Correct Answer: B. I and III only

Correct Answer: A. Rutherford

Correct Answer: C) SO₂

Explanation

Answer: Option C - SO\(_2\)

Sulfur dioxide (SO\(_2\)) is a versatile gas that can act as both a reducing agent and an oxidizing agent, depending on the reaction conditions and the substances involved. In this case, SO\(_2\) can behave as a reducing agent towards chlorine (Cl\(_2\)) and as an oxidizing agent toward hydrogen sulfide (H\(_2\)S).

When SO\(_2\) reacts with chlorine, it reduces the chlorine by donating electrons, as shown in the equation:

In this reaction, SO\(_2\) donates two electrons to Cl\(_2\), converting it to SOCl\(_2\) (thionyl chloride).

On the other hand, when SO\(_2\) reacts with hydrogen sulfide, it acts as an oxidizing agent by accepting electrons from H\(_2\)S, as demonstrated in the following equation:

In this reaction, SO\(_2\) accepts two electrons from H\(_2\)S, leading to the formation of sulfur (S) and water (H\(_2\)O).

Therefore, SO\(_2\) can behave as both a reducing and oxidizing agent, depending on the reaction and the substances involved.

Explanation

(a) Definition of Electrochemical Cell: Electrochemical cell is defined as a device/cell which converts chemical energy to electrical energy/or electricity is generated by a chemical change.

(b) Aluminium can be prepared commercially by the application of electrolysis:

(i) Name the: I. Electrolyte used in the process: (i) Aluminium oxide/(Molten) alumina. II. Bauxite. III. Carbon electrodes/graphite electrodes.

(ii) Reasons why cryolite, NaAl F\(_6\) is added to the electrolyte: –As a solvent. –Helps reduce the melting point of alumina (Al\(_2\)O\(_3\)). –To increase the conductivity of electrolyte.

(c)(i) Gaseous fuels produced from coke: –Producer gas. –Water gas.

(c)(ii) Water gas.

(iii) Reason for your answer in(c)(ii): The producer gas contains about 67%/More non-combustible nitrogen and 33% less combustible carbon (II) oxide while water gas contains equal amount of combustible carbon (II) oxide and hydrogen.

(iv) Balance equation for each production of gaseous fuel: Producer gas. 2C\(_{(s)}\) + O2\(_{2(g)}\) + 4N\(_{2(g)}\) --> 2CO\(_{(g)}\) +4N\(_{2(g)}\). Water gas: C\(_{(s)}\) + H\(_{2}\)O —> CO\(_{(g)}\) + H\(_{2(g)}\)

(d)(i) What would be observed when:

I. Chlorine gas is bubbled through aqueous sodium iodide. –Colourless solution turn reddish brown.

II. Chlorine gas is passed over heated iron in a hard glass tube. –Grey silvery solid turns brown.

III. Aqueous silver trioxonitrate (V) is added to aqueous sodium bromide. –Cream/(pale) yellow precipitate.

(ii) Chemical equation for reactions in (d)(i):

(i) Cl\(_2\) +2Nal ---> 2NaCl + l\(_2\)

(ii) 2Fe +3Cl\(_{2}\) —> 2FeCl\(_{2}\)

(iii) AgNO\(_{3}\) +NaBr —> NaNO\(_3\) + AgBr.

Correct Answer: B) 15.0 atm

Explanation

P1V1 = P2V2 P2 = 10

Correct Answer: A) 0.30 mol

Explanation

In this question, we have a solution containing 0.20 moles of NaBr and 0.20 moles of MgBr\(_2\) in 2.0 dm\(^3\) of water. We are asked to find out how many moles of Pb(NO\(_3\))\(_2\) must be added in order to precipitate all the bromide as insoluble PbBr\(_2\). To solve this problem, we need to use the concept of stoichiometry, which is the relationship between the amounts of reactants and products in a chemical reaction. In this case, we know that Pb(NO\(_3\))\(_2\) reacts with NaBr and MgBr\(_2\) to form PbBr\(_2\), which is insoluble and precipitates out of the solution. The balanced chemical equation for this reaction is: Pb(NO\(_3\))\(_2\) + 2NaBr + MgBr\(_2\) ? PbBr\(_2\) + 2NaNO\(_3\) + Mg(NO\(_3\))\(_2\) From this equation, we can see that 1 mole of Pb(NO\(_3\))\(_2\) reacts with 2 moles of NaBr and 1 mole of MgBr\(_2\) to produce 1 mole of PbBr\(_2\). Therefore, to precipitate all the bromide as PbBr\(_2\), we need to add an amount of Pb(NO\(_3\))\(_2\) that is equal to the total number of moles of NaBr and MgBr\(_2\) in the solution. The total number of moles of NaBr and MgBr\(_2\) in the solution is: 0.20 mol (NaBr) + 0.20 mol (MgBr\(_2\)) = 0.40 mol Therefore, we need to add 0.40 moles of Pb(NO\(_3\))\(_2\) to precipitate all the bromide as PbBr\(_2\). The correct answer is option A: 0.30 mol. This answer is incorrect given the explanation. The correct answer should be option D: 0.40 mol.

Correct Answer: D. SO₂ and NO

Correct Answer: A. Nitrogen from liquid air

Correct Answer: A. manufacture of ammonia

Correct Answer: C. negative