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How many ml of 50 percent dextrose solution and how many ml of water are needed to prepare 4.5 L of a 1 percent solution?
Y= 444.4, x=844.4
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How many mL of a 20 percent solution are needed to prepare 500 mL of a 5 percent solution?
125 ml 500(ml) * 0.05 = 25 25 / 0.20 = 125
How much of a 4 percent solution is needed for a 100mg dose?
2.5ml
Mary is experimenting with chemicals in the laboratory She mixes a solution that is 10 percent acid with a solution that is 25 percent acid How much of the 10 percent acid solution will be needed to m?
16 2/3 liters
A chemist needs 90 milliliters of a 77 percent solution but has only 75 percent and 84 percent solutions available Find how many milliliters of each that should be mixed to get the desired solution?
Let v be the capacity in milliliters of the 75% solution required then (90 - v) is the required capacity of the 84% solution needed. 75/100v + 84/100(90 - v) = 77/100 x 90 75v + 84(90 - v) = 77 x 90 75v + 7560 - 84v = 6930 9v = 7560 - 6930 = 630 v = 70 : therefore (90 - v) = 20 The mixed solution requires 70ml of the 75% solution and 20ml of the 84% solution to create 90ml of 77% solution.
85 percent is 120 percent percent of what number?
To find a percentage you need to firstly divide the number (120) by 100. 120 / 100 = 1.2 Then you need to times this number by the percent needed (85). 1.2 x 85 = 102 Therefore 85% of 120 is 102.
How many ml are needed to prepare 1 liter of 2.5 percent dextrose solution?
25
How many grams of dextrose are needed to prepare 0.5l of d5w?
To prepare 0.5L of D5W (5% dextrose in water), you need 25 grams of dextrose. This is because 5% of 0.5L is 25 grams.
How many ml of 25percent dextrose and needed to prepare 500 ml of 40percent dextrose if you are to make 40percent dextrose from 25percent dextrose and 60percent dextrose?
Oh, what a lovely question! To prepare 500 ml of 40% dextrose, you'll need to mix 200 ml of 25% dextrose with 300 ml of 60% dextrose. By blending these two solutions together, you can create the desired 40% dextrose solution with care and precision. Just remember to mix gently and enjoy the process!
How many ml of 50 percent dextrose solution are needed to obtain 25 g of dextrose?
694 mL of glucose solution 0,2 M are needed.
How many mL of a 20 percent solution are needed to prepare 500 mL of a 5 percent solution?
125 ml 500(ml) * 0.05 = 25 25 / 0.20 = 125
How do you prepare 6n ammonium hydroxide from 30 percent solution?
To prepare a 6N ammonium hydroxide solution from a 30% solution, you would first calculate the molarity of the 30% solution. Then, you would dilute the 30% solution with water to reach the desired concentration. Calculate the volume of the 30% solution needed and dilute it with water to reach the final volume for a 6N concentration.
How much solute is needed to prepare 150mL of a 40.0 percent lithium nitrate solution?
To prepare a 40.0% lithium nitrate solution, 60mL of the solution would be solvent (water) and 90mL would be solute (lithium nitrate). Therefore, you would need 90mL of lithium nitrate to prepare 150mL of the 40.0% solution.
What volume of 24 percent citric acid is needed to prepare eight 3 ounce bottles of 10 percent citric acid solution?
To prepare the 10% citric acid solution, you will need to dilute the 24% citric acid solution. The amount needed can be calculated using the formula for dilution: C1V1 = C2V2. Given that the final volume is 8 bottles x 3 ounces = 24 ounces and the concentration is 10%, the volume of 24% citric acid needed is (10%)(24 ounces) / 24% = 10 ounces.
How can you prepare 1N H2SO4 from H2SO4 95 percent?
To prepare 1N H2SO4 from 95% H2SO4, you would first need to dilute the 95% H2SO4 with water by adding the appropriate amount of water to achieve the desired concentration. To calculate the volume of 95% H2SO4 needed to make 1N solution, you need to use the formula: (Normality of stock solution) * (Volume of stock solution) = (Normality of diluted solution) * (Volume of diluted solution). Adjust the volumes accordingly to prepare the desired 1N solution.
A chemist has one solution that is 80 percent acid and another solution that is 30 percent acid How much of the second 30 percent solution is needed to make a 400 L solution that is 62 percent acid?
144liters
How much water must be added to 425 g of formaldehyde to prepare a 40.0 percent by mass solution of formaldehyde?
To calculate the amount of water needed, first determine the mass of formaldehyde in the final solution. Since the solution is 40.0% formaldehyde by mass, the mass of formaldehyde is 40.0% of the final solution mass. Subtract the mass of formaldehyde in 425 g from this value to find the mass of water needed to prepare the 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 5?
50liters
