Question 511886: How many liters of a 40% and 80% acids must be mixed in order to make 20 liters of a 55% acid solution?
A 12 liter mixture is 60% alcohol. if 8 liters of a 25% alcohol mixture are added, what percent mixture will be formed.
Answer by Maths68(1474)   (Show Source):
You can put this solution on YOUR website! (1)
How many liters of a 40% and 80% acids must be mixed in order to make 20 liters of a 55% acid solution?
Solution A
Amount = x
Concentration =40% = 0.4
Solution B
Amount = 20-x
Concentration =80% = 0.8
Resultant Solution
Amount =20
Concentration =55%=0.55
[Amount Solution A * Concentration A] + [Amount Solution B * Concentration of B] = Amount of Resultant * Concentration of resultant
(0.4)x+(0.8)(20-x)=(20)(0.55)
0.4x+16-0.8x=11
0.4x-0.8x=11-16
-0.4x=-5
-0.4x/-0.4=-5/-0.4
x=12.5
Solution A
Amount = x =12.5 liters
Concentration =40% = 0.4
Solution B
Amount = 20-x =20-12.5=7.5 liters
Concentration =80% = 0.8
Result
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12.5 liters of a 40% and 7.5 liters of 80% acids must be mixed in order to make 20 liters of a 55% acid solution
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(2)
A 12 liter mixture is 60% alcohol. if 8 liters of a 25% alcohol mixture are added, what percent mixture will be formed.
Solution A
Amount =12 liters
Concentration =60% = 0.6
Solution B
Amount = 8 liters
Concentration =25% = 0.25
Resultant Solution
Amount =(8+12)=20 liters
Concentration =x
[Amount Solution A * Concentration A] + [Amount Solution B * Concentration of B] = Amount of Resultant * Concentration of resultant
(12*0.6)+(8)(0.25)=20x
7.2+2=20x
9.2=20x
9.2/20=20x/20
0.46=x
x=0.46
x=46/100= 46%
Result
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If we add 12 liter 60% alcohol solution to 8 liters of a 25% alcohol solution, a mixture of 20 liters of 46% alcohol will be formed.
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