For which of the following equilibria does `"k"_("eq") = ["o"_2]`? a. o2(l) o2(g) b. 2o3(g) 3o2(g) c. 2h2o(l) 2h2(g) + o2(g) d. 2hg(s) +o2(g) 2hgo(s)

For plato users

the answer is a. O2(l) O2(g)

hope this helps!

For a hypothetical reaction:

aA ↔ bB

the equilibrium constant is given as:

Keq = [B]^b/[A]^a

activity of pure solids and liquids = 1

For the given reactions:

A) O2(l) ↔ O2(g)

Keq = [O₂]

B) 2O3(g) ↔ 3O2(g)

Keq = [O₂]³/[O₃]²

C) 2H2O(l) ↔2H2(g) + O2(g)

Keq = [H₂]²[O₂]

D) 2Hg(s) + O2(g) ↔ 2HgO(s)

Keq = [Hg]²[O₂]

Ans: A

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.

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