0
What else can I help you with?
How many gallons of a 50 percent antifreeze solution most be mixed with 90 gallons 10 percent antifreeze to get a mixture that is 40 percent antifreeze?
Multiply These Equals What you mix 50 .10+x 5+50x Plus what you have 90 .10 9 Equals what you need 40 x 40x 5+50x+9=40x 14+50x=40x 10x=14 x=1.4 140%
How many gallons of 80 percent antifreeze solution must be mixed with 100 gallons of 10 percent antifreeze to get a mixture that is 70 percent antifreeze?
600 gallons. To solve this think about the amount of antifreeze in the solution. When 10% antifreeze is added to x times as much 80% antifreeze, the resultant percentage antifreeze will be: (10 + 80x)/(1 + x) So to obtain a 70% solution, x will need to solve: (10 + 80x)/(1 + x) = 70 ⇒ 10 + 80x = 70 + 70x ⇒ 10x = 60 ⇒ x = 6 So to 100 gallons of 10% antifreeze solution, 100 x 6 = 600 gallons of 80% antifreeze solution will be needed to make it a 70% solution. Consider adding 100 gallons of 10% antifreeze and 100 gallons of 80% antifreeze together and then taking half the resultant solution, that is so that you have 100 gallons of the mixture; the amount of antifreeze in this 100 gallons is (10 gallons + 80 gallons) ÷ 2 = 45 gallons, that is a 45% mixture. Now consider adding 100 gallons of 10% antifreeze and 200 gallons of 80% antifreeze together and then taking a third of the resultant solution, that is so that you have 100 gallons of the mixture; the amount of antifreeze in this 100 gallons is (10 gallons + 2 x 80 gallons) ÷ 3 = 562/3 gallons, that is a 562/3% mixture. If x times as much of the 80% antifreeze is added to the If 1 gallon of 10% antifreeze was added together with 80% antifreeze, the resultant mixture will still be (10+90)/2 = 45%. So if x times as much of the 80% Consider the amount of antifreeze in the solution. When two solutions of the same volume is added, the amount of antifreeze in the same volume is half the new amount. Consider the ratio of anti-freeze to water in each of the solutions in fraction form of antifreeze/water: 10% is 10/90 70% is 70/30 80% is 80/20 When two ratios are added together, the amount of antifreeze in the solution is added together and the amount of water is added together to give some odd fraction maths: 10% + 80% = (10+80)/(90+20) = 90/110 =
How many quarts of pure antifreeze must be added to 6 quarts of a 10 percent antifreeze solution to obtain a 20 percent antifreeze solution?
To find out how many quarts of pure antifreeze must be added to 6 quarts of a 10% antifreeze solution to obtain a 20% solution, we can set up the equation. The initial amount of antifreeze in the solution is (0.10 \times 6 = 0.6) quarts. Let (x) be the amount of pure antifreeze to add. The final solution will be (6 + x) quarts, and we need the total antifreeze to equal (0.20(6 + x)). Setting up the equation (0.6 + x = 0.20(6 + x)) and solving gives (x = 1.2) quarts. Thus, 1.2 quarts of pure antifreeze must be added.
How many gallons do you need to fill up a 500 gallon propane tank to 60 percent?
60% * 500 = 300three hundred gallons (of liquid - the gas before being liquefied - takes up MUCH more volume)
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 45 percent antifreeze?
To create a 45% antifreeze solution from 1 gallon (128 ounces) of pure antifreeze, you need to determine how much of the solution should consist of antifreeze. Since you want 45% antifreeze, you can set up the equation: 0.45 * (128 + x) = 128, where x is the amount of water added. Solving for x, you find that you need to add approximately 104 ounces of water to achieve a 45% antifreeze solution.
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.
How many gallons of a solution that is 75 antifreeze must be mixed with 4 gallons of a 30 solution to obtain a mixture that is 50 antifreeze?
You will need 3.2 gallons.
How many gallons of a 50 percent antifreeze solution most be mixed with 90 gallons 10 percent antifreeze to get a mixture that is 40 percent antifreeze?
