Mintnote Science
Edexcel IGCSE Chemistry
⚗ Core Practicals Revision
IGCSE Chemistry
Core Practicals
Visual step-by-step methods and active recall flashcards for all 12 Edexcel IGCSE Chemistry core practicals.
12 Core Practicals
Visual Methods
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Core Practical 01
Solubility
Spec ref: 1.3 · Investigate solubility at different temperatures
Purpose: To find how the solubility of a solid (e.g. potassium nitrate) changes with temperature and construct a solubility curve.
Step-by-Step Method
- 1Weigh a known mass of solid (e.g. 10 g KNO₃) into a boiling tube
- 2Add 10 cm³ of distilled water to the boiling tube
- 3Place the boiling tube in a hot water bath and heat, stirring until all solid dissolves
- 4Remove from water bath and allow to cool slowly while stirring continuously
- 5Record the temperature at which the first crystals appear — this is the saturation temperature⚠ Watch closely — crystals appear suddenly
- 6Repeat with different masses of solid (e.g. 15 g, 20 g, 25 g) to get results at different temperatures
- 7Calculate solubility (g per 100 g water) and plot solubility against temperature
Key Information
Variables
| Independent | Temperature of solution |
| Dependent | Mass of solid that dissolves |
| Control | Volume of water, type of solid |
Key Result
Solubility of most solids increases with temperature. The solubility curve shows this relationship.
Calculating Solubility
If 12 g KNO₃ dissolves in 20 g water at 50°C:
Solubility = (12 ÷ 20) × 100 = 60 g per 100 g water
Solubility = (12 ÷ 20) × 100 = 60 g per 100 g water
Key Definition
Saturated solution: a solution in which no more solute can dissolve at that temperature
Always stir continuously while cooling — this gives a more accurate crystallisation point
Never heat above the required temperature — it changes the equilibrium and gives inaccurate results
Active Recall Flashcards
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Q — Define
What is a saturated solution?
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Answer
A solution in which no more solute can dissolve at that temperature
Q — Recall
How does solubility of most solids change as temperature increases?
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Answer
Solubility increases as temperature increases
Q — Method
Why must you stir continuously while the solution cools?
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Answer
To give a more accurate crystallisation point — ensures even cooling and prevents missing the moment crystals first appear
Q — Apply
A saturated KNO₃ solution at 80°C is cooled to 20°C. What happens?
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Answer
Crystals form — the solution contains more dissolved solid than can remain dissolved at 20°C
Core Practical 02
Chromatography
Spec ref: 1.7 · Investigate mixtures using paper chromatography
Purpose: To separate and identify the components of a mixture of dyes or inks using paper chromatography and calculate Rf values.
Step-by-Step Method
- 1Draw a pencil baseline 2 cm from the bottom of the chromatography paperMust be pencil — ink would run with the solvent
- 2Place small spots of each sample on the baseline using a capillary tube. Allow to dry between applications
- 3Pour solvent (e.g. water or ethanol) into the beaker to a depth of 1 cm⚠ Solvent level MUST be below the baseline
- 4Place paper in the solvent — do not let spots touch the solvent directly
- 5Wait until the solvent front is near the top of the paper
- 6Remove paper immediately and mark the solvent front with a pencil⚠ Do this instantly — solvent evaporates quickly
- 7Measure distances and calculate Rf values for each spot
Key Information
Rf Formula
Rf = distance travelled by spot ÷ distance travelled by solvent front
Rf is always between 0 and 1
Interpreting Results
| Pure substance | Produces one spot |
| Mixture | Produces more than one spot |
| Identification | Match Rf to reference values |
Mark the solvent front in pencil immediately on removal — if you wait, it evaporates and Rf cannot be calculated
Rf values only compare if the SAME solvent and SAME paper are used
If spots overlap: make the initial spots smaller / use less concentrated samples. Overlapping spots cannot be identified.
Active Recall Flashcards
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Q — Recall
Why is the baseline drawn in pencil, not pen?
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Answer
Pencil is insoluble in the solvent — ink would run and interfere with the results
Q — Recall
Why must the solvent be below the baseline?
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Answer
If above the baseline, the spots would dissolve directly into the solvent instead of being carried up the paper
Q — Calculate
Spot travels 4.8 cm. Solvent front travels 8.0 cm. Calculate Rf.
