13.5 gallons
The 50% acid solution will contain 4.5 gallons of acid & 4.5 gallons of water.
So let X be the number of gallons of pure acid to add. You can set up an equation to solve for X:
(4.5 + X )
------------- = 0.80 (80% solution)
(9 + X)
The top part of the fraction is the amount of acid in the new 80% solution & the bottom part will be the total amount of solution you have. Cross multiply and solve for x.
(9+X)*(0.80) = (4.5+X)*(1)
7.2 + 0.8X = 4.5 + X
7.2 - 4.5 = 1X - 0.8X
2.7 = 0.2X
X = 2.7/0.2 = 13.5
25 gallons
x=45
50 gallons @ 3% must be added.
50% acid in a 6 gallon solution means that 3 gallons are acid. 9 gallons more acid will give you a total of 12 gallons of acid in a 15 gallon solution. 12 is 80% of 15.
x = 10% solution y = 35% solution .1x+.35y=.20*12 x+y=12 y=12-x .1x+.35(12-x)=.20*12 .1x+4.2-.35x=2.4 -.25x=-1.8 .25x=1.8 x=7.2 gallons y=4.8 gallons
10 gallons..
25 gallons
x=45
50 gallons @ 3% must be added.
Make an equation: x(.10) = 50gal(.15) Solve algebraically: x = 75 gal
pH less than 7
50% acid in a 6 gallon solution means that 3 gallons are acid. 9 gallons more acid will give you a total of 12 gallons of acid in a 15 gallon solution. 12 is 80% of 15.
614
630
70gallons
132 gallons 16 : 84 of 60 gallons = 9.6 : 50.4 gallons 80 : 20 of 132 gallons = 105.6 : 26.4 gallons added = 115.2 : 76.8 gallons = 1.5:1 ( 60:40)
tv=tv, so 0,25x50=0,1v; v=12,5/0,1; v=125ml of water. So the solution must have 125ml of water for the title to be 10%, then we must add 125-50ml(75ml) of water to it.