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tprost
@tprost
July 2022
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При сгорании некоторой массы углеводорода получено 7,7 г CO2 и 3,6 г H2O
. Определите формулу
углеводорода
и его
массу
.
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Радость25
(x+y/4)*O2+СхHу=x*СO2+(y/2)*H2O
x*(12+2*16)/7,7 = y/2*(2*1+16)/3,6
x*(12+2*16)*36 = y/2*(2*1+16)*77
x*(44)*36 = y/2*(18)*77
x*4*2 = y/2*1*7
x*16 = y*7
C7Н16
16 votes
Thanks 28
IUV
Verified answer
Решение во вложении:
13 votes
Thanks 12
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Answers & Comments
x*(12+2*16)/7,7 = y/2*(2*1+16)/3,6
x*(12+2*16)*36 = y/2*(2*1+16)*77
x*(44)*36 = y/2*(18)*77
x*4*2 = y/2*1*7
x*16 = y*7
C7Н16
Verified answer
Решение во вложении: