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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?
Molarity = moles of solute/Liters of solution Molarity = 5 moles solute/4.5 Liters of solution = 1 M solution ==========
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?
Molarity = moles of solute/Liters of solution Molarity = 5 moles solute/4.5 Liters of solution = 1 M solution ==========
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.
