Question

how many helium atoms are contained in each of the following number of moles? be sure each of your answer entries has the correct number of significant figures.
note: reference the fundamental constants table for additional information.
part 1 of 3
6.60 mol he = \square he atoms
part 2 of 3
0.350 mol he = \square he atoms

Explanation:

Part 1 of 3

Step 1: Recall Avogadro's number

Avogadro's number is \( 6.022 \times 10^{23} \) atoms/mol. To find the number of atoms, we multiply the number of moles by Avogadro's number.

Step 2: Calculate the number of He atoms

Given \( 6.60 \) mol He, the number of atoms \( N = n \times N_A \), where \( n = 6.60 \) mol and \( N_A = 6.022 \times 10^{23} \) atoms/mol.
\[
N = 6.60 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol}
\]
\[
N = 6.60 \times 6.022 \times 10^{23}
\]
\[
N \approx 3.97452 \times 10^{24}
\]
Rounding to three significant figures (since 6.60 has three), we get \( 3.97 \times 10^{24} \) (or using the calculator's precision, maybe \( 3.97 \times 10^{24} \) or more accurately \( 3.97452\times 10^{24}\approx 3.97\times 10^{24}\) when considering significant figures). But let's do the exact multiplication: \( 6.60\times6.022 = 39.7452\), so \( 39.7452\times 10^{23}=3.97452\times 10^{24}\approx 3.97\times 10^{24}\) (or if we use \( 6.02\times 10^{23}\) for simplicity, \( 6.60\times6.02\times 10^{23}= 39.732\times 10^{23}=3.9732\times 10^{24}\approx 3.97\times 10^{24}\)).

Wait, actually, let's check the significant figures. 6.60 has three, Avogadro's number is often taken as \( 6.022\times 10^{23} \) (four significant figures), so the result should have three. So \( 6.60\times6.022\times 10^{23}= 39.7452\times 10^{23}=3.97452\times 10^{24}\approx 3.97\times 10^{24}\) (or \( 3.97\times 10^{24} \) when rounded to three significant figures). But maybe the system expects using \( 6.02\times 10^{23} \), let's recalculate: \( 6.60\times6.02 = 39.732 \), so \( 39.732\times 10^{23}=3.9732\times 10^{24}\approx 3.97\times 10^{24}\).

Alternatively, using a calculator: \( 6.60 \times 6.02214076\times 10^{23} = 6.60\times6.02214076 = 39.746129016 \), so \( 39.746129016\times 10^{23}=3.9746129016\times 10^{24}\approx 3.97\times 10^{24}\) (three significant figures).

Step 1: Recall Avogadro's number

Avogadro's number is \( 6.022 \times 10^{23} \) atoms/mol. To find the number of atoms, we multiply the number of moles by Avogadro's number.

Step 2: Calculate the number of He atoms

Given \( 0.350 \) mol He, the number of atoms \( N = n \times N_A \), where \( n = 0.350 \) mol and \( N_A = 6.022 \times 10^{23} \) atoms/mol.
\[
N = 0.350 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol}
\]
\[
N = 0.350 \times 6.022 \times 10^{23}
\]
\[
N \approx 2.1077 \times 10^{23}
\]
Rounding to three significant figures (since 0.350 has three), we get \( 2.11 \times 10^{23} \) (or more accurately, \( 2.1077\times 10^{23}\approx 2.11\times 10^{23}\) with three significant figures).

Answer:

\( 3.97 \times 10^{24} \) (or in the box, if we consider the format, maybe \( 3.97\times10^{24} \), but let's check the exact calculation. Wait, maybe the problem expects using \( 6.02\times 10^{23} \), so \( 6.60\times6.02\times 10^{23}= 39.732\times 10^{23}=3.9732\times 10^{24}\approx 3.97\times 10^{24}\). So the answer is \( 3.97 \times 10^{24} \) (or \( 3.97\times10^{24} \)).

Part 2 of 3