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Which particles conduct electricity in molten lead (II) bromide and lead metal? Explain melting point differences for ionic and covalent chlorides.

Chemistry

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5. Which particles conduct electricity in:

a) Molten lead (II) bromide
✅ Correct. The particles are ions (Pb2+\text{Pb}^{2+} and Br−\text{Br}^{-}).

b) Lead metal
✅ Correct. The particles are delocalized electrons.

6. The information in the table below relates to the physical properties of the chlorides of certain elements.

a) Explain why melting points of NaCl\text{NaCl} and MgCl2\text{MgCl}_2 are higher than those of SiCl4\text{SiCl}_4, PCl3\text{PCl}_3 and SCl2\text{SCl}_2 chlorides in the table above.
NaCl\text{NaCl} and MgCl2\text{MgCl}_2 are ionic compounds with strong electrostatic forces of attraction between oppositely charged ions in a giant ionic lattice structure. A large amount of energy is required to overcome these strong forces, resulting in high melting points.
SiCl4\text{SiCl}_4, PCl3\text{PCl}_3, and SCl2\text{SCl}_2 are covalent compounds that exist as simple discrete molecules. The forces between these molecules are weak intermolecular forces (van der Waals forces). Only a small amount of energy is required to overcome these weak forces, leading to low melting points.

7. 400 cm3400\text{ cm}^3 of a gas D diffuses through a porous plug in 50 seconds, while 600 cm3600\text{ cm}^3 of oxygen gas diffuses from the same apparatus in 30 seconds. Calculate the relative molecular mass of gas D. (O=16\text{O} = 16)

  1. Calculate the rate of diffusion for gas D and oxygen.

    Rate of diffusion = Volume diffusedTime taken\frac{\text{Volume diffused}}{\text{Time taken}}

    RateD=400 cm350 s=8 cm3/sRate_D = \frac{400\text{ cm}^3}{50\text{ s}} = 8\text{ cm}^3/\text{s} RateO2=600 cm330 s=20 cm3/sRate_{O_2} = \frac{600\text{ cm}^3}{30\text{ s}} = 20\text{ cm}^3/\text{s}
  2. Determine the relative molecular mass of oxygen gas.

    Relative atomic mass of O = 16.
    Relative molecular mass of O2=2×16=32\text{O}_2 = 2 \times 16 = 32.

  3. Apply Graham's Law of Diffusion.

    Graham's Law states that the rate of diffusion of a gas is inversely proportional to the square root of its relative molecular mass.

    RateDRateO2=MO2MD\frac{Rate_D}{Rate_{O_2}} = \sqrt{\frac{M_{O_2}}{M_D}}

    Substitute the known values:

    820=32MD\frac{8}{20} = \sqrt{\frac{32}{M_D}} 0.4=32MD0.4 = \sqrt{\frac{32}{M_D}}
  4. Square both sides of the equation.

    (0.4)2=32MD(0.4)^2 = \frac{32}{M_D} 0.16=32MD0.16 = \frac{32}{M_D}
  5. Solve for MDM_D.

    MD=320.16M_D = \frac{32}{0.16} MD=200\boxed{M_D = 200}

    8. Study the flow chart below and answer the questions the follow.

    a) White precipitate A: Calcium carbonate (CaCO3\text{CaCO}_3)

    b) Solution B: Calcium hydrogencarbonate (Ca(HCO3)2\text{Ca(HCO}_3)_2)

    c) Gas C: Carbon(IV) oxide (CO2\text{CO}_2)

    9. Classify the following processes as either chemical or physical.

    i) Heating Ammonium Chloride: Chemical change

    ii) Obtaining kerosene from crude oil: Physical change

    iii) Souring of milk: Chemical change

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