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in ce raport molar se afla acidul azotic si acidul clorhidric stiind ca amestecul contine 8,116 % azot ?

Răspuns :

MHNO3 = 63g/mol
MHCl = 36,5g/mol

63x+36,5y..............14x
100............................8,116

511,3x+296,234y = 1400x => 888,692x = 296,234y => x/y = 1/3