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A chemist has one solution that is 60 percent chlorinated and another solution that is 40 percent chlorinated How much of the first 60 percent solution is needed to make a 100 L solution that is 50?
To create a 50% chlorinated solution from the 60% and 40% solutions, the chemist will need to mix the two in equal amounts. Therefore, 50 L of the 60% solution and 50 L of the 40% solution are needed to make a 100 L solution that is 50% chlorinated.
What is the molarity of a solution with 5mol solute in 4.5L solution?
The molarity of the solution can be calculated by dividing the moles of solute by the liters of solution. In this case, it would be 5 moles / 4.5 liters = 1.11 M.
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.
What is the molarity of a solution consisting of 0.250 mol NaCl dissolved in 2.25 L of solution?
The molarity of the solution is calculated by dividing the moles of solute (0.250 mol NaCl) by the liters of solution (2.25 L). Molarity = moles of solute / liters of solution Molarity = 0.250 mol / 2.25 L = 0.111 M
A solution of 30 percent acid and a solution of 60 percent acid are to be mixed to produce 50 liters of 57 percent acid How many liters of each solution should be used?
t = number of liters of 30% acid solution s = number of liters of 60% acid solution t+s=57 .30t+.60s=.50*57 t=57-s .30(57-s)+.60s=.50*57 .30*57 -.30s +.60s = .50*57 .30s = .50*57 - .30*57 = .20*57 s = .20*57/.30 = 38 liters of 30% solution t = 57 - s = 57-38 = 19 liters of 60% solution
A chemist wants to make a 10 solution of fertilizer. how much water and how much of a 30 solution should the chemist mix to get 30 L of a 10 solution?
30 liters of a 10 % solution of fertilizer has .1(30) = 3 liters of fertilizer 1 liter of 30% solution has .3 liter of fertilizer 10 liters of 30% solution has 3 liters of fertilizer so, the chemist needs 10 liters of the 30% solution and 20 liters of water to make 30 liters of a 10% solution.
A chemist is making 200 L of a solution that is 62 percent acid He is mixing an 80 percent acid solution with a 30 percent acid solution How much of the 80 percent acid solution will he use?
The chemist will use 100 liters of the 80% acid solution and 100 liters of the 30% acid solution to make a 200-liter solution that is 62% acid. The amount of acid in the 80% solution will be 0.8 * 100 = 80 liters, and in the 30% solution, it will be 0.3 * 100 = 30 liters.
A chemist has one solution that is 60 percent chlorinated and another that is 40 percent chlorinated How much of each solution is needed to make a 100 liter solution that is 50 percent chlorinated?
50 Liters of the 60% solution.
A chemist has one solution that is 60 percent chlorinated and another solution that is 40 percent chlorinated How much of the first 60 percent solution is needed to make a 100 L solution that is 50?
To create a 50% chlorinated solution from the 60% and 40% solutions, the chemist will need to mix the two in equal amounts. Therefore, 50 L of the 60% solution and 50 L of the 40% solution are needed to make a 100 L solution that is 50% chlorinated.
A beaker contains 625 milliliters of water A chemist pours 25 liters of the water in a solution How many milliliters are left in the beaker?
33ml
A chemist has 2 solutions of h2so4 One 40 concentration and the other 25 solution How many liters must be mixed to make 129 liters of 31 solution?
For a detailed answer visit: algebra.com/algebra/homework/word/mixtures/Mixture_Word_Problems.faq.question.786696.html
A chemist has one solution that is 80 percent acid and another solution that is 30 percent acid How much of the first 80 percent solution is needed to make a 400 L solution that is 62 percent acid?
To create a 400 L solution that is 62% acid, you would need 200 L of the 80% acid solution and 200 L of the 30% acid solution. This would result in a final solution with the desired concentration.
A chemist has one solution which is 50 percent acid and a second solution which is 25 percent acid how much of each should be mixed to make a 10 liters of a 40 percent acid solution?
6 litres of 50% + 4 litres of 25%
How many liters of a 90 percent acid solution must be mixed with a 15 percent acid solution to get 600 L of a 80 percent solution?
Mixing 80 liters of 15% solution and 520 liters of 90% solution will give 600 liters of 80% solution.
What is the molarity of a solution with 5mol solute in 4.5L solution?
The molarity of the solution can be calculated by dividing the moles of solute by the liters of solution. In this case, it would be 5 moles / 4.5 liters = 1.11 M.
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 30 solution should be added to a 90 solution to make 18 liters of a 75 solution?
4.5 litres of a 30% solution to the appropriate quantity of the 90% solution.
