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To solve this problem, we need to determine the number of atoms of Barium (Ba), Phosphorus (P), and Oxygen (O) in 0.5 moles of Barium Phosphate, which has the chemical formula \( \text{Ba}_3(\text{PO}_4)_2 \). Here's a step-by-step solution to find the number of atoms for each element:
1. Identify the composition of Barium Phosphate \(\text{Ba}_3(\text{PO}_4)_2 \):
- Barium Phosphate contains:
- 3 atoms of Barium (Ba)
- 2 units of phosphate ion (\(\text{PO}_4\)).
- In each phosphate ion (\(\text{PO}_4\)), there are:
- 1 atom of Phosphorus (P)
- 4 atoms of Oxygen (O)
2. Calculate the number of atoms in 1 mole of Barium Phosphate:
- Barium (Ba):
- There are 3 atoms of Barium per formula unit of Barium Phosphate.
- Phosphorus (P):
- There are 2 phosphate ions per formula unit.
- Each phosphate ion contains 1 atom of Phosphorus.
- Therefore, there are \(2 \times 1 = 2\) atoms of Phosphorus per formula unit of Barium Phosphate.
- Oxygen (O):
- There are 2 phosphate ions per formula unit.
- Each phosphate ion contains 4 atoms of Oxygen.
- Therefore, there are \(2 \times 4 = 8\) atoms of Oxygen per formula unit of Barium Phosphate.
3. Factor in Avogadro's number:
- Avogadro's number (\(N_A\)) is \(6.022 \times 10^{23}\) atoms/mol, which means that 1 mole of any substance contains \(6.022 \times 10^{23}\) representative particles (in this case, atoms).
4. Calculate the number of atoms in 1 mole of Barium Phosphate:
- Barium (Ba) atoms in 1 mole:
- \(3 \times 6.022 \times 10^{23} = 1.8066 \times 10^{24}\) atoms
- Phosphorus (P) atoms in 1 mole:
- \(2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24}\) atoms
- Oxygen (O) atoms in 1 mole:
- \(8 \times 6.022 \times 10^{23} = 4.8176 \times 10^{24}\) atoms
5. Adjust for 0.5 moles of Barium Phosphate:
- Since the given quantity is 0.5 moles, multiply each of the above results by 0.5:
- Barium (Ba) atoms in 0.5 moles:
- \(1.8066 \times 10^{24} \times 0.5 = 9.033 \times 10^{23}\) atoms
- Phosphorus (P) atoms in 0.5 moles:
- \(1.2044 \times 10^{24} \times 0.5 = 6.022 \times 10^{23}\) atoms
- Oxygen (O) atoms in 0.5 moles:
- \(4.8176 \times 10^{24} \times 0.5 = 2.4088 \times 10^{24}\) atoms
Thus, in 0.5 moles of Barium Phosphate, there are:
- \(9.033 \times 10^{23}\) atoms of Barium (Ba)
- \(6.022 \times 10^{23}\) atoms of Phosphorus (P)
- [tex]\(2.4088 \times 10^{24}\)[/tex] atoms of Oxygen (O)
1. Identify the composition of Barium Phosphate \(\text{Ba}_3(\text{PO}_4)_2 \):
- Barium Phosphate contains:
- 3 atoms of Barium (Ba)
- 2 units of phosphate ion (\(\text{PO}_4\)).
- In each phosphate ion (\(\text{PO}_4\)), there are:
- 1 atom of Phosphorus (P)
- 4 atoms of Oxygen (O)
2. Calculate the number of atoms in 1 mole of Barium Phosphate:
- Barium (Ba):
- There are 3 atoms of Barium per formula unit of Barium Phosphate.
- Phosphorus (P):
- There are 2 phosphate ions per formula unit.
- Each phosphate ion contains 1 atom of Phosphorus.
- Therefore, there are \(2 \times 1 = 2\) atoms of Phosphorus per formula unit of Barium Phosphate.
- Oxygen (O):
- There are 2 phosphate ions per formula unit.
- Each phosphate ion contains 4 atoms of Oxygen.
- Therefore, there are \(2 \times 4 = 8\) atoms of Oxygen per formula unit of Barium Phosphate.
3. Factor in Avogadro's number:
- Avogadro's number (\(N_A\)) is \(6.022 \times 10^{23}\) atoms/mol, which means that 1 mole of any substance contains \(6.022 \times 10^{23}\) representative particles (in this case, atoms).
4. Calculate the number of atoms in 1 mole of Barium Phosphate:
- Barium (Ba) atoms in 1 mole:
- \(3 \times 6.022 \times 10^{23} = 1.8066 \times 10^{24}\) atoms
- Phosphorus (P) atoms in 1 mole:
- \(2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24}\) atoms
- Oxygen (O) atoms in 1 mole:
- \(8 \times 6.022 \times 10^{23} = 4.8176 \times 10^{24}\) atoms
5. Adjust for 0.5 moles of Barium Phosphate:
- Since the given quantity is 0.5 moles, multiply each of the above results by 0.5:
- Barium (Ba) atoms in 0.5 moles:
- \(1.8066 \times 10^{24} \times 0.5 = 9.033 \times 10^{23}\) atoms
- Phosphorus (P) atoms in 0.5 moles:
- \(1.2044 \times 10^{24} \times 0.5 = 6.022 \times 10^{23}\) atoms
- Oxygen (O) atoms in 0.5 moles:
- \(4.8176 \times 10^{24} \times 0.5 = 2.4088 \times 10^{24}\) atoms
Thus, in 0.5 moles of Barium Phosphate, there are:
- \(9.033 \times 10^{23}\) atoms of Barium (Ba)
- \(6.022 \times 10^{23}\) atoms of Phosphorus (P)
- [tex]\(2.4088 \times 10^{24}\)[/tex] atoms of Oxygen (O)
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