Answer :

There are approximately 6.83 x 10^22 atoms of oxygen in 39.3 moles of Cr3(PO4)2. To determine the number of atoms of oxygen (O) in 39.3 moles of Cr3(PO4)2, we need to use Avogadro's number and the mole-to-atom conversion factor.

First, we calculate the molar mass of Cr3(PO4)2.

Cr (chromium) has a molar mass of 52 g/mol, P (phosphorus) has a molar mass of 31 g/mol, and O (oxygen) has a molar mass of 16 g/mol. We multiply the molar mass of each element by its subscript and add them up:

3 x 52 g/mol (Cr) + 2 x (31 g/mol (P) + 4 x 16 g/mol (O)) = 3 x 52 g/mol + 2 x 31 g/mol + 8 x 16 g/mol = 156 g/mol + 62 g/mol + 128 g/mol = 346 g/mol.

Next, we use Avogadro's number, which is 6.022 x 10^23 atoms/mol.

To find the number of moles of oxygen, we divide the given mass (39.3 moles) by the molar mass of Cr3(PO4)2 (346 g/mol):

39.3 moles / 346 g/mol = 0.1136 moles.

Finally, we multiply the number of moles of oxygen by Avogadro's number to find the number of atoms:

0.1136 moles x 6.022 x 10^23 atoms/mol = 6.83 x 10^22 atoms.

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