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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?
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 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.
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?
Let x be the liters of the 30% acid solution and y be the liters of the 60% acid solution. We can set up a system of equations: x + y = 50 (total liters) and 0.3x + 0.6y = 0.57*50 (acid content). Solving this system of equations, we find that x = 20 liters of the 30% acid solution and y = 30 liters of the 60% acid solution.
What is the method for making mordant black 2 indicator?
To make Mordant Black 2 indicator, dissolve Mordant Black 2 powder in a suitable solvent, such as water or alcohol, to the desired concentration. The indicator solution can then be used for various analytical purposes, such as detecting the presence of cations in solution based on color changes.
When a small amount of white powder was shaken with distilled water in a test tube powder sank bottom when drops of universal indicator were added it turn red?
The white powder is likely to be a base as it sank to the bottom when mixed with distilled water. The red color observed after adding universal indicator indicates that the solution became acidic. This suggests that the white powder was an insoluble base that released a basic solution when mixed with water, which turned acidic upon adding the indicator.
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 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 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 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 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 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 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 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.
