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anonymous

  • one year ago

How many grams of nitrogen are present in 46.34 g ammonium nitrate?

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  1. anonymous
    • one year ago
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    mmonium nitrate is NH4NO3 moleculear weight is 80 g / mole nitrogen weight per mole is 28 g / mole % N by weight in a mole = 100 * 28 /80 = 35% 46.34 g amonium nitrate is 46.34*0.35= 16.22 g N

  2. anonymous
    • one year ago
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    @lily768 From where you get the 80

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