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​:
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:​
Reducing agent is a substance that is oxidized.
Oxidation state of Zn increases from 0 to +2, whereas the oxidation state of N decreases from +5 to +4.
The oxidation state of H remains +1.
The oxidation state of O remains -2.

Explanation:
In a primary alcohol, the OH group is bonded to a CH2 group. The alcohol group initially reacts with KMnO4​ to form an aldehyde (CHO) group, as in option B.
However, the aldehyde is further oxidzed to a carboxylic acid (COOH) group unless it is removed by distillation. Here, the solution is heated under reflux, so the carboxylic acid is the major product.

Explanation:​
Assuming "x" to be the oxidation number of N:

• In nitrate NO3− , x +3 (-2) = -1. Therefore, x = +5
• In nitrite NO2− , x +2 (-2) = -1. Therefore, x = +3

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.

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:​
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.

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​:
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.

Explanation:​
The HPO^4− ion contains two negatively charged oxygen atoms which can bond to hydrogen ions, producing phosphoric acid (H3PO^4​).
The HPO^4− ion can therefore react with two H+ ions.
This question is moderately difficult as the student must recognise that H^+ ions bond to the negatively charged oxygen atoms, or that HPO^4− is derived from H3PO^4​.

Explanation:​
Water accepts a hydrogen ion to produce H3O^+. This question is easy as the formation of H3O^+ is covered explicitly in the syllabus

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

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:​
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:​

Acids react with carbonates to produce a salt, water, and carbon dioxide.

Explanation:
Use the mnemonic CNAP; cathode negative, anode positive for the polarity in electrolytic cells.

Explanation:​
In equation 1, the oxidation state of Cl decreases from +5 to 0 (reduced), and the oxidation state of Iodine increases from 0 to +5 ( oxidized).
In equation 2, the oxidation state of Cl increases from 0 to +1 in NaClO (oxidized), and also decreases from 0 to -1 in NaCl (reduced).

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​:
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.

A

Explanation:​
NaOH is a base so reacts with a hydrogen ion. HNO3​, nitric acid, releases a hydrogen ion through the breaking (heterolysis) of the O-H bond.
This question is moderately difficult as the student must recognise that NaOH is a base and HNO3​ an acid and correctly interpret a structural formula that may be unfamiliar.

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​:
Statement II is incorrect as acids produce solutions with pH lower than 7.