Simple equality will do here as you have three things and require a fourth.
(X L)(2.5 M NaOH) = (2.0 L)(0.75 M NaOH)
2.5X = 1.5
X = 0.60 Liters needed
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how many milliliters of 2.25M NaOH is needed to make 125mL of 0.75M NaOH solution
This depends on your experiment.
The answer is 20,15 mL.
The concentration of a solution is typically given in terms of the volume of solution, in liters.
Concentration of NaOH = 0.025 M = 0.025 Moles per Litre of SolutionVolume of Solution required = 5.00LWe can say therefore that:Number of Moles of NaOH needed to prepare the solution= Concentration of NaOH * Volume of Solution requiredTherefore:Number of Moles of NaOH needed to prepare the solution= 0.025M * 5.00L= 0.125molesFrom this we can say that 0.125 moles of NaOH are needed to prepare a 5.00 L solution with a concentration of 0.025M of NaOH.
400 mls would require 40g of glucose for a 10% solution and thus 20g for a 5% solution.
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
To prepare 6 nM ammonium hydroxide a 30 percent solution you need to know the volume of the 30 percent solution that you have and the volume of 6nM solution you would like to make. Then use the following formula: C1V1 = C2V2 where C = concentration in moles/Liter and V = volume in liters.
This depends on your experiment.
The answer is 20,15 mL.
The concentration of a solution is typically given in terms of the volume of solution, in liters.
Concentration of NaOH = 0.025 M = 0.025 Moles per Litre of SolutionVolume of Solution required = 5.00LWe can say therefore that:Number of Moles of NaOH needed to prepare the solution= Concentration of NaOH * Volume of Solution requiredTherefore:Number of Moles of NaOH needed to prepare the solution= 0.025M * 5.00L= 0.125molesFrom this we can say that 0.125 moles of NaOH are needed to prepare a 5.00 L solution with a concentration of 0.025M of NaOH.
Yes, by décreasing the volume tenfold.
42.8 grams of KIO3 and add H2O until the final homogeneous solution has a volume of 2.00 liters
400 mls would require 40g of glucose for a 10% solution and thus 20g for a 5% solution.
The answer is 15,039 g hydrogen chloride (HCl).
Molarity = moles of solute/Liters of solutionOr, for our purposes....,Liters of solution (volume) = moles of solute/MolarityVolume (liters) = 0.150 moles HCl/4.00 M HCl= 0.0375 liters = 37.5 milliliters======================
7 liters of a 20% acid solution consists of 1.4 liters of acid (20% of the total volume) mixed with 5.6 liters of water (80% of the total volume). The amount of acid isn't going to change in the new solution. You are just going to add enough water to make it a 10% solution instead of a 20% solution. So it will be more dilute. That means that 1.4 liters of acid will represent 1/10 of the volume of the new solution. So the total volume of the new solution will be 10 x 1.4 or 14 liters. The amount of water in the new solution will be 14 - 1.4 = 12.6 liters. That is a difference of 12.6 - 5.6 = 7 liters from the amount of water you started with. So you need to add 7 liters of water to the original 20% solution to make it a 10% solution. This makes sense because if you double the amount of the mixture from 7 liters to 14 liters and the amount of acid is unchanged, the solution will be half as strong.