Answer
.06 g
Work Step by Step
We find:
$$ \frac{250*10^{-3} g}{1} \frac{mol \ molecules}{387.31 g} \frac{3 \ mol \ P}{mol \ molecules} \frac{30.97 g}{mol \ P} = .06 g$$
Chemistry: The Molecular Science (5th Edition)
by Moore, John W.; Stanitski, Conrad L.
Published by
Cengage Learning
ISBN 10:
1285199049
ISBN 13:
978-1-28519-904-7
Chapter 2 - Chemical Compounds - Questions for Review and Thought - Topical Questions - Page 90d: 77c
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