Final 4-Hour Sprint: Chemistry First Aid Kit

SPRINT 1: Acids, Bases & Salts / QA (Focus: 60 mins)

Tactical Goal: Master the core SOPs (Salt Prep, QA) and the "Explain Why" scoring phrases.

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1. Pasir Ris Secondary School, Paper 2, Question 8

Salt Preparation (Excess Method) Stoichiometry Integration

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📋 Full Question:

(a) A student prepares crystals of copper(II) chloride by adding an excess of copper(II) carbonate to 50.0 cm³ of 2.00 mol/dm³ hydrochloric acid.

(i) Write a balanced chemical equation, including state symbols...

(ii) Explain why the student add an excess of copper(II) carbonate?

(iii) Describe how the student would obtain pure and dry crystals...

(b) (i) Calculate the number of moles of copper(II) chloride...

(ii) Hence, calculate the maximum mass...

💡 Why this question? It's the perfect all-in-one. It tests equation writing, the #1 "Explain Why" rationale, the full crystallisation SOP, and links directly to a standard stoichiometry calculation. Mastering this one question covers 50% of this topic.

Answer Template:

(a)(i) Write a balanced chemical equation...

CuCO₃(s) + 2HCl(aq) → CuCl₂(aq) + H₂O(l) + CO₂(g)

Marks awarded for: Correct formulae, Correct balancing and state symbols

(a)(ii) Explain why the student add an excess...

"To ensure that all the hydrochloric acid is completely reacted. If any acid remains, it will be an impurity in the final salt solution, making the crystals impure."

(a)(iii) Describe how the student would obtain pure and dry crystals...

"Heat the filtrate in an evaporating dish until it becomes saturated."
"Allow the saturated solution to cool slowly at room temperature for crystals to form."
"Filter to collect the crystals, wash them with a small amount of cold distilled water, and dry them between pieces of filter paper."

(b)(i) Calculate the number of moles of copper(II) chloride...

Step 1 (Moles of Known):

Moles of HCl = C × V = 2.00 mol/dm³ × (50.0 / 1000) dm³ = 0.100 mol

Step 2 & 3 (Mole Ratio & Moles of Unknown):

From the equation, mole ratio of HCl : CuCl₂ is 2 : 1

Therefore, moles of CuCl₂ = 0.100 / 2 = 0.0500 mol

(b)(ii) Hence, calculate the maximum mass...

Step 4 (Convert to Answer):

Mᵣ of CuCl₂ = 64 + 2(35.5) = 135

Mass of CuCl₂ = moles × Mᵣ = 0.0500 mol × 135 g/mol = 6.75 g

2. Bedok South Secondary School, Paper 2, Section C, Question 9

Qualitative Analysis (Flowchart Deduction)

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📋 Full Question:

A green solution, A, is divided into three portions...

(i) ...aqueous sodium hydroxide is added... A green precipitate, B, is formed, which is insoluble in excess.

(ii) ...dilute nitric acid is added, followed by aqueous barium nitrate. A white precipitate, C, is formed.

(iii) ...dilute nitric acid is added, followed by aqueous silver nitrate. No visible reaction occurs.

(a) Identify the cation and anion present in solution A.

(b) Identify substances B and C.

(c) Write the ionic equation for the formation of precipitate C.

💡 Why this question? It is a classic, high-stakes QA flowchart. It forces you to activate your memory of precipitate colours (Fe²⁺), anion tests (SO₄²⁻ present, Cl⁻ absent), and link them in a logical chain.

Answer Template:

(a) Identify the cation and anion present in solution A.

Detective Work:

Clue 1: "A green solution" → Suspect is Fe²⁺
Clue 2: "green precipitate, insoluble in excess" → Confirms cation is Fe²⁺
Clue 3: "white precipitate with barium nitrate" → Confirms anion is SO₄²⁻
Clue 4: "No reaction with silver nitrate" → Confirms Cl⁻ is absent

Cation: Iron(II) ion or Fe²⁺
Anion: Sulfate ion or SO₄²⁻

(b) Identify substances B and C.

Substance B: Iron(II) hydroxide or Fe(OH)₂
Substance C: Barium sulfate or BaSO₄

(c) Write the ionic equation for the formation of precipitate C.

Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

Marks for: Correct ions, Correct formula and state symbols

SPRINT 2: Stoichiometry & The Mole Concept (Focus: 60 mins)

Tactical Goal: Solidify the calculation SOPs for both standard and limiting reactant problems.

