1st Year Chemistry — MCQs & Practice Questions

In order of increasing proton (atomic) number, arranged in periods (rows) and groups (columns) of similar properties.

Reactivity increases down Group I because the outer electron is further from the nucleus and more easily lost.

Group VII atoms all have 7 outer electrons, so they react similarly (gaining/sharing one electron). Noble gases have a full outer shell, so they have no tendency to gain, lose or share electrons and are unreactive.

1. Reactivity decreases down the group[1]
2. The outer shell is further from the nucleus / more shielded[1]
3. So it is harder to attract/gain an electron[1]
4. Chlorine is more reactive than bromine, so it displaces bromine[1]
5. The solution turns orange (bromine formed)[1]
6. Cl₂ + 2KBr → 2KCl + Br₂[1]

Reactivity falls down the group (outer shell further/shielded, harder to gain e⁻); chlorine displaces bromine: Cl₂ + 2KBr → 2KCl + Br₂.

Reactivity increases down the group because the outer electron is further from the nucleus and more shielded, so it is lost more easily.

1. Melting point rises from Na to a peak at Si, then falls sharply to Ar[1]
2. Silicon is a giant covalent (macromolecular) structure[1]
3. so a huge number of strong covalent bonds must be broken — very high melting point[1]
4. First ionisation energy generally increases across the period[1]
5. because nuclear charge increases while shielding stays similar[1]
6. so the outer electron is held more tightly and needs more energy to remove[1]

Melting point rises to a giant-covalent peak at Si then falls to molecular Ar; first ionisation energy rises (more nuclear charge, similar shielding).

Proton: charge +1, mass 1. Neutron: charge 0, mass 1. Electron: charge −1, mass negligible (about 1/1840).

Atoms of the same element (same number of protons / same proton number) that have different numbers of neutrons, and so different mass numbers.

Protons = 17, electrons = 17 (neutral atom), neutrons = 35 − 17 = 18.

1. Both isotopes have the same number of electrons / the same electronic configuration (2,8,7)[1]
2. Chemical properties depend on the (outer) electrons, so they react identically[1]
3. Ar = (35 × 75 + 37 × 25) ÷ 100sets up weighted mean[1]
4. Ar = (2625 + 925)/100 = 35.5[1]
5. 7 electrons in the outer shell[1]
6. Gains 1 electron, so the chloride ion is Cl⁻ (charge −1)[1]

Same electron arrangement → identical chemistry; Ar = 35.5; 7 outer electrons, so Cl⁻.

Aᵣ = (10 × 20 + 11 × 80) ÷ 100 = (200 + 880) ÷ 100 = 1080 ÷ 100 = 10.8.

1. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²[1]
2. electron count totals 20[1]
3. Ca²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ (loses the two 4s electrons)[1]
4. The first two electrons are removed from the outer 4s sub-shell[1]
5. The third is removed from the 3p, a shell closer to the nucleus[1]
6. which is held much more strongly, so far more energy is needed — showing two electrons in the outer shell[1]

Ca = 1s²2s²2p⁶3s²3p⁶4s²; Ca²⁺ = 1s²2s²2p⁶3s²3p⁶; the jump after 2 electrons shows only two electrons in the outer shell.

Sodium loses one electron to chlorine; sodium becomes Na⁺ and chlorine becomes Cl⁻, and the oppositely charged ions attract by strong electrostatic forces.

It has a giant ionic lattice; the many strong electrostatic forces between oppositely charged ions need a large amount of energy to overcome.

Methane has weak intermolecular forces between molecules, which need little energy to overcome. Diamond is a giant covalent structure with strong covalent bonds throughout, needing very large energy to break.

1. Magnesium loses 2 electrons to oxygen, forming Mg²⁺ and O²⁻[1]
2. The oppositely charged ions attract by strong electrostatic forces[1]
3. When molten, the ions are free to move and carry charge[1]
4. When solid, the ions are fixed in the lattice and cannot move[1]
5. CO₂ is simple molecular with weak intermolecular forces[1]
6. Little energy is needed to separate the molecules, so it is a gas[1]
7. MgO is a giant ionic lattice with strong forces needing much energy, so it is a high-melting solid[1]

Mg→Mg²⁺ + O→O²⁻ (electrostatic attraction); MgO conducts molten (mobile ions) not solid (fixed ions); CO₂ gas (weak intermolecular forces) vs MgO solid (giant lattice, strong forces).

Down the group the atoms have more electrons, so the van der Waals (London) forces between them are stronger and need more energy to overcome, raising the boiling point.

1. Methane is tetrahedral[1]
2. with a bond angle of 109.5°[1]
3. because four bonding pairs repel to get as far apart as possible[1]
4. Water molecules form hydrogen bonds (O–H, H bonded to O)[1]
5. methane has only weak van der Waals forces[1]
6. hydrogen bonds are much stronger, so water needs far more energy to boil[1]

Methane: tetrahedral, 109.5°, four repelling bonding pairs; water boils higher because of strong hydrogen bonding vs methane's weak van der Waals forces.

Mr of CH₄ = 12 + 4 = 16. Moles = mass ÷ Mr = 8.0 ÷ 16 = 0.50 mol.

The sum of the relative atomic masses of all the atoms shown in the molecular formula of a substance.

4 Al + 3 O₂ → 2 Al₂O₃.

1. Mr(CaCO₃) = 40 + 12 + 48 = 100[1]
2. Moles CaCO₃ = 25 ÷ 100 = 0.25 mol[1]
3. Moles CaO = 0.25 (1:1 ratio); Mr(CaO) = 56[1]
4. Mass CaO = 0.25 × 56 = 14 g[1]
5. Moles CO₂ = 0.25 (1:1 ratio)[1]
6. Volume = moles × 24 = 0.25 × 24[1]
7. Volume CO₂ = 6.0 dm³[1]

Mr = 100; 14 g CaO; 6.0 dm³ CO₂.

V = 500 ÷ 1000 = 0.50 dm³. c = n ÷ V = 0.20 ÷ 0.50 = 0.40 mol/dm³.

1. Mr(Fe₂O₃) = 112 + 48 = 160[1]
2. Moles Fe₂O₃ = 320 ÷ 160 = 2 (tonne-moles)[1]
3. Moles Fe = 2 × 2 = 4; Ar(Fe) = 56[1]
4. Mass Fe = 4 × 56 = 224 tonnes[1]
5. Desired product mass in equation = 2 × 56 = 112; total product mass = 112 + 3 × 44 = 244[1]
6. Atom economy = (112 ÷ 244) × 100[1]
7. = 45.9% (≈ 46%)[1]

Mr = 160; 224 tonnes of iron; atom economy ≈ 46%.

At constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure: PV = constant (P₁V₁ = P₂V₂).

P₁V₁ = P₂V₂ → 100 × 500 = 250 × V₂ → V₂ = 50000 ÷ 250 = 200 cm³.

Gas particles have negligible volume; there are no forces between them; collisions are perfectly elastic; they move randomly (any two).

1. P = pressure, V = volume, n = number of moles[1]
2. R = gas constant, T = temperature in kelvin[1]
3. V = nRT ÷ P = (0.50 × 8.31 × 300) ÷ 100000[1]
4. = 1246.5 ÷ 100000[1]
5. V = 0.0125 m³ (12.5 dm³)[1]
6. Low pressure (and/or high temperature)[1]

P,V,n,R,T defined; V ≈ 0.0125 m³; ideal behaviour at low pressure / high temperature.

Mole fraction of He = 0.25 ÷ (0.25 + 0.75) = 0.25. Partial pressure = 0.25 × 200 = 50 kPa.

1. Rate of diffusion is inversely proportional to the square root of the molar mass[1]
2. rate(H₂)/rate(O₂) = √(32/2) = √16[1]
3. = 4 times faster[1]
4. At high pressure the molecules are close, so their own volume and the forces between them matter[1]
5. at low temperature they move slowly, so intermolecular attractions have a bigger effect — both break the ideal assumptions[1]

Graham's law: rate ∝ 1/√M; hydrogen diffuses 4× faster; real gases deviate at high P/low T where molecular volume and intermolecular forces matter.

