IGCSE Chemistry Answers #

1. Answer: Carbon

2. Answer: Mole

3. Answer: Avogadro’s constant

4. Answer: Relative atomic mass

5. Answer: $\text{I}_2 + \text{Cl}_2 \rightarrow 2\text{ICl}$

6. Question: Element Y has two isotopes ⁶⁹Y and ⁷¹Y in ratio 3:2. Calculate relative atomic mass.

   Formula: $A_r = \frac{(\% \text{ of isotope A} \times \text{mass of isotope A}) + (\% \text{ of isotope B} \times \text{mass of isotope B})}{100}$

   Answer:

   First convert ratio to percentages: 3:2 = 60%:40%

   $A_r = \frac{(60 \times 69) + (40 \times 71)}{100}$

   = $\frac{4140 + 2840}{100} = \frac{6980}{100} = 69.8$

   Element Y is Gallium

7. Answer:

   1. Bromide ($\text{Br}^-$)
   2. Potassium bromide

   3. Question: Calculate mass of calcium ions in 250 cm³ of water.

      Formula: $\frac{\text{mass in sample}}{\text{volume of sample}} = \frac{\text{mass needed}}{\text{new volume}}$

      Mass in 1000 cm³ = 44 mg

      Mass in 250 cm³ = $\frac{44 \times 250}{1000} = 11$ mg

   4. $\text{F} + \text{e}^- \rightarrow \text{F}^-$

8. Question: Complete table to calculate relative formula mass of UO₂F₂.

   Answer:

   Element	Number of atoms	Relative atomic mass	Total mass
   Uranium	1	238	238
   Oxygen	2	16	32
   Fluorine	2	19	38

   Relative formula mass = 238 + 32 + 38 = 308

9. 	Electrons	Neutrons	Protons
   $\text{^{17}O}$	8	9	8
   $\text{^{59}Co}^{2+}$	25	32	27

10. Answer:

    1. $100.00 – 100.00 = 0.00$%
    2. Oxygen

11. Question: Calculate maximum mass of ammonium nitrate from 820g calcium nitrate.

    Formula: Number of moles = $\frac{\text{mass}}{\text{M}_r}$

    Answer:

    (a) Number of moles = $\frac{820}{164} = 5.0$ mol

    (b) From equation: 1 mol Ca(NO₃)₂ → 2 mol NH₄NO₃

    So: 5.0 mol Ca(NO₃)₂ → 10.0 mol NH₄NO₃

12. Question: Calculate Mr of NH₄NO₃ and maximum mass produced.

    Formula: Mass = number of moles × $M_r$

    Answer:

    (c) $M_r$ of NH₄NO₃ = 14 + 4 + 14 + 48 = 80

    (d) Maximum mass = 10.0 × 80 = 800 g

13. Question: Determine value of x in FeSO₄•xH₂O with mass 278g.

    Answer:

    Mass of FeSO₄ = 56 + 32 + 64 = 152 g

    Mass of H₂O = 18x g

    Total mass = 152 + 18x = 278

    18x = 278 – 152 = 126

    x = 7

14. Answer: $\text{K} + \text{Cl}_2 \rightarrow 2\text{KCl}$

15. Answer: $(\text{NH}_4)_2\text{SO}_4 + 2\text{NaOH} \rightarrow 2\text{NH}_3 + 2\text{H}_2\text{O} + \text{Na}_2\text{SO}_4$

16. Question: Calculate number of molecules in 15g of ethane.

    Formula: Number of molecules = moles × $6.02 \times 10^{23}$

    Answer:

    $M_r$ of C₂H₆ = (2 × 12) + (6 × 1) = 30

    Number of moles = $\frac{15}{30} = 0.5$ mol

    Number of molecules = $0.5 \times 6.02 \times 10^{23} = 3.01 \times 10^{23}$

17. Question: Calculate percentage by mass of nitrogen in NH₄NO₃.

    Formula: Percentage = $\frac{\text{mass of element}}{\text{M}_r} \times 100$

    Answer:

    $M_r$ of NH₄NO₃ = 14 + 4 + 14 + 48 = 80

    Mass of nitrogen = 14 + 14 = 28

    Percentage = $\frac{28}{80} \times 100 = 35$%

18. Answer:

    1. Unsaturated means contains double or triple bonds between carbon atoms
    2. $\text{C}_3\text{H}_8 \rightarrow \text{C}_3\text{H}_6 + \text{H}_2$
    3. Propene

19. Answer: $\text{Cu} + 2\text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{SO}_2 + 2\text{H}_2\text{O}$

20. Answer:

    1. Name	Methanal	Ethanal	Propanal	Butanal
       Molecular formula	CH₂O	C₂H₄O	C₃H₆O	C₄H₈O

    2. $\text{C}_n\text{H}_{2n}\text{O}$

21. Answer: $2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$

22. Answer: $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$

23. Answer: $\text{H}_2\text{SO}_4 + 2\text{NH}_3 \rightarrow (\text{NH}_4)_2\text{SO}_4$

24. Answer:

    1. NH₄⁺
    2. PH₄I

25. Answer: $2\text{Al} + \text{Fe}_2\text{O}_3 \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe}$

26. Answer: $\text{Ca}_3\text{P}_2 + 6\text{H}_2\text{O} \rightarrow 3\text{Ca(OH)}_2 + 2\text{PH}_3$

27. Answer: $\text{P}_4\text{O}_{10} + 12\text{NaOH} \rightarrow 4\text{Na}_3\text{PO}_4 + 6\text{H}_2\text{O}$

28. Answer:

    Fe + Sn²⁺ → Fe²⁺ + Sn

    Sn + Cu²⁺ → no reaction

29. Answer: $2\text{K}_{(s)} + \text{Br}_2_{(l)} \rightarrow 2\text{KBr}_{(s)}$

30. Answer:

    1. Excess of oxygen means more oxygen than needed for complete combustion
    2. Volume of oxygen used = 150 – 25 = 125 cm³

    3. Volume ratio	1	:	5	:	3

    4. C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O
       Formula = C₅H₁₂

31. Answer: $2\text{Cu(NO}_3)_2 \rightarrow 2\text{CuO} + 4\text{NO}_2 + \text{O}_2$

32. Answer:

    1. $6\text{Li} + \text{N}_2 \rightarrow 2\text{Li}_3\text{N}$

33. Answer:

    1. (i) Chloride (Cl⁻)
       (ii) Na⁺ and NO₃⁻
       (iii) Mass = $\frac{6 \times 250}{1000} = 1.5$ mg

34. Type of atom	Number	Relative atomic mass	Total
    Copper	1	64	64
    Nitrogen	4	14	56
    Hydrogen	16	1	16
    Oxygen	2	16	32

    Relative molecular mass = 64 + 56 + 16 + 32 = 168

35. Question: Calculate mass of magnesium needed to produce 1.24g of MgF₂.

    Formula: Use simple proportion

    Answer:

    Given: 2.40g Mg makes 6.20g MgF₂

    For 1.24g MgF₂: Mass of Mg = $\frac{2.40 \times 1.24}{6.20} = 0.48$ g

36. Answer: $6\text{Li} + \text{N}_2 \rightarrow 2\text{Li}_3\text{N}$

37. Question: Find x in Na₂SO₄•xH₂O from given data.

    Formula: Find mole ratio of H₂O to Na₂SO₄

    Answer:

    Moles of Na₂SO₄ = $\frac{0.71}{142} = 0.005$ mol

    Mass of H₂O lost = 1.61 – 0.71 = 0.90 g

    Moles of H₂O = $\frac{0.90}{18} = 0.05$ mol

    Ratio = $\frac{0.05}{0.005} = 10$, so x = 10

38. Answer: $\text{AgNO}_3_{(aq)} + \text{NaI}_{(aq)} \rightarrow \text{AgI}_{(s)} + \text{NaNO}_3_{(aq)}$

39. Answer: $2\text{Zn(NO}_3)_2 \cdot 6\text{H}_2\text{O} \rightarrow 2\text{ZnO} + 4\text{NO}_2 + \text{O}_2 + 12\text{H}_2\text{O}$

40. Answer: $2\text{Na} + \text{F}_2 \rightarrow 2\text{NaF}$

41. Question: Calculate volume of CO₂ from 12.6g NaHCO₃.

    Formula: Volume = moles × 24 (at room temperature)

    Answer:

    $M_r$ of NaHCO₃ = 23 + 1 + 12 + 48 = 84

    Moles of NaHCO₃ = $\frac{12.6}{84} = 0.15$ mol

    From equation: 2 NaHCO₃ → 1 CO₂

    So 0.15 mol NaHCO₃ makes 0.075 mol CO₂

    Volume = 0.075 × 24 = 1.8 dm³