Consider the following data. no2(g) equilibrium reaction arrow identifying the requirement for light no(g) + o(g) k = 6.8 ✕ 10-49 o3(g) + no(g) equilibrium reaction arrow no2(g) + o2(g) k = 5.8 ✕ 10-34 calculate a value for the equilibrium constant for the reaction below. (hint: when reactions are added together, the equilibrium expressions are multiplied.)

Explanation:

Reactions:

With:

With:

To achive the reaction we combine those two reactions:

The equilibrium constant:

6.4 cm (max) - 25 cm (min)

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.