am facut cate ceva , 4 si inceputul de la
CnH2n-6---ardere----> nCo2 + (2n-6)/2H2O
se da: 1mol hidrocarbura
m= 352gCo2--> niu= 352g/44g/mol= 8mol CO2---> 8mol C
m= 90gH20---> niu=90g/18g/mol= 5molH20---> 10molH
1mol hidrocarbura are 8C si 10H---> C6H10
-IZOMERI:orto, meta si para metil toluen(xileni) ,etil benzen
9.formula generala CnH2n-6
100g hidroc...........8,69gH
14n-6.......................2n-6
200n-600=121,66n-52,14--- n=7---> toluen
r1;toluen +O---> acid benzoic
r2:toluen + CH3cL---> ETIL BENZEN
R3: toluen +2Cl2---toluen dublu substituit la catena laterala--> clorura de benziliden
