x = amount of acid to add.
y = final volume.
5 gal + x = y
original amount of acid + acid added = final amount of acid
(5 X .2) + x = O.5y
Subtract the second equation from the first one.
5 - 1 + x - x = y - 0.5y
4 = 0.5y
8 = y
Therefore the final volume is 8 gallons.
5 gal + x = 8
x = 3 gal. the amount of pure acid to add.
Check the answer
5 X .2 = 1 gal of acid in original solution.
3 gallons added = 4 gallons total acid in solution.
4 gallons total acid in final solution of 8 gallons total solution = 50% acid.
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.