0
How much of water should be add to obtain a solution that is twenty percent antifreeze?
To obtain a solution that is twenty percent antifreeze, you would add 4 parts water to 1 part antifreeze. This means that for every 1 part of antifreeze, you would add 4 parts of water. This would result in a total of 5 parts of solution, with 1 part being antifreeze and 4 parts being water, achieving a solution that is twenty percent antifreeze.
Add your answer:
Earn +20 pts
How many gallons of a 40 percent antifreeze solution should be mixed with a 10 percent solution to make 50 gallons of a 30 percent solution?
Let the amount of pure alcohol be x, so the amount of 25% solution would be x + 5. So we have:x + 0.15(5) = 0.25(x + 5)x + 0.75 = 0.25x + 1.25x - 0.25x = 1.25 - 0.750.75x = 0.5x = 0.5/0.75 = 0.666... = 6/9 = 1/3 of the literAdded noteto deal with the so called 'dilution contraction' of total volumeIf it were % by MASS ( %m/m), it's quite easy to do (based on the 'Mass Conservation Law). You calculate with mass (kg) and mass-% (%m/m) i.s.o. volume (L) and vol% (%v/v).However if the meaning was: % by Volume ( %v/v) then calculation appears to become quite complicated, but not impossible if you know at least the density values of all solutions (original 100%v/v or 15%v/v and final 25%v/v).DO NOT use: (orig. volume) + (added volume) = final volume, as done above, if exact figures are necessary.It's only a rule of thump, an approximation. This is because fluids can contract on mixing at dilution. There is no rule such as: conservation of volume.Your case: 0.33 L + 5 L (is not equal but) < 5.33 L final solution.
How many milliliter of a 20 percent and a 65 percent solutions should you mix together to make 500 milliliter of a 45 percent solution?
Let x be the amount of the 20% solution and y be the amount of the 65% solution. The total volume equation is: x + y = 500. The total amount of the solute equation is: 0.20x + 0.65y = 0.45(500). Solve these two equations to find the amounts of each solution needed.
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.
A chemist needs 4 liters of a 50 percent salt solution All she has available is a 20 percent salt solution and a 70 percent salt solution How much of each of the two solutions should she mix to obtain?
To achieve a 50% salt solution with 4 liters total volume, the chemist needs 2 liters of each of the 20% and 70% solutions. This is because the resulting solution will be a combination of the two strengths in equal amounts, leading to an overall concentration of 50%.
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.
How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60 percent antifreeze?
2 gallons.
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 80 percent antifreeze?
0.25 gallons of water (or 1 quart)
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 35 and 8203 antifreeze?
Approx 1.86 gallons.
How many liters of a 20 percent solution of acid should be added to 10 liters of a 30 percent solution of acid to obtain a 25 percent solution?
10 liters.
How much water should be added to 30 gallons of a solution that is 70 percent antifreeze in order to get a mixture that is 60 percent antifreeze?
First off do not use tap water. Use only distilled water. You will need to add 3 gallons of distilled water to the solution to get 60% antifreeze and 40% distilled water or a 60/40 mix.
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 80 percent solution must be mixed with 100 gallons of a 15 percent antifreeze solution to get a mixture that is 70 percent antifreeze?
For these types of problems, I recommend finding the initial amount of antifreeze. To find this value, multiply the percentage of antifreeze (15%) by the amount of solution (100 gallons). This gets:0.15*100=15 gallons of antifreeze*Now we need make an equation that represents this problem. We let x be the number of gallons of 80% solution.(.80x+15)/(x+100)=0.70The top of the equation represents the amount of antifreeze, while the bottom is the total volume of the solution. If we divide these values, we should get the final concentration %, .70.Now we just need to manipulate the equation algebraically:.80x+15=.70x+70.10x=55x=550 gallons of 80% solution*If the concentration of a solution is given in percentage, it is probably referring to mass, not volume. Because of this, the values for the volume of antifreeze are actually untrue. But because the density of both substances remains the same, the final answer should be correct.
How much of an 18 percent solution of sulfuric acid should be added to 360 ml of a 10 percent solution to obtain a 15 percent solution?
To determine the amount of 18% solution to add: Let x = volume of 18% solution to be added. 0.18x + 0.10(360) = 0.15(x + 360) Solving for x, you would need to add 75 ml of the 18% solution to the 360 ml of 10% solution to obtain a 15% solution.
How much water should be added to 15 g of salt solution to obtain 15 percent salt solution?
Dissolve 15 g salt in 100 mL water.
What mix should you use for antifreeze in a car?
50 percent coolant, 50 percent water
How much of a ten percent solution should be added to two cups of twenty percent solution to make a twelve percent solution?
2%
