Asked by Mike
What mass of pure silver would you obtain from 85.0 g of AgNO3?
I got 53.55 g is that correct?
I got 53.55 g is that correct?
Answers
Answered by
Damon
Ag = 108 g/mol
N =14 g/mol
O3 = 3*16 = 48 g/mol
so AgNO3 = 170 g/mol of which 108 is silver
(108/170)(85) = 54, close enough
N =14 g/mol
O3 = 3*16 = 48 g/mol
so AgNO3 = 170 g/mol of which 108 is silver
(108/170)(85) = 54, close enough
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