How many atoms of phosphorus are in 7.00 mol of copper(II) phosphate?. Created by rigrace21. chemistry-us

Answer: The number of phosphorus atoms in given amount of copper(II) phosphate is Explanation:We are given:Moles of copper(II) phosphate = 7.00 mol1 mole of copper(II) phosphate contains 3 moles of copper, 2 moles of phosphorus and 8 moles of oxygen atomsMoles of phosphorus in copper(II) phosphate = According to the mole concept:1 mole of a compound contains number of particlesSo, 7.00 moles of copper(II) phosphate will contain = number of phosphorus atoms.Hence, the number of phosphorus atoms in given amount of copper(II) phosphate is

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

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rigrace21

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

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BatteringRam

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