Chemical Reactions and Equations

The reaction is \(\mathrm{CaCO_3(s) \rightarrow CaO(s) + CO_2(g)}\). In an intact closed container, no matter, including \(\mathrm{CO_2}\), leaves the system, so the total mass of the container and all its contents remains unchanged. Option B could apply to an open container, where escaping \(\mathrm{CO_2}\) can make the measured mass decrease.

The law of conservation of mass applies to the complete system. In an open container, a produced gas may escape, so it is not included when the container and its remaining contents are weighed later. Therefore, the measured mass appears lower, but total mass has not been destroyed. Option B is incorrect because mass is not destroyed in a chemical reaction.

Subscripts specify the fixed number of atoms of each element in a compound. For example, changing H₂O to H₂O₂ changes water into hydrogen peroxide, so it changes the substance itself. Only coefficients should be changed when balancing because they change the number of molecules or moles without changing the chemical formula.

The product is \(2MgO\). Each formula unit of \(MgO\) contains one oxygen atom, and the coefficient 2 represents two formula units. Therefore, the total number of oxygen atoms is \(2 \times 1 = 2\). Although \(O_2\) also has two oxygen atoms, it is on the reactant side; atoms on the product side are counted from \(2MgO\).

A thin layer of magnesium oxide forms on magnesium when it is exposed to air. This layer can prevent direct contact between magnesium and oxygen and hinder burning. Rubbing with sandpaper removes the oxide layer, exposing fresh magnesium so that it burns readily. The purpose is not to make the ribbon thinner.

Magnesium reacts with oxygen in air to form white magnesium oxide. The balanced equation is \(2Mg + O_2 \rightarrow 2MgO\). This is a combination reaction because two reactants form one product. Magnesium chloride would require chlorine, which is not present in this reaction.

On thermal decomposition, calcium carbonate forms solid calcium oxide and carbon dioxide gas: \(\mathrm{CaCO_3(s) \xrightarrow{\Delta} CaO(s) + CO_2(g)}\). Therefore, the solid product is calcium oxide. Calcium hydroxide is formed when calcium oxide reacts with water, not by heating calcium carbonate alone.

Calcium carbonate undergoes thermal decomposition on heating to form calcium oxide and carbon dioxide: \(\mathrm{CaCO_3 \xrightarrow{\Delta} CaO + CO_2}\). The released \(\mathrm{CO_2}\) reacts with lime water, \(\mathrm{Ca(OH)_2}\), forming insoluble calcium carbonate: \(\mathrm{CO_2 + Ca(OH)_2 \rightarrow CaCO_3 \downarrow + H_2O}\). This precipitate makes the lime water appear milky. Oxygen does not form this precipitate with lime water.

The thermal decomposition reaction is:
\[
2\mathrm{Pb(NO_3)_2} \rightarrow 2\mathrm{PbO} + 4\mathrm{NO_2} + \mathrm{O_2}
\]
Here,
\(\mathrm{NO_2}\)
is the brown gas, whereas
\(\mathrm{O_2}\)
is colourless. Therefore, the brown colour identifies
\(\mathrm{NO_2}\)
, not oxygen. Oxygen is indeed formed, but it is confirmed by rekindling a glowing splint, not by observing brown fumes.

Silver chloride undergoes photodecomposition in sunlight: \(2\mathrm{AgCl}(s)\xrightarrow{\text{sunlight}}2\mathrm{Ag}(s)+\mathrm{Cl}_2(g)\). The silver metal produced is grey, so silver chloride gradually appears grey. Although chlorine gas is also formed, it is not responsible for the grey colour.

Silver bromide is photosensitive. In light, it undergoes photodecomposition: \(2\mathrm{AgBr}\xrightarrow{\text{light}}2\mathrm{Ag}+\mathrm{Br}_2\). A dark-coloured bottle reduces the light reaching the compound and therefore helps prevent this decomposition. It is not used to increase thermal decomposition or solubility in water.

The balanced equation for electrolysis of water is \(2H_2O(l) \rightarrow 2H_2(g) + O_2(g)\). At the same temperature and pressure, gas volumes are proportional to their numbers of moles. Therefore, the volume ratio of hydrogen to oxygen is \(2:1\). The ratio \(1:2\) would represent oxygen to hydrogen, not the order asked in the question.

In the electrolysis of water, hydrogen gas is produced at the cathode and oxygen gas at the anode. From \(2H_2O \rightarrow 2H_2 + O_2\), the volume ratio of hydrogen to oxygen is \(2:1\). Therefore, the greater volume of gas, hydrogen, is produced at the cathode. Oxygen is produced at the anode, but its volume is half that of hydrogen.

