дано
n(CO2) = 0.6 mol
n(H2O) =0.3 mol
D(возд ) = 2.69
--------------------
CxHy-?
M(CxHy) = D(возд) * 29 = 2.69*29 = 78.01 g/mol
n(H) = 2n(H2O) = 2*0.3 = 0.6 mol
C:H = 0.6 : 0.6 = 6 : 6
C6H6 - БЕНЗОЛ
ответ БЕНЗОЛ
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Verified answer
дано
n(CO2) = 0.6 mol
n(H2O) =0.3 mol
D(возд ) = 2.69
--------------------
CxHy-?
M(CxHy) = D(возд) * 29 = 2.69*29 = 78.01 g/mol
n(H) = 2n(H2O) = 2*0.3 = 0.6 mol
C:H = 0.6 : 0.6 = 6 : 6
C6H6 - БЕНЗОЛ
ответ БЕНЗОЛ