A chemist studying the properties of photographic emulsions needed to prepare 500 mL of 0.167 M AgNO3(aq).....?
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Answer:
moles AgNO3 = 0.500 L x 0.167 M=0.0835 mass AgNO3 = 0.0835 x 169.87 g/mol=14.2 g
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Other answers
moles AgNO3 = 0.500 L x 0.167 M=0.0835 mass AgNO3 = 0.0835 x 169.87 g/mol=14.2 g
Dr.A
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