If you needed to make 100 ml of a 0.2 m fruit drink solution from the 1.0 m fruit drink solution, how would you do it? (hint: use msvs = mdvd to find the amount of concentrated solution you need, then add water to reach 100 ml.) show your work.

We take 20 ml of 1.0 M fruit drink and add 80 ml of water to get 100 ml of 0.2 M solution.

Explanation:

According to the neutralization law,

where,

= molarity of stock solution = 0.2 M

= volume of stock solution = 100 ml

= molarity of resulting solution = 1.0 M

= volume of resulting solution = ?

Now put all the given values in the above law, we get the volume of resulting solution.

Therefore, the volume of 1.0 M required is 20 ml.

We take 20 ml of 1.0 M fruit drink and add 80 ml of water to get 100 ml of 0.2 M solution.

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consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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