## Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall

# Chapter 1 - Sections 1.1-1.8 - Exercises - Problems by Topic - Page 39: 60b

#### Answer

$1228g/L = 1228 \times 10^{-3} g/mL$ and $1228 \times 10^{-9} kg/ML$

#### Work Step by Step

1)$1228\frac{g}{L}\times \frac{L}{1000mL} = 1.228g/mL = 1228 \times 10^{-3} g/mL$ 2) $1228\frac{g}{L} \times\frac{kg}{1000g} \frac{10^{6}L}{ML} = 1228 \times 10^{-9} kg/ML$

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