1)массовая доля кальция в соединении равна 40%, углерода-12% кислорода-48%. выведите формулу данного соединения.
2)объемная доля кислорода в воздухе равна 21%. какой объем воздуха необходим для получения 35 л кислорода?
3) в 180г волы расстворили 20 г сахара. определите массовую долю сахарав полученном расстворе.
4) при выпаривании 80 г расствора получили 4 г соли. определите массовую долю вещества в исходном расстворе.
5) массовая доля примесей в известняке составляет 8%. найдите массу примесей в 350 г известняка.
Предмет:
ХимияАвтор:
azkabanbrcyЗадача №1
w(Ca) = 40%
w(C) = 12%
w(O) = 48%
CaxCyOz
Пусть у нас есть 100 грамм вещества, тогда, в растчете на 100 грамм вещества получим:m(Ca) = 40г
m(C) = 12г
m(O) = 48г
Дело за малым, рассчитаем количественные соотношения, т.е. рассчитаем количества вещесвта каждого атома в молекуле:
n(Ca) = m(Ca)/M(Ca) = = 40г/40г/моль = 1моль
n(C) = m(C)/M(C) = 12г/12г/моль = 1моль
n(O) = m(O)/M(O) = 48г/16г/моль = 3моль
n(Ca):n(C):n(O) = 1:1:3, значит формула в-ва:
CaCO3
Задача №2
V(возд) = V(кислорода)/q/100% = 35л/21%/100% = 35л/0,21 = 166,6л
Дадача №3
Рассчитаем массу раствора:
m(р-ра) = m(р.в) + m(р-ля) = 20г + 180г = 200г
Рассчитаем массовую долю:
w = m(р.в.)/m(р-ра)*100% = 20г/200г/*1005 = 10%
Задача №4
m(р-ра) = 80г
m(р.в.) = 4г
w = m(р.в.)m(р-ра)*100% = 4г/80г100% = 5%
Задача №5
m(ч.в.) = w*m(смеси) = 8%/100%*350г = 0,08*320г = 28г
Автор:
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