2 câu trả lời
$\text{n$_{O2}$=1,26.10$^{21}$:6.10$^{23}$=2,1.10$^{-3}$(mol)}$
$\text{⇒m$_{O2}$=2,1.10$^{-3}$.32=0,0672(g)}$
$\text{$n_{O_2}$ = $\dfrac{A}{N}$ = $\dfrac{1,26. 10^{21}}{6. 10^{23}}$ = $\dfrac{21}{10000}$ mol }$
$\rightarrow$ $\text{$m_{O_2}$ = $\dfrac{21}{10000}$ . 32 = 0,0672 gam}$
$\text{#Kudo}$