Question

1.11
1.12
Dau coroană.

Answer 1

1.11.Maer(gaz)=28,8g/mol

dgaz=M/Mgaz => M=dgaz*Mgaz=1,557*28,9=45g/mol

C=24/12=2 moli C ||:1 => 2 atomi C

H=7/1=7 moli H ||:1 => 7 atomi H

N=14/14 = 1 mol N ||:1 => 1 atom N

F.B => (C2H7N)n

F.M => 12n*2+7n*1+14n=45 => 45n=45 => n=1  => C2H7N

CaHbOcNdXe

NE=[(2a+2)-(b-d+e)]/2 = [(2*2+2)-(7-1)]/2 = 6-6/2 = 0 => nu are leg. duble

Formula plana => CH3-NH-CH3 -> dimetilamina

c) MC2H7N=45g/mol

In 45g C2H7N.....................................14g N
xg........................................................2,8g N

x=45*2,8/14=9g 

d) pV=m/M*R*T => m=pVM/RT = 2*1312*45/0,082*320 = 118080/26,24 = 4500g = 4,5 Kg C2H7N

1.12.

pV=nRT => n=pV/RT=6*12,3/0,082*300 = 73,8/24,6 = 3 moli CO2

1 mol CO2.........................12g C
3 moli CO2......................x=36g C

18g H2O.............................2gH
54g H2O.............................x=6g H

n=m/M=36/12=3 moli C ||:3 => 1 atom C

n=m/M=6/1=6 moli H || : 3 => 4 atomi H

CH4 - metanul

b) MCH4=16g/mol

%C=12/16*100=75%

%H=25%

c) ro=M/Vm=16/22,4=0,714g/L


Nu sunt sigur la 1,12..