Peptization can be defined as a process of converting a precipitate into colloidal sol by shaking it with dispersion medium in the presence of small amount of electrolyte.The electrolyte used in this process is known as a peptizing agent.
Express the relation among cell constant, the resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solution related to its conductivity?
The conductivity (k) of the solution in a cell is the reciprocal of its resistivity.
The quantity 1/2 is cell constant.
L --> Distance between 2 electrodes
a --> Area of cross section
R --> Resistance
The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9S cm2mol-1. Calculate the conductivity of the conductivity of this solution.
A reaction is of second order with respect to a reactant. How is its rate affected if the concentration of the reactant is (i) doubled (ii) reduced to half?
Let the concentration of the reactant be [A] = a
Rate of reaction, R = k [A]2 = ka2
(i) If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be
R = k (2a)2
Therefore, the rate of the reaction would increase by 4 times.
(ii) If the concentration of the reactant is reduced to half, i.e., then the rate of the reaction would be
Therefore, the rate of the reaction would be reduced to
How is copper extracted from a low-grade ore of it?
Copper can be obtained from low grade from low through the process of leaching using acid or bacteria (leaching processes in which ore is treated with a suitable reagent which dissolves ore but not the impurities).
Explain the role of each of the following:
(i) NaCN in the extraction of silver
(ii) SiO2 in the extraction of copper
(i) The roasted ore of gold is leached with a solution of sodium cyanide in the presence of oxygen for many days. The role of NaCN in this process is to dissolve the gold to form an aurocyanide complex, from which the metal is obtained by displacement.
4Au + 8NaCN + 2H2O + O2 ---> 4 Na [Au(CN)2] + 4KOH
2Na[Au(CN)2] + Zn ---> Na2[Zn(CN)4] + 2Au
(ii) Copper matte contains Cu2S and FeS. In the blast furnace, copper matte is added with powdered coke and silica. The oxidation of ore takes place in this process. As a result, cuprous oxide and ferrous oxide are produced. The role of silica in this process is to remove the iron oxide obtained as ‘slag’. FeO combines with silica (flux) to form iron silicate, FeSiO3(slag).
FeO + SiO2 FeSiO2
Explain the following facts giving appropriate reason in each case:
(i) NF3 is an exothermic compound whereas NCl3 is not.
(ii) All the bonds in SF4 are not equivalent.
(i) As we move down the group 17, the size of the atom increases from fluorine to chlorine. The instability of NCl3 is due to the weak NCl bond. This is due to the large difference in the size of nitrogen and chlorine atoms. On the other hand, atoms of both nitrogen (75 pm) and fluorine (72 pm) are small sized. Thus, bonding in NF3 is quite strong and it is an exothermic compound.
(ii)SF4 has four bonded atoms and one lone pair so SF4 has sp3d hydridisation and thus have trigonal bipyramid structure in which one axial position is occupied by a lone pair of electrons. This lone pair finds a position that minimizes the number of 900 repulsion it has with bonding electron pairs. This results in two types of angles.
Equatorial bond angle F - S - F° (LP-BP repulsion >BP – BP repulsion)
Axial bond angle F - S - F<90Â°
What is the basicity of H3PO2 acid and why?
Basicity of H3PO2 depends on upon the number of ionizable -OH groups present in the molecule. That is the number of hydrogen attached to the electronegative atom oxygen.
H3PO2 has one ionizable –OH group, thus its basicity is 1.
The structure of H3PO2 is as follows:
What are n-types semiconductors?
The semiconductor whose increased conductivity is a result of negatively-charged electrons is called an n-type semiconductor. These are generated when the crystal of a group 14 element such as Si or Ge is doped with a group 15 element such as P or As.