Key Takeaways

1. Chemical Equations & Balancing

  • Balancing Equations:
    • Ensure the number of atoms of each element is equal on both sides.
    • Example:
      2CuCO3(s)+C(s)→2Cu(s)+3CO2(g)2CuCO3​(s)+C(s)→2Cu(s)+3CO2​(g)
      • Tip: Balance metals first, then non-metals, and finally oxygen/hydrogen.
  • State Symbols:
    • Use (s) = solid, (l) = liquid, (g) = gas, (aq) = aqueous.
    • Example: In 2NH3(aq)+H2SO4(aq)→(NH4)2SO4(aq)2NH3​(aq)+H2​SO4​(aq)→(NH4​)2​SO4​(aq), ammonia is dissolved (aq).

2. Moles & Formula Mass

  • Formula Mass (Mᵣ):
    • Sum of atomic masses in a compound.
    • Example:
      CuCO3=63.5 (Cu)+12 (C)+3×16 (O)=123.5 g/molCuCO3​=63.5(Cu)+12(C)+3×16(O)=123.5g/mol
    • Common Mistake: Forgetting to multiply oxygen atoms by 3.
  • Mole Calculations:
    • Use n=mass (g)Mrn=Mr​mass (g)​.
    • Example: 1 mole of CO₂ = 12 + (2 × 16) = 44 g.

3. Practical Work: Making Salts

  • Steps for Soluble Salts (e.g., CuSO₄):
    1. React excess metal oxide (CuO) with acid (H₂SO₄).
    2. Filter to remove unreacted CuO.
    3. Evaporate filtrate to crystallisation point.
    4. Dry crystals using filter paper.
    • Tip: Excess reactant ensures all acid is neutralised.
  • Precipitation Reactions:
    • Example:
      Na2SO4(aq)+Ca(NO3)2(aq)→CaSO4(s)+2NaNO3(aq)Na2​SO4​(aq)+Ca(NO3​)2​(aq)→CaSO4​(s)+2NaNO3​(aq)
    • Filter, wash, and dry the precipitate.

4. Indicators & pH

  • Red Cabbage Indicator:
    • Colours: Red (acid), Purple (neutral), Blue/Green (alkali).
    • Calibration: Test with known acids (vinegar) and alkalis (oven cleaner).
  • Universal Indicator:
    • pH Scale:pH0–6 (Acid)7 (Neutral)8–14 (Alkali)ColourRed → YellowGreenBlue → Purple

5. Reversible Reactions & Equilibrium

  • Dynamic Equilibrium:
    • Forward and reverse reactions occur at equal rates.
    • Example:
      CuSO4(s)+5H2O(l)⇌CuSO4⋅5H2O(s)CuSO4​(s)+5H2​O(l)⇌CuSO4​⋅5H2​O(s)
    • Le Chatelier’s Principle: Increasing temperature shifts equilibrium to favour endothermic direction.

6. Catalysts

  • Role: Speed up reactions without being used up.
    • Example: MnO₂ in H2O2→H2O+O2H2​O2​→H2​O+O2​.
    • Effect: Lowers activation energy, increasing successful collisions.

7. Calculations

  • Heat Energy (q = mcΔT):
    • Example: Mixing 25 cm³ HCl + 25 cm³ NaOH (total mass = 50 g):
      q=50 g×4.2 J/g°C×7 °C=1470 Jq=50g×4.2J/g°C×7°C=1470J.
  • Titration Calculations:
    • Use C1V1=C2V2C1​V1​=C2​V2​ for 1:1 ratios.
    • Example:
      Moles of NaOH=0.1 mol/dm3×23.51000=0.00235 molMoles of NaOH=0.1mol/dm3×100023.5​=0.00235mol.
  • Percentage Yield:
    • Percentage Yield=Actual YieldTheoretical Yield×100Percentage Yield=Theoretical YieldActual Yield​×100.

8. Ionic Equations

  • Simplified Neutralisation:
    H+(aq)+OH−(aq)→H2O(l)H+(aq)+OH−(aq)→H2​O(l)
    • Tip: Cancel spectator ions (e.g., Na⁺, Cl⁻).

9. Tests for Ions

  • Sulfate Ions:
    • Add HCl (to remove CO₃²⁻), then BaCl₂ → white precipitate (BaSO₄).
  • Chloride Ions:
    • Add HNO₃, then AgNO₃ → white precipitate (AgCl).

