Answer:
[Cl2] equilibrium = 0.0089 M
Explanation:
<u>Given:</u>
[SbCl5] = 0 M
[SbCl3] = [Cl2] = 0.0546 M
Kc = 1.7*10^-3
<u>To determine:</u>
The equilibrium concentration of Cl2
<u>Calculation:</u>
Set-up an ICE table for the given reaction:

I 0 0.0546 0.0546
C +x -x -x
E x (0.0546-x) (0.0546-x)
![Kc = \frac{[SbCl3][Cl2]}{[SbCl5]}\\\\1.7*10^{-3} =\frac{(0.0546-x)^{2} }{x} \\\\x = 0.0457 M](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BSbCl3%5D%5BCl2%5D%7D%7B%5BSbCl5%5D%7D%5C%5C%5C%5C1.7%2A10%5E%7B-3%7D%20%3D%5Cfrac%7B%280.0546-x%29%5E%7B2%7D%20%7D%7Bx%7D%20%5C%5C%5C%5Cx%20%3D%200.0457%20M)
The equilibrium concentration of Cl2 is:
= 0.0546-x = 0.0546-0.0457 = 0.0089 M
There are several ways of expressing concentration of solution. Few of them are listed below
1) mass percentage
2) volume percentage
3) Molarity
4) Normality
5) Molality
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It implies that, 3ml H2O2 is present in 100 ml of solution
Thus, 400 ml of solution would contain 4 X 3 = 12 ml H2O2
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what are the statements that you are able to choose from or do you just have to know?