Question
how many moles of copper are in 4.567x10^28 atoms of copper
Answers
GPT-4o mini
To calculate the number of moles of copper in \( 4.567 \times 10^{28} \) atoms, we use Avogadro's number, which is approximately \( 6.022 \times 10^{23} \) atoms/mole.
The formula to calculate the number of moles (\( n \)) is given by:
\[
n = \frac{\text{number of atoms}}{\text{Avogadro's number}}
\]
Now, substituting the values into the formula:
\[
n = \frac{4.567 \times 10^{28} \text{ atoms}}{6.022 \times 10^{23} \text{ atoms/mole}}
\]
Calculating this gives:
\[
n \approx \frac{4.567 \times 10^{28}}{6.022 \times 10^{23}} \approx 7.59 \times 10^4 \text{ moles}
\]
Thus, there are approximately \( 7.59 \times 10^4 \) moles of copper in \( 4.567 \times 10^{28} \) atoms of copper.
The formula to calculate the number of moles (\( n \)) is given by:
\[
n = \frac{\text{number of atoms}}{\text{Avogadro's number}}
\]
Now, substituting the values into the formula:
\[
n = \frac{4.567 \times 10^{28} \text{ atoms}}{6.022 \times 10^{23} \text{ atoms/mole}}
\]
Calculating this gives:
\[
n \approx \frac{4.567 \times 10^{28}}{6.022 \times 10^{23}} \approx 7.59 \times 10^4 \text{ moles}
\]
Thus, there are approximately \( 7.59 \times 10^4 \) moles of copper in \( 4.567 \times 10^{28} \) atoms of copper.