QUESTION 1 How many grams of Fe(NO2)3 in 0.943 mol? 95.0 g 1299 O 136 9...

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Answer 1

Answer #1

- formular weight of Fe(NO2) = x atomic wt of Fe + 3xatomic wt of N + 6x atomic out of o = (2x 56g)+(3 x 14g) + (6 x 169) =

Answer 2

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