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  • 04-02-2020
  • Chemistry
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Calculate the moles of O atoms in 0.658 g of Mg(NO3)2.

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Tringa0 Tringa0
  • 04-02-2020

Answer:

There are 0.0267 moles of oxygen in 0.658 grams of magnesium nitrate.

Explanation:

Mass of magnesium nitrate = m = 0.658 g

Molar mass of magnesium nitrate = M = 148 g/mol

Moles of magnesium nitrate = n

[tex]n=\frac{m}{M}=\frac{0.658 g}{148 g/mol}=0.004446 mol[/tex]

1 mole of magnesium nitrate has 6 moles of oxygen atoms. Then 0.004446 moles magnesium nitrate will have :

[tex]6\times 0.004446 mol=0.026676 mol\approx 0.0267[/tex]

There are 0.0267 moles of oxygen in 0.658 grams of magnesium nitrate.

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