10. Rate of Reaction

  • Factors:
    • Temperature, concentration, surface area, catalysts.
    • Graphs: Steeper slope = faster rate.
    • Example: Powdered CaCO₃ reacts faster than lumps.

Key Tips for Success

  1. Show Working: Even simple steps gain method marks.
  2. Check Units: Convert cm³ → dm³ by dividing by 1000.
  3. Balance Equations First: Avoid errors in mole ratios.
  4. Use Excess Reactant: Ensures all limiting reactant is used.
  5. Revise Chemical Tests: Memorise sulfate/chloride tests.

50 GCSE Chemistry Questions

Section A: Chemical Equations & Balancing

  1. Balance the equation:
    CuCO3(s)+C(s)→Cu(s)+CO2(g)CuCO3​(s)+C(s)→Cu(s)+CO2​(g)
  2. State the products when calcium oxide reacts with water.
  3. Write the balanced equation for the reaction between zinc and sulfuric acid.

Section B: Moles & Formula Mass

  1. Calculate the formula mass of ammonium sulfate, (NH4)2SO4(NH4​)2​SO4​.
  2. How many moles are in 9.22 g of lead iodide (Mr=461Mr​=461)?
  3. Calculate the mass of carbon dioxide produced when 1 mole of copper carbonate decomposes.

Section C: Practical Work (Salts)

  1. Describe the steps to make pure dry copper sulfate crystals from copper oxide and sulfuric acid.
  2. Explain why excess copper oxide is used in the reaction with sulfuric acid.
  3. What is a precipitate? Give an example reaction.

Section D: Indicators & pH

  1. How would you calibrate a red cabbage indicator?
  2. State the colour of universal indicator at pH 9.
  3. Why does ammonia solution turn red litmus blue?

Section E: Reversible Reactions & Equilibrium

  1. Define dynamic equilibrium.
  2. Explain how increasing temperature affects the equilibrium in 2SO2(g)+O2(g)⇌2SO3(g)2SO2​(g)+O2​(g)⇌2SO3​(g).

Section F: Catalysts

  1. What is the role of manganese(IV) oxide in the decomposition of hydrogen peroxide?
  2. How does a catalyst affect activation energy?

Section G: Calculations

  1. Calculate the heat energy released when 50 g of solution heats up by 7°C (c=4.2 J/g°Cc=4.2J/g°C).
  2. A student neutralises 25 cm³ of 0.1 mol/dm³ HCl with NaOH. Calculate the moles of HCl used.
  3. If 0.95 g of calcium sulfate is made (theoretical yield = 1.36 g), calculate the percentage yield.

Section H: Ionic Equations

  1. Write the ionic equation for the neutralisation of HCl with NaOH.

Section I: Tests for Ions

  1. Describe the test for sulfate ions.
  2. How would you confirm the presence of chloride ions in a solution?

Section J: Rate of Reaction

  1. Explain why powdered marble reacts faster than lumps with HCl.
  2. Sketch a graph showing how concentration affects reaction rate.

