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DJKLHy
@DJKLHy
July 2022
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Какой объем углекислого газа образуется при сгорании 200 мг этанола.(p=0.8г/см^3)
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KotoFey328
M(C2H5OH) = 200 *0.8 = 160 г.
2C2H5OH + O2 -> 2CO2 + 2H2O
160г/88г = xл/44.8л
x=160*44.8/88 = 80л
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Answers & Comments
2C2H5OH + O2 -> 2CO2 + 2H2O
160г/88г = xл/44.8л
x=160*44.8/88 = 80л