Chemistry (7th Edition)

Published by Pearson
ISBN 10: 0321943171
ISBN 13: 978-0-32194-317-0

Chapter 1 - Chemical Tools: Experimentation and Measurement - Section Problems - Page 29: 50

Answer

- There are $10^9$ $pg$ in 1 mg and $3.5 \times 10^4$ $pg$ in $35$ $ng$.

Work Step by Step

1 mg $ = 10^{-3}g \times \frac{1 pg}{10^{-12}g} = 10^9pg$ - There are $10^9$ $pg$ in 1 mg. $35 ng \times \frac{10^{-9}g}{1ng} \times \frac{1pg}{10^{-12}} = 3.5 \times 10^4$ $pg$ - There are $3.5 \times 10^4$ $pg$ in $35$ $ng$.
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