Certainly! Let's work through the problem step by step:
### Step 1: Determine the molar masses.
First, we need to know the molar masses of Na₂CO₃ and He.
- Molar mass of Na₂CO₃ (sodium carbonate):
- Sodium (Na): 23.00 g/mol × 2 = 46.00 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol × 3 = 48.00 g/mol
Adding these together,
[tex]\[
46.00 + 12.01 + 48.00 = 106.01 \text{ g/mol}
\][/tex]
- Molar mass of Helium (He):
[tex]\[
4.00 \text{ g/mol}
\][/tex]
### Step 2: Calculate the number of moles.
Next, using the given masses, we can find the number of moles for each compound.
- Moles of Na₂CO₃:
Using the formula:
[tex]\[
\text{Moles} = \frac{\text{Given mass (g)}}{\text{Molar mass (g/mol)}}
\][/tex]
[tex]\[
\text{Moles of Na₂CO₃} = \frac{5.3 \text{ g}}{106.01 \text{ g/mol}} \approx 0.0500 \text{ moles}
\][/tex]
- Moles of He:
[tex]\[
\text{Moles of He} = \frac{52 \text{ g}}{4.00 \text{ g/mol}} = 13.0 \text{ moles}
\][/tex]
### Step 3: Utilize Avogadro's number to find the number of atoms.
Avogadro's number is [tex]\(6.022 \times 10^{23}\)[/tex] atoms per mole, which will help us convert moles to atoms.
- Number of atoms in Na₂CO₃:
Sodium carbonate has 6 atoms per molecule (2 Na, 1 C, 3 O).
The total number of atoms:
[tex]\[
\text{Atoms of Na₂CO₃} = 0.0500 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mole} \times 6
\][/tex]
[tex]\[
= 1.8068 \times 10^{23} \text{ atoms}
\][/tex]
- Number of atoms in He:
Helium (He), being monoatomic, means each mole is just Avogadro's number.
[tex]\[
\text{Atoms of He} = 13.0 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mole}
\][/tex]
[tex]\[
= 7.8286 \times 10^{24} \text{ atoms}
\][/tex]
### Final Result:
- Moles of Na₂CO₃: [tex]\(0.0500 \text{ moles}\)[/tex]
- Moles of He: [tex]\(13.0 \text{ moles}\)[/tex]
- Number of atoms in Na₂CO₃: [tex]\(1.8068 \times 10^{23} \text{ atoms}\)[/tex]
- Number of atoms in He: [tex]\(7.8286 \times 10^{24} \text{ atoms}\)[/tex]
