Asked by Anonymous
You're asked to prepare 125mL of .0321 M AgNO3. How many grams would you need of a smaple known to be 99.81% AgNO3 by mass?
Answers
Answered by
bobpursley
masspureAgNO3=molmassAgNO3*volumeinLiter*molarity
MassAgNO3= massabove/.9981
MassAgNO3= massabove/.9981
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