2007-10-30 06:27:04 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Atomic weights: H=1 N=14 HN3=43 Let the solution be called S.
4.38gHN3/408mLS x 1molHN3/43gHN3 x 1000mLS/1LS = 0.250 mole HN3 per L, which is the molarity.
Was that the question?
2007-10-30 06:33:27 · answer #1 · answered by steve_geo1 7 · 0⤊ 0⤋
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2016-05-26 02:37:50 · answer #2 · answered by ute 3 · 0⤊ 0⤋
mol HN3 = 4.38g/MM; MM = 43.0g/mol
M = mol/V = 0.102mol/0.408L =0.250M
2007-10-30 06:36:16 · answer #3 · answered by papastolte 6 · 0⤊ 0⤋
http://chemistry.about.com/
2007-10-30 09:18:01 · answer #4 · answered by novabludag 2 · 0⤊ 1⤋