felipe0387 12/09/2017 Olá!!!1 mol de qualquer gás nas CNTP ocupa um volume de 22,4L:1 mol 22,4L 6,02•10²³ partículasM(SO3) = 32+3•16 = 80g/mol80g --- 1 mol --- 22,4L --- 6,02•10²³6,02•10²³ 22,4L1,2•10²⁴ XX = 44,65L
felipe0387
Olá!!!
1 mol de qualquer gás nas CNTP ocupa um volume de 22,4L:
1 mol 22,4L 6,02•10²³ partículas
M(SO3) = 32+3•16 = 80g/mol
80g --- 1 mol --- 22,4L --- 6,02•10²³
6,02•10²³ 22,4L
1,2•10²⁴ X
X = 44,65L