The percent concentration of a solution can be calculated from; mass of solute /mass of solution * 100. The mass of the solute here is 8.1 g.
<h3>What is concentration?</h3>
The term concentration refers to the amount of solute presnt in a solution. There are many ways of expressing concentration such as molarity, molality and percentage.
Here;
mass of solution = 230.5 g
Percent of solute = 3.5 %
3.5 = x/ 230.5 * 100
3.5 = 100x/230.5
230.5(3.5) = 100x
x = 230.5(3.5) /100
x = 8.1 g
Learn more about percent concentration: brainly.com/question/202460?
Answer: theres no image or claim
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
MgCl2 because it is the only option in which a metal appears with a nonmetal. In this case, the metal transfers electrons to the nonmental because the metal has a lower ionization energy.
