Ustal wzór węglowodoru o masie cząsteczkowej 44u zawierającego 18% wodoru.
44u - 100%
x - 18 %
CxHy
m = 12x + 1y = 44u
100 - 18 = 82 [%] - C
82% * 44 = 12x
36,08 = 12x
x = 3
44 - 3*12 = 44 - 36 = 8 - H
wzór - C3H8 - propan
m= 44u 44=12x+1y 44u - 100% x=44*82/100 = 36 u
C-12u x - 82% 36/12 = 3
H-1u
44=12*3+1y
44-36=1y
8=y
wzór : C3H8
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CxHy
m = 12x + 1y = 44u
100 - 18 = 82 [%] - C
82% * 44 = 12x
36,08 = 12x
x = 3
44 - 3*12 = 44 - 36 = 8 - H
wzór - C3H8 - propan
m= 44u 44=12x+1y 44u - 100% x=44*82/100 = 36 u
C-12u x - 82% 36/12 = 3
H-1u
44=12*3+1y
44-36=1y
8=y
wzór : C3H8