Question

Calculeaza densitatea hidrogenului si a oxigenului in conditii normale
R: ph2= 0,089 g/L; po2=1,428g/L

Answer 1

$V_m=22,4l$

$1 \ mol \ H_2=2A_H=2*1=2g$

$\rho_{H_2}=\frac{m_{H_2}}{V_m}=\frac{2g}{22,4L}=0,089g/L$

$1 \ mol \ O_2=2A_O=2*16=32g$

$\rho_{O_2}=\frac{m_{O_2}}{V_m}=\frac{32g}{22,4L}=1,428g/L$