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RonSwaggerOne
@RonSwaggerOne
July 2022
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Сколько молей кислорода (O2) требуется для получения 242 г окиси магния?
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Pom1dorka2
Verified answer
2Mg + O2 = 2MgO
n(MgO) = m/M = 242г / 40г/моль = 6,05 моль
n(O2)= 3, 025 моль
2 votes
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Answers & Comments
Verified answer
2Mg + O2 = 2MgOn(MgO) = m/M = 242г / 40г/моль = 6,05 моль
n(O2)= 3, 025 моль