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CFCs (Chlorofluorocarbons) have a lifetime in the atmosphere of about 20 to 100 years, and consequently one free chlorine atom from a CFC molecule can do a lot of damage, destroying ozone molecules for a long time

Ethene is the most important organic chemical, by weight, that is manufactured. It is an important raw material for the production of plastics and antifreeze solutions as well as solvents.

An example of benzene addition reaction is the formation of cyclohexane by a mix of benzene vapour and hydrogen passed over nickel catalyst; while chlorobenzene is formed by substitution reaction.

The bonding between a non-metal and a metal is an ionic bonding. In metal hydrides, Hydrogen takes on the ionic charge of -1, hence forming an ionic bond with the metals.

Given the molecular formula, \(CH_{2}\). The empirical formula will be \((CH_{2})n\)= 42 \((12+(2\times1))n =42\) ⇒ \(14n =42\) n = 3, hence the empirical formula is \(C_{3}H_{6}\)

No expanded working has been attached for this question yet. The verified answer is Option A.

Ethene can be gotten from the dehydration of ethanol in the presence of excess \(H_{2}SO_{4}\) or \(H_{3}PO_{4}\).

The equilibrium position is shifted to the left with a decrease in pressure in this system because the number of gaseous molecules on the right is less. (Le-Chatelier's principle).

The preferential discharge of ions is influenced by: - The position of the ions in the electrochemical series - The concentration of the ions in the electrolyte - The nature of the electrode

\(_{12}Mg = 2, 8, 2\) \(= 1s^{2} 2s^{2} 2p^{6} 3s^{2}\)

Stainless steel comprises iron (Fe), chromium (Cr), manganese (Mn), silicon (Si), carbon (C) and, in many cases, significant amounts of nickel (Ni) and molybdenum (Mo).

\(1mole = 6.02 \times 10^{23} ions\) \(0.02moldm^{-3} = 0.02 \times 6.02 \times 10^{23}\) =\(0.1204 \times 10^{23} = 1.204 \times 10^{22}\) \(\approxeq 1.2 \times 10^{22}\)

In order to extinguish a petrol fire, you have to cut off the oxygen. This can be done by using carbon dioxide gas to dilute the oxygen available and stop the burning.

Using n = cv where n = no of mole, c = molar concentration(mol/dm3) and v = volume( dm3) volume = 500\(cm^{3}\) = 0.5dm3, c = 0.005\(moldm^{-3}\) number of mole(n) = c x v = 0.5 X 0.005 = 0.0025mol. but n = \(\frac{mass}{ molar mass}\) = 0.0025 = \(\frac{mass}{ 98}\) (where 98g/mol is the molar mass of H2SO4) mass of H2SO4 = 0.0025 x 98 = 0.245g.

The easiest and most drastic way to change the boiling point is by tampering with the pressure above your water. To lower the boiling point, you need simply to decrease its external pressure.