0
How many ounces of pure water must be added to 50 ounces at a 15 percent saline solution to make a saline solution that is 10 percent salt?
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
What else can I help you with?
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.
A mixture contains ninety-two ounces of glycol and water and is forty-five percent glycol. If the mixture is to be weakened to thirty percent how much water must be added?
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.
How much water must be added to 12 L of a 40 percent solution of alcohol to obtain a 30 percent solution?
4 litres
How much of a 50 percent solution should be added to 1 liter to make a 25 percent solution?
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.
A car's radiator has a capacity of 15 quarts and it currently contains 12 quarts of a 30 percent antifreeze solution. How many quarts of pure antifreeze must be added to strengthen the solution 40 per?
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.
How many ounces of 20 percent saline solution must be added to 40 ounces of a 5 percent saline solution to make a 10 percent saline solution?
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
What is Sodium chloride 0.9 percent?
This is an isotonic saline solution; 9 g/L sodium chloride solution in water with added glucose.
How much water must be added to a gallon of 10 percent saline solution to create a 2 percent solution?
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!!:)
What 0.18 percent saline mean?
Saline is basically water with common salt (Sodium Chloride) added. Therefore - 1 litre of saline has 0.18 percent salt.
Does a normal saline solution needs to be added a neutralizer?
Not necessary
What does liquid water turn in to when heat is added?
A saline solution is formed.
When NaCl is added to water the solution will have what?
The chemical and physical properties are changed.
How do you make a percent solution of saline for nasal irrigation?
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.
What type of solution is salt on water?
When salt is added to water, it forms a homogeneous solution, meaning the salt particles dissolve evenly throughout the water. This type of solution is called a saline solution.
How much pure acid must be added to 12 ounces of a 40 percent acid solution in order to produce 60 percent acid solution?
To find the amount of pure acid to add, set up an equation based on the amount of acid in the original solution and the final solution. Let x be the amount of pure acid to add. The amount of acid in the original solution is 0.4 × 12 = 4.8 ounces. The amount of acid in the final solution is 0.6 × (12 + x) = 4.8 + 0.6x ounces. Therefore, to get a 60% acid solution, you would need to add 10.67 ounces of pure acid.
How much of a ten percent solution should be added to two cups of twenty percent solution to make a twelve percent solution?
2%
How much water must be added to 1 liter of a 5 percent saline solution to get a 2 percent saline solution?
At dilution always true:Volume*concentration = amount of solute (= constant, not changing)So 1 (litre) * 5 (%) => 5
