1. 1000ml.......1molHCl
250ml..........mol/4 ----> m= 36,5/4gHCl trebuie sa se gaseasc in masa solutiei de conc36,5%
36,5/100= 36,5/4xms----. .>ms=25g--> Vs=25g/1,18g/ml
2. NH4Cl + NaOH= NH4OH+ NaCl
20/100= md/ 200x1,225----> md=40x1,225 g NaOH
revenind la ecuatie si calculand masele molare,rezulta:
40g NaOH.....53,5NH4Cl
40x1,225.......m= 1,225x53,5g clorura de amoniu Termina calculele si nu mai puneti cate 2 probleme la o tema !