Asked by Enoch
At 603 K, Kc for the reaction NH4Cl <===> NH3 + HCl is 0.011. What is the equilibrium concentration of NH3 if the initial concentration of NH4Cl is 0.500 M. Please help me with this question because i got 2 answers so prove me.
Answers
Answered by
McDonald Warea
206
Answered by
DrBob222
Why not show your work for the two answers you have so we can find the error. That's a lot easier than us working the problem and letting you look for the error.
......NH4Cl(g) --> NH3(g) + HCl(g)
I......0.5..........0........0
C......-x...........x........x
E.....0.5-x.........x........x
Kc = 0.011 = (x)(x)/(0.50x)
Solve for x.
......NH4Cl(g) --> NH3(g) + HCl(g)
I......0.5..........0........0
C......-x...........x........x
E.....0.5-x.........x........x
Kc = 0.011 = (x)(x)/(0.50x)
Solve for x.
Answered by
LIJAPE JAMES
Please answer me with the working out...
Answered by
Terrence Kasau
please can you help me with the correct answer
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