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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 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 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 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 70 percent antifreeze solution should be mixed with 60 gallons of 10 percent antifreeze to get a mixture that is 60 percent antifreeze?
.70x+.10(60)=.60(x+60) let x= number of gallon at 70% .7x+6=.6x+36 .1x=30 x=300
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 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 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 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 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 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 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 12 percent indicator solution must be mixed with a 20 percent indicator solution to get 10 gallons of a 14 percent solution?
Let x represent the amount of 12% solution and (10-x) represent the amount of 20% solution. The equation to solve is: 0.12x + 0.20(10-x) = 0.14(10). Solving for x gives x = 4, so you need 4 gallons of the 12% solution and 6 gallons of the 20% solution to make 10 gallons of the 14% solution.
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?
Let x represent the gallons of 10% ammonia solution. The total volume of the mixture is x + 50 gallons. The equation for the mixture is: 0.10x + 0.30(50) = 0.15(x + 50). Solving this equation gives x = 50 gallons of the 10% ammonia solution needed.
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?
Let x be the gallons of the 80% acid solution needed. The amount of acid in the 80% solution is 0.8x, and the amount in the 16% solution is 0.16*60=9.6. We want a total of (x+60) gallons of solution with 60% acid, so we have the equation 0.8x + 9.6 = 0.6(x+60). Solving for x gives x = 24 gallons.
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 much water should be mixed with 2 gallons of a 50 percent acid solution in order to get an 2 percent acid solution?
To get a 2% acid solution, you need to dilute the 50% acid solution with water. Since the final volume is 2 gallons, you will need to mix 2 gallons of water with the 2 gallons of 50% acid solution to get a 2% acid solution.