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Answer
Rf = 4.8 ÷ 8.0 = 0.60
Q — Interpret
A chromatogram shows 4 spots for a food dye. What does this tell you?
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Answer
The food dye is a mixture of 4 components / is not a pure substance
Q — Explain
Why must you mark the solvent front immediately?
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Answer
The solvent evaporates quickly after removal — the front disappears and Rf cannot be calculated
Core Practical 03
Combustion & Reduction
Spec ref: 2.22–2.24 · Combustion of Mg and reduction of copper oxide
Purpose: Part A — investigate the combustion of magnesium and confirm mass changes. Part B — reduce copper oxide using hydrogen gas.
Part A — Combustion of Magnesium
- 1Weigh a clean crucible + lid. Record mass.
- 2Add coiled magnesium ribbon. Weigh crucible + lid + Mg.
- 3Heat strongly with lid on, lifting the lid briefly at intervals⚠ Lid prevents MgO smoke escaping — but must allow air in
- 4Continue until no further reaction — all Mg burned to white MgO
- 5Cool completely, then reweigh. Mass should have increased.
Part B — Reduction of Copper Oxide
- 1Place black copper oxide powder in a hard glass tube
- 2Pass hydrogen gas through the tube to displace all airTest escaping H₂ — squeaky pop confirms no air
- 3Heat strongly while continuing to pass hydrogen gas
- 4Observe: black CuO → pink/orange copper metal; water condenses at cool end
- 5Keep H₂ flowing while cooling — prevents re-oxidation of copper⚠ Critical step — do not turn off H₂ until cool
Key Equations
2Mg + O₂ → 2MgO (combustion — mass increases)
CuO + H₂ → Cu + H₂O (reduction — black → pink/orange)
In calculations: mass of oxygen gained = (final mass) − (mass of Mg before). Always set out three masses first.
The most common error: forgetting to lift the lid allows MgO smoke to escape → loss of mass → wrong results
Safety: never look directly at burning magnesium — the intense white light can permanently damage eyes. Use eye protection at all times.
Active Recall Flashcards
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Q — Explain
Why is the lid lifted briefly during Mg combustion?
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Answer
To let air in (so Mg can react with O₂) but prevent MgO smoke escaping — escaping smoke would cause a loss of mass
Q — Recall
What observation shows copper oxide has been reduced?
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Answer
The black solid turns pink/orange — copper metal has formed
Q — Explain
Why must H₂ keep flowing while the tube cools?
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Answer
If H₂ stops, air enters the hot tube and re-oxidises the copper back to black copper oxide
Q — Identify
In CuO + H₂ → Cu + H₂O, which is the reducing agent?
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Answer
H₂ (hydrogen) is the reducing agent — it removes oxygen from CuO (and is itself oxidised to H₂O)
Q — Safety
State one hazard of using H₂ and one precaution.
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Answer
Hazard: H₂ is highly flammable/explosive with air. Precaution: expel all air before heating; test escaping gas with a lit splint first
Core Practical 04
Acid-Metal Reactions
Spec ref: 2.19–2.20 · Compare reactions of metals with dilute acids
Purpose: To investigate and compare how different metals react with dilute acids, measuring the rate of hydrogen gas production.
Step-by-Step Method
- 1Measure 20 cm³ of dilute HCl into a conical flask using a measuring cylinder
- 2Weigh a measured mass of metal (e.g. 0.5 g). Record mass.
- 3Connect the flask to a gas syringe or inverted measuring cylinder over water
- 4Add metal to acid and immediately start the stopwatch
- 5Record volume of H₂ collected every 30 seconds for 5 minutes
- 6Repeat with Mg, Zn, and Fe — keeping all other variables the same
- 7Plot volume of gas vs time for each metal on the same axes
Key Information
Expected Rate Order
Mg > Zn > Fe > Cu (no reaction)
Variables
| Independent | Type of metal |
| Dependent | Volume of H₂ per unit time |
| Control | Volume of acid, concentration, mass of metal, temperature |
Alternative Method — Mass Loss
Place flask on a balance, add metal and acid. Record mass every 30 s — mass decreases as H₂ escapes. Faster mass loss = more reactive metal.
