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MrFoxD
@MrFoxD
October 2021
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Какой объем кислорода потребуется доя сгорания 4,8г магния?
Химия, 8 класс
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melissa17
2Mg+O2=2MgO
n(Mg)=4.8/24=0.2 моль
0.2 =x
2 = 1
x=0.1
n(O2)=0.1 моль
V(O2)=0.1*22.4=2.24 л
1 votes
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Answers & Comments
n(Mg)=4.8/24=0.2 моль
0.2 =x
2 = 1
x=0.1
n(O2)=0.1 моль
V(O2)=0.1*22.4=2.24 л