Question

дано: w(c) =40% , w(H) = 6,7%, Mr (Cx HyOz) = 60​

Answer 1

Решение:

w(O) = 100 - 40 - 6,7 = 53,3 %

кол-во атомов:

C = (M(CxHyOz) * w(C))/M(C) = (60*0,4)/12 = 2

H = (60*0,067)/1 = 4

O = (60*0,533)/16 = 2

CH3COOH