There are 10 nanomoles in a 10 micromolar stock solution due to the conversion factor: 1 micromolar = 1,000 nanomoles/mL.
You need 6,9 mL stock solution.
This is a saturated solution.
A solution can have one or more solvents. In a binary solution, there are two solvents, and in a ternary solution, there are three solvents, and so on. Each solvent plays a role in dissolving the solute and determining the properties of the solution.
No, there can be many solutes in a solution, but only one solvent.
1 M is equal to 1000 mM. Calculating molarity is important in biochemical and molecular experiments. Normally 10 X or 20X stock solution are prepared from which the working solutions are diluted as per the need of the concentration (in mM)
You need 6,9 mL stock solution.
To make 1 liter of a 1000 ppm solution from a 1000 ppm stock solution, you would need 1 ml of the stock solution. This is because 1 ml of the 1000 ppm stock solution contains 1000 parts of solute in 1 million parts of solution, which is equivalent to 1 liter.
To make a 0.25 M solution of ammonium sulfate from a stock solution of 6 M, you would need to dilute the stock solution. The dilution equation is C1V1 = C2V2 where C1 and V1 are the concentration and volume of the stock solution, and C2 and V2 are the concentration and volume of the final solution. You would need to set up this equation to calculate the volume of the stock solution needed and then convert that volume to grams using the molar mass of ammonium sulfate.
You would need 50 mL of the 2.0 M NaBr solution to make 200 mL of 0.50 M NaBr solution. This can be calculated using the formula: (C1V1) = (C2V2), where C1 = concentration of stock solution, V1 = volume of stock solution, C2 = final concentration, and V2 = final volume.
To prepare 1000 ml of 0.02 M NaCl solution, you would need 40 ml of 5 M NaCl solution, which you can calculate using the formula C1V1 = C2V2, where C1 is the concentration of the stock solution, V1 is the volume of the stock solution needed, C2 is the desired concentration, and V2 is the final volume. The dilution factor in this case would be 25, as you are diluting the 5 M solution 25 times to achieve the desired 0.02 M concentration.
Too many unknowns in your question. Is this 3% by mass or by volume? Does the quantity of final solution matter? IE do you need 100 ml, 1 liter or 5000ml. What is the density of the hydrogen peroxide? (needed for a volume % problem) Assuming you mean 3% by mass, then that means 3 g of hydrogen peroxide in 100 g of solution. 300 micromolar = 3 x 10-4 molar. Assuming you want to make one liter then you need 3 x 10-4 moles of peroxide. The molar mass of peroxide is 34 g/mole. 34 g/mole x 3 x 10-4 moles = 1.02 x 10-2 grams of peroxide 1.02 x 10-2 grams / .03 = 0.34 grams of the original solution. Weigh (mass) accurately 0.34 g of the original solution in a 1 liter volumetric flask. Add distilled water until the total volume is 1 liter.
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To prepare 0.500 L of 0.470 M HNO3 solution, you would need 103.6 mL of the 11.4 M stock solution. This can be calculated using the dilution formula: M1V1 = M2V2, where M1 is the concentration of the stock solution, V1 is the volume of stock solution needed, M2 is the final concentration, and V2 is the final volume.
If i understand the question correctly, basically you want to make your standard 1/5 the conc of the stock. Therefore, 300ml stock diluted up with solvent
Two, the Dhaka Stock Exchange and the Chittagong Stock Exchange.
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There were about 30 Stock Market crashes in history.