Water molecules form hydrogen bonds, which are relatively strong intermolecular forces requiring extra energy to overcome, raising the boiling point.

A crystalline solid has a regular, ordered arrangement of particles (e.g. NaCl); an amorphous solid has a random, disordered arrangement (e.g. glass).

Van der Waals (London dispersion) forces are weakest, then permanent dipole–dipole forces, then hydrogen bonding is strongest.

1. The number of electrons increases down the group[1]
2. so van der Waals forces get stronger, raising the boiling point[1]
3. Water molecules form hydrogen bonds[1]
4. which are much stronger than the van der Waals forces in the others, needing more energy[1]
5. Van der Waals (London dispersion) forces[1]
6. are present in all molecules / all four hydrides[1]

Bigger molecules → stronger van der Waals forces (H₂S→H₂Te); water is anomalous due to hydrogen bonding; van der Waals forces are present in all.

Water molecules form hydrogen bonds (H bonded to O), which are strong intermolecular forces; methane has only weak van der Waals forces. Far more energy is needed to overcome water's hydrogen bonds, so it boils much higher.

1. In ice, hydrogen bonds hold the molecules in a regular open lattice[1]
2. with more empty space between molecules than in the liquid[1]
3. so a given mass takes up more volume — lower density — and ice floats[1]
4. A liquid boils when its vapour pressure equals the external pressure[1]
5. at altitude the atmospheric pressure is lower[1]
6. so a lower temperature is enough for the vapour pressure to match it[1]

Ice floats because hydrogen bonds form an open lattice with extra empty space; water boils lower at altitude because the lower air pressure is matched at a lower vapour pressure.

The heat energy change for a reaction measured at standard conditions (298 K, 1 atm), with substances in their standard states.

The total enthalpy change of a reaction is independent of the route taken, provided the initial and final conditions are the same. It lets us calculate enthalpy changes that cannot be measured directly.

q = mcΔT, where q = heat energy, m = mass of solution, c = specific heat capacity, ΔT = temperature change.

1. q = mcΔT = 100 × 4.2 × 30[1]
2. = 12600 J = 12.6 kJ[1]
3. Moles ethanol = 0.50 ÷ 46 = 0.0109 mol[1]
4. ΔH = energy ÷ moles = 12.6 ÷ 0.0109[1]
5. = 1156 kJ/mol (magnitude)[1]
6. ΔH is negative (exothermic): ≈ −1160 kJ/mol[1]
7. Heat is lost to the surroundings / incomplete combustion[1]

q = 12.6 kJ; 0.0109 mol; ΔH ≈ −1160 kJ/mol; lower magnitude due to heat loss / incomplete combustion.

Bonds broken = 436 + 242 = 678 kJ. Bonds made = 2 × 431 = 862 kJ. ΔH = 678 − 862 = −184 kJ/mol (exothermic).

1. The energy released when gaseous ions[1]
2. combine to form one mole of solid ionic lattice[1]
3. MgO contains Mg²⁺ and O²⁻ ions with 2+ and 2− charges[1]
4. NaCl contains only 1+ and 1− ions[1]
5. the ions in MgO are also smaller[1]
6. higher charges and smaller radii give stronger electrostatic attraction, so a much larger (more exothermic) lattice energy[1]

Lattice energy = energy released forming a mole of lattice from gaseous ions; MgO's higher charges and smaller ions give much stronger attraction than NaCl.

The change in concentration of a reactant or product per unit time.

The power to which the concentration of that reactant is raised in the rate equation.

First order in A, second order in B, overall third order. Doubling [B] multiplies the rate by 2² = 4.

1. k increases with temperature[1]
2. Particles have more energy and collide more often[1]
3. and a greater fraction exceed the activation energy, so more collisions are successful[1]
4. The rate-determining step is the slowest step in a multi-step reaction[1]
5. It controls the overall rate[1]
6. s⁻¹[1]

k rises with temperature (more successful collisions); rate-determining step = the slowest step, which limits overall rate; first-order k has units s⁻¹.

