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427v2
@427v2
July 2022
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Какой объём воздуха необходим для окисления 3,5 г лития?
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Alexei78
Дано
m(Li)=3.5 g
------------------
V(возд)-?
3.5 X
4Li+O2-->2Li2O M(Li)=7 g/mol
4*7 22.4
X=3.5*22.4 / 28 = 2.8L
V(возд) = V(O2)*100% / 21% = 2.8*100 / 21 = 13.3L
ответ 13.3 л
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Answers & Comments
m(Li)=3.5 g
------------------
V(возд)-?
3.5 X
4Li+O2-->2Li2O M(Li)=7 g/mol
4*7 22.4
X=3.5*22.4 / 28 = 2.8L
V(возд) = V(O2)*100% / 21% = 2.8*100 / 21 = 13.3L
ответ 13.3 л