can some one please help me please fast fast

label Chemistry
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Carbonyl bromide, COBr2, decomposes to CO and Br2 by the following simple, gas-phase, reaction: 
COBr2(g) ↔ Br2(g) + CO(g)

If a 2.0 liter flask is found to contain 0.093 mol of COBr2 at equilibrium, what will be the equilibrium concentrations of CO and Br2 in the same flask? The equilibrium constant for this reaction is Kc= 0.190. Enter your answer in scientific notation

May 6th, 2015

DATA:

Moles of CoBr2 = (0.093 mol*1000)/2000 = 0.0465M

Keq = 0.190

[Co]=?

[Br2]=?

SOLUTION:

Keq = [Br2][Co]/[CoBr2]

Suppose,

X= [Br2]

As,

[Br2]=[Co]

So,

X=x

Put the values in equation;

0.190= (x*x)/0.0465

X2=0.0088

X= 0.0938M

S0, the molar concentration of [Br2] and [Co] is 0.0938M


May 6th, 2015

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May 6th, 2015
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