Overall order = 2 + 1 = 3 (third order). Doubling [A] multiplies the rate by 2² = 4.

1. The slowest step in a multi-step reaction[1]
2. which controls (limits) the overall rate[1]
3. A catalyst provides an alternative pathway with lower activation energy[1]
4. so a greater fraction of collisions are successful, raising k[1]
5. Higher temperature gives molecules more energy[1]
6. so more exceed the activation energy — more successful collisions and a larger k[1]

Rate-determining step = slowest step controlling the rate; catalyst and higher temperature both raise k by increasing the fraction of successful (Ea-exceeding) collisions.

Kc = [NH₃]² ÷ ([N₂][H₂]³).

A large Kc means the equilibrium lies to the right — the mixture contains mostly products.

Increasing temperature shifts the equilibrium in the endothermic (backward) direction, so the yield of product falls and Kc decreases.

1. Higher pressure increases the yield of SO₃[1]
2. The equilibrium shifts to the side with fewer gas molecules[1]
3. 3 gas molecules on the left, 2 on the right, so it shifts right[1]
4. A catalyst does not change the yield / position of equilibrium[1]
5. It speeds up the forward and backward reactions equally (equilibrium reached faster)[1]
6. Increasing temperature decreases Kc[1]

Higher pressure → more SO₃ (fewer gas moles on right); catalyst → no change in yield (faster only); higher temperature → Kc decreases.

Kc = [SO₃]² ÷ ([SO₂]² [O₂]).

1. Higher pressure increases the yield of ammonia[1]
2. the system shifts to the side with fewer gas molecules[1]
3. 4 gas molecules on the left, 2 on the right, so it shifts right[1]
4. Increasing temperature decreases Kc[1]
5. the forward reaction is exothermic, so higher temperature favours the endothermic reverse[1]
6. less product forms, so Kc falls[1]

Higher pressure → more NH₃ (fewer gas moles on the right); higher temperature → lower Kc (favours the endothermic reverse).

Acids produce hydrogen ions, H⁺, in aqueous solution; their pH is below 7.

A strong acid is fully ionised in water; a weak acid is only partly ionised. It is about degree of ionisation, not concentration.

Hydrochloric acid + sodium hydroxide → sodium chloride + water. HCl + NaOH → NaCl + H₂O.

1. CuO + H₂SO₄ → CuSO₄ + H₂O (correct formulae)[1]
2. Equation balanced[1]
3. Warm the acid and add excess copper(II) oxide, then filter off the unreacted solid[1]
4. Heat the filtrate to evaporate some water / to the point of crystallisation[1]
5. Leave to cool so crystals form[1]
6. Filter and dry the crystals (e.g. between filter paper)[1]
7. Excess ensures all the acid reacts, and the excess solid can simply be filtered off[1]

CuO + H₂SO₄ → CuSO₄ + H₂O; react with excess base, filter, evaporate to crystallisation, cool, filter, dry; excess guarantees all acid used and is easily removed.

pH = −log₁₀(1 × 10⁻³) = 3. pH 3 is below 7, so the solution is acidic.

1. CuO + H₂SO₄ → CuSO₄ + H₂O (correct formulae)[1]
2. balanced[1]
3. Warm the acid and add copper(II) oxide in excess until no more reacts[1]
4. Filter to remove the excess copper(II) oxide[1]
5. Heat the filtrate to evaporate water to the point of crystallisation, then leave to cool[1]
6. Filter off the crystals and dry them (e.g. between filter paper)[1]

CuO + H₂SO₄ → CuSO₄ + H₂O; react with excess base, filter, evaporate to crystallisation, cool, filter, dry.

Oxidation is an increase in oxidation number; reduction is a decrease in oxidation number.

The voltage of that half-cell measured against a standard hydrogen electrode under standard conditions (298 K, 1 mol dm⁻³, 1 atm).

EMF = E°(cathode) − E°(anode) = +0.34 − (−0.76) = +1.10 V. The zinc electrode is the negative terminal (it is oxidised).

1. Zn → Zn²⁺ + 2e⁻[1]
2. This is oxidation (loss of electrons)[1]
3. The salt bridge completes the circuit[1]
4. and allows ions to flow to balance the charge in each half-cell[1]
5. Electrons flow from zinc to copper[1]
6. because zinc is more easily oxidised / has the more negative electrode potential[1]

Zn → Zn²⁺ + 2e⁻ (oxidation); salt bridge completes circuit and balances charge; electrons flow zinc → copper (zinc more easily oxidised).

Silver has the more positive E°, so it is reduced (cathode) and copper is oxidised (anode). E°cell = 0.80 − 0.34 = +0.46 V.

1. The zinc electrode is negative[1]
2. Zn → Zn²⁺ + 2e⁻ (oxidation)[1]
3. E°cell = E°(cathode) − E°(anode) = 0.34 − (−0.76)[1]
4. = +1.10 V[1]
5. Electrons flow from zinc to copper in the external circuit[1]
6. because zinc has the more negative electrode potential, so it is oxidised and releases electrons[1]

Zinc electrode negative (Zn → Zn²⁺ + 2e⁻); EMF = +1.10 V; electrons flow zinc → copper (zinc more easily oxidised).

A family of organic compounds with the same general formula and similar chemical properties, where each member differs from the next by CH₂.

Add bromine water: an alkene decolourises it (orange to colourless); an alkane does not (stays orange).

Many ethene monomers (with C=C double bonds) join together, with no other product, to form a long-chain polymer. Repeat unit: –(CH₂–CH₂)–.

1. C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂[1]
2. correct formulae / balanced[1]
3. Steam / water, ~300 °C, ~60 atm, phosphoric acid catalyst[1]
4. C₂H₄ + H₂O → C₂H₅OH[1]
5. balanced equation[1]
6. It is a continuous process / faster / gives purer ethanol (any one)[1]

Fermentation: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂; hydration: C₂H₄ + H₂O → C₂H₅OH (steam, ~300 °C, ~60 atm, H₃PO₄); hydration is continuous/faster/purer.

Add bromine water: ethene (an alkene) decolourises it from orange to colourless; ethane (an alkane) leaves it orange (no reaction).

1. Cracking breaks long-chain hydrocarbons into shorter ones[1]
2. using heat and a catalyst[1]
3. to make more useful small molecules (fuels) and alkenes for plastics[1]
4. Many ethene monomers (with C=C double bonds) join together[1]
5. the double bonds open and link into a long chain, with no other product[1]
6. The product is poly(ethene)[1]

Cracking breaks long hydrocarbons (heat + catalyst) into useful short molecules and alkenes; addition polymerisation joins many ethene monomers into poly(ethene).

Free-radical substitution. Initiation: Cl₂ → 2Cl• (UV). Propagation: Cl• + CH₄ → •CH₃ + HCl, then •CH₃ + Cl₂ → CH₃Cl + Cl•.

1. The mechanism is electrophilic addition[1]
2. the electron-rich C=C double bond attracts the electrophile (the δ+ H of HBr)[1]
3. The H can add to either carbon, giving either a secondary or a primary carbocation[1]
4. the secondary carbocation is more stable than the primary[1]
5. so the H adds to the terminal (CH₂) carbon (Markovnikov's rule)[1]
6. giving mainly 2-bromobutane[1]

Electrophilic addition; the C=C attracts the electrophile; via the more stable secondary carbocation the major product is 2-bromobutane.

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂, using yeast at about 30 °C in the absence of air. An advantage is that it uses a renewable resource (sugar/plant crops) rather than crude oil.

1. Oxidise the ethanol with acidified potassium dichromate (warm/reflux)[1]
2. (or ferment ethanol in air) to give ethanoic acid[1]
3. The magnesium fizzes / dissolves, giving off hydrogen gas[1]
4. Mg + 2CH₃COOH → (CH₃COO)₂Mg + H₂[1]
5. Ethyl ethanoate is formed, plus water[1]
6. using a concentrated sulfuric acid catalyst[1]

Oxidise ethanol → ethanoic acid; with magnesium it fizzes giving H₂ (Mg + 2CH₃COOH → (CH₃COO)₂Mg + H₂); with ethanol it makes ethyl ethanoate + water (conc. H₂SO₄ catalyst).

