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How much of an 18 percent solution of sulfuric acid should be added to 360 ml of a 10 percent solution to obtain a 15 percent solution?
To determine the amount of 18% solution to add: Let x = volume of 18% solution to be added. 0.18x + 0.10(360) = 0.15(x + 360) Solving for x, you would need to add 75 ml of the 18% solution to the 360 ml of 10% solution to obtain a 15% solution.
If you have a 300ml of a 15 percent hydrochloric acid solution in dilute the solution to 1000ml with sterile water what percent should appear on the label?
The concentration of the diluted solution will be 15(300/1000) = 4.5 %, if the percent is expressed on a weight/volume basis.
What should the minimum percent of Isopropyl Alcohol be to prevent contamination of an IV?
The minimum recommended concentration of Isopropyl Alcohol to prevent contamination of an IV site is typically 70%. Below this concentration, it may not effectively kill bacteria and other microorganisms. Therefore, it is important to use a solution that contains at least 70% Isopropyl Alcohol to ensure proper disinfection.
How many times should you dilute 0.1n hcl to obtain 0.01n hcl?
You would need to dilute the 0.1N HCl solution with distilled water in a 1:9 ratio (1 part HCl solution to 9 parts distilled water) once in order to obtain a 0.01N HCl solution.
Is there a solution of alcohol and iodine?
Yes, a solution of alcohol and iodine is known as tincture of iodine. It is commonly used as an antiseptic for disinfecting wounds and skin before surgery. However, it should be used with caution as it can cause skin irritation and staining.
What fraction of 10 percent alcohol should be mixed with 70 percent alcohol to obtain 50 percent alcohol?
2/3 of 70% and 1/3 of 10%
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.
How many ml of water should be added to 95 percent ethyl alcohol to make 1 liter of a 30 percent ethyl alcohol solution?
684 ml
How many ounces of the 50 percent alcohol solution should be mixed with the 1 percent alcohol solution to make 8 oz of the 20 percent alcohol solution?
Let x = ounces of 50% solution, and y = ounces of 1% solution. So that we have: 0.5x + 0.01y = 8(0.2) which is a linear equation in two variables, meaning there are infinitely many choices of mixing those solutions.
How much of an 18 percent solution of sulfuric acid should be added to 360 ml of a 10 percent solution to obtain a 15 percent solution?
To determine the amount of 18% solution to add: Let x = volume of 18% solution to be added. 0.18x + 0.10(360) = 0.15(x + 360) Solving for x, you would need to add 75 ml of the 18% solution to the 360 ml of 10% solution to obtain a 15% solution.
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60 percent antifreeze?
2 gallons.
How much water should be added to 15 g of salt solution to obtain 15 percent salt solution?
Dissolve 15 g salt in 100 mL water.
How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
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 of water should be add to obtain a solution that is twenty percent antifreeze?
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.
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 80 percent antifreeze?
0.25 gallons of water (or 1 quart)
How many liters of a 20 percent solution of saline should a nurse mix with 10 liters of a 50 percent solution to obtain a mixture that is 40 percent saline?
Let x = the amount of 20% solution Let x + 10 = the amount of the final solution. So we have: (.20)x + (.50)(10) = (.40)(x + 10) .20x + 5 = .40x + 4 .20x = 1 x = 5 liters of 20% solution of saline.
