
How would you prepare 250 mL of a 0.250 M solution of fluoride ions from solid CaF2? A)Calculate the mass of CaF2.
What volume of 6.00 M NaOH solution is required to prepare 750.0 mL of a 0.250 M NaOH solution? Explain how you would prepare this solution.
how would you prepare a stock solution of 0.100 M solution of MgCl2 in 500 mL? Using the stock solution, how would you make a 0.078 M solution in 25.00 mL?
1. How many grams of solute would you use to prepare 1.50 L of 0.250 M glucose, C6H12O6? 2. What volume of 0.250 M H2SO4 is needed to react with 50.0 mL of 0.100 M NaOH?
An aqueous solution contains 0.486 M ammonia (NH3). How many mL of 0.250 M hydroiodic acid would have to be added to 150 mL of this solution in order to prepare a buffer with a pH of 9.160.
How many mL of 0.250 M BaCl2 would you need to precipitate all the sulfate ion present in 38.1 mL of 0.197 M H2SO4 solution?
How many mL of 0.250 M BaCl2 would you need to precipitate all the sulfate ion present in 38.1 mL of 0.197 M H2SO4 solution? What is the concentration (M) of the chloride ion remaining in the resulting solution?
How many grams of potassium bromide are needed to prepare 300.0 mL of a solution that is 2.0%w/v? Do NOT include any units, just answer with the NUMBER of grams, for example: 4.2 or 13 or 220. Type your response What is the new concentration, if solvent is added to 20.0 mL of a solution that is 4.0%w/v, until the total volume of the solution is 100.0 mL? (DO NOT include any units with your answer.) Type your response 25...
how many mL of a 3.0 M solution must be used in order to make 300.0 ML of 0,25M solution
3. 300.0 mL of a 2.500 M solution X was mixed with 300.0 mL of a 2.500 M solution Y in a calorimeter. Both of the solutions were at the same temperature initially. Determine the heat of neutralization (kJ/mole) if the temperature goes from 12.0 °C to 29.0 °C. Assume a molar ratio of 1:1. The specific heat of the solution is 4.281 J/g °C. The density of the mixture is 1.050 g/mL. 4. Sketch a crude time-temperature graph for...