Solution

d-orbital have four lobes and two nodes and opposite lobes have the same sign

Solution

The electrolysis of pure alumina faces some difficulties. Pure alumina is a bad conductor of electricity. The fusion temperature of pure alumina is about 2000^oC and at this temperature when electrolysis is carried out on the fused mass, the metal formed vaporizes, as the boiling point of aluminium is 1800^oC. These difficulties are overcome by using a mixture containing alumina, cryolite (Na₃,AlF₆), and fluorspar (CaF₂)

Thus the conductivity is increased and the fusion temperature is lowered.

Solution

Acetone would distinguish cis-cyclopenta−1,2−diol from the trans-isomer.
With cis isomer, acetone forms bicylic ketal. With trans isomer, formation of bicyclic ketal is not possible due to presence of angle strain.

Solution

Stability depends on enthalpy of formation more the negative value of enthalpy of formation more stability

Solution: 

If the solubilities are S₁, S₂, S₃ respectively, then the K_sp values for them are
For A₂B, K_sp = 4 S31 [A₂B ? 2A⁺ + B^2-]
For AB, K_sp = S22 [AB ? A⁺ + B^-]
For AB₃, K_sp = 27S43 [AB₃ ? A³⁺ + 3B^-]
Since the value of K_sp is same
∴ The correct order of solubilities is
AB > A₂B > AB₃.

Solution

Galena (PbS) are sulphide ores. Cassiterite SnO₂ (oxide ore), and Magnetite Fe₃O₄ (oxide ore) are not sulphide ore. The froth flotation process is used to concentrate sulphide ores, based on preferential wetting properties with frosting agent and water.

Solution

The number of moles of organic compound $$\displaystyle = \frac {2.68}{134} = 0.2 $$ mol.
The number of moles of silver iodide $$\displaystyle  = \frac {14.08}{234.77} = 0.06 $$
Thus, 0.2 moles of organic compound gives 0.6 moles of $$AgI$$.
1 mole of organic compound will give 3 moles of $$AgI$$.
Hence, the number of methoxy groups present in the organic compound is 3.

SOLUTION

Rate of S_N?2 reaction ∝  (1/ steric overcrowding in transition state)

So, the reactivity of CH₃?Cl will be the most. 

And as the steric hinderance increases rate of S_N?2 will decrease. And hinderance increases with crowding on the central atom.

Solution

One liter 1N solution of KMnO4? corresponds to 1 g equivalent. In acid medium for oxidation, the equivalent weight of KMnO₄? is one fifth of its molecular mass. It is 15 / 58?=31.6.
Hence, 1 g equivalent of KMnO₄? corresponds to 31.6 g.

Hence, in order to prepare one liter 1N solution of KMnO₄?, 31.6 grams of KMnO4? are required, if the solution to be used in acid medium for oxidation.