Gas Test
Lit splint → squeaky pop confirms hydrogen
When asked to “justify using data” — always quote specific numbers with units from the table
Copper produces NO gas — it is below hydrogen in the reactivity series
Active Recall Flashcards
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Q — Recall
What gas is produced when metals react with dilute acids? How do you test for it?
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Answer
Hydrogen gas — test with a lit splint → squeaky pop
Q — Explain
Why does copper produce no gas with dilute HCl?
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Answer
Copper is below hydrogen in the reactivity series — it is not reactive enough to displace hydrogen from the acid
Q — Recall
State three variables that must be controlled.
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Answer
Volume of acid · concentration of acid · mass of metal · temperature (any three)
Q — Apply
Mg stops producing gas at 120 s but Zn continues. Why?
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Answer
Mg is completely used up at 120 s — no reactant remains. Zn is less reactive so still reacting at 120 s
Core Practical 05
Preparation of Copper Sulfate
Spec ref: 2.35 · Prepare hydrated copper sulfate crystals
Purpose: To prepare pure, dry hydrated copper sulfate crystals (CuSO₄·5H₂O) by reacting excess copper oxide with dilute sulfuric acid.
Step-by-Step Method
- 1Warm 25 cm³ of dilute sulfuric acid gently in a beaker (do not boil)
- 2Add excess black copper oxide powder in small amounts, stirring after each additionStop adding CuO when no more dissolves
- 3Filter the hot mixture through filter paper in a funnel — excess CuO is residue; blue CuSO₄ solution passes through as filtrateCuO excess ensures all acid is used up
- 4Heat the filtrate gently to evaporate — stop when crystals begin to form on a glass rod dipped in the solution
- 5Leave to cool slowly — blue crystals of CuSO₄·5H₂O form
- 6Filter crystals, pat dry with filter paper, leave in warm place to complete drying
Key Information
Equation
CuO + H₂SO₄ → CuSO₄ + H₂O
Why Excess CuO?
Ensures all the acid is used up — any remaining acid would contaminate the crystals. Excess CuO is removed by filtration.
Observations
• Black CuO dissolves → solution turns blue
• Blue crystals form on cooling
• Blue crystals form on cooling
Never boil the filtrate — overheating drives off water of crystallisation, giving white anhydrous CuSO₄ (not the desired product)
“Filter to remove impurities” = 0 marks. Say specifically: excess copper oxide is removed by filtration
Active Recall Flashcards
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Q — Explain
Why is excess CuO added rather than the exact amount?
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Answer
To ensure all the acid is used up — remaining acid would contaminate the crystals. Excess CuO is removed by filtration.
Q — Method
How is excess copper oxide removed from the solution?
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Answer
Filtration — excess CuO stays on filter paper; copper sulfate solution passes through as filtrate
Q — Predict
What happens if the filtrate is heated too strongly?
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Answer
White anhydrous CuSO₄ powder forms — not the desired blue hydrated crystals. Water of crystallisation is driven off.
Q — Recall
What are the two observations when CuO is added to warm H₂SO₄?
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Answer
Black CuO dissolves · solution turns blue
Core Practical 06
Preparation of Lead Sulfate
Spec ref: 2.38 · Prepare lead sulfate by precipitation
Purpose: To prepare a sample of dry lead sulfate using a precipitation reaction — used for insoluble salts that cannot be made by crystallisation.
Step-by-Step Method
- 1Measure 25 cm³ of lead(II) nitrate solution into a beaker
- 2Add 25 cm³ of sodium sulfate solution — white precipitate forms immediately
- 3Stir to ensure complete precipitation
- 4Filter — lead sulfate precipitate collected on filter paper as residue
- 5Wash residue with distilled water (three times) to remove soluble NaNO₃ impurities⚠ Washing is essential — don’t skip this step
- 6Dry at low temperature in an oven or leave in a warm place
Key Information
Full Equation
Pb(NO₃)₂(aq) + Na₂SO₄(aq) → PbSO₄(s) + 2NaNO₃(aq)
Ionic Equation
Pb²⁺(aq) + SO₄²⁻(aq) → PbSO₄(s)
Why precipitation?
Lead sulfate is insoluble — it cannot be made by crystallisation. Precipitation is used for all insoluble salts.
