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(Solved): The equilibrium constant, K_(c), for the following reaction is 1.80\times 10^(-4) at 298K. NH_(4)HS( ...
The equilibrium constant,
K_(c), for the following reaction is
1.80\times 10^(-4)at
298K.
NH_(4)HS(s)⇌NH_(3)(g)+H_(2)S(g)This reaction is
◻favored at equilibrium. Enter PRODUCT or REACTANT. The concentrations of
NH_(3)and
H_(2)Swill be
◻at equilibrium. Enter HIGH or LOW.
◻
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