Correct Answer: C. Z

Correct Answer: B. 6.02 x 10²²

Explanation

(a)(i) Two industrial uses of hydrogen: —filling of weather balloons. —manufacture of plastics. —manufacture of ammonia by the Haber process. —manufacture of methanol. —oil refinery process/hydrocracking —used in fuel cells. —manufacture of margarine. —in oxy-hydrogen flame (for welding and cutting metals). —as reducing agent in production of metals, e.g. Cu and.Pb from their oxides. —conversion of coal to crude oil. —(liquid) hydrogen is used as rocket fuel/gaseous fuel (Any 2).

(ii) Mg(HCO\(_3\))\(_{2(aq)}\) —> MgCO\(_3\)\(_{(s)}\) + H\(_2\)O\(_{(l)}\) + Cu\(_3\).

In the equation above: (1) the type of hardness of water being removed is temporary hardness of water. (2) two disadvantges of hardness of water are: —scales/furrinab of kettles/pipes causing blockage leading to power wastage. —soap is wasted not suitable for laundry. —not suitable for use in tanning, textiles and paper production (Any 2).

(b)(i) Disadvatnages of using graphite electrodes in the extraction of aluminium by electrolysis: At high temperature oxygen gas produced at the anode reacts with the graphite electrodes to form carbon(IV) oxide.

(ii) Balanced equation of reaction between calcium oxide and water to form slaked lime: CaO + H\(_2\)O \(\to\) Ca(OH)\(_2\)

(iii) One use of slaked lime: —used to treat acidic soil. —used to make mortar/plasters/cement/hold bricks. —in manufacture of glass. —purification of water/removing hardness of water. —manufacture of white wash. —to test for CO\(_2\). —production of bleaching powder: —to recover or produce ammonia from ammonium chloride used in solvay process.

(c)(i) Saponification is the alkaline hydrolysis of fats/oils to produce soap and propane —1,2,3-triol

(ii) Raw materials needed for the manufacture of soap: —oil/fat. —alkali (1) (sodium hydroxide/potassium hydroxide

(iii) Main by-product obtained from the manufacture of soap is propane —1,2,3-triol.

(d) Extraction of iron in the blast furnace using iron-ore, coke and limestone as raw materials with the aid of chemical equations..

(i) Bottom of furnace: The O\(_2\) in the hot air combines with carbon(coke) to produce carbon (IV) oxide which is reduced to carbon (II) oxide.

C + O\(_2\) \(\to\) CO\(_{2(g)}\)

CO\(_{2(g)}\) + C \(\to\) 2CO\(_{(g)}\)

(ii) Middle of the furnace: The calcium carbonate CaCO\(_3\) decomposes under great heat to form CO\(_2\) and CaO and CaCO\(_3\) decomposes to remove the impurities SiO\(_2\)

CaCO\(_3\) —> CaO\(_s\) + CO\(_{2(g)}\)

CaO + SiO\(_2\) --> CaSiO\(_3\).

(iii) Top of furnace: Iron (III) oxide is reduced to metallic iron Fe\(_2\)O\(_2\) + 3CO 2Fe + 3CO\(_2\)

OR Fe\(_3\)O\(_4\) + 4CO —> 3Fe + 4CO\(_2\).

Correct Answer: A) 10s

Explanation

In order to explain this question, we need to understand the concept of diffusion and how it relates to the molecular weight of gases. Diffusion is the process of a substance spreading out to evenly occupy its container or environment. In the case of gases, lighter gases diffuse faster than heavier gases under the same conditions.

Graham's law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it can be represented as:

\[\frac{Rate_1}{Rate_2} = \sqrt{\frac{Molar\ Mass_2}{Molar\ Mass_1}}\]

Here, we are given that methane (CH₄) diffuses in 20 seconds, and we have to find the time taken for the same volume of sulfur (IV) oxide (SO₂) to diffuse under the same conditions. First, we need to calculate the molar mass of both gases.

The molar mass of methane (CH₄): C = 12, H = 1 (4 hydrogen atoms) = 4, so the molar mass of CH₄ = 12 + 4 = 16 g/mol.

The molar mass of sulfur (IV) oxide (SO₂): S = 32, O = 16 (2 oxygen atoms) = 32, so the molar mass of SO₂ = 32 + 32 = 64 g/mol.

Now, applying Graham's law of diffusion:

\[\frac{Rate_{CH_4}}{Rate_{SO_2}} = \sqrt{\frac{Molar\ Mass_{SO_2}}{Molar\ Mass_{CH_4}}}\]

Let the time taken for SO₂ to diffuse be x seconds. Since rate is inversely proportional to time, we can write:

\[\frac{20}{x} = \sqrt{\frac{64}{16}}\]

Solving for x:

\[x = 20 \times \sqrt{\frac{16}{64}} = 20 \times \frac{1}{2} = 10\]

So, it will take 10 seconds for the same volume of sulfur (IV) oxide to diffuse under the same conditions.

Correct Answer: A. carbon

Correct Answer: A) coatings of pipes

Correct Answer: D. Hydrogen bond

Correct Answer: A. X₃N

Correct Answer: B) Ethyne

Correct Answer: D. it contains more H₃O^- than OH^-