What this diagnostic is for
Use this page to find which Unit 4 topics to revisit: one question per topic, two for the larger Topics 4.5 and 4.7. Answer every question before opening any answer, then use the table below.
These are original Marlbridge practice questions, not past exam questions. They are not calibrated, so your total gives no predicted score. A wrong answer points to a topic; it is not a grade.
Calculator and data. Scientific calculator allowed. Molar masses: H 1.008, C 12.01, O 16.00, Na 22.99, Al 26.98, Cl 35.45, Fe 55.85 g mol⁻¹. R = 0.08206 L atm mol⁻¹ K⁻¹ and T(K) = T(°C) + 273.15. All sodium, potassium, ammonium and nitrate salts are soluble in water.
Question 1 (multiple choice · 4.1)
Which of these is a physical change?
- (A) A copper roof slowly turns green when left outdoors for many years
- (B) Milk left in a warm room turns sour and forms lumps
- (C) Iron filings are stirred into sulfur powder at room temperature
- (D) Vinegar is poured onto baking soda and the mixture fizzes
Answer and explanation
Answer: (C). Stirring two solids together makes a mixture; a magnet could still pull the iron out. No new substance forms.
- (A) A new green solid forms on the copper: a new substance.
- (B) Souring makes new substances, which is why the taste and texture change.
- (D) The fizzing is a new gas (carbon dioxide) made from two substances that were not gases.
If you missed this: read “Physical changes” and “Evidence of a chemical change” in the Topic 4.1 study guide.
Question 2 (multiple choice · 4.2)
Magnesium hydroxide, Mg(OH)₂, is insoluble in water. Solid magnesium hydroxide is added to excess dilute nitric acid, HNO₃, and it dissolves. Which is the net ionic equation?
- (A) Mg(OH)₂(s) + 2H⁺(aq) → Mg²⁺(aq) + 2H₂O(l)
- (B) H⁺(aq) + OH⁻(aq) → H₂O(l)
- (C) Mg(OH)₂(s) + 2HNO₃(aq) → Mg(NO₃)₂(aq) + 2H₂O(l)
- (D) Mg²⁺(aq) + 2NO₃⁻(aq) → Mg(NO₃)₂(aq)
Answer and explanation
Answer: (A). Nitric acid is a strong acid, so it is written as H⁺(aq) and NO₃⁻(aq). The solid stays as one formula. Nitrate, unchanged on both sides, is the spectator. Check: 1 Mg, 2 O and 4 H on each side; charge +2 on each side.
- (B) splits the solid into ions; that fits a soluble hydroxide such as NaOH.
- (C) is the molecular equation. It still contains the spectator nitrate.
- (D) shows only spectators; Mg(NO₃)₂(aq) is just its ions in water.
If you missed this: read “Three ways to write a reaction in solution” and Worked example 3 in the Topic 4.2 study guide.
Question 3 (multiple choice · 4.3)
Nitrogen and hydrogen react: N₂(g) + 3H₂(g) → 2NH₃(g). A “before” box holds 5 N₂ molecules and 9 H₂ molecules. The reaction goes to completion. Which “after” box is consistent with the equation?
- (A) 6 NH₃ molecules and 2 N₂ molecules
- (B) 10 NH₃ molecules
- (C) 6 NH₃ molecules only
- (D) 6 NH₃ molecules and 2 H₂ molecules
Answer and explanation
Answer: (A). The 9 H₂ react with 3 N₂ (ratio 3 : 1) to make 6 NH₃. Two N₂ are left over, unchanged. Atom check: before, 10 N and 18 H; after, 6 + 4 = 10 N and 18 H.
- (B) assumes all 5 N₂ react: 30 H atoms after, from 18 before.
- (C) loses the leftover N₂: only 6 N atoms after, from 10 before.
- (D) swaps the excess reactant: 22 H atoms but only 6 N atoms after.
If you missed this: read “The rules a consistent diagram follows” and Worked example 1 in the Topic 4.3 study guide.
Question 4 (multiple choice · 4.4)
Solid sodium chloride can be melted (a physical change). The molten salt can then be electrolysed into sodium metal and chlorine gas (a chemical change). Which statement best explains the difference?
- (A) Electrolysis breaks ionic bonding and forms new metallic and covalent bonds; melting leaves the same Na⁺ and Cl⁻ ions, only less ordered.
- (B) Melting does not affect the attractions between the ions at all; only electrolysis does.
- (C) Electrolysis needs more energy, and any process that needs a lot of energy is chemical.
- (D) Melting can be reversed by cooling, and any reversible change is physical.
Answer and explanation
Answer: (A). Liquid NaCl still contains Na⁺ and Cl⁻ in a 1 : 1 ratio: still sodium chloride. Electrolysis makes Na (metallic bonding) and Cl₂ (a covalent bond): new substances.
- (B) Melting does overcome some ionic attractions, so the ions can move.
- (C) Energy size is a clue; the test is whether new substances form.
- (D) Reversibility is not the test either; many reactions can be reversed.
If you missed this: read “Two kinds of attraction, two kinds of change” and “When the line is blurred” in the Topic 4.4 study guide.
Question 5 (multiple choice · 4.5)
In a blast furnace, iron(III) oxide is reduced by carbon monoxide: Fe₂O₃(s) + 3CO(g) → 2Fe(l) + 3CO₂(g). What mass of iron can be made from 40.0 g of Fe₂O₃, with excess CO?
- (A) 14.0 g
- (B) 28.0 g
- (C) 35.0 g
- (D) 80.0 g
Answer and explanation
Answer: (B).
- M(Fe₂O₃) = 2(55.85) + 3(16.00) = 159.70 g mol⁻¹, so n = 40.0 ÷ 159.70 = 0.2505 mol.
- Mole ratio: 2 mol Fe per 1 mol Fe₂O₃, so n(Fe) = 0.5009 mol.
- Mass: 0.5009 × 55.85 = 27.98 g, which is 28.0 g.
- (A) uses a 1 : 1 ratio and misses the 2 in front of Fe.
- (C) uses a wrong molar mass, 127.70 g mol⁻¹, with only one oxygen in the formula.
- (D) treats the coefficient as a mass ratio (40.0 × 2): more iron than oxide, which is impossible.
If you missed this: read “Why a balanced equation lets you predict amounts” and Worked example 1 in the Topic 4.5 study guide.
Question 6 (short answer · 4.5)
A student adds 0.270 g of aluminium foil to 20.0 mL of 1.20 M hydrochloric acid at 25.0 °C and 1.00 atm:
2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)
(a) Identify the limiting reactant. Show your reasoning. (b) Calculate the volume of hydrogen gas that forms. (c) Calculate the mass of the excess reactant that is left over.
Worked answer
(a) Moles of Al: 0.270 ÷ 26.98 = 0.010007 mol. Moles of HCl: 1.20 × 0.0200 L = 0.0240 mol. Divide by the coefficients: Al 0.010007 ÷ 2 = 0.00500; HCl 0.0240 ÷ 6 = 0.00400. The smaller value is for HCl, so HCl is limiting (all the Al would need 0.0300 mol).
(b) n(H₂) = 0.0240 mol HCl × (3 mol H₂ / 6 mol HCl) = 0.0120 mol. V = nRT / P = (0.0120)(0.08206)(298.15) ÷ 1.00 = 0.294 L (294 mL).
(c) Al used = 0.0240 × (2 / 6) = 0.00800 mol. Al left = 0.010007 − 0.00800 = 0.002007 mol, which is 0.002007 × 26.98 = 0.0542 g of aluminium.
Common slips: calling Al limiting because it has fewer moles (predicts 0.367 L); using 25.0 instead of 298.15 K.
If you missed this: read “Limiting reactant, excess reactant and yield” and Worked example 2 in the Topic 4.5 study guide.
Question 7 (multiple choice · 4.6)
A student dissolves 0.212 g of pure sodium carbonate in water and titrates it with hydrochloric acid:
Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)
The equivalence point is reached after 20.00 mL of acid. What is the concentration of the hydrochloric acid?
- (A) 0.0500 M
- (B) 0.1000 M
- (C) 0.2000 M
- (D) 0.000200 M
Answer and explanation
Answer: (C). n(Na₂CO₃) = 0.212 ÷ 105.99 = 0.002000 mol, so n(HCl) = 2 × 0.002000 = 0.004000 mol. Concentration = 0.004000 ÷ 0.02000 L = 0.2000 M.
- (A) divides by 2 instead of multiplying.
- (B) assumes a 1 : 1 ratio.
- (D) divides by 20.00 mL instead of 0.02000 L.
If you missed this: read “Finding the equivalence point from amounts” and Worked example 1 in the Topic 4.6 study guide.
Question 8 (multiple choice · 4.7)
Which of these is an acid-base reaction?
- (A) 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)
- (B) NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)
- (C) BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)
- (D) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
Answer and explanation
Answer: (B). No oxidation number changes. Water passes a proton to ammonia: acid-base.
- (A) makes a hydroxide, but Na goes from 0 to +1 and H from +1 to 0: redox.
- (C) Ba²⁺ and SO₄²⁻ form an insoluble solid: precipitation.
- (D) is combustion, a kind of redox.
If you missed this: read “A decision path you can use every time” in the Topic 4.7 study guide.
Question 9 (multiple choice · 4.7)
In which species does nitrogen have an oxidation number of +3?
- (A) NO₂
- (B) N₂O
- (C) NaNO₂
- (D) NH₄⁺
Answer and explanation
Answer: (C). In NaNO₂: +1 + N + 2(−2) = 0, so N = +3.
- (A) NO₂: N = +4. The subscript is not the oxidation number.
- (B) N₂O: 2N + (−2) = 0, so each N is +1.
- (D) NH₄⁺: N + 4 = +1, so N = −3 (the sum equals the ion’s charge).
If you missed this: read “Oxidation numbers: bookkeeping for electrons” in the Topic 4.7 study guide.
Question 10 (multiple choice · 4.8)
HCOOH(aq) + CN⁻(aq) → HCOO⁻(aq) + HCN(aq)
Which two species form a conjugate acid-base pair?
- (A) HCOOH and HCN
- (B) HCOOH and HCOO⁻
- (C) CN⁻ and HCOO⁻
- (D) HCOOH and CN⁻
Answer and explanation
Answer: (B). HCOOH donates a proton and becomes HCOO⁻: one H⁺ and a charge of 1 apart. The other pair is CN⁻ and HCN.
- (A) Both are acids, from different pairs.
- (C) Both are bases.
- (D) These are the reacting acid and base, not a pair.
If you missed this: read “Conjugate acid-base pairs” and Worked example 1 in the Topic 4.8 study guide.
Question 11 (short answer · 4.9)
Copper metal dissolves in dilute nitric acid. The products are Cu²⁺ ions and nitrogen monoxide gas, NO.
(a) Give the oxidation number of nitrogen in NO₃⁻ and in NO. (b) Write the balanced oxidation and reduction half-reactions in acidic solution. (c) Combine them into the balanced net ionic equation, and show that it conserves charge.
Worked answer
(a) NO₃⁻: N + 3(−2) = −1, so N = +5. NO: N = +2. Nitrogen gains 3 electrons: nitrate is reduced.
(b) Oxidation: Cu(s) → Cu²⁺(aq) + 2e⁻. Reduction: balance O with 2H₂O, H with 4H⁺, then charge (+3 left, 0 right) with 3e⁻: NO₃⁻(aq) + 4H⁺(aq) + 3e⁻ → NO(g) + 2H₂O(l).
(c) Multiply the oxidation by 3 and the reduction by 2 (6e⁻ each), add, and cancel the electrons:
3Cu(s) + 2NO₃⁻(aq) + 8H⁺(aq) → 3Cu²⁺(aq) + 2NO(g) + 4H₂O(l)
Charge: left 2(−1) + 8(+1) = +6; right 3(+2) = +6. Atoms: 3 Cu, 2 N, 6 O and 8 H on each side.
Common slips: electrons left in the final equation; using 5 electrons (2 + 3) instead of 6.
If you missed this: read “The half-reaction method” and Worked example 1 in the Topic 4.9 study guide.
Your next step
Tick each question you got wrong. A short answer counts as missed if any part was wrong.
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 4.1 Introduction for Reactions | 1 | Topic 4.1 study guide |
| 4.2 Net Ionic Equations | 2 | Topic 4.2 study guide |
| 4.3 Representations of Reactions | 3 | Topic 4.3 study guide |
| 4.4 Physical and Chemical Changes | 4 | Topic 4.4 study guide |
| 4.5 Stoichiometry | 5, 6 | Topic 4.5 study guide |
| 4.6 Introduction to Titration | 7 | Topic 4.6 study guide |
| 4.7 Types of Chemical Reactions | 8, 9 | Topic 4.7 study guide |
| 4.8 Introduction to Acid-Base Reactions | 10 | Topic 4.8 study guide |
| 4.9 Oxidation-Reduction (Redox) Reactions | 11 | Topic 4.9 study guide |
How to use your result
- Look at topics, not the total.
- Read every explanation. If you guessed, count the question as missed.
- Fix gaps in order. Topics 4.1 to 4.4 cover what a reaction is and how to write it; 4.5 and 4.6 count moles; 4.7 to 4.9 sort reactions by what is transferred. Start with the earliest topic you missed in each group.
- Then practise with each topic’s practice questions, then the Unit 4 mixed review, where questions combine topics.