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Pharmacist mixed a 20 percent solution with a 30 percent solution to obtain 100 liters of a 24 percent solution How much of the 20 percent solution did the pharmacist use in the mixture?
10
How many gallons of a 9 percent salt solution must be mixed with 25 gallons of a 31 percent solution to obtain a 20 percent solution?
25 gallons
How many gallons of 3 percent salt solution must be mixed with 50 gallons of 7 percent salt solution to obtain a 5 percent salt solution?
50 gallons @ 3% must be added.
Is it possible to mix a 10 percent acid solution and a 40 percent acid solution to obtain a 60 percent acid solution?
No. The reulting concentration (percent) must be between the two components. So, with the two acids you are mixing, you cannot get an acid that is less than 10% or more than 40%
How many quartz of pure antifreeze must be added to 6 quartz of a 10 percent antifreezesolution to obtain a 20 percent antifreeze solution?
4.2 quarts
A pharmacist mixed a 20 percent solution with a 30 percent solution to obtain 100 liters of a 24 percent solution How much of the 20 percent solution did the pharmacist use in the mixture?
A pharmacist mixed a 20 percent solution with a 30 percent solution to obtain 100 liters of a 24 percent solution. How much of the 20 percent solution did the pharmacist use in the mixture (in liters).
Pharmacist mixed a 20 percent solution with a 30 percent solution to obtain 100 liters of a 24 percent solution How much of the 20 percent solution did the pharmacist use in the mixture?
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 liters of a 40 percent acid solution must be added to 100 liters of a 25 percent solution to obtain a 30 percent solution?
Suppose L litres are required.@ 40% it will contain 0.4L of the active ingredient.Then total volume of mixture = L + 100 litresand volume of active ingredient = 0.4L + 25 litresThe strength is (0.4L + 25)/(L + 100) = 30% = 0.3(0.4L + 25) = 0.3*(L + 100)0.4L + 25 = 0.3*L + 300.1L = 5L = 50 litres.
How many liters of water must be added to 7 liters of a 20 percent acid solution to obtain a 10 percent acid solution?
7 liters of a 20% acid solution consists of 1.4 liters of acid (20% of the total volume) mixed with 5.6 liters of water (80% of the total volume). The amount of acid isn't going to change in the new solution. You are just going to add enough water to make it a 10% solution instead of a 20% solution. So it will be more dilute. That means that 1.4 liters of acid will represent 1/10 of the volume of the new solution. So the total volume of the new solution will be 10 x 1.4 or 14 liters. The amount of water in the new solution will be 14 - 1.4 = 12.6 liters. That is a difference of 12.6 - 5.6 = 7 liters from the amount of water you started with. So you need to add 7 liters of water to the original 20% solution to make it a 10% solution. This makes sense because if you double the amount of the mixture from 7 liters to 14 liters and the amount of acid is unchanged, the solution will be half as strong.
A bottle of commercial concentrated aqueous ammonia is labeled 28.89 percent NH3 by mass and density is 0.8960 grams mL. What is the molarity of the ammonia solution?
To find the molarity, first calculate the moles of NH3 in 100g of the solution using the mass percent. Then convert the volume of the solution (1 mL) to liters. Finally, divide moles by liters to obtain the molarity.
How many liters would you need to get 0.5moles if you had 0.1m solution?
To find out how many liters of a 0.1 M solution are needed to obtain 0.5 moles, you can use the formula: [ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} ] Rearranging this gives: [ \text{liters of solution} = \frac{\text{moles of solute}}{\text{Molarity (M)}} ] Substituting in the values: [ \text{liters of solution} = \frac{0.5 \text{ moles}}{0.1 \text{ M}} = 5 \text{ liters} ] Therefore, you would need 5 liters of a 0.1 M solution to obtain 0.5 moles.
How many gallons of a 9 percent salt solution must be mixed with 25 gallons of a 31 percent solution to obtain a 20 percent solution?
25 gallons
How do you obtain copper sulfate crystals from a mixture of copper sulfate and sand?
To obtain copper sulfate crystals from a mixture with sand, you can dissolve the mixture in water. The copper sulfate will dissolve, while the sand will not. You can then filter the solution to separate the sand from the copper sulfate solution. By evaporating the water from the copper sulfate solution, you can obtain copper sulfate crystals.
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 many gallons of 3 percent salt solution must be mixed with 50 gallons of 7 percent salt solution to obtain a 5 percent salt solution?
50 gallons @ 3% must be added.
How many liters of a 5.0 percent glucose solution would you need to obtain 75 g of glucose?
To calculate the amount of glucose in the solution, use the formula: mass = volume × concentration Rearrange the formula to solve for volume: volume = mass / concentration Substitute the values to find: volume = 75 g / 0.05 = 1500 mL = 1.5 L Therefore, you would need 1.5 liters of the 5.0% glucose solution to obtain 75 g of glucose.
