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The equilibrium constant for the following reaction is 78:      A   +   B    <---->            2 C    

The equilibrium constant for the following reaction is 78:  

   A   +   B    <---->            2 C

        If 0.80 mol of A and 0.80 mol of B in a 1.000 L reaction vessel is allowed to reach equilibrium, what is the concentration of C?  

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Answer #1

keam =78. A+B ka when A and B are in eam with c then we can write keam = [.]² [c] = concentration of a [A16] [B] - - » 78 & [​​​​​​I hope you will understand this. Thank you...

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