yes it is a solution
How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
solution
6 litres of 40% and 2 litres of 80%.
You need 1 1/3 quarts of pure antifreeze.
0.6 of a pint.
6 litres of the 30% and 4 litres of 60%.
During the summer your engine would actually run hotter because antifreeze has a lower specific heat than does water. (That means a pure antifreeze solution will act as a better insulator than water---less heat will be transferred through the pure antifreeze solution.) Even during the winter, a 100 percent antifreeze solution isn't a good idea. A pure antifreeze solution actually provides less protection against freezing than does a mix solution.
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.
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.
Nothing, you have a solution of antifreeze. However if you then cool the solution the freezing point will be lower than that of pure water.
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.
4 gallons Let x be the amount of antifreeze needed to be added. We know that the total amount of antifreeze in the new solution must equal the amount of antifreeze in the old solution + x: .40*(x+12)=x+.20*12 .60x=2.40 x=4 gallons