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I) The compound is CH3-CH(CH3)-CH3.

Chemistry

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  1. a)

i) The compound is CH3−CH(CH3)−CH3\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_3.
Name: 2-methylpropane

ii) The compound is CH3−CH2−C≡C−CH3\text{CH}_3-\text{CH}_2-\text{C}\equiv\text{C}-\text{CH}_3.
Name: Pent-2-yne

b) To distinguish between 2-methylpropane (an alkane) and pent-2-yne (an alkyne):
Add bromine water (reddish-brown).

  • Pent-2-yne (alkyne) will decolorize the bromine water because it is unsaturated and undergoes an addition reaction.
  • 2-methylpropane (alkane) will not decolorize the bromine water because it is saturated and does not readily undergo addition reactions.
  1. a) Gay-Lussac's Law of combining volumes states that when gases react, they do so in volumes that bear a simple whole-number ratio to one another and to the volumes of the gaseous products, provided that all volumes are measured at the same temperature and pressure.

b) Step 1: Write the balanced chemical equation for the combustion of carbon (II) oxide.

2CO(g)+O2(g)→2CO2(g)\text{2CO}(g) + \text{O}_2(g) \rightarrow \text{2CO}_2(g)
  1. Use the mole ratio (volume ratio for gases) from the balanced equation.

    From the equation, 2 volumes of CO produce 2 volumes of CO2\text{CO}_2.

  2. Calculate the volume of gas formed.

    Since the ratio is 2:22:2, 20 cm320 \text{ cm}^3 of CO will produce 20 cm320 \text{ cm}^3 of CO2\text{CO}_2.
    The volume of gas formed is 20 cm3\boxed{\text{20 cm}^3}.

    1. a) The products formed when a candle burns in air are carbon dioxide and water.

    b)

    i) One observation made in the combustion tube is that the silvery-white magnesium ribbon turns into a white solid.

    ii) Gas Y is Hydrogen.

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