A 1.00 M sample HI is placed in a 1-L vessel at 460°C, and the reaction system is allowed to come to equilibrium. The Hl partially decomposes, forming Hz and I2. What is the equilibrium
concentration of HI if the equilibrium constant is 52.8?
H2(g) + 12(g) = 2 HI(g) at 460°C?

[HI] = 0.784M

[I₂] = 0.108M

[H₂] = 0.108M

Explanation:

Based on the equilibrium reaction:

H₂(g) + I₂(g) ⇄ 2HI(g)

The equilibrium constant, K, is:

K = 52.8 = [HI]² / [H₂] [I₂]

Where [] are equilibrium concentrations of each gas.

As initial concentration of HI is 1.00M, the equilibrium concentrations of the gases is:

[HI] = 1.00M - 2X

[I₂] = X

[H₂] = X

Replacing:

52.8 = [1.00-2X]² / [X] [X]

52.8X² = 4X² - 4X + 1

0 = -48.8X² - 4X + 1

Solving for X:

X = -0.1899M. False solution, there is no negative concentrations

X = 0.108M. Right solution.

Replacing, equilibrium concentrations are:

[HI] = 1.00M - 2*0.108M

it's c (engineer's do their work with mechanical stuff not on any living being)

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.