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This depends on the desired pH, volume of solution, initial pH, etc.
it is very easy to prepare working solution from a stock solution we use the formula for this purpose which is: C1V1 = C2V2 C1 is the concentration of the stock solution V1 required volume from the stock solution C2 concentration of the working solution V2 volume of the working solution
It depends on the Molarity of the solution. if the (mols x molar mass)/volume> 1, its more dense.
25/1.05 = 23.81 ml (rounded)
A reaction of ammonia and water make ammonium hydroxide.I think ammonium hydroxide is a weak base. it has most of ammonia molecule because ammonia has a property of nucleophilic and same thing is done by the ammonium hydroxide.
You cannot make a solution of AgCl, it is an insoluble salt
You cannot make a solution of AgCl, it is an insoluble salt
By definition, a 0.61M sodium nitrate solution contains 0.61 moles of sodium nitrate per liter, which is equivalent to 0.61 mmol/ml. Therefore, the volume of this solution required to contain 400mmol is 400/0.61 or 6.6 X 102 ml, to the justified number of significant digits.
The answer is 252,05 mL.
how many moles are there in 56.0 grams of silver nitrate?
This depends on the desired pH, volume of solution, initial pH, etc.
It depends on the Molarity of the solution. if the (mols x molar mass)/volume> 1, its more dense.
it is very easy to prepare working solution from a stock solution we use the formula for this purpose which is: C1V1 = C2V2 C1 is the concentration of the stock solution V1 required volume from the stock solution C2 concentration of the working solution V2 volume of the working solution
Look up molecular mass of lead nitrate in the periodic table. Formula of lead nitrate is Pb(NO3)2 Molecular mass is 331.2 gm/mole A one molar solution has 1 mole per liter of solution. 13.6 grams/331.2 gm/mole = 0.041 moles 0.041 moles/0.15 molar = .274 liters This is not exactly correct, because when you add a solute to a solvent, the volume of the solution may be more or less than the original volume of solvent, depending upon the interaction of the solvent and the solute. Actually, the final volume of solution should be 0.274 liters. The most accurate way is to dissolve the solute is slightly less than the calculated amount of solvent needed, and then adjust the volume to the final amount after the solute is completely dissolved.
The answer is 20,15 mL.
This depends on your experiment.
400 mls would require 40g of glucose for a 10% solution and thus 20g for a 5% solution.