Short Answer Questions

The salts formed and their solubility when the given reactants are mixed is listed in the table below:

From the table, we can see that silver nitrate and zinc chloride form silver chloride (AgCl) which is an insoluble precipitate, while for all others, no precipitate is formed.

(a) The red litmus paper turned blue, which indicates that the solution is basic (alkaline) in nature. Hydroxyl ions (OH^-) are responsible for turning red litmus paper blue.

(b) To prevent back suction of water and to increase area of absorption of Ammonia gas.

Raising the temperature to 600°C would decrease the yield of ammonia because the forward reaction is exothermic, and increasing temperature shifts the equilibrium toward the endothermic reverse reaction, reducing ammonia formation.

Chloromethane or methyl chloride

(a) Complete combustion of butene (C₄H₈) produces four moles of carbon dioxide and four moles of water only.

C₄H₈ + 6O₂ ⟶ 4CO₂ + 4H₂O

(b) Structural formula of butene (C₄H₈) is shown below:

When barium chloride solution is added to the acid prepared by the contact process i.e., sulphuric acid (H₂SO₄), a white precipitate of barium sulphate (BaSO₄) is formed, which is insoluble in water.

H₂SO₄ + BaCl₂ ⟶ BaSO₄ + 2HCl

(a) Equation for the catalytic oxidation of ammonia is given below:

$4\text{NH}_3 + 5\text{O}_2 \xrightarrow[800 ^{\circ} \text{C}]{\text{Pt.}} \text{4NO} + \text{6H}_{2}\text{O}\uparrow + \Delta$

(b) The platinum (catalyst) continues to glow even after the heating is discontinued because the catalytic oxidation of ammonia is an exothermic reaction.

(a)

$\text{C}_{12}\text{H}_{22}\text{O}_{11}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{12C}\text{(s)} + \text{11H}_2\text{O}$

(b) Conc. H₂SO₄ acts as dehydrating agent. It readily removes chemically combined water molecules.

(a) A freshly prepared ferrous sulphate solution is used because on exposure to the atmosphere, it is oxidised to ferric sulphate, which will not give the brown ring test.

(b) FeSO₄.NO
[Nitroso Ferrous sulphate, a brown compound]

Given,
Mass of KClO₃ = 367.5 g
Molar mass of KClO₃ = 39 (K) + 35.5 (Cl) + 3×16 (O) = 122.5 g/mol
Molar mass of KCl = 39 (K) + 35.5 (Cl) = 74.5

$$\begin{matrix} 2\text{KClO}_3 & \xrightarrow{MnO_2} & 2\text{KCl} & + & 3\text{O}_2 \\ 2[39 + 35.5 & & 2[39 & & 3(22.4) \\ + 3(16)] & & + 35.5] & & = 67.2 \text{ lit.} \\ = 245 \text{ g} & & = 149 \text{g} \\ \end{matrix}$$

(a) 245 g of KClO₃ produces 67.2 lit. of O₂

∴ 367.5g of KClO₃ will give,

= 67.2 x $\frac{367.5}{245}$ = 100.8 l of O₂

∴ The volume of the colourless and odourless gas O₂ liberated during the experiment is 100.8 liters.

(b) 245 g of KClO₃ produces 149 g of KCl

∴ 367.5g of KClO₃ produces,

= 149 x $\frac{367.5}{245}$ = 223.5g of KCl

∴ The weight of the KCl left behind as a residue in the test tube is 223.5g.

Two reasons for using Duralumin instead of Aluminium is:

1. Duralumin is light and strong, while aluminium is light and weak. It is also unaffected by moist air. Duralumin has a strength up to six times greater than pure aluminium.
2. It is corrosion-resistant and has high tensile strength.

(a) Carbon (C): Atomic mass = 12 g/mol
Mass = 5 moles x 12 g/mol = 60 g

Sodium (Na): Atomic mass = 23 g/mol
Mass = 5 moles x 23 g/mol=115 g

∴ Krish’s container is heavier

(b) 1 mole = 6.022 x 10²³ atoms

Both the containers have 5 moles of substance.

5 moles x 6.022 x 10²³ = 3.011 x 10²⁴ atoms

∴ Both the containers have the same number of atoms.

(a) Most reactive element is B

(b) Least reactive element is C

Explanation:

Above reactions are displacement reaction. In first and second reactions B is displacing A and C from their oxide and sulphate respectively. In third reaction, A displaces C from its oxide. Considering all three reaction, B is more reactive than A and C.
C is less reactive, and is not displacing either A and B in above reactions.

(a) The conductivity of PQ₂ is due to the presence of free ions.

(b) During electrolysis of aqueous PQ₂, if thickening of the cathode and thinning of the anode is observed, the anode material will be metal P.

Explanation:

PQ₂ is an ionic compound made of metal P and nonmetal Q, where P is likely a divalent metal cation (P²⁺)

(a) Ions, ionic compounds conduct electricity in molten or aqueous forms due to the movement of ions.

(b) Thickening of the cathode means metal P²⁺ ions gain electrons (reduction) and deposit as solid metal P. Thinning of the anode means anode is dissolving, which only happens if the anode is made of the same metal (P).

If the anode were graphite, it would not dissolve, but might produce gas instead. So, the anode material must be metal P

The incorrect steps are:

1. Step 2: Froth floatation instead of leaching.
2. Step 3: Roasting instead of calcination.

The corrected flowchart is shown below:

(a) Nickle

(b) The complete chemical equation for the conversion of C₂H₄ to C₂H₆ (Ethane) is shown below:

$\underset{\text{ ethene}}{\text{C}_2\text{H}_4} + \text{H}_2 \xrightarrow[\text{Nickle}]{200^{\circ}\text{C}} \underset{\text{ethane}}{\text{C}_2\text{H}_6}$

The compound is Propan-1-ol.

The structure of the compound is given below: