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14.0 g x 1 mole/187.56 g = 0.0746 moles0.0746 moles Cu(NO3)2 x 6 moles O/mole Cu(NO3)2 = 0.448 moles O atoms

0.448 moles O atoms x 6.02x10^23 atoms/mole = 2.70x10^23 atoms of oxygen

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Q: How many oxygen atoms are present in a 14.0 g sample of Cu(NO3)2?
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