The ionic equation includes only Pb²⁺ and SO₄²⁻ — spectator ions (Na⁺ and NO₃⁻) are omitted
Washing the precipitate is essential — NaNO₃ would otherwise contaminate the lead sulfate product
Active Recall Flashcards
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Q — Recall
Name the two solutions mixed to make lead sulfate.
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Answer
Lead(II) nitrate solution + sodium sulfate solution
Q — Explain
Why is the precipitate washed with distilled water?
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Answer
To remove soluble impurities — sodium nitrate (NaNO₃) — which would otherwise contaminate the product
Q — Write
Write the ionic equation for the precipitation reaction.
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Answer
Pb²⁺(aq) + SO₄²⁻(aq) → PbSO₄(s)
(Na⁺ and NO₃⁻ are spectator ions — omit them)
(Na⁺ and NO₃⁻ are spectator ions — omit them)
Q — Explain
Why is precipitation used rather than crystallisation?
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Answer
Lead sulfate is insoluble — it cannot be dissolved and crystallised. Precipitation forms it directly as a solid.
Core Practical 07
Energy Changes — Exothermic & Endothermic
Spec ref: 3.1–3.5 · Investigate temperature changes in reactions
Purpose: To measure temperature changes in chemical reactions and classify them as exothermic (temperature rises) or endothermic (temperature falls).
Method — Solution Reactions
- 1Measure a fixed volume of solution into a polystyrene cupPolystyrene cup = good insulator → less heat loss
- 2Record the initial temperature (T₁) with a thermometer
- 3Add the second reagent (solution or solid) and stir immediately
- 4Record the maximum temperature (T₂ for exothermic) or minimum temperature (T₂ for endothermic)
- 5Calculate ΔT = T₂ − T₁
- 6Calculate energy change using q = mcΔT
Key Information
Key Formula
q = m × c × ΔT
q = energy (J) · m = mass of solution (g)
c = 4.2 J g⁻¹ °C⁻¹ · ΔT = temperature change (°C)
c = 4.2 J g⁻¹ °C⁻¹ · ΔT = temperature change (°C)
For mixed solutions: m = total volume
If mixing 25 cm³ + 25 cm³, use m = 50 g (assuming density = 1 g/cm³)
m = TOTAL volume of both solutions combined — the most common error is using only one volume
Exothermic: ΔT is positive (temperature rises). Endothermic: ΔT is negative (temperature falls)
Temperature-Time Graph Shape
Exothermic: flat line → sharp rise when reagent added → gradual fall as heat lost to surroundings.
Endothermic: flat line → sharp fall → gradual return toward room temperature.
Endothermic: flat line → sharp fall → gradual return toward room temperature.
Active Recall Flashcards
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Q — Recall
What is the formula for calculating energy change?
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Answer
q = m × c × ΔT
c for water = 4.2 J g⁻¹ °C⁻¹
c for water = 4.2 J g⁻¹ °C⁻¹
Q — Calculate
25 cm³ NaOH + 25 cm³ HCl. Temperature rises from 19.5°C to 26.3°C. Calculate q.
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Answer
m = 50 g · ΔT = 6.8°C
q = 50 × 4.2 × 6.8 = 1428 J
q = 50 × 4.2 × 6.8 = 1428 J
Q — Explain
Why is a polystyrene cup used instead of a glass beaker?
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Answer
Polystyrene is a good insulator — reduces heat loss to surroundings, giving more accurate temperature readings
Q — Classify
Classify: combustion of methane, photosynthesis, dissolving NH₄NO₃
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Answer
Combustion: exothermic · Photosynthesis: endothermic · Dissolving NH₄NO₃: endothermic
Q — Improve
Suggest two reasons why actual energy is lower than theoretical value in combustion experiments.
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Answer
Heat lost to surroundings / air · Incomplete combustion of fuel · Not all heat transferred to water
Core Practical 08
Electrolysis
Spec ref: 1.46–1.60C · Investigate electrolysis of solutions
Purpose: To investigate the products of electrolysis of different solutions and identify gases produced at each electrode.
