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

Published by Prentice Hall

# Chapter 2 - Sections 2.1-2.9 - Exercises - Problems by Topic: 90c

#### Answer

$111\ kg\ Mn$

#### Work Step by Step

1. Use Avogadro's number to convert from atoms to moles of manganese. Then use the molar mass of manganese to convert from moles to grams of manganese. Finally, convert grams into kilograms. 2. $1.22\times10^{27}\ atoms\ Mn\times\frac{1\ mol\ Mn}{6.022\times 10^{23}\ atoms\ Mn}\times\frac{54.94\ g \ Mn}{1\ mol\ Mn}\times\frac{1\ kg\ Mn}{1000\ g\ Mn}= 111\ kg\ Mn$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.