Reactivity decreases down the group because the outer shell is further from the nucleus and more shielded, so it is harder to attract an electron.

Damp red litmus paper turns blue in ammonia gas (ammonia is alkaline).

Chlorine is more reactive (a stronger oxidising agent) than bromine, so it removes electrons from bromide ions: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂.

1. Sulfur dioxide and nitrogen oxides are released when fossil fuels burn[1]
2. They dissolve in rainwater[1]
3. forming acidic solutions (sulfurous/sulfuric and nitric acids)[1]
4. SO₂ + H₂O → H₂SO₃[1]
5. Damages plants/forests, or acidifies lakes killing fish[1]
6. or corrodes buildings/limestone and metals (any valid effect)[1]

SO₂/NOₓ from burning fuels dissolve in rain forming acids; SO₂ + H₂O → H₂SO₃; acid rain damages forests, lakes, buildings.

With potassium bromide the solution turns orange as bromine is displaced; with potassium iodide it turns brown as iodine is displaced. Chlorine is more reactive than both. Equation: Cl₂ + 2KBr → 2KCl + Br₂.

1. Halogens react by gaining one electron[1]
2. down the group the outer shell is further from the nucleus and more shielded[1]
3. so an electron is attracted/gained less easily — reactivity falls[1]
4. The inert pair effect is the reluctance of the outer s-electron pair to bond in heavier p-block elements[1]
5. so they favour an oxidation state two lower than the group maximum[1]
6. lead is more stable as Pb²⁺ than Pb⁴⁺[1]

Halogen reactivity falls down the group (electron gained less easily); the inert pair effect makes heavier elements like lead favour the lower (+2) state.

Reactivity decreases down the group because the outer shell is further from the nucleus and more shielded, so it is harder to attract an electron.

Damp red litmus paper turns blue in ammonia gas (ammonia is alkaline).

Chlorine is more reactive (a stronger oxidising agent) than bromine, so it removes electrons from bromide ions: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂.

1. Sulfur dioxide and nitrogen oxides are released when fossil fuels burn[1]
2. They dissolve in rainwater[1]
3. forming acidic solutions (sulfurous/sulfuric and nitric acids)[1]
4. SO₂ + H₂O → H₂SO₃[1]
5. Damages plants/forests, or acidifies lakes killing fish[1]
6. or corrodes buildings/limestone and metals (any valid effect)[1]

SO₂/NOₓ from burning fuels dissolve in rain forming acids; SO₂ + H₂O → H₂SO₃; acid rain damages forests, lakes, buildings.

With potassium bromide the solution turns orange as bromine is displaced; with potassium iodide it turns brown as iodine is displaced. Chlorine is more reactive than both. Equation: Cl₂ + 2KBr → 2KCl + Br₂.

1. Halogens react by gaining one electron[1]
2. down the group the outer shell is further from the nucleus and more shielded[1]
3. so an electron is attracted/gained less easily — reactivity falls[1]
4. The inert pair effect is the reluctance of the outer s-electron pair to bond in heavier p-block elements[1]
5. so they favour an oxidation state two lower than the group maximum[1]
6. lead is more stable as Pb²⁺ than Pb⁴⁺[1]

Halogen reactivity falls down the group (electron gained less easily); the inert pair effect makes heavier elements like lead favour the lower (+2) state.

About 78% nitrogen and about 21% oxygen (the remaining ~1% is mostly argon with a little carbon dioxide).

Add anhydrous copper(II) sulfate — it turns from white to blue in the presence of water (or blue cobalt chloride paper turns pink).

Carbon dioxide is a greenhouse gas; it absorbs and re-emits heat (infrared radiation) that would otherwise escape, so the extra CO₂ traps more heat and raises average global temperatures.

1. Water (moisture)[1]
2. Oxygen (air)[1]
3. Galvanising coats the iron with a layer of zinc[1]
4. The zinc is more reactive than iron, so it corrodes/loses electrons in preference (sacrificial protection)[1]
5. so the iron is protected even where the coating is scratched[1]
6. Painting / oiling / greasing / coating with plastic (any one)[1]

Water + oxygen needed; galvanising = zinc coating that sacrificially protects (zinc more reactive) even if scratched; also painting/oiling.

Sulfur impurities in the coal burn to sulfur dioxide, which dissolves in rainwater to form an acidic solution (sulfurous/sulfuric acid). This acid rain damages plants and forests, acidifies lakes killing fish, or corrodes limestone buildings and metals.

1. Carbon monoxide forms by incomplete combustion[1]
2. when there is not enough oxygen to make CO₂[1]
3. it is toxic because it prevents the blood from carrying oxygen[1]
4. The converter provides a catalyst surface where pollutants react[1]
5. carbon monoxide is oxidised to carbon dioxide[1]
6. 2CO + O₂ → 2CO₂ (and NOₓ is reduced to N₂)[1]

CO from incomplete combustion, toxic (stops blood carrying O₂); catalytic converter oxidises CO to CO₂ (2CO + O₂ → 2CO₂) and reduces NOₓ to N₂.

By simple distillation: heat the solution so water boils off, then condense the vapour back to liquid; the salt remains behind.

A pure substance produces a single spot on the chromatogram; a mixture separates into two or more spots.

The Rf value is the ratio of the distance moved by the spot to the distance moved by the solvent front. Rf = distance moved by spot ÷ distance moved by solvent.

1. Add water and stir to dissolve the copper(II) sulfate[1]
2. Filter — sand is the residue in the filter paper[1]
3. Wash and dry the sand[1]
4. Heat the filtrate to evaporate water to the point of crystallisation[1]
5. Leave to cool so crystals form[1]
6. Filter off and dry the crystals[1]

Dissolve, filter (sand = residue), wash and dry it; evaporate the filtrate to crystallisation, cool, filter and dry the crystals.

Dip a clean wire in the salt and hold it in a blue flame: sodium gives a yellow flame, potassium gives a lilac flame.

1. Dissolve the solid and add sodium hydroxide solution[1]
2. a green precipitate forms[1]
3. confirming the iron(II) ion, Fe²⁺[1]
4. To a fresh sample add dilute (hydrochloric or nitric) acid, then barium chloride solution[1]
5. a white precipitate forms[1]
6. confirming the sulfate ion, SO₄²⁻[1]

Fe²⁺: NaOH gives a green precipitate. SO₄²⁻: acid + barium chloride gives a white precipitate.

By simple distillation: heat the solution so water boils off, then condense the vapour back to liquid; the salt remains behind.

A pure substance produces a single spot on the chromatogram; a mixture separates into two or more spots.

The Rf value is the ratio of the distance moved by the spot to the distance moved by the solvent front. Rf = distance moved by spot ÷ distance moved by solvent.

1. Add water and stir to dissolve the copper(II) sulfate[1]
2. Filter — sand is the residue in the filter paper[1]
3. Wash and dry the sand[1]
4. Heat the filtrate to evaporate water to the point of crystallisation[1]
5. Leave to cool so crystals form[1]
6. Filter off and dry the crystals[1]

Dissolve, filter (sand = residue), wash and dry it; evaporate the filtrate to crystallisation, cool, filter and dry the crystals.

Dip a clean wire in the salt and hold it in a blue flame: sodium gives a yellow flame, potassium gives a lilac flame.

1. Dissolve the solid and add sodium hydroxide solution[1]
2. a green precipitate forms[1]
3. confirming the iron(II) ion, Fe²⁺[1]
4. To a fresh sample add dilute (hydrochloric or nitric) acid, then barium chloride solution[1]
5. a white precipitate forms[1]
6. confirming the sulfate ion, SO₄²⁻[1]

Fe²⁺: NaOH gives a green precipitate. SO₄²⁻: acid + barium chloride gives a white precipitate.