## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 4 - Exercises - Page 189: 66

#### Answer

125 mL of $0.150$ M $Li_2S$

#### Work Step by Step

1. Use the molarities and other information as conversion factors: $$125 \space mL \space Co(NO_3)_2 \times \frac{1 \space L}{1000 \space mL} \times \frac{ 0.150 \space mol \space Co(NO_3)_2 }{1 \space L \space Co(NO_3)_2 } \times \frac{ 1 \space mol \space Li_2S }{ 1 \space mol \space Co(NO_3)_2 } \times \frac{1 \space L}{ 0.150 \space mol \space Li_2S } \times \frac{1000 \space mL}{1 \space L} = 125 \space mL$$

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