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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.

Q: How much of water should be add to obtain a solution that is twenty percent antifreeze?

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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.

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.

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.

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%.

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.

Related questions

How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?

2 gallons.

0.25 gallons of water (or 1 quart)

Approx 1.86 gallons.

10 liters.

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.

.70x+.10(60)=.60(x+60) let x= number of gallon at 70% .7x+6=.6x+36 .1x=30 x=300

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.

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.

Dissolve 15 g salt in 100 mL water.

50 percent coolant, 50 percent water

2%