Răspunsuri

2014-03-10T22:08:44+02:00
%O = (masa oxigen/masa molara compus)*100
masa molara = 2* A aluminiu + 3 * A oxigen = 2*27 + 3*16=54+48=102g/mol
%O=(3*16/102)*100=>%O=47,059%
2014-03-10T22:40:18+02:00
Masa moleculara Al2O3=2*27+3*16= 57+48=102g/mol
in 102 g Al2O3 ...... 48 g O2
in 100 g Al2O3 ...... x g O2
x= 48*100 supra 102 = 4800 supra 102 = aproximativ 47,1% O2
2 4 2