What is the oxidation number for chromium in each of these compounds. show your work. a. k2cro4 b. cr2o3

A. + 6.

B. + 3.

Explanation:

A. K₂CrO₄:

We have that the sum of the oxidation no. of different elements in the compound multiplied by its no. is equal to the overall charge of the compound.

∵ 2(oxidation no. of K) + (oxidation no. of Cr) + 4(oxidation no. of O) = 0.

oxidation no. of K = + 1, oxidation no. of O = - 2.

∴ 2(+ 1) + (oxidation no. of Cr) +4(- 2) = 0.

∴ 2 + (oxidation no. of Cr) - 8 = 0.

∴ (oxidation no. of Cr) - 6 = 0.

∴ oxidation no. of Cr = + 6.

B. Cr₂O₃:

∵ 2(oxidation no. of Cr) + 3(oxidation no. of O) = 0.

oxidation no. of O = - 2.

∴ 2(oxidation no. of Cr) + 3(- 2) = 0.

∴ 2(oxidation no. of Cr) - 6 = 0.

∴ 2(oxidation no. of Cr) = + 6.

∴ (oxidation no. of Cr) = + 6/2 = + 3.

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%