In $58 \cdot g$ $58*g$ of $ammonium nitrate$ $"ammonium nitrate"$ , what mass of nitrogen is present?

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In $58 \cdot g$ $58*g$ of $ammonium nitrate$ $"ammonium nitrate"$ , what mass of nitrogen is present?

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Answer 1 · 2017-05-31T04:50:57

Approx. $20 \cdot g$ $20*g$ .

Explanation:

We work out a molar quantity of $ammonium nitrate:$ $"ammonium nitrate:"$

$Moles of ammonia nitrate = \frac{58.0 \cdot g}{80.05 \cdot g \cdot m o l^{- 1}} = 0.725 \cdot m o l$ $"Moles of ammonia nitrate"=(58.0*g)/(80.05*g*mol^-1)=0.725*mol$ .

And we take this molar quantity, and multiply it by the appropriately modified molar mass of nitrogen...........

$0.725 \cdot m o l \times 2 \times 14.01 \cdot g \cdot m o l^{- 1} = ? ? \cdot g$ $0.725*molxx2xx14.01*g*mol^-1=??*g$

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In $58 \cdot g$ $58*g$ of $ammonium nitrate$ $"ammonium nitrate"$ , what mass of nitrogen is present?

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Explanation:

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In 58*g58⋅g58*g of "ammonium nitrate"ammonium nitrate"ammonium nitrate", what mass of nitrogen is present?

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In $58 \cdot g$ $58*g$ of $ammonium nitrate$ $"ammonium nitrate"$ , what mass of nitrogen is present?