A) 0.00358 M
B) 0.0358 M
C) 0.358 M
D) 3.58 M
Correct Answer: C
Solution :
\[{{K}_{c}}=\frac{{{[N{{H}_{3}}]}^{2}}}{[{{N}_{2}}]\,\,{{[{{H}_{2}}]}^{3}}}\] \[2.37\times {{10}^{-3}}=\frac{{{x}^{2}}}{[2]\,\,{{[3]}^{3}}}={{x}^{2}}=0.12798\] x = 0.358 M.You need to login to perform this action.
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