0.1 N iodide would be 0.1 moles of the iodide salt (e.g. KI) per liter of solution. For 500 ml, you would need 0.05 moles of the iodide salt. You need to state the salt (KI, NaI, LiI, etc.) in order to determine the actual mass required.
You would need to dilute the 6M acetic acid solution by adding the appropriate volume of water. To prepare 500 mL of 1M solution, you would need to take (1/6)th of the volume of the 6M solution, which is (1/6) x 500 mL = 83.33 mL of the 6M solution. Dilute this with water to reach a final volume of 500 mL.
125 ml 500(ml) * 0.05 = 25 25 / 0.20 = 125
You need 116,88 g dried and pure sodium chloride.
The pH of a solution containing 3.0 g of HI in 500 ml of water is approximately 1.7.
To prepare a 500 ppm (parts per million) solution of lead nitrate, you would need to dissolve 500 grams of lead nitrate in 1 million grams (1,000,000 grams) of solution. The specific weight of lead nitrate would depend on its molar mass and density. It is more common to express concentrations in terms of molarity (moles per liter) rather than ppm for making solutions.
To prepare 500 mL of a 0.5 M solution of acetic acid, you would need to take 100 mL of the 2.5 M stock solution and dilute it to 500 mL. This is because the molarity equation (M1V1 = M2V2) can be used to calculate the volume of stock solution needed for dilution.
To make a 500 ml solution of 3% H2O2 from a 50% solution, you would need to dilute the 50% solution with water. You would add 325 ml of water to 175 ml of the 50% H2O2 solution to achieve a final volume of 500 ml with a concentration of 3% H2O2.
To prepare a weight/volume (w/v) solution of calcium chloride in 500 mL of water, you first need to decide the desired concentration (e.g., 1%, 5%, etc.). For instance, to create a 1% w/v solution, you would dissolve 5 grams of calcium chloride in enough water to make a total volume of 500 mL. Adjust the amount of calcium chloride accordingly based on the specific concentration you want.
Weighting half value of 23+1+12+(3x16) = 84/2 = 42 gram NaHCO3 gives you 0.5 mole of it, to be dissolved in about 200 mL and there after to be filled up to 500 mL (= 500 cm3) will result in a 0.5 mol / 500 mL = 0.5 mol / 0.5 L = 1 mol/L = 1M NaHCO3
Lactated Ringer's solution typically comes in premade volumes like 500 mL or 1000 mL. However, you can request a healthcare provider or pharmacist to prepare a custom 0.45 Lactated Ringer's solution by diluting the standard concentration with sterile water or saline.
To prepare 3% hydrogen peroxide from 30% hydrogen peroxide, you need to dilute the 30% solution with water. The formula to calculate the dilution is C1V1 = C2V2, where C1 is the initial concentration (30%), V1 is the volume of the initial solution, C2 is the final concentration (3%), and V2 is the final volume of the solution you want to prepare. By plugging in the values, you can determine the volume of 30% hydrogen peroxide needed and the volume of water needed to achieve a 3% solution.
A water solution of this medicine is obtained.