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How much of water should be add to obtain a solution that is twenty percent antifreeze?
Wiki User
∙ 15y ago
80% water
Wiki User
∙ 15y ago
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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?
222.223 ml @ 20% solution = 44.444 ml 277.777 ml @ 65% solution = 180.555 ml total = 225 ml out of 500 ml = 45%
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?
You will need more data about all densities (in kg/Litre)and you must be sure of using mass% = (g solute)/(100 g solution)Solve two equations for both X and Y:4*d50*50 = X*d20*20 + Y*d70*70 (based on salt mass balance in diff. sol'n.)4*d50 = X*d20 + Y*d70 (based on solutions mass balance)In which:dm = density of the 'm'% salt solution in kg/Litre)X and Y = volume of the 20% and 70% salt solutions respectively
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.
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 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 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?
600ml.
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 water should be mixed to 12ml of alcohol to obtain a 12 percent of alcohol solution?
88 ml
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
