At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.100 M [N2]=[O2]=0.100 M and [ NO ] = 0.600 M . [NO]=0.600 M. N 2 ( g ) + O 2 ( g ) − ⇀ ↽ − 2 NO ( g ) N2(g)+O2(g)↽−−⇀2NO(g) If more NO NO is added, bringing its concentration to 0.900 M, 0.900 M, what will the final concentration of NO NO be after equilibrium is re‑established?

The final concentration of NO be after equilibrium is re‑established is 0.825 M.

Explanation:

The given balanced chemical equation is as follows.

Now, equilibrium constant for this reaction will be as follows.

It is given that concentrations at the equilibrium are:

    = 0.1 M and [NO] = 0.6 M

Therefore, the value of is as follows.

                     = 36.0

NO concentration of 0.9 M is added to the system. So,

Initial:      0.1        0.1          0.9

Change:    x         x             2x

Equibm:(0.1 + x)  (0.1 + x)   (0.9 - 2x)

Now, we will find the value of x as follows.

         36.0 =

          x = 0.0375

Therefore, final concentration of NO after the equilibrium that is re-established is as follows.

                     0.9 - 2x

                  =

                  = 0.9 - 0.075

                  = 0.825 M

Therefore, we can conclude that the final concentration of NO be after equilibrium is re‑established is 0.825 M.

the answer is na (sodium)