Solution Found!
Calculate each of the following: a. number of Li atoms in 4.5 moles of Li b. number of
Chapter 0, Problem 7.4(choose chapter or problem)
Calculate each of the following:
a. number of \(\mathrm{Li}\) atoms in 4.5 moles of \(\mathrm{Li}\)
b. number of \(\mathrm{CO}_{2}\) molecules in 0.0180 mole of \(\mathrm{CO}_{2}\)
c. moles of \(\mathrm{Cu}\) in \(7.8 \times 10^{21}\) atoms of \(\mathrm{Cu}\)
d. moles of \(\mathrm{C}_{2} \mathrm{H}_{6}\) in \(3.75 \times 10^{23}\) molecules of \(\mathrm{C}_{2} \mathrm{H}_{6}\)
Questions & Answers
QUESTION:
Calculate each of the following:
a. number of \(\mathrm{Li}\) atoms in 4.5 moles of \(\mathrm{Li}\)
b. number of \(\mathrm{CO}_{2}\) molecules in 0.0180 mole of \(\mathrm{CO}_{2}\)
c. moles of \(\mathrm{Cu}\) in \(7.8 \times 10^{21}\) atoms of \(\mathrm{Cu}\)
d. moles of \(\mathrm{C}_{2} \mathrm{H}_{6}\) in \(3.75 \times 10^{23}\) molecules of \(\mathrm{C}_{2} \mathrm{H}_{6}\)
ANSWER:Step 1 of 4
(a) Given - Moles of Li = 4.5
1 mole of an element contains Avogadro's number of atoms.
\(1 \text { mole of } \mathrm{Li}=6.02 \times 10^{23} \mathrm{Li} \text { atoms }\)
The conversion factor will be ; \(\frac{6.02\times10^{23}\mathrm{\ Li}\text{ atoms }}{1\mathrm{~mol\ }\mathrm{Li}}\)
Let’s calculate the number of Li atoms in 4.5 moles.
\(4.5 \mathrm{~mol} \times \frac{6.02 \times 10^{23} \mathrm{Li} \text { atoms }}{1 \mathrm{~mol} \mathrm{Li}}=27.1 \times 10^{23} \mathrm{Li} \text { atoms }\)
Therefore, \(27.1 \times 10^{23} \mathrm{Li}\) atoms present in 4.5 moles.