if heat of precipitation of AgCl=-36k.j/mole
find heat of precipitation of 287 gm of AgCl knowing that (Ag=108 . CL35.5)
1 answer
36 kJ/mol = (287 g molar mass AgCl) = ?
find heat of precipitation of 287 gm of AgCl knowing that (Ag=108 . CL35.5)
1 answer