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karinazasimenk
@karinazasimenk
August 2022
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Какой объём (н.у) занимает 4,816*10 в 23 степени молекул аммиака NH3
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k1598
∨= V\ Vm = N\Na
V = Vm * N\ Na = 22.4 л\моль * 4,816 * 10^23 / 6,02*10^23 моль^-1= 17,92 л
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Answers & Comments
V = Vm * N\ Na = 22.4 л\моль * 4,816 * 10^23 / 6,02*10^23 моль^-1= 17,92 л