3 years ago

How many grams of potassium nitrate (KNO3) are required to prepare 0.250L of a 0.700 M

Chemistry

1 answer:

CaHeK987 [17]3 years ago

6 0

Answer:

17.78g

Explanation:

m/M = C×V

m/101.102= 0.7× 0.25

m= 101.102×0.7×0.25= 17.78g

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