Step-by-Step Method
- 1Set up the electrolysis cell with two carbon electrodes connected to a d.c. power supply
- 2Fill the cell with the chosen electrolyte (e.g. dilute H₂SO₄, CuSO₄ solution, or brine)
- 3Place inverted test tubes filled with the electrolyte over each electrode to collect gas
- 4Switch on the power supply and wait until a reasonable volume of gas has been collected
- 5Test each gas to identify it — see key information panel
- 6Observe any colour changes in the solution during electrolysis
- 7Repeat with different electrolytes and record all observations
Products Summary
| Electrolyte | Cathode | Anode |
| Dilute H₂SO₄ | H₂ | O₂ |
| CuSO₄ (carbon) | Cu | O₂ |
| CuSO₄ (copper) | Cu | Cu dissolves |
| Brine (conc.) | H₂ | Cl₂ |
Gas Tests
Cathode H₂Lit splint → squeaky pop
Anode O₂Glowing splint relights
Anode Cl₂Bleaches damp litmus paper
CuSO₄ ratioH₂:O₂ volume = 2:1
Copper Purification (Spec 1.56–1.57)
| Anode | Impure copper — dissolves: Cu → Cu²⁺ + 2e⁻ |
| Cathode | Pure copper — deposited: Cu²⁺ + 2e⁻ → Cu |
| Solution | Stays blue — Cu²⁺ replaced as fast as removed |
| Anode sludge | Impurities (Ag, Au) fall off — do not dissolve |
With copper electrodes in CuSO₄: solution stays blue (anode dissolves, replacing Cu²⁺). With carbon electrodes: solution fades as Cu²⁺ is used up
Copper purification is separately examined — know both electrode equations and why the solution stays blue
Active Recall Flashcards
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Q — Recall
What is the test for chlorine gas? State the observation.
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Answer
Hold damp litmus paper near the gas — it is bleached / turns white
Q — Explain
Why does CuSO₄ solution stay blue when copper electrodes are used?
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Answer
The copper anode dissolves, releasing Cu²⁺ into solution at the same rate Cu²⁺ is deposited at the cathode — concentration stays constant
Q — Recall
Why are carbon electrodes used instead of metals?
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Answer
Carbon is inert — it does not react with the electrolyte or dissolve, giving uncontaminated results
Q — Explain
Why is twice the volume of gas collected at the cathode vs anode in dilute H₂SO₄?
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Answer
From the equations, H₂:O₂ molar ratio = 2:1 — twice as many moles of H₂ are produced per mole of O₂
Q — Recall
Write the electrode equations for electrolysis of brine.
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Answer
Cathode: 2H⁺ + 2e⁻ → H₂
Anode: 2Cl⁻ → Cl₂ + 2e⁻
Anode: 2Cl⁻ → Cl₂ + 2e⁻
Q — Purification
In copper purification, why does the CuSO₄ solution stay blue throughout?
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Answer
The impure copper anode dissolves, releasing Cu²⁺ at the same rate Cu²⁺ is deposited at the cathode — concentration stays constant
Core Practical 09
Rusting
Spec ref: 2.28 · Investigate conditions necessary for iron to rust
Purpose: To determine which conditions (air/oxygen and water) are necessary for iron to rust by using a controlled experiment with five test tubes.
Step-by-Step Method
- 1Set up 3 test tubes, each containing a clean iron nail
- 2Tube 1: add a little water (air + water present — control)
- 3Tube 2: add silica gel to absorb all moisture (dry air only — no water)Seal the top of the tube
- 4Tube 3: add boiled (deaerated) water + seal with oil layer (water only — no air)Boiling removes dissolved oxygen
- 5Leave all tubes for one week at room temperature
- 6Observe each tube and record whether rusting has occurred
Expected Results
| Tube 1 (air + water) | Rusts — both conditions present |
| Tube 2 (dry air only) | No rust — no water |
| Tube 3 (water, no air) | No rust — no oxygen |
Conclusion
Both oxygen AND water are needed for iron to rust.
Rust Prevention Methods
Painting · Greasing · Galvanising (Zn coating)
Sacrificial protection · Stainless steel · Electroplating
Sacrificial protection · Stainless steel · Electroplating
For galvanising — zinc corrodes sacrificially even when scratched because Zn is MORE reactive than Fe
Active Recall Flashcards
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Q — Recall
State the two conditions required for iron to rust.
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Answer
Oxygen (from air) AND water — both must be present
Q — Explain
Why does the nail in boiled water + oil NOT rust?
