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The equilibrium constant K, for the reaction below is is 7.755x10 CD+E 1st attempt See Periodic Table Part 1 (1point) The ini
Part 3 (1 point) Feedback See Hint What is the equilibrium concentration of E? O 1.978 x10M
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Answer #1

                C----------------> D      +   E

I             1.111*10^-2       1.111*10^-2           1.111*10^-2

C              +x                           -x                      -x

E            1.111*10^-2+x       1.111*10^-2 -x     1.111*10^-2 -x

        Kc   = [D][E]/[C]

       7.755*10^-5   = (1.111*10^-2-x)(1.111*10^-2-x)/(1.111*10^-2 + x)

     7.755*10^-5*(1.111*10^-2 + x)    = (1.111*10^-2-x)(1.111*10^-2-x)

             x   = 0.00984

[C]      = 1.111*10^-2 +0.00984   = 0.02095M

[D]   = 1.111*10^-2 -0.00984       = 0.00127M

[E]    = 1.111*10^-2 -0.00984       = 0.00127M

                    

            

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