n = CV = (2.5 M)(0.200 L) = 0.5 mol m = nM = (0.5 mol)(39.1 g/mol) = 19.55 g Assuming a 2.5 M KOH solution. Mass used should be proportionally larger if more than one potassium ion are in the compound being dissolved. e.g. 39.1 g should be used if making a 2.5 M K2SO4 solution.
To make a 1% aqueous solution of potassium hydroxide, you would mix 1 gram of potassium hydroxide with 99 grams of water (for a total of 100 grams solution). This would give you a solution where 1% of the total weight is potassium hydroxide.
A solution that is refered to as a percentage of something (like potassium hydroxide, KOH) refers to the mass of the solute compared to the total solution, so a 5% KOH solution would be 5g KOH + 95g H2O, and the 5g KOH would be 5% of the 100g total of the solution.
To make a 10 percent aqueous solution of potassium hydroxide, you would dissolve 10 grams of potassium hydroxide in enough water to make a total solution volume of 100 mL. This solution would be considered a 10 percent concentration by weight. Be cautious when handling potassium hydroxide as it is a caustic substance.
To make a potassium thiocyanate solution, first dissolve potassium thiocyanate powder in distilled water. Stir the mixture until the powder is completely dissolved. The resulting solution can be used for various chemical reactions or experiments.
To make a potassium thiocyanate solution, simply dissolve potassium thiocyanate powder in distilled water until the desired concentration is achieved. Stir the solution until the powder is completely dissolved. Take proper safety precautions as potassium thiocyanate can be harmful if mishandled.
290 grams
At 60 degrees Celsius, the solubility of potassium chlorate (KClO₃) in water is approximately 7.2 grams per 100 grams of water. To create a saturated solution in 200 grams of water, you would need about 14.4 grams of potassium chlorate (7.2 g/100 g water x 200 g water = 14.4 g KClO₃). Thus, 14.4 grams of potassium chlorate would be required for saturation at this temperature.
To make a 1% aqueous solution of potassium hydroxide, you would mix 1 gram of potassium hydroxide with 99 grams of water (for a total of 100 grams solution). This would give you a solution where 1% of the total weight is potassium hydroxide.
To make a 10% KI solution, dissolve 10 grams of potassium iodide (KI) in 90 grams of water, for a total of 100 grams of solution. This will give you a 10% weight/volume (w/v) solution of KI.
A solution that is refered to as a percentage of something (like potassium hydroxide, KOH) refers to the mass of the solute compared to the total solution, so a 5% KOH solution would be 5g KOH + 95g H2O, and the 5g KOH would be 5% of the 100g total of the solution.
To make a 0.25N K2CrO4 solution, you need to first calculate the molecular weight of K2CrO4 (potassium chromate). Then, determine the grams of K2CrO4 needed to make the desired volume of solution at a concentration of 0.25N. Dissolve this amount of K2CrO4 in the required volume of solvent, usually water, to make the final solution.
To calculate the grams needed to mix a 0.10 M potassium fluoride solution, you would first find the molar mass of KF (58.10 g/mol). Then use the formula Molarity = moles/Liter to find the moles of KF needed (moles = Molarity x Volume). Finally, convert moles to grams using the molar mass: grams = moles x molar mass. In this case, you would need 5.81 grams of KF to make a 0.10 M solution in 1.0 liter.
Calculate the mass (in grams) of sodium sulfide that is needed to make 360ml of a 0.50 mol/L solution
To find the grams of potassium chlorate needed, you would first calculate the moles of potassium chlorate using the molarity and volume provided. Then, use the molar mass of potassium chlorate to convert moles to grams. So, the calculation would be: moles = Molarity x Volume (in liters), and then grams = moles x molar mass of potassium chlorate.
To make a 2 molar solution of hydrochloric acid, you would need to know the volume of the solution you want to make. Once you have the volume, you can use the molarity formula (M = moles of solute / liters of solution) to calculate the grams of hydrochloric acid needed.
The answer is 7,5g.
To calculate the total amount of sodium chloride needed for a 13 L solution at 4 grams per liter, multiply the concentration by the volume of the solution: 4 grams/L x 13 L = 52 grams of sodium chloride. Therefore, you will need 52 grams of sodium chloride to make the 13 L solution.