NOT: 15 ounces to make 10%
Correct answer:
Twenty ounces
Calculus:
Let it be M ounces. Then:
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!!:)
Saline is basically water with common salt (Sodium Chloride) added. Therefore - 1 litre of saline has 0.18 percent salt.
At dilution always true:Volume*concentration = amount of solute (= constant, not changing)So 1 (litre) * 5 (%) => 5
add 4 parts water per part solution
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.
Add 25 oz of pure waterLet 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 + 0.90X1X - 0.90X = 45 - 42.50.10X = 2.5X = 25
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