Помогите пожалуйстаСколько молекул при нормальных условиях займут 112 л озона О3 - №5915764

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Сколько молекул при нормальных условиях займут 112 л озона О3
- Предмет:
  Химия
- Автор:
  emeliaaqnv
- 5 лет назад

Ответы 1

n(O3) = 112 л / 22,4 л / моль = 5 молькол-во молекул = 5 моль * 6,02 * 10 ^23 = 30,1 * 10^23
- Автор:
  mitchell26
- 5 лет назад
- 0
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