A stock solution with a concentration of 6.53 M is used to make 250 mL of a 1.38 M solution. How many milliliters of stock solution will you need?

The answer to your question is 52 ml

Explanation:

Data

Volume 1 = ?

Concentration 1 = 6.63M

Volume 2 = 250 ml

Concentration 2 = 1.38 M

To solve this problem use the dilution equation.

Formula

                Volume 1 x Concentration 1 = Volume 2 x Concentration 2

-Solve for Volume 1

                Volume 1 = Volume 2 x Concentration 2 / Concentration 1

-Substitution

                 Volume 1 = (250 x 1.38) / 6.63

-Simplification

                  Volume 1 = 345 / 6.63

-Result

                 Volume 1 = 52.04 ml

-Conclusion

I need 52 ml of the stock solution.

the correct name for the n3 ion is azide. azide is a polyatomic anion with a -1 charge and is written as n3 -1. it consists of three nitrogen atoms bonded together. hope i . let me know if you need anything else.