Question

Ex 236 cu explicatii!!!

Answer 1

28,7 g AgCl, M.AgCl = 143,5 g/mol

143,5 g AgCl ...................... 35,5g Cl

28,7g AgCl ..................... m = 7,1 g Cl

=> masa de C+H = 9,9-7,1 = 2,8 g

daca compusul clorurat s-a obtinut prin aditie hidrocarbura poate fi alchena sau alchina

nr.moli = m/M.compus => M.compus = m/n = 9,9/0,1 = 99 g/mol

notam compusul CaHbClc

9,9g compus ...................... 7,1g Cl

99g ....................................... 35,5c

=> c = 2

=> CaHbCl2 = 99 => 12a+b = 99-71 = 28

=> a > sau = cu 2

=> a = 2 => b = 4 => DA

=> hidorcarbura va fi C2H4 iar compusul C2H4Cl2

=> A.