Making a Formalin solution in a container (HCHO 37.5%, MEOH 12%)?
968613
If you want to make 1 Liter of this final solution, you would need to take 721 mL of your 52% formaldehyde solution, add 120 mL of methanol to it and bring the final volume up to 1 Liter with water.  You should be able to scale this up or down to make any particular volume you need.\n\nWhat isn't coming out right in your calculations?
1079987
No safe way to do it.  Distillation perhaps, but that's not safe as sulphate fumes are not that good for you.  \n\nGrab the stock bottle and make up a fresh solution.
500934
Combine 19 parts of solvent with 1 part solute.  Boom, 5% solution.  \n\nIf you have an already diluted substance, just figure out the ratio and try to make it 1/20.  For example, if you have a 50% solution of NaCl and water, add two parts (2*50% is 100% or 1 part) of it to 18 parts water to make a 5% NaCl and H20 solution.
481495
1. Perform vacuum distillation.  Place the solution in a vacuum flask; apply vacuum (removing air from vessel); apply heat and "boil" the water off.  Collect the condensate and measure its volume.  Weigh the remaining sugar that is left behind in the vacuum flask.\n  -or -\n2. There may be a way to measure the refractive index of the sugar solution.\n-or-\n3. Heat a sample of the solution to determine the boiling point elevation; if you have a table that compares molarity vs. boiling point elevation, you may be able to determine the molarity of the solution by knowing at what temp. the solution boils.\n\n4. You may be able to extrapolate by weighing exactly one cubic centimeter (cc) (volume) of the solution at 25 degrees C.  There may be a table comparing weight of a 1 cc solution at that temp (as the density of 1 cc of water is different than a 1cc sugar solution.
712851
Dissolve 3.2 g of KMnO4 in water to make 500 mL final volume. This will be 0.04M KMnO4 solution.
844407
Easy, the molecular weight of HCl is 36.461 gr per mole. \n\nIf your trying to reach 4 molar solution then you need 4 x 36.461g HCL diluted to 1 liter. you need 145.844g HCl in the liter\n\nBut since your probably using 32-38 % HCl (actual amount might be on the bottle) you need to multiply you'll need to adjust for inactives by assuming your active molar mass is ~35% of 36.461 or 12.761g \n\nin order to make 1 liter of 4 molar solution you'll need 416.71g \n\n36.461/12.761 * 145.844 = 416.71\n\nassuming it is 35 % active. \n\nYou should always standardize your solution after mixing to determine final activity!!!!!
327703
The amount is diluted by a factor of 25.  So the resulting solution would have a concentration of 11.6M divided by 25 or \nor 0.464M.
242296
22.49 L = 2 moles of HNO3 will react with one  mole of K2HPO4\n\nyou only have .0014 mol  of K2HPO4 available\n\nSO, YOU WILL NEED 31.45 mls of HNO3 to react with .0014mol of K2HPO4
868179
2Rb + 2 H20 --> 2RbOH + H2
1073643
the easiest way is to use a pH paper to test the pH. and just compare the reading with the one given with the test kit. \n\nOr you may do some neutralisation. And then calculate with the  amount of alkali you use for neutralisation. By that you can calculate the concentration of hydroxide ions in the alkali that you use. Then calculate the pOH for the alkali you use. pH=14 - pOH.\neg. \n1mol of NaOH gives you 1 mole of OH ions.\nso the pOH = -log1 = 0\npH = 14 - 0 = 14\n\nJust some rough guide. Hopefully it helps...Good luck.
776525
Since you are pretty, here's how:\n\n1. Compute for the number of moles NaOH needed to