Explanation:​
A conjugate acid-base pair differ by a single hydrogen ion. Carbonic acid and carbonate (choice A) differ by two hydrogen ions so are not a conjugate acid-base pair. Boric acid and its hydroxide adduct (choice B) differ by a hydroxide ion and ammonia and hydroxylamine (choice D) by an oxygen atom.
This question is categorised as very hard as the student must recall the definition of a conjugate acid-base pair, interpret the structural formulas of unfamiliar molecules and identify a conjugate acid and base without the aid of a reaction equation.

Explanation:​
C3H8​ is a saturated alkane and C3H6​ is an unsaturated alkene. The test of unsaturation using Br2(aq) can demonstrate the presence or absence of a carbon-carbon double bond.
In the presence of a double bond, Br2(aq) can add across the double bond in the alkene producing a dibromo-alkane, and a change of color from brown to colorless will be observed.
In the case of the absence of a double bond (saturated hydrocarbon), no color change would occur.

Explanation​:
Step 1 is a free radical substitution, converting the alkane to a chloroalkane and producing HCl as a by-product.
Step 2 is a nucleophilic substitution, in which the C−Cl bond breaks and is replaced by a hydroxyl (C−OH) group; the products are an alcohol and sodium chloride.
Step 3 is an oxidation. To prevent the aldehyde from being oxidised further to a carboxylic acid, the product must be removed by distillation as it forms.

Explanation:​
Spontaneous reactions occur in voltaic cells converting chemical energy to electrical energy, whereas, non-spontaneous reactions occur in electrolytic cells requiring electrical energy and converting this to chemical energy.
Electrolytic cells typically consist of one reaction vessel containing the aqueous of molten electrolyte. No salt bridge is therefore required.

Neutralization involves acid and base reacting to form a salt and water. Reaction A does not show those criteria to be categorized as a neutralization reaction.

Explanation:​
Students must compare ways of breaking bonds. They must recall that homolytic fissions form free radicals (species with unpaired electrons) as shared electrons split once the bond breaks.
In contrast, heterolytic fissions produce positively and negatively charged particles (ions) as both shared electrons go to just one of the atoms in the bond.

This question is categorized as hard. Students must recall the ways of breaking bonds, distinguish homolytic and heterolytic fission, and understand that free radicals have unpaired electrons.

Explanation:​
A heterolytic fission involves the break up of a bond so that both electrons go to one of the atoms. This process produces ions, as shown below.

The heterolytic fission requested occurs like this:

The species produced are ions, (CH3)3C^+ and Br^−.

This question is categorized as hard. The student must interpret the condensed formula given, recall how heterolytic fissions occur in that specific bond, and predict the product of this process.

Explanation:
Non-spontaneous reactions take place in electrolytic cells using an external power supply.
The positive electrode is the anode where oxidation takes place.
The aqueous metal ion (Mx^+) is reduced at the cathode to produce the solid metal.

Explanation:​
Least reactive metal is the strongest oxidizing agent/ weakest reducing agent.
Z is the most reactive metal (can displace both X and Y), followed by Y(can only displace X), and lastly, X (cannot displace any).

Explanation​:
The positive ions (cations) move in the direction of the cathode as the positive Cu2+ ions are used up.
Zn being more reactive than copper, Zn(s) is the anode where oxidation occurs, and Cu(s) is the cathode where reduction of Cu2+(aq) occur

Explanation​:
NaOH is a strong base and so would release a high concentration of OH^− ions in water. Choices A and C are incorrect as HCl and H2SO4 are strong acids and would give low $\mathrm{pH}$s. $\mathrm{NaCl}$ is a neutral salt.

Explanation:​
A secondary alcohol is a compound in which the carbon attached to the hydroxyl group is also attached to two other carbons.
Upon its oxidation, a removal of 2H atoms occurs and a ketone is formed.

Explanation:​
Oxidation involves the loss of electrons and an increase in oxidation state.

Explanation​:
In this nucleophilic substitution reaction, the hydroxide donates a lone pair to break the C-Br bond and form a new C-O bond. The C atom of compound B accepts the electron pair, so is an electrophile.
The compound is a secondary bromoalkane, as the C atom of the C-Br bond is attached to two C atoms.

Explanation:
H2SO4 is a strong acid and fully ionizes its first proton leading to a high H+ concentration.
HSO4− is a weak acid that only partially dissociates to produce $.$ [SO42−] is, therefore, low and not equal to HSO4−

Explanation:​
$\mathrm{pH}$ = $-$.
For concentrations of H+ greater than 1 mol dm^−3, this would result in a negative $\mathrm{pH}$ value.

Explanation:
NaOH is a strong base and has a high pH,  NH3,​ is a weak base with a lower $\mathrm{pH}$ than NaOH. CH3COOH​ is a weak acid, $\mathrm{HCl}$ is a strong acid and so has the lowest pH.

Explanation:​

Metals and acids react to produce a salt and hydrogen gas. The formulae of the ions are Mg2+ and NO3−, therefore for a neutral salt, two nitrate ions are required for each magnesium ion.

Explanation​:
Zn2+(l) gains electrons and is reduced to Zn(s) at the cathode. Cl−(l) loses electrons and is oxidised to Cl2(g) at the anode.

Explanation:​
A base accepts a hydrogen ion, whereas an acid donates it. The base in the forward and backward reactions are therefore the species which gain hydrogen ions during the reaction: NH3 and HSO4^−, respectively.
This question is moderately difficult as the student must recall the definition of a base and identify specific examples of bases in a reaction equation.

Explanation:​
A nucleophile is a species that is electron-rich and an electron donor. An electrophile is an electron deficient species.

Explanation:​
The conjugate base of HCl is the compound produced when a hydrogen ion is removed. The conjugate base is therefore the chloride ion, Cl^−.
This question is hard as the student must link the names of chemicals to their formula and identify a conjugate base without the aid of an explicit reaction equation.

Explanation​:
More reactive metals are more likely to lose electrons (be oxidised) and so, Zn(s) will lose electrons to produce Zn^2+(aq) .
Cu^2+(aq)will gain electrons forming Cu(s) leading to a decrease in Cu^2+(aq) concentration and a paler blue colour of the solution.
Electrons flow from the anode(Zn) to the cathode (Cu).

Explanation:​
A tenfold decrease in [H+], leads to a $\mathrm{pH}$ increase of 1.