Answer:
there are **1.06 x 10^19** Ba atoms in 3.50 x 10^3 mg of barium carbonate, BaCO3.
Step-by-step explanation:
To calculate the number of Ba atoms in 3.50 x 10^3 mg of barium carbonate, BaCO3, we can use the following steps:
1. Convert the mass of BaCO3 to grams.
2. Calculate the number of moles of BaCO3.
3. Calculate the number of moles of Ba.
4. Multiply the number of moles of Ba by Avogadro's number to get the number of Ba atoms.
**Step 1: Convert the mass of BaCO3 to grams.**
```
3.50 x 10^3 mg * (1 g / 1000 mg) = 3.50 x 10^-2 g
```
**Step 2: Calculate the number of moles of BaCO3.**
```
Moles of BaCO3 = Mass of BaCO3 / Molar mass of BaCO3
```
The molar mass of BaCO3 is 197.34 g/mol.
```
Moles of BaCO3 = 3.50 x 10^-2 g / 197.34 g/mol = 1.76 x 10^-4 mol
```
**Step 3: Calculate the number of moles of Ba.**
```
Moles of Ba = Moles of BaCO3 * (Moles of Ba / Moles of BaCO3)
```
There is 1 mole of Ba per mole of BaCO3.
```
Moles of Ba = 1.76 x 10^-4 mol * (1 mol Ba / 1 mol BaCO3) = 1.76 x 10^-4 mol
```
**Step 4: Multiply the number of moles of Ba by Avogadro's number to get the number of Ba atoms.**
```
Number of Ba atoms = Moles of Ba * Avogadro's number
```
Avogadro's number is 6.022 x 10^23 atoms/mol.
```
Number of Ba atoms = 1.76 x 10^-4 mol * 6.022 x 10^23 atoms/mol = 1.06 x 10^19 atoms
```
Therefore, there are **1.06 x 10^19** Ba atoms in 3.50 x 10^3 mg of barium carbonate, BaCO3.