Смешали два раствора массой 2кг и 8кг и получили раствор с концентрацией 12%.найти концентрацию 1 и 2 раствора? дано: масса 1 раствора-8кг масса 2 раствора-2кг концентрация полученного раствора-12% найти: концентрацию 1 и 2

Дано:
m₁(p-pa) = 8кг
m₂(p-pa) = 2кг
w₃=12%
Найти:
w₁, w₂
Решение:
Для решения нужно узнать сколько грамм вещества содержится в смешанном растворе, разделить полученную массу вещества поровну между исходными растворами и поделить на массу их растворов:
m₃(в-ва) = m₃(p-pa)*w₃ = (2000г + 8000г) * 0,12 = 1200г
Т.е. по 600г вещества должно быть в первом и втором растворе. Мы нашли m₁(в-ва) и m₂(в-ва). Ищем концентрацию.

% = 7,5%
% = 30%
ответ: смешали растворы концентрацией w₁ = 7,5% и w₂ = 30%.

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ответ:

дано:

m(картофеля)=1620 кг

w((с6н10o5)n)=20%

¥(c6h12o6)=75%

найти: m(c6h12o6)=?

решение :

находим массу крахмала в картофеле :

m((c6h10o5)n)=w*m(картофеля)/100= 20*1620/100=324 кг

находим молярную массу крахмала :

m((c6h10o5)n)=(12*6+1*10+16*5)*n=(82+80)n=162n кг/кмоль

находим количество вещества крахмала:

n((c6h10o5)n)=m/m=324/162n= 2/n кмоль

составляем уравнение реакции :

(c6h10o5)n+nh2o=> nc6h12o6

составляем пропорцию , в которой за х берем количество вещества глюкозы:

x кмоль /n кмоль = 2/n кмоль/1 кмоль

отсюда:

x/n=2/n

получаем:

x=2n/n=2 кмоль

находим молярную массу глюкозы:

m(c6h12o6)=12*6+1*12+16*6=72+12+96=72+108=180 кг/кмоль

находим теоретическую массу глюкозы :

m(теор)(c6h12o6)=n*m=2*180=360 кг

находим практическую массу глюкозы :

m(практ)(c6h12o6)=¥*m(теор)/100=75*360/100=270 кг

ответ: 270 кг

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