Răspuns:
Explicație:
6,73 L 4,48 L 4,5 g
CₓHyO + (2x+(y/2)-1/2) O₂ ----> xCO₂ + y/2H₂O
(4x+y-2/4)*22,4L 22,4x L 9y (18*(y/2) ) g
[tex]\frac{6,73}{\frac{4x+y-2}{4} *22,4 } = \frac{4,48}{22,4x} =\frac{4,5}{9y} \\\\\frac{6,73}{\frac{4x+y-2}{4} *22,4 } = \frac{4,48}{22,4x} \\\\<=> \frac{6,73}{\frac{4x+y-2}{4}} = \frac{4,48}{x} \\\\\\<=> \frac{4*6,73}{4x+y-2} = \frac{4,48}{x} \\\\<=> 4x+y-2=6x <=> y=2x+2\\\\ \frac{6,73}{\frac{4x+y-2}{4} *22,4 } = \frac{4,5}{9y}\\\\\frac{4*6,73}{(4x+y-2) *22,4 }= \frac{4,5}{9y}\\\\\\4x+y-2= 2,4y <=>4x-2=1,4y[/tex]
y=2x+2
4x-2=1,4y => 4x-2=1,4*(2x+2)
<=> 2*(2x-1)=1,4*2*(x+1)
<=> 2x-1= 1,4x+1,4 <=> 0,6x=2,4 => x=4
y=2x+2=8+2=10
=> F.ch.M C₄H₁₀O
NE= 2*4+2-10/2=0 ( compus saturat)
d.butanolul e raspunsul corect ( CH₃-CH₂-CH₂-CH₂-OH)
