ms = 600 g , c% = 60% , c% = md x 100 / ms => md = ms x c% / 100 = 360 g HNO3
Mhno3 = 1 x1 + 14 x 1 + 16 x 3 = 63 g/mol
=> la 63 gr ................ avem.......... 3 x 16 g O
360 gr .............. vom avea......... a g O => a = 274,28 g O
n = nr moli = mo2 / Ao2 = N / Na => N = mo2 x Na / Ao2 = 274,28 x 6,022 x 10la23 / 16 => N = 103,23 x 10la23 atomi O