Multiply These Equals What you mix 50 .10+x 5+50x Plus what you have 90 .10 9 Equals what you need 40 x 40x 5+50x+9=40x 14+50x=40x 10x=14 x=1.4 140%
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 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 many gallons of antifreeze does a 2001 peterbilt hold?
Need to know model and engine in order to answer this.
How many gallons of 80 percent antifreeze solution must be mixed with 100 gallons of 10 percent antifreeze to get a mixture that is 70 percent antifreeze?
600 gallons. To solve this think about the amount of antifreeze in the solution. When 10% antifreeze is added to x times as much 80% antifreeze, the resultant percentage antifreeze will be: (10 + 80x)/(1 + x) So to obtain a 70% solution, x will need to solve: (10 + 80x)/(1 + x) = 70 ⇒ 10 + 80x = 70 + 70x ⇒ 10x = 60 ⇒ x = 6 So to 100 gallons of 10% antifreeze solution, 100 x 6 = 600 gallons of 80% antifreeze solution will be needed to make it a 70% solution. Consider adding 100 gallons of 10% antifreeze and 100 gallons of 80% antifreeze together and then taking half the resultant solution, that is so that you have 100 gallons of the mixture; the amount of antifreeze in this 100 gallons is (10 gallons + 80 gallons) ÷ 2 = 45 gallons, that is a 45% mixture. Now consider adding 100 gallons of 10% antifreeze and 200 gallons of 80% antifreeze together and then taking a third of the resultant solution, that is so that you have 100 gallons of the mixture; the amount of antifreeze in this 100 gallons is (10 gallons + 2 x 80 gallons) ÷ 3 = 562/3 gallons, that is a 562/3% mixture. If x times as much of the 80% antifreeze is added to the If 1 gallon of 10% antifreeze was added together with 80% antifreeze, the resultant mixture will still be (10+90)/2 = 45%. So if x times as much of the 80% Consider the amount of antifreeze in the solution. When two solutions of the same volume is added, the amount of antifreeze in the same volume is half the new amount. Consider the ratio of anti-freeze to water in each of the solutions in fraction form of antifreeze/water: 10% is 10/90 70% is 70/30 80% is 80/20 When two ratios are added together, the amount of antifreeze in the solution is added together and the amount of water is added together to give some odd fraction maths: 10% + 80% = (10+80)/(90+20) = 90/110 =
How many gallons of antifreeze does a 2001 Lincoln town car hold or need?
According to the Owner Guide the cooling system capacity is 15.8 quarts It came from the factory with a 50 / 50 mix of antifreeze and distilled water So , 7.9 quarts or slightly under ( 2 gallons ) of antifreeze
How much antifreeze do you need?
How much coolant does your car or truck hold? You want to have one half pure antifreeze and one half water. If you want to mix your own and if your car holds 4 gallons of coolant then you need 2 gallons of distilled water and 2 gallons of antifreeze. Or you can buy the kind of antifreeze that is already mixed called "PRE-MIXED COOLANT" at most part stores and just fill it up.
Is their a special kind of antifreeze for 1998 Lincoln navigator and how many gallons does it need?
According to the 1998 Lincoln Navigator Owner Guide : It came from the factory with a 50 / 50 mix of distilled water and GREEN color antifreeze ( meeting Ford specification ESE-M97B44-A ) The engine cooling system capacity is 20.8 U.S. quarts , so you would need : ( 10.4 quarts / 2.6 gallons of antifreeze ) for a 50 / 50 mix * Ford states not to exceed 60 % antifreeze
How many gallons of antifreeze do you need for your 8 cylinder motor?
50% Antifreeze and 50% water mixed together. The winter months you need a higher ratio of antifreeze to water and in summer you need the 1/2 and 1/2 ration. It is not suggested to run pure Antifreeze in your car or truck. You can find pre mixed in the jugs at most Wal-marts/Kmarts/Auto Parts stores.
How many grams of active ingredient will you need to prepare 2.5 gallons of a 45.6 percent solution?
4314.9 grams