Section K: Titrations

  1. In a titration, 23.5 cm³ of 0.1 mol/dm³ H2SO4H2​SO4​ neutralises 25 cm³ NaOH. Calculate the NaOH concentration.

Section L: Bond Energy

  1. Use bond energies to calculate the enthalpy change for 2H2O→2H2+O22H2​O→2H2​+O2​.

Answers

  1. Balanced equation:
    2CuCO3(s)+C(s)→2Cu(s)+3CO2(g)2CuCO3​(s)+C(s)→2Cu(s)+3CO2​(g)
  2. Products: Calcium hydroxide:
    CaO(s)+H2O(l)→Ca(OH)2(s)CaO(s)+H2​O(l)→Ca(OH)2​(s)
  3. Zinc reaction:
    Zn(s)+H2SO4(aq)→ZnSO4(aq)+H2(g)Zn(s)+H2​SO4​(aq)→ZnSO4​(aq)+H2​(g)
  4. Formula mass of (NH4)2SO4(NH4​)2​SO4​:
    (2×18)+32+(4×16)=132 g/mol(2×18)+32+(4×16)=132g/mol
  5. Moles of PbI₂:
    n=9.22461=0.02 moln=4619.22​=0.02mol
  6. Mass of CO₂:
    From 1 mole of CuCO3CuCO3​, 3 moles of CO₂ are produced.
    Mass = 3×44=132 g3×44=132g.
  7. Steps for CuSO₄ crystals:
    • Add excess CuO to H₂SO₄.
    • Filter to remove unreacted CuO.
    • Evaporate filtrate to crystallisation point.
    • Dry crystals using filter paper.
  8. Excess CuO: Ensures all acid is neutralised.
  9. Precipitate: Insoluble solid formed in a reaction, e.g., CaSO4(s)CaSO4​(s).
  10. Calibrate indicator: Test with vinegar (acid), oven cleaner (alkali), and water (neutral).
  11. pH 9 colour: Blue (universal indicator).
  12. Ammonia solution: Contains OH−OH− ions, which turn litmus blue.
  13. Dynamic equilibrium: Forward and reverse reactions occur at equal rates in a closed system.
  14. Temperature effect: Increasing temperature shifts equilibrium to favour endothermic direction (left for exothermic reactions).
  15. MnO₂ role: Catalyses decomposition of H2O2H2​O2​ into H2OH2​O and O2O2​.
  16. Activation energy: Catalyst provides an alternative pathway with lower activation energy.
  17. Heat energy:
    q=50×4.2×7=1470 Jq=50×4.2×7=1470J
  18. Moles of HCl:
    n=0.1×251000=0.0025 moln=0.1×100025​=0.0025mol
  19. Percentage yield:
    0.951.36×100=69.9%1.360.95​×100=69.9%
  20. Ionic equation:
    H+(aq)+OH−(aq)→H2O(l)H+(aq)+OH−(aq)→H2​O(l)
  21. Sulfate test: Add HCl (to remove carbonates), then BaCl₂ → white precipitate (BaSO4BaSO4​).
  22. Chloride test: Add AgNO3AgNO3​ → white precipitate (AgClAgCl).
  23. Surface area: Powder increases surface area → more frequent collisions.
  24. Graph: Steeper curve for higher concentration.
  25. NaOH concentration:
    Moles of H2SO4=0.1×23.51000=0.00235 molH2​SO4​=0.1×100023.5​=0.00235mol.
    Moles of NaOH = 2×0.00235=0.0047 mol2×0.00235=0.0047mol.
    Concentration = 0.00470.025=0.188 mol/dm30.0250.0047​=0.188mol/dm3.
  26. Bond energy calculation:
  • Energy to break bonds: 4×464=1856 kJ4×464=1856kJ.
  • Energy released forming bonds: 2×436+498=1370 kJ2×436+498=1370kJ.
  • Enthalpy change: 1856−1370=+486 kJ1856−1370=+486kJ (endothermic).

Section M: Data Analysis & Graphs

  1. The table below shows the volume of oxygen produced in the decomposition of hydrogen peroxide with different masses of MnO₂ catalyst.
Time (hours)Volume of O₂ (cm³) with 1g MnO₂Volume of O₂ (cm³) with 2g MnO₂
2495505
10990985

Question: What conclusion can be drawn about the effect of catalyst mass on reaction rate?

  1. A graph shows the loss of mass over time when marble chips react with HCl. Question: Which curve (steep or gradual) represents smaller marble chips? Explain.
  2. Question: Why does the line on a rate vs. concentration graph not pass through all plotted points?
  3. Question: A student’s experiment shows a 35°C temperature rise using 3.5g CaO. Calculate the moles of CaO used (Mr=56Mr​=56).
  4. Question: If 0.0455 moles of Ca(OH)₂ are made (theoretical = 0.0625 mol), calculate the percentage yield.

Section N: Enthalpy & Bond Energy

  1. Question: Calculate the enthalpy change for 2H2O→2H2+O22H2​O→2H2​+O2​ using bond energies:
  • H-H = 436 kJ/mol, O-H = 464 kJ/mol, O=O = 498 kJ/mol.
  1. Question: Burning ethanol releases 1470 kJ. If 0.92g ethanol (Mr=46Mr​=46) is burned, calculate the enthalpy change per mole.
  2. Question: Dissolving 8g NH₄NO₃ in water causes a temperature drop from 19°C to 11°C. Calculate the heat energy absorbed (c=4.2 J/g°Cc=4.2J/g°C).
  3. Question: Define ΔHΔH and explain why experimental values may differ from theoretical values.

Section O: Equilibrium & Le Chatelier

  1. Question: For N2(g)+3H2(g)⇌2NH3(g)N2​(g)+3H2​(g)⇌2NH3​(g) (exothermic), predict the effect of:
    a) Increasing pressure
    b) Adding a catalyst
  2. Question: At 450°C, sulfur trioxide yield decreases. Explain why industries still use this temperature.
  3. Question: If ammonia is removed from the equilibrium mixture, what happens to the yield?
  4. Question: What does the ⇌⇌ symbol indicate about a reaction?

