A mixture initially contains A, B, and C in the following concentrations: [A] = 0.500 M , [B] = 0.800 M , and [C] = 0.350 M . The following reaction occurs and equilibrium is established: A+2B⇌C At equilibrium, [A] = 0.330 M and [C] = 0.520 M . Calculate the value of the equilibrium constant, Kc.

The value of the equilibrium constant is 7.447.

Explanation:

A  +   2B  ⇌  C

Initially

0.500 M   0.800 M   0.350 M

At equilibrium :

(0.500-x) M (0.800-2x)M     (0.350+x)M

Concentration at equilibrium:

[A] = 0.330 M = (0.500-x)M

x = 0.17 M

[B] = (0.800-2x)M  = 0.800 - 2  0.17 M = 0.46 M

[C] = 0.520 M

The expression of an equilibrium constant will be give as:

The value of the equilibrium constant is 7.447.

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.