[BT]Nguyên tử.

N

nangtien_lonton

Đặt %[TEX]2H[/TEX] = x%

Có [TEX]\frac{(100-x)+2x}{100}=1,008[/TEX]

=> x = 0,8

1 ml H2O ~ 1 g H2O ~ [TEX]\frac{1}{18}mol H2O[/TEX] ~ [TEX]\frac{1}{9}mol H[/TEX]

=> n_2H = [TEX]\frac{1}{9}.0,8%[/TEX] = 8,88 . 10^(-4) mol

=> 5,35 . 10^20 ntử
 
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