Alkanes

Carbon and hydrogen have very similar electronegativities, so C-H bonds are almost non-polar and C-C bonds are non-polar; there is no partial charge for a nucleophile or electrophile to attack. Both bonds are also strong (about 410 and 350 kJ mol⁻¹), so a lot of energy is needed to break them. The bonds are strong, not weak, and they are not highly polar. Alkanes are saturated: they contain only single bonds, so there is no C=C. Each carbon in an alkane is sp³ hybridised (tetrahedral, 109.5°), and all four electrons are used in sigma bonds; sp² describes alkene carbons.

Hexane is saturated and has no C=C, so there is no electron-rich pi bond for bromine to add to; in the dark the orange colour stays. (Substitution by bromine needs UV light and is slow.) Decolourising bromine water is the positive test for an alkene; thinking that alkanes do this is a classic error. Turning purple is not a bromine observation at all; purple is the colour of manganate(VII). No precipitate forms: any bromoalkane would be a liquid, and there is no reaction here anyway.

Carbon: 10 = 8 + 2, so X has two carbons. Hydrogen: 22 = 18 + 4, so X has four hydrogens: X is C₂H₄, ethene. Cracking always produces at least one alkene, because there are not enough hydrogens for every product to be saturated. C₂H₆ would need 24 H on the left. C₂H₂ would leave 2 H unaccounted for. Two CH₄ molecules need 8 H, so the equation would not balance.

In the converter, toxic carbon monoxide is oxidised and nitrogen monoxide is reduced at the same time, giving CO₂ and N₂, both non-toxic. Unburnt hydrocarbons are also oxidised to CO₂ and H₂O. The reverse reaction would produce pollutants, not remove them. Soot (C) is a product of incomplete combustion, but the converter is not designed to burn it, and this equation removes only one pollutant. Forming NO₂ from NO creates another pollutant (it is a cause of acid rain and photochemical smog).

Boiling point depends on instantaneous dipole-induced dipole forces. Hexane has six carbons (more electrons than the C₅ alkanes) and is unbranched, so its molecules have the greatest contact area: it has the strongest forces and highest boiling point (69 °C). 2,2-dimethylbutane is an isomer of hexane with the same number of electrons, but its compact, branched shape reduces contact, so its boiling point is lower (50 °C). Pentane has fewer electrons (36 °C). 2-methylbutane has both fewer electrons and branching, giving the lowest boiling point of the four (28 °C).

For one C₈H₁₈: 8 C give 8CO₂ and 18 H give 9H₂O. Oxygen atoms on the right = 16 + 9 = 25, so 12.5 O₂ are needed. Doubling to remove the half gives 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O, so b = 25. 12.5 is correct per mole of octane but the question asks for whole-number coefficients. 17 comes from 8 + 9, adding the numbers of product molecules instead of the oxygen atoms. 50 doubles the oxygen atoms twice, counting atoms rather than O₂ molecules.

n(octane) = 1000 / 114.0 = 8.77 mol. Each mole gives 8 mol CO₂, so n(CO₂) = 70.2 mol, and mass = 70.2 × 44.0 = 3088 g = 3.09 kg. 0.386 kg uses a 1 : 1 ratio, ignoring the eight carbons per molecule. 6.18 kg takes the 16 from the equation written for 2C₈H₁₈ without halving it. 1.42 kg is the mass of water produced (9 mol H₂O per mole, M_r 18.0), not carbon dioxide.

From C₃H₈ + 5O₂ → 3CO₂ + 4H₂O, 20.0 cm³ of propane uses 100 cm³ of O₂ and makes 60.0 cm³ of CO₂. Oxygen left over = 150 − 100 = 50.0 cm³. Water is a liquid at room temperature, so its volume is negligible. Total gas = 60.0 + 50.0 = 110 cm³. 60.0 cm³ forgets the excess oxygen. 190 cm³ counts 80 cm³ of water as a gas. 170 cm³ is just the starting volume, assuming no change in the number of gas molecules.