Respuesta :
Answer:
20.4 moles of [tex]PBr_{3}[/tex] contain [tex]3.68\times 10^{25}[/tex] bromine atoms
Explanation:
1 molecule of [tex]PBr_{3}[/tex contains one P atom and three Br atoms
So 1 mol of [tex]PBr_{3}[/tex contains 3 moles of Br.
We know that 1 mol = [tex]6.023\times 10^{23}[/tex] molecules
So, there are [tex](3\times 6.023\times 10^{23})[/tex] atoms of Br in 1 mol of [tex]PBr_{3}[/tex]
Hence [tex]3.68\times 10^{25}[/tex] bromine atoms are present in [tex]\frac{3.68\times 10^{25}}{3\times 6.023\times 10^{23}}[/tex] moles of [tex]PBr_{3}[/tex] or 20.4 moles of [tex]PBr_{3}[/tex]
The moles of phosphorus bromide with 3.68 × 10²⁵ bromine atoms have been 20.36 mol.
The moles have been defined as the mass of the compound with respect to the molecular mass of the compound. The number of atoms in a mole of the compound has been equivalent to the Avogadro number.
The Avogadro number has been a constant that has a value of 6.023 [tex]\rm \times\;10^2^3[/tex] atoms.
The given phosphorus bromide has 3 bromine atoms in a molecule. The mass of a molecule has been equivalent to the molar mass of the compound that has been 270.69 g/mol.
1 mole = 6.023 [tex]\rm \times\;10^2^3[/tex] atoms
1 mole [tex]\rm PBr_3[/tex] = 3 times bromine
1 mole [tex]\rm PBr_3[/tex] = 3 × 6.023 [tex]\rm \times\;10^2^3[/tex] Bromine atoms
1 mole [tex]\rm PBr_3[/tex] = 18.069 [tex]\rm \times\;10^2^3[/tex] Bromine atoms
The moles of phosphorus bromide with 3.68 × 10²⁵ bromine atoms has been:
18.069 [tex]\rm \times\;10^2^3[/tex] Bromine atoms = 1 mole [tex]\rm PBr_3[/tex]
3.68 × 10²⁵ bromine atoms = [tex]\rm \dfrac{1}{18.069\;\times\;10^2^3} \;\times\;3.68\;\times\;10^2^5[/tex] mol
= 20.36 mol.
The moles of phosphorus bromide with 3.68 × 10²⁵ bromine atoms have been 20.36 mol.
For more information about the atoms in a mole compound, refer to the link:
https://brainly.com/question/1445383
