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Evgesh-ka [11]
3 years ago
14

How many atoms of phosphorus are in 7.00 mol of copper(II) phosphate?

Chemistry
1 answer:
Rama09 [41]3 years ago
3 0

<u>Answer:</u> The number of phosphorus atoms in given amount of copper(II) phosphate is 8.431\times 10^{24}

<u>Explanation:</u>

We are given:

Moles of copper(II) phosphate Cu_3(PO_4)_2 = 7.00 mol

1 mole of copper(II) phosphate contains 3 moles of copper, 2 moles of phosphorus and 8 moles of oxygen atoms

Moles of phosphorus in copper(II) phosphate = (2\times 7.00mol

According to the mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of particles

So, 7.00 moles of copper(II) phosphate will contain = (2\times 7\times 6.022\times 10^{23}=8.431\times 10^{24} number of phosphorus atoms.

Hence, the number of phosphorus atoms in given amount of copper(II) phosphate is 8.431\times 10^{24}

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