дано
W(C) = 27.3%
M(CxOy) =44 g/mol
CxOy - ?
W(O) = 100% - W(C) = 100 - 27.3 = 72.7% Ar(C) = 12 a.e.m
n(C) = 27.3 / 12 = 2.275 mol
Ar(O) = 16 a.e.m.
n(O) = 72.7 / 16 = 4.543 mol
n(C) : n(O)= 2.275 : 4.543 = 1: 2
CO2
ответ CO2
Объяснение:
дано
W(C) = 27.3%
M(CxOy) =44 g/mol
CxOy - ?
W(O) = 100% - W(C) = 100 - 27.3 = 72.7% Ar(C) = 12 a.e.m
n(C) = 27.3 / 12 = 2.275 mol
Ar(O) = 16 a.e.m.
n(O) = 72.7 / 16 = 4.543 mol
n(C) : n(O)= 2.275 : 4.543 = 1: 2
CO2
ответ CO2
Объяснение: