Answer: Concentration of
in the equilibrium mixture is 0.31 M
Explanation:
Equilibrium concentration of
= 0.729 M
The given balanced equilibrium reaction is,

Initial conc. x 0 0
At eqm. conc. (x-2y) M (y) M (3y) M
The expression for equilibrium constant for this reaction will be:
3y = 0.729 M
y = 0.243 M
![K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5By%5D%5Ctimes%20%5B3y%5D%5E3%7D%7B%5Bx-2y%5D%5E2%7D)
Now put all the given values in this expression, we get :



concentration of
in the equilibrium mixture = 
Thus concentration of
in the equilibrium mixture is 0.31 M
<span>In a mole of anything, there are 6.023 x 10^23 units. So, in 3.9 moles of sulfur, there are 3.9 * 6.023 x 10^23 = 23 x 10^23 = 2.3 x 10^24 atoms (keeping only 2 sig figs). Hope I help!!
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It can hold 4 minutes more electrons.