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Problem 2 SolutionIn 80.042 g (1 mole) of ammonium nitrate (bomb city!), there are 28.014 grams of nitrogen, 47.997 g of oxygen and the rest is hydrogen. a. What percent of ammonium nitrate is nitrogen? b. What percent of ammonium nitrate is oxygen? c. What percent of ammonium nitrate is hydrogen? (a) 34.999% (b) 59.965% (c) 5.036% Solution Steps for Part (a):What percent of ammonium nitrate is nitrogen?
Answer:We'll fill in the amounts that we know, and use the clues from the statement:What% of 80.042 is 28.014 x * 80.042 = 28.014 dividing by 80.042 gives x = .34998688. Since these are obviously measurements, we will round to 5 digits: 34.999% Solution Steps for Part (b):What percent of ammonium nitrate is oxygen?
Answer:We'll use the clues from the statement:What% of 80.042 is 47.997 x * 80.042 = 47.997 dividing by 80.042 gives x = .59964769. Since these are obviously measurements, we will round to 5 digits: 59.965%. Notice that the units on 80.042 and 47.997 cancel out when we divide. Percents really are dimensionless! Solution Steps for Part (c):What percent of ammonium nitrate is hydrogen?
Answer:Let's take the easy way out:100% total  34.999% nitrogen  59.965% oxygen = 5.036% hydrogen

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