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10 gallons..

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Q: How many gallons of a 12 percent indicator solution must be mixed with a 20 percent indicator solution to get 10 gallons of a 14 percent solution?
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How many gallons of a 80 percent acid solution must be mixed with 60 gallons of a 16 percent solution to obtain a solution that is 60 percent?

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)


A solution of 30 percent acid and a solution of 60 percent acid are to be mixed to produce 50 liters of 57 percent acid How many liters of each solution should be used?

t = number of liters of 30% acid solution s = number of liters of 60% acid solution t+s=57 .30t+.60s=.50*57 t=57-s .30(57-s)+.60s=.50*57 .30*57 -.30s +.60s = .50*57 .30s = .50*57 - .30*57 = .20*57 s = .20*57/.30 = 38 liters of 30% solution t = 57 - s = 57-38 = 19 liters of 60% solution


How much water should be mixed with 2 gallons of a 50 percent acid solution in order to get an 2 percent acid solution?

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.


What is the method for making mordant black 2 indicator?

mordant black 2 is used as an indicator in comleometric titrations, usuall against Di Na EDTA Solution. the indicator is mixed with NaCl/ KI in appropriate quantity...


If 5 liters of a 4 percent silver iodide solution must be mixed with a 10 percent solution to get a 6 percent solution How many liters of the 10 percent solution are needed?

2.5 is the answer....5(.04)+.10(x)= .06(5+x).2+.1x=.3+.06x.04x=.1or 4x=10 so x =2.5

Related questions

How many gallons of a 9 percent salt solution must be mixed with 25 gallons of a 31 percent solution to obtain a 20 percent solution?

25 gallons


How many gallons of 3 percent salt solution must be mixed with 50 gallons of 7 percent salt solution to obtain a 5 percent salt solution?

50 gallons @ 3% must be added.


How many gallons of a 90 percent antifreeze solution must be mixed with 80 gallons of 25 percent antifreeze to get a mixture that is 80 percent antifreeze?

614


How many gallons of a 90 percent antifreze solution must be mixed with 90 gallons of 10 percent antifreeze to get a mixture that is 80 percent antifreeze?

630


How many gallons of a 50 percent antifreeze solution must be mixed with 70 gallons of 30 percent antifreeze to get a mixture that is 40 percent antifreeze?

70gallons


How many gallons of a 80 percent acid solution must be mixed with 60 gallons of a 16 percent solution to obtain a solution that is 60 percent?

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)


How much pure acid should be mixed with 5 gallons of 70 percent acid solution in order to get a 90 percent acid solution?

x=45


How many gallons of a 10 percent ammonia solution should be mixed with 50 gallons of a 30 percent ammonia solution to make a 15 percent ammonia solution?

Make an equation: x(.10) = 50gal(.15) Solve algebraically: x = 75 gal


How much pure acid should be mixed with 2 gallons of a 50 percent acid solution in order to get an 80 percent acid solution?

pH less than 7


How much pure acid should be mixed with 6 gallons of a 50 percent acid solution in order to get an 80 percent acid solution?

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.


How much of 10 percent acid Solution must be mixed with 35 percent acid solution mixed to get 12 Gallos of 20 percent solution?

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


How many gallons of a 80 percent antifreeze solution must be mixed with 70 gallons of 20 percent antifreeze to get a mixture that is 70 percent antifreeze?

70 gallons of 20% solution contains 70*0.2 = 14 gallons of antifreeze. Suppose you need G gallons of the 80% antifreeze solution. This will contain 0.8*G gallons of antifreeze. Total volume of solution = G + 70 gallons Volume of antifreeze required in this solutions to make it a 70% solution is 0.7*(G + 70) = 0.7G + 49 gallons. Volume of antifreeze = 14 + 0.8G gallons So 0.7G + 49 = 14 + 0.8G 0.7G + 35 = 0.8G 35 = 0.1G 350 = G Answer: 350 gallons.