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murielcoleman
@murielcoleman
July 2022
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Вычислите объем воздуха, который нужен для сжигания 12,8 г метанола
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Luussa
Hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
1 votes
Thanks 1
marygia
Verified answer
12,8г х
2CH3OH+3O2->2CO2+4H2O
m=64г V=67,2л
Решаем пропорцией 12,8/64=х/67,2
х=13,44л
13,44-21%
х-100%
х=64л
2 votes
Thanks 1
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Answers & Comments
Verified answer
12,8г х2CH3OH+3O2->2CO2+4H2O
m=64г V=67,2л
Решаем пропорцией 12,8/64=х/67,2
х=13,44л
13,44-21%
х-100%
х=64л