Question

cineva putin ajutor cu ex 4 si 7 ? :)

Answer 1

4)
dN2=Mhidrocarbura/MN2
Mhidrocarbura=dN2*MN2
Mhidrocarbura=1,9285*28=54g/mol
notam formula: CxHy
MCxHy=12x+y=54 (I)
12x/y=8/1 => y=12x/8=1,5x (II)
(II) -> (I): 12x+1,5x=13,5x=54 => x=4
y=1,5*4=6
f.m: C4H6
7)
notam formula: CxHy
MCxHy=12x+y g/mol
fiind gazoasa => d=M/Vm => M=Vm*d=22,4*1,25=28g/mol
12x+y=28 (I)
28g..............100%
12x............85,71%
y................14,29%
12x=28*85,71/100=24 => x=2
y=28*14,29/100=4
f.m: C2H4