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Chemistry 14 Online
OpenStudy (anonymous):

How many GRAMS of oxygen are present in 5.50E22 molecules of dinitrogen tetraoxide ?

OpenStudy (kc_kennylau):

5.50E22 molecules of dinitrogen tetraoxide = 2.20E23 atoms of oxygen = 16/6.022E23*2.20E23 grams = 5.845 grams

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