Add 25 oz of pure water
Let X be the volume of pure water to add. Then the total amount of pure water you add plus the total amount of water in the 15% solution will equal the total amount of water in the final 10% solution.
So,
X + 50*(1-0.15) = (50 + X) * (1-0.10)
X + 50*(0.85) = (50 + X)*(0.90)
X + 42.5 = 45 + 0.90X
1X - 0.90X = 45 - 42.5
0.10X = 2.5
X = 25
10 liters.
The original mixture contains 41.4 ounces of glycol. for this to be 30 percent of the mixture, the total mixture must be 138 ounces, so 46 ounces of water must be added.
4 litres
Mix this 50% solution in equal quantities with water(?) to halve it's strength. So use 1 litre of the 50% solution and 1 litre of water of that's what you are diluting it with.
The original solution has 3.6 quarts of antifreeze in it. The equation then becomes (3.6 + x)/(12 + x) = 0.40, where x is the amount of antifreeze added. X is then equal to 2.
NOT: 15 ounces to make 10%Correct answer:Twenty ouncesCalculus:Let it be M ounces. Then:20%*(M ounce) + 5%*(40 ounce) = 10%*(M+40 ounce)20*M + 200 = 10*M + 400(20-10)*M = 400-20010M = 200M = 20
This is an isotonic saline solution; 9 g/L sodium chloride solution in water with added glucose.
6 ounces
There is 1 gallon of 10% solution...There is 0% saline in water...The equation is (0.1)(1) + 0x = 0.02 (x+1)...Answer is: 4 gallons. Hope I helped!!:)
Not necessary
Saline is basically water with common salt (Sodium Chloride) added. Therefore - 1 litre of saline has 0.18 percent salt.
The chemical and physical properties are changed.
A saline solution is formed.
A one percent saline solution (roughly isotonic) can be made by adding 1 teaspoon of salt to a litre of drinking water. The salt should not contain added iodine or anti-caking agents. Which means that kosher or pickling salt are appropriate but table salt isn't. The saline should be heated to a luke warm temperature before use.
The density of 40ml of saline solution in a 50 ml beaker is 1.0046g/mL. The density will vary based upon the concentration of the salt added to the solution.
About 80ml of water must be added to 40ml of a 25 percent by weight solution to make a 2 percent by weight solution.
At dilution always true:Volume*concentration = amount of solute (= constant, not changing)So 1 (litre) * 5 (%) => 5