Section P: Titration Calculations

  1. Question: 29.85 cm³ of H₂SO₄ neutralises 40 cm³ of 1.5 mol/dm³ NaOH. Calculate the H₂SO₄ concentration.
  2. Question: In a titration, initial burette reading = 1.15 cm³, final = 40.60 cm³. Calculate the volume of acid used.
  3. Question: A student titrates 25 cm³ Li₂CO₃ with 0.1 mol/dm³ HCl. Average titre = 38.25 cm³. Calculate moles of Li₂CO₃.

Section Q: Percentage Yield & Error

  1. Question: A student makes 0.95g CaSO₄ (theoretical = 1.36g). Calculate percentage yield.
  2. Question: Why might percentage yield exceed 100%?
  3. Question: A reaction produces 36.12g BaSO₄ (theoretical = 58.3g). Calculate percentage yield.

Section R: Practical Methods

  1. Question: Describe how to obtain dry BaSO₄ precipitate from a mixture.
  2. Question: Explain why nitric acid is used instead of sulfuric acid to make CO₂ from CaCO₃.
  3. Question: Why is excess sodium hydroxide used in making Na₂SO₄?

Section S: Rate of Reaction

  1. Question: Sketch a graph showing how temperature affects reaction rate.
  2. Question: Explain why increasing concentration increases reaction rate.

Detailed Answers

  1. Conclusion: Doubling the catalyst mass has minimal effect on final yield but speeds up initial reaction rate.
  2. Answer: Steeper curve = smaller chips (greater surface area increases collision frequency).
  3. Answer: Random errors in measurements cause scatter; the line shows the trend.
  4. Moles of CaO:
    n=3.556=0.0625 moln=563.5​=0.0625mol
  5. Percentage yield:
    0.04550.0625×100=72.8%0.06250.0455​×100=72.8%
  6. Enthalpy change:
  • Bonds broken: 4×464=1856 kJ4×464=1856kJ
  • Bonds formed: 2×436+498=1370 kJ2×436+498=1370kJ
  • ΔH=+486 kJΔH=+486kJ (endothermic).
  1. Enthalpy per mole:
    Moles of ethanol = 0.9246=0.02 mol460.92​=0.02mol
    ΔH=14700.02=73,500 kJ/molΔH=0.021470​=73,500kJ/mol
  2. Heat absorbed:
    Mass = 25g (water) + 8g = 33g
    q=33×4.2×8=1108.8 Jq=33×4.2×8=1108.8J
  3. Definition: ΔHΔH = enthalpy change. Experimental losses (heat, incomplete reactions) cause lower yields.
  4. Equilibrium shifts:
    a) Higher pressure favours NH₃ (fewer gas molecules).
    b) Catalyst speeds up both forward/reverse reactions equally – no shift.
  5. Industrial choice: Lower yield at 450°C is offset by faster reaction rate.
  6. Yield increase: System shifts right to produce more NH₃.
  7. Reversible reaction: Reaction can proceed in both directions.
  8. H₂SO₄ concentration:
    Moles NaOH = 1.5×0.04=0.06 mol1.5×0.04=0.06mol
    Moles H₂SO₄ = 0.06/2=0.03 mol0.06/2=0.03mol
    C=0.030.02985=1.01 mol/dm3C=0.029850.03​=1.01mol/dm3
  9. Volume used:
    40.60−1.15=39.45 cm340.60−1.15=39.45cm3
  10. Moles Li₂CO₃:
    Moles HCl = 0.1×0.03825=0.003825 mol0.1×0.03825=0.003825mol
    Moles Li₂CO₃ = 0.003825/2=0.00191 mol0.003825/2=0.00191mol
  11. Percentage yield:
    0.951.36×100=69.9%1.360.95​×100=69.9%
  12. Yield >100%: Impurities, incomplete drying, or measurement errors.
  13. Percentage yield:
    36.1258.3×100=61.9%58.336.12​×100=61.9%
  14. Dry BaSO₄: Filter, wash with distilled water, dry between filter paper.
  15. Acid choice: Sulfuric acid forms insoluble CaSO₄, stopping the reaction.
  16. Excess NaOH: Ensures all acid is neutralised for pure Na₂SO₄.
  17. Graph: Curve shifts left (steeper) with higher temperature.
  18. Explanation: More particles per unit volume → more frequent collisions.