Pure water contains very few ions and is therefore a poor conductor of electricity. Dilute sulphuric acid supplies mobile ions, allowing current to pass through the solution and enabling electrolysis of water. Therefore, option A is correct. Adding an acid does not make the solution neutral; it makes it acidic.

The reaction is \(2H_2O \rightarrow 2H_2 + O_2\). At the same temperature and pressure, the gas volumes are in the ratio of their coefficients, so \(H_2:O_2 = 2:1\). Therefore, for 8 mL of oxygen, the volume of hydrogen is \(2 \times 8 = 16\) mL. The 8 mL option is incorrect because the volumes of the two gases are not equal.

Iron is more reactive than copper, so it displaces copper from copper sulphate solution. The displaced copper deposits on the iron nail as a reddish-brown coating. The reaction is \(\mathrm{Fe + CuSO_4 \rightarrow FeSO_4 + Cu}\). Iron oxide in option C is associated with rusting, whereas the coating in this reaction is copper.

The blue colour of aqueous copper sulphate solution is mainly due to hydrated copper(II) ions, \(\mathrm{Cu^{2+}}\). When the concentration of these ions decreases, the blue colour becomes less intense and the solution appears paler. Sulphate ions, \(\mathrm{SO_4^{2-}}\), do not cause the blue colour, so option B is not the correct ionic reason. Exam tip: For solution colour changes, first identify the ion responsible for the colour.

In a displacement reaction, a more reactive metal displaces a less reactive metal from its salt solution. For example, zinc displaces copper from copper(II) sulfate solution: \(\mathrm{Zn + CuSO_4 \rightarrow ZnSO_4 + Cu}\). Therefore, zinc is more reactive than copper. Option B is incorrect because copper cannot displace zinc from a zinc salt solution.

A more reactive metal displaces a less reactive metal from a solution of its salt. Since A displaces B, A is more reactive than B. Since B displaces C, B is more reactive than C. Thus, the reactivity order is A > B > C, so metal A is the most reactive. Option B is more reactive than C, but not than A.

In \(\mathrm{BaCl_2 + Na_2SO_4 \rightarrow BaSO_4\downarrow + 2NaCl}\), the ions of the two compounds exchange partners. \(\mathrm{Ba^{2+}}\) combines with \(\mathrm{SO_4^{2-}}\) to form an insoluble precipitate of \(\mathrm{BaSO_4}\), while \(\mathrm{Na^+}\) and \(\mathrm{Cl^-}\) form \(\mathrm{NaCl}\). Therefore, this is a double displacement reaction. Option A describes single displacement, in which a free element replaces an element in a compound. Exam tip: The general pattern \(\mathrm{AB + CD \rightarrow AD + CB}\) indicates a double displacement reaction.

In a precipitation reaction, ions from two aqueous solutions combine to form a solid that is insoluble in water. For example,
\(\mathrm{BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq)}\).
\(\mathrm{BaSO_4}\) separates from the solution as an insoluble white solid, so it is a precipitate. In contrast, the soluble substance described in option B would remain in a clear solution rather than form a precipitate.

The reaction is \(\mathrm{CuO + H_2 \rightarrow Cu + H_2O}\). Oxygen is removed from \(\mathrm{CuO}\) and combines with hydrogen to form water. Removal of oxygen is reduction, so copper(II) oxide undergoes reduction. In contrast, hydrogen gains oxygen and is oxidised.

The reaction is \(\mathrm{CuO + H_2 \rightarrow Cu + H_2O}\). Hydrogen gains oxygen from copper oxide to form water. As gain of oxygen is oxidation, hydrogen is oxidised. Copper oxide, on the other hand, loses oxygen to form copper and is therefore reduced. Exam tip: In oxygen-based redox questions, the substance that gains oxygen is oxidised.

Based on oxygen, oxidation is the addition of oxygen to a substance, whereas reduction is the removal of oxygen from a substance. For example, when copper gains oxygen, copper oxide forms; when oxygen is removed from copper oxide, copper is obtained. Therefore, option A is correct. Option B interchanges the definitions of oxidation and reduction.

Rusting requires iron to be in contact with oxygen and water or moisture. Paint and oil form a protective coating on the iron surface, so air and moisture cannot reach the metal. Therefore, option A is correct. Raising the temperature does not create a barrier between iron and oxygen or moisture.