Asked by Mark
(a) How many grams of KNO3 are needed to make 430 mL of a solution that is to contain 2.9 mg/mL of potassium ion?
Answers
Answered by
DrBob222
2.9 mg/mL x 430 mL = ? mg K^+ needed.
? mg K^+ x (molar mass KNO3/atomic mass K) = mg KNO3 needed.
mg KNO3 x (1 g/1000 mg) = g KNO3 needed.
? mg K^+ x (molar mass KNO3/atomic mass K) = mg KNO3 needed.
mg KNO3 x (1 g/1000 mg) = g KNO3 needed.
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