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1. Broadrick Secondary School, Paper 3, Question 8

Stoichiometry (Limiting Reactant)

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📋 Full Question:

Iron(II) carbonate reacts with dilute hydrochloric acid... FeCO₃ + 2HCl → ...

2.32 g of iron(II) carbonate was reacted with 50 cm³ of 0.10 mol/dm³ of dilute hydrochloric acid.

(a) Calculate the number of moles of iron(II) carbonate present.

(b) Calculate the number of moles of dilute hydrochloric acid present.

(c) (i) Identify the limiting reactant... Show your reasoning.

(ii) Calculate the volume of carbon dioxide produced...

💡 Why this question? This is the quintessential Limiting Reactant problem. It forces you to execute the full SOP: calculate moles of both reactants, compare them using the mole ratio, identify the L.R., and then use the L.R. for the final product calculation.

Answer Template:

(a) Calculate the number of moles of iron(II) carbonate...

Mᵣ of FeCO₃ = 56 + 12 + 3(16) = 116

Moles of FeCO₃ = mass / Mᵣ = 2.32 g / 116 g/mol = 0.0200 mol

(b) Calculate the number of moles of dilute hydrochloric acid...

Moles of HCl = C × V = 0.10 mol/dm³ × (50 / 1000) dm³ = 0.00500 mol

(c)(i) Identify the limiting reactant...

Step 1 (Mole Ratio): From the equation FeCO₃ + 2HCl → ..., the mole ratio FeCO₃ : HCl is 1 : 2

Step 2 (Compare): To react with 0.0200 mol of FeCO₃, moles of HCl needed = 2 × 0.0200 = 0.0400 mol

Step 3 (Judge): We only have 0.00500 mol of HCl, which is less than the 0.0400 mol needed.

Conclusion: Hydrochloric acid (HCl) is the limiting reactant.

(c)(ii) Calculate the volume of carbon dioxide produced...

Step 1 (Mole Ratio): From the equation, mole ratio HCl : CO₂ is 2 : 1

Step 2 (Moles of Product): Moles of CO₂ produced = Moles of HCl / 2 = 0.00500 / 2 = 0.00250 mol

Step 3 (Convert to Answer): Volume of CO₂ = moles × 24 = 0.00250 mol × 24 dm³/mol = 0.0600 dm³

2. Meridian Secondary School, Paper 2, Section C, Question 8(d,e,f)

Stoichiometry (Multi-step with unit conversion)

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📋 Full Question:

...Each antacid tablet contains 600 mg of calcium carbonate.

(d) Calculate the relative formula mass of calcium carbonate.

(e) ...Calculate the number of moles of hydrochloric acid that would have reacted with 1.5 antacid tablets.

(f) Calculate the volume of gas produced...

💡 Why this question? It tests your ability to handle non-standard units (mg), work with real-world numbers (1.5 tablets), and execute a multi-step calculation chain. It's a great test of procedural accuracy.

Key Skills:

Unit conversion: mg to g
Working with fractional amounts (1.5 tablets)
Multi-step stoichiometry chain
Gas volume calculations at STP

SPRINT 3: Chemical Bonding & Structure (Focus: 60 mins)

Tactical Goal: Master the "Golden Templates" for explaining properties and the detailed rules for dot-and-cross diagrams.

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1. Chua Chu Kang Secondary School, Paper 2, Question 10

Ionic vs Covalent Comparison

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📋 Full Question:

Chlorine is a non-metallic element.

(a) State the type of bond... Explain your answer.

(b) Draw 'dot and cross' diagrams for sodium with chlorine and hydrogen with chlorine.

(c) Describe the differences in melting points... Explain, in terms of bonding, why this is so.

(d) Explain why both substances... can conduct electricity when dissolved in water.

💡 Why this question? It's the ultimate comparison question. It forces you to construct diagrams for both ionic and covalent substances and then immediately apply the corresponding "Golden Templates" to explain the differences in their properties (melting point, conductivity). It covers 80% of this topic in a single question.

Key Concepts:

Ionic bonding: Metal + Non-metal → Electron transfer → High melting point, conducts when molten/aqueous
Covalent bonding: Non-metal + Non-metal → Electron sharing → Lower melting point, does not conduct (unless acidic)
Golden Templates: Always mention "strong ionic/covalent bonds" and "large amount of energy"

2. Pasir Ris Secondary School, Paper 2, Question 3

Ionic Properties Focus

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📋 Full Question:

Magnesium chloride has a high melting point and is a conductor of electricity when molten.

