At 350°c, keq = 1.67 × 10-2 for the reversible reaction 2hi (g) ⇌ h2 (g) + i2 (g). what is the concentration of hi at equilibrium if [ h2 ] is 2.44 × 10-3 m and [ i2 ] is 7.18 × 10-5 m?

According to the reversible reaction equation:

2Hi(g) ↔ H2(g) + i2(g)

and when Keq is the concentration of the products / the concentration of the reactants.

Keq = [H2][i2]/[Hi]^2

when we have Keq = 1.67 x 10^-2

[H2] = 2.44 x 10^-3

[i2] = 7.18 x 10^-5

so, by substitution:

1.67 x 10^-2 = (2.44 x 10^-3)*(7.18x10^-5)/[Hi]^2

∴[Hi] = 0.0033 M

Answer : option a) 84explanation :

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.

Answercreating graphs and performing calculations;