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dilute twice in volume
By using the formula V1 x N1 = V2 x N2 Taking V1= 250 ml; N1= 0.35M; N2= 5.7M. V2 = volume of 5.7M needed to dilute V2 = V1 x N1 N2 = 250 x 0.35 = 15.35ml 5.7
Dilute solution contain small amounts of solute for a certain volume of solvent.
Dilute a measured volume of the 100 mM solution with 19 times its own volume of pure water to produce a 5mM solution.
Some people actually can measure liquid volume with sound. This is because they know how long it takes sound to travel through a certain amount of liquid.
You calculate 0.05 percent of the volume of the base liquid, then add that amount of whatever you want to dilute in this quantity.
dilute twice in volume
The answer depends on the dilution factor and if the sulfuric acid was 100% to start.
=0.142857.. of the original volume.
You need to add an amount of solvent, such as water, to dilute it but you didn't specify the volume of the starting solution.
It is used to measure small volume of liquid to be added where needed .
By using the formula V1 x N1 = V2 x N2 Taking V1= 250 ml; N1= 0.35M; N2= 5.7M. V2 = volume of 5.7M needed to dilute V2 = V1 x N1 N2 = 250 x 0.35 = 15.35ml 5.7
20 volume is 6% solution. To make it 3% solution just add same volume of water to the original 6% solution and you have double volume of 3% solution.
DEPENDS HOW MUCH MOTHER LIQUID YOU START WITH. EXAMPLE, 100 MLS WITH 35 GRMS OF ADDITIVE (subtract the volume of displacement) would be a 35 % dilution. In theory you would add 15% if mother liquid was 100 Mil's, (minus its displacement,) this is what a volumetric flask is for,,, hope this helped.. if Ir's not to important just figure your total liquid and add 15% of your whatever
Yes, it is possible.
The concentration of the diluted solution will be 15(300/1000) = 4.5 %, if the percent is expressed on a weight/volume basis.
Determine the concentration desired and then pour the needed amount of water into beaker. Add the the acid volume to this beaker full of water.