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Answer
Boiling removed dissolved oxygen; oil prevents oxygen re-entering. No oxygen present → one condition missing → no rusting
Q — Explain
Why does galvanised iron not rust even when scratched?
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Answer
Zinc is more reactive than iron — it corrodes sacrificially in preference to the iron underneath
Q — Explain
Why does stainless steel not rust even though it contains iron?
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Answer
Chromium forms a thin, impermeable oxide layer on the surface — prevents O₂ and water reaching the iron underneath
Core Practical 10
Rates of Reaction
Spec ref: 3.9–3.16 · Investigate factors affecting the rate of reaction
Purpose: To investigate how concentration, temperature, surface area and catalysts affect the rate of the reaction between marble chips and hydrochloric acid.
Method A — Gas Volume (CaCO₃ + HCl)
- 1Measure 30 cm³ of dilute HCl into a conical flask
- 2Weigh 2 g of marble chips. Add to flask and immediately seal with a bung connected to a gas syringe
- 3Start stopwatch when marble chips added
- 4Record volume of CO₂ in the gas syringe every 30 seconds
- 5Repeat changing ONE variable at a time (concentration, temperature, chip size)
- 6Plot volume vs time — steeper initial gradient = faster rate
Method B — Mass Loss
Place flask on a balance. Add CaCO₃ and HCl — leave flask open. Record mass every 30 s. Mass decreases as CO₂ escapes. Faster mass loss = faster rate. Plot mass vs time.
Method C — Disappearing Cross (Na₂S₂O₃ + HCl)
Reaction: Na₂S₂O₃ + 2HCl → 2NaCl + SO₂ + S + H₂O
Place flask on paper with a cross. Add Na₂S₂O₃ and HCl. Time until cross disappears through cloudy sulfur precipitate.
Rate = 1 ÷ time | Shorter time = faster rate
Place flask on paper with a cross. Add Na₂S₂O₃ and HCl. Time until cross disappears through cloudy sulfur precipitate.
Rate = 1 ÷ time | Shorter time = faster rate
Disappearing cross: subjective endpoint — different observers judge differently. State this as a limitation.
Factors & Explanations
| Concentration ↑ | Rate ↑ — more frequent collisions |
| Temperature ↑ | Rate ↑ — more frequent AND more energetic collisions; more particles exceed Ea |
| Surface area ↑ | Rate ↑ — more particles exposed, more collisions |
| Catalyst added | Rate ↑ — lower activation energy pathway |
Equation
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
Temperature: must state TWO things — more frequent collisions AND more particles exceed activation energy. One point = 1 mark, not 2.
All curves level off at the same final volume if the same mass of marble chips is used — final volume depends on amount of reactant, not rate
Active Recall Flashcards
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Q — Explain (2 marks)
Explain why increasing temperature increases the rate of reaction.
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Answer
Particles move faster → more frequent collisions. More particles have energy ≥ activation energy → greater proportion of successful collisions
Q — Explain
Why does marble powder react faster than marble chips?
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Answer
Powder has a much larger surface area — more particles exposed, more collisions per second with acid
Q — Interpret
Three curves reach the same final volume but at different rates. What does this mean?
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Answer
Same amount of reactant was used in all three (same final volume). Only the rate differed — the limiting reactant determines final volume, not rate
Q — Recall
State one limitation of the disappearing cross method.
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Answer
Subjective — different observers judge different endpoints. The point at which the cross “disappears” varies between people.
Q — Explain
How does a catalyst increase the rate of reaction?
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Answer
Provides an alternative reaction pathway with a lower activation energy — more particles have sufficient energy to react
Q — Method
Describe the disappearing cross method. What is measured and how is rate calculated?
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Answer
Na₂S₂O₃ + HCl → cloudy sulfur. Time until cross disappears through flask. Rate = 1 ÷ time. Shorter time = faster rate.
Core Practical 11
Catalytic Decomposition
Spec ref: 3.13 · Investigate catalytic decomposition of hydrogen peroxide
Purpose: To compare the effectiveness of different catalysts on the rate of decomposition of hydrogen peroxide (H₂O₂).
Step-by-Step Method
- 1Measure 20 cm³ of hydrogen peroxide into a conical flask connected to a gas syringe
- 2Weigh 0.5 g of catalyst (e.g. MnO₂). Record mass.
- 3Add catalyst to the flask, seal immediately and start the stopwatch
- 4Record volume of O₂ every 30 seconds for 5 minutes
- 5Repeat with different catalysts (Fe₂O₃, raw potato/liver) keeping mass constant
- 6Recover and reweigh catalyst at end — confirm mass unchanged
Key Information
Equation
2H₂O₂ → 2H₂O + O₂
Gas Test
Hold a glowing splint to the gas — it relights → confirms oxygen
After the experiment
Recover and reweigh catalyst — mass is unchanged (catalyst is not consumed)
Both curves reach the same final volume — because the same amount of H₂O₂ was used. The catalyst only affects the RATE, not the total amount of O₂ produced.
Catalyst mass stays the same at end — this is a guaranteed exam question. Students who write “it decreases” are confusing catalyst with reactant.
Active Recall Flashcards
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Q — Recall
Write the balanced equation for the decomposition of hydrogen peroxide.
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Answer
2H₂O₂ → 2H₂O + O₂
Q — Recall
What happens to the catalyst mass at the end of the experiment?
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Answer
Mass stays the same / unchanged — a catalyst is not consumed in the reaction; it is regenerated
Q — Explain
Why do both catalyst curves reach the same final volume of O₂?
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Answer
Same volume of H₂O₂ was used — the total O₂ depends on the amount of H₂O₂, not the catalyst. Catalyst only changes rate.
Q — Explain
Why does raw potato cause H₂O₂ to decompose?
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Answer
Raw potato contains the enzyme catalase — a biological catalyst that decomposes H₂O₂ by lowering the activation energy
Core Practical 12
Preparation of Ethyl Ethanoate
Spec ref: 4.43C · Prepare a sample of an ester
Purpose: To prepare a sample of ethyl ethanoate by reacting ethanol with ethanoic acid in the presence of a concentrated sulfuric acid catalyst.
Step-by-Step Method
- 1Mix ethanol and ethanoic acid in a round-bottomed flask or test tube
- 2Add a few drops of concentrated sulfuric acid as catalyst. Handle with extreme care.
- 3Heat gently using a water bath — do NOT use a Bunsen burner⚠ Both reagents are flammable — no naked flame
- 4Pour the mixture carefully into sodium carbonate solution in a separating funnel
- 5Shake gently — CO₂ is produced (effervescence). Release gas pressure regularly.
- 6Allow layers to separate — ester floats as upper layer; aqueous layer below
- 7Run off lower aqueous layer and collect the ester (upper layer)
- 8Identify the product by its distinctive fruity smell
Key Information
Equation (reversible!)
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O
Why Na₂CO₃ solution is used
Neutralises unreacted ethanoic acid → removes it
Neutralises H₂SO₄ catalyst → removes it
Dissolves ethanol → removes it
Leaves insoluble ester as upper layer
Neutralises H₂SO₄ catalyst → removes it
Dissolves ethanol → removes it
Leaves insoluble ester as upper layer
Ester layer
Upper layer — ester is less dense than water
Key Observation
When mixture added to Na₂CO₃ solution: effervescence observed — CO₂ gas produced as Na₂CO₃ neutralises the acid.
Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂
Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂
The reversible sign ⇌ is required in the equation — writing → instead loses a mark
Never heat with a Bunsen burner — ethanol and ethanoic acid are both flammable
Active Recall Flashcards
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Q — Recall
Name the catalyst used and explain why a water bath is used.
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Answer
Catalyst: concentrated H₂SO₄. Water bath used because ethanol and ethanoic acid are flammable — no naked flame
Q — Explain
Why is sodium carbonate solution used to wash the product?
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Answer
Neutralises and removes unreacted ethanoic acid, the H₂SO₄ catalyst, and dissolves ethanol — leaving pure ester
Q — Explain
Which layer contains the ester and why?
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Answer
Upper layer — ethyl ethanoate is less dense than water and immiscible (does not mix) with the aqueous layer
Q — Recall
What observation identifies the product as an ester?
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Answer
A distinctive fruity smell — esters are volatile compounds with characteristic fruity odours
Q — Write
Write the equation for preparation of ethyl ethanoate. What sign is essential?
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Answer
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O
The reversible sign ⇌ is essential — writing → loses a mark
The reversible sign ⇌ is essential — writing → loses a mark
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