(i) Name the type of chemical bonding...

(ii) Draw a 'dot and cross' diagram for magnesium chloride...

(iii) Explain why magnesium chloride is a conductor of electricity when molten but not in the solid state.

(b) ...Explain the difference in their boiling points (vs Chlorine)...

💡 Why this question? It is a laser-focused test of the ionic bonding toolkit: drawing the diagram (with correct charges/brackets for a Group 2 metal) and explaining conductivity using the "mobile ions" keyword.

Critical Keywords:

For conductivity: "mobile ions" / "free to move" / "carry charge"
For melting point: "strong ionic bonds" / "large amount of energy to overcome"
For diagram: Always include charges, brackets for polyatomic ions, 2+ for Group 2 metals

SPRINT 4: Redox & Reactivity + Final Review (Focus: 45 mins)

Tactical Goal: Activate your knowledge of the reactivity series and the core redox definitions.

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1. Deyi Secondary School, Paper 2, Section B, Question 10(b)

Redox & Reactivity (Deducing Reactivity Series)

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Three experiments were carried out to find the order of reactivity of three metals. The metals used were zinc, tin and unknown metal Q.

(i) Suggest the name of metal Q.

(iii) List the three metals in the order of increasing reactivity.

(iv) Write a chemical equation for the reaction involving zinc and tin(II) chloride.

💡 Why this question? It is the best single question for testing your ability to apply the displacement rule and deduce the reactivity series from experimental evidence, which is a core C-level skill.

Key Concepts:

Displacement Rule: A more reactive metal can displace a less reactive metal from its salt solution
Reactivity Series: Use experimental results to arrange metals in order
Evidence Analysis: If metal X displaces metal Y, then X is more reactive than Y

2. Pasir Ris Secondary School, Paper 2, Question 4

Redox (Identifying Agents)

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📋 Full Question:

In each of these redox equations, identify the oxidising agent and the reducing agent.

(a) Zn + Fe²⁺ → Zn²⁺ + Fe

(b) 2Cu₂O + C → 4Cu + CO₂

💡 Why this question? It's a direct, no-frills test of your ability to apply the OIL RIG principle and the "Agent" model. It's a quick and effective way to activate this fundamental concept.

Answer Template:

(a) Zn + Fe²⁺ → Zn²⁺ + Fe

Analysis:

Zn → Zn²⁺ (loses electrons → oxidation → Zn is reducing agent)
Fe²⁺ → Fe (gains electrons → reduction → Fe²⁺ is oxidising agent)

Oxidising agent: Fe²⁺ or iron(II) ion
Reducing agent: Zn or zinc

(b) 2Cu₂O + C → 4Cu + CO₂

Analysis:

Cu₂O → Cu (copper reduced from +1 to 0 → Cu₂O is oxidising agent)
C → CO₂ (carbon oxidized from 0 to +4 → C is reducing agent)

Oxidising agent: Cu₂O or copper(I) oxide
Reducing agent: C or carbon

🔑 Remember: OIL RIG

Oxidation Is Loss (of electrons)
Reduction Is Gain (of electrons)

The substance that gets oxidized is the reducing agent
The substance that gets reduced is the oxidising agent

🧪 Final 4-Hour Sprint: The Golden Questions "First Aid Kit"

SPRINT 1: Acids, Bases & Salts / QA (Focus: 60 mins)

1. Pasir Ris Secondary School, Paper 2, Question 8

📋 Full Question:

Answer Template:

2. Bedok South Secondary School, Paper 2, Section C, Question 9

📋 Full Question:

Answer Template:

SPRINT 2: Stoichiometry & The Mole Concept (Focus: 60 mins)

1. Broadrick Secondary School, Paper 3, Question 8

📋 Full Question:

Answer Template:

2. Meridian Secondary School, Paper 2, Section C, Question 8(d,e,f)

📋 Full Question:

Key Skills:

SPRINT 3: Chemical Bonding & Structure (Focus: 60 mins)

1. Chua Chu Kang Secondary School, Paper 2, Question 10

📋 Full Question:

Key Concepts:

2. Pasir Ris Secondary School, Paper 2, Question 3

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Critical Keywords:

SPRINT 4: Redox & Reactivity + Final Review (Focus: 45 mins)

1. Deyi Secondary School, Paper 2, Section B, Question 10(b)

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Key Concepts:

2. Pasir Ris Secondary School, Paper 2, Question 4

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Answer Template: