Question

Cate duble legaturi sunt in compusii A si B cunoscand ca
-pentru a hidrogena 2,24L (cn) compus gazos A, sunt necesari 0,492L H₂ masurati la 27 grade C si 5 atm
-pentru bromurarea completa a 3,4g compus B sunt necesare 400g solutie 2% Br₂
Vaporii substantei B au densitatea 3,33g/L la 1,5 atm si 100 grade C

Answer 1

2,24/22,4 = 0.1 moli compus A.
Se foloseste relatia : p*V=niu*R*T => niu = p*V/R*T => niu = 5*0.492/0.082*(27+273)=2.46/24.6 = 0.1 moli H2
0.1 moli A .... 0.1 moli H2
 1 mol      .....      x       => x=1 mol H2. => A are o singura legatura dubla.

c=md*100/ms => md = c*ms/100 = 2*400 / 100 =8 grame Br2.
d=m/v => v=m/d
p*v = niu*R*T =>p*m/d = m/M *R*T Se simplifica m => p/d=R*T/M => M=d*R*T/p
M=3.33*0.082*373/1.5≈68
3.4g B ..........8 g Br2
68 g B ...........x*160   => x*160 = 68*8/3.4 <=> x*160 = 160 => x =1, adica 1 mol de Br2 aditionat. => B are o singura legatura dubla.