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1. How many milliliters of a 0.184 M NaNO3 solution contain 0.113 moles of NaNO3?ans:614 mL

Jodi Brooks

in Chemistry

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Ashley Howard on November 7, 2018

C=0.184 mol/Ln=0.113 moln=CVV=n/CV=0.113/0.184=0.614 L = 614 mL


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