain your answer.
Acidified potassium manganate(VII) is used to detect reducing agents.
State the colour change when it reacts with a reducing agent.
Aqueous potassium iodide is used to detect oxidising agents. State the colour change.
In the electrolysis of molten lead(II) bromide, write the ionic half-equation at each electrode and state which is oxidation.
Explain why the reaction between hydrochloric acid and sodium hydroxide is not a redox reaction.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | loss of electrons; increase in oxidation number loss of electrons B1 increase in oxidation number B1 | 2 |
| (b)(i) | +3; 0 +3 B1 0 B1 | 2 |
| (ii) | aluminium; it is oxidised / loses electrons aluminium B1 it gains oxygen / loses electrons / its oxidation number increases (0 to +3) B1 | 2 |
| (c)(i) | purple to colourless purple to colourless B1 | 1 |
| (ii) | colourless to brown colourless to (yellow-)brown B1 | 1 |
| (d) | Pb²⁺ + 2e⁻ → Pb; 2Br⁻ → Br₂ + 2e⁻; anode reaction is oxidation cathode: Pb²⁺ + 2e⁻ → Pb B1 anode: 2Br⁻ → Br₂ + 2e⁻ B1 balanced with electrons B1 oxidation occurs at the anode (bromide ions lose electrons) B1 | 4 |
| (e) | no change in oxidation numbers no element changes its oxidation number / no electrons transferred B1 | 1 |
A student wants to prepare pure, dry crystals of copper(II) sulfate from copper(II) oxide and dilute sulfuric acid.
Describe the method.
Explain why excess copper(II) oxide is used.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | excess base, filter, evaporate, crystallise warm the acid and add copper(II) oxide until it is in excess (no more dissolves) B1 filter to remove excess oxide B1 heat the filtrate to evaporate some water (to crystallisation point) B1 leave to cool and crystallise B1 filter/pat dry crystals with filter paper B1 | 5 |
| (b) | ensure all acid reacts to make sure all the acid is used up B1 | 1 |
Write balanced symbol equations for these reactions, including state symbols.
Magnesium reacts with dilute hydrochloric acid to form magnesium chloride and hydrogen.
Methane burns completely in oxygen.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) correct formulae and balancing B1 correct state symbols B1 | 2 |
| (b) | CH₄ + 2O₂ → CO₂ + 2H₂O correct formulae B1 balanced B1 | 2 |
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