J.R. S. answered 05/07/19
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2.00 L x 2.00 mol/L = 4 moles needed
4 moles x 63 g/mole = 252 g
Estefany M.
asked 05/07/19How many grams of solute is needed to prepare a solution of 2.00 L of 2.00 M nitric acid, HNO3, solution?
J.R. S. answered 05/07/19
Ph.D. University Professor with 10+ years Tutoring Experience
2.00 L x 2.00 mol/L = 4 moles needed
4 moles x 63 g/mole = 252 g
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