A concentration cell consists of two Sn/Sn2+ half-cells. The solution in one half-cell A is 0.13 M Sn(NO3)2 and is 0.87 M Sn(NO3
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Answer:
47.68 mL
Explanation:
In this case, we have a <u>dilution problem.</u> So, we have to start with the dilution equation:
We have to remember that in a dilution procedure we go from a <u>higher concentration to a lower one</u>. With this in mind, We have to identify the <u>concentration values</u>:
The higher concentration is C1 and the lower concentration is C2. Now, we can identify the <u>volume values</u>:
The V2 value has <u>"mL"</u> units, so V1 would have <u>"mL"</u> units also. Now, we can include all the values into the equation and <u>solve for "V1"</u>, so:
So, we have to take 47.68 mL of the 6 M and add 139.31 mL of water (187-47.68) to obtain a solution with a final concentration of 1.53 M.
I hope it helps!