Class 12th

8. Separation of two sulfide ores can be accomplished by adjusting the proportion of oil to water or by using depressants. In the case of an ore containing ZnS and PbS, for example, the depressant NaCN is used. It forms a complex with ZnS and prevents it from coming into contact with froth, but PbS remains in contact with froth.

7. The reduction is the removal of oxygen. Hydrogen and carbon both form oxides that are easy to remove both water and carbon dioxide which are produced as a gas at hot reactions. Although carbon and hydrogen are better reducing agents, they are not used to reduce metallic oxides at high temperatures because at high temperatures carbon and hydrogen react with metals to form carbide and hydrides respectively.

6. Both processes have two basic requirements:

(i) The metal should react with an available reagent to form a volatile compound.

(ii) The volatile compound should be easily decomposable, allowing for easy metal recovery.

5. Low-grade copper ores contain a small percentage of copper that is 0.27% copper. By hydrometallurgy, low-grade copper is leached by treating with acid when copper metal goes into solution as copper ions. The solution containing copper ions is treated with scrap iron or hydrogen gas. Since iron or hydrogen is more reactive than copper so they reduced copper ions from solution to copper metal.

26.  (ii) Free valence electrons 

4. Iron oxides are mixed with limestone and coke in order to decompose carbonates and oxidize sulfides. This mixture is then fed into the blast furnace. Many reactions take place in different temperature ranges of the blast furnace. By oxidizing impurities from cast iron in a reverberatory furnace which is lined by hematite, wrought iron or malleable iron is obtained. This is the purest form of commercial iron. This can be shown with the following reaction:

Fe2O₃ + 3C → 2Fe + 3CO.

Sulfur, silicone and phosphorus are oxidized and passed into the slag and removed as impurities when limestone is added as a flux to it.

3. From Ellingham diagram, we know that ΔG^° _{(C, CO)}  < G^° _{(Fe, Fe)}, the following reactions are:

C+ $\frac{1}{2}$ O₂→CO

2Fe+O₂→2FeO

Therefore, FeO can be reduced to Fe by coke. 

25. Option (i) Very low value of electrical conductivity in the molten state is correct since in molten state, the ionic solids possess free ions which help in the conduction of electricity. Due to the availability of these free ions the electrical conductivity of ionic compounds in molten state is found to be very high. 

2. Extraction of chlorine from brine is an oxidation method. The reactions involved are:

2Cl⁻ + 2H₂O → 2OH⁻ + H₂ + Cl₂

For this reaction, the value of ΔG°=+422 kJ, which is positive. Using the formula ΔG°=−nE°F, we get a negative value of E° =−2.2 V.

Since,  E°=−2.2 V, it would naturally require an external voltage greater than 2.2 V for the reaction to occur.

24. Option (iii) (C) and (D) is correct since Iodine is a non-polar molecular solid which acts as an insulator while H₂O is a hydrogen bonded molecular solid which also acts as an insulator. Whereas Mg is a metal and acts as a conductor in solid and in its molten state and TiO is also an ionic solid which acts as an insulator in solid state but acts as a conductor in molten and in aqueous solutions.