222.223 ml @ 20% solution = 44.444 ml
277.777 ml @ 65% solution = 180.555 ml
total = 225 ml out of 500 ml = 45%
Let x be the amount of the 20% solution and y be the amount of the 65% solution. The total volume equation is: x + y = 500. The total amount of the solute equation is: 0.20x + 0.65y = 0.45(500). Solve these two equations to find the amounts of each solution needed.
A 2% salt solution is hypotonic compared to a 4% salt solution because it has a lower concentration of salt. In osmosis, water flows from hypotonic to hypertonic solutions, so in this case, water would flow from the 2% solution to the 4% solution to try to equalize the concentrations.
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.
Osmosis would occur from the 15 percent sugar solution to the 25 percent sugar solution since the concentration of water is higher in the 15 percent solution. Water molecules will move across the semipermeable membrane to equalize the concentration of sugar on both sides.
Chemists typically use percent by weight or molarity to prepare and describe solutions because these measures are more accurate and reflective of the actual concentration of solute in the solution. Percent by volume can fluctuate with temperature changes and can be affected by differences in the volumes of the solute and solvent, making it less precise for analytical purposes.
To achieve a 50% salt solution with 4 liters total volume, the chemist needs 2 liters of each of the 20% and 70% solutions. This is because the resulting solution will be a combination of the two strengths in equal amounts, leading to an overall concentration of 50%.
A 3 percent solution is 1.5 times as strong as a 2 percent solution.
yes it is isotonic solution.
The percent volume of ethanol in a solution is the volume of ethanol divided by the total volume of the solution, multiplied by 100. It is commonly used to express the concentration of ethanol in alcoholic beverages or solutions.
Let Q be the quantity of 80% solution required then (300 - Q) is the quantity of 30% solution as together the two solutions must equal 300. Then, [80 x Q] + [30 x (300 - Q)] = [40 x 300] 80Q + 9000 - 30Q = 12000 50Q = 3000 Q = 60...........which means (300 - Q) = 240 60ml of 80% acid solution + 240ml of 30% solution produces 300ml of 40% 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.
A 2% salt solution is hypotonic compared to a 4% salt solution because it has a lower concentration of salt. In osmosis, water flows from hypotonic to hypertonic solutions, so in this case, water would flow from the 2% solution to the 4% solution to try to equalize the concentrations.
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.
times it together
Molarity: the concentration of a solution in moles of solute per liter of solution. Molality: the concentration of a solution in moles of solute per kilogram of solvent. Mass percent: the percentage of the total mass of a solution that is contributed by the solute. Volume percent: the percentage of the total volume of a solution that is contributed by the solute. Parts per million (ppm): the concentration of a solution in parts per million by weight.
58.1 ml
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).
You will have to assume that the 2 % is a volume fraction, then the volume of copper sulfate in the solution would be 11.5 milliliter(575 ml*(0.02). If it were a weight fraction, then you would have to have more information on the solution density.