Предмет: Химия, автор: davidara

К 160 г 5% р-ра HNO3 добавим 20 г H2O. Определите массовую долю азотной к-ты в полученном р-ре.

Ответы

Автор ответа: mrvladimir2
1
ω= 160*0,05/(160+20) = 0,0444 = 4,44%
Автор ответа: braslavliliya
0
Дано m(раств.HNO3)=160 . массовая доля=5%=0,05 m (H2O)=20 г. Найдём массу азотной кислоты) по формуле мас.доля=m(вещества)\m (раствора) отсюда m(вещ.)=0,05*160=8г   массовая доля полученного раствора=8\(160+20)=0,044=4,4%
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…………………………………………………………………………………………………………………………………………………………

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