Question

Va rog frumos 39, multumesc de ajutor!

Answer 1

31/.

a)

1 mol CO2 ........ 22,4 L  ............. 44 g/mol ....... 12g C

                       33,6 L ....................................... x = 18 g C

=> 21,6 - 18 = 3,6 g H

C / H = 18 / 3,6 = 5 / 1

b)

C = 5/12 = 0,4167

H = 1/1 = 1  : 0,4167

=> (CH2,4) = Fb

c)

dN2 = miu.hc./miuN2

=> miu.hc. = dN2xmiu.N2 = 2,571x28 = 72 g/mol

=> hc. = CxHy => 12x+y

21,6 g hidroc. ............. 18g C ........... 3,6g H

72g ............................ 12x .................. y

=> x = 5  ,  y = 12

=> Fm = C5H12

d)

       CH3

CH3-C-CH3

        CH3

N.E. = 2a+2-b/2 = (2x5+2-12)/2 = 0