Question

найти процентную долю кислорода H2O​

Answer 1

w = Ar(O)*100%/Mr(H2O) = 16 * 100% / 18 = 89%

Answer 2

$\sf M_r(H_2O) = 2A_r(H)+A_r(O) = 2*1 + 16 = 2 +16 = 18.$

$\sf \omega(O) = \frac{A_r(O)}{M_r(H_2O)}=\frac{16}{18} \approx 0,889\approx 88,9\%.$

Ответ : ≈ 88,9%.