Koliko je atoma broma supstucijom uvedeno u molekulu toluena ako je procenat broma 46,78 ?
C6H5CH3+Br2-------->C6H5Br+CH3Br
46.78%:80*x=100%:80*x+92-x
80*x*100=(79*x+92)*46.78
8000*x=3695.62*x+4303.76
8000*x-3695.62*x=4303.76
4304.38*x=4303.76
x=4303.76/4304.38=1 atom Br2
powered by Drupal
organska hemija
C6H5CH3+Br2-------->C6H5Br+CH3Br
46.78%:80*x=100%:80*x+92-x
80*x*100=(79*x+92)*46.78
8000*x=3695.62*x+4303.76
8000*x-3695.62*x=4303.76
4304.38*x=4303.76
x=4303.76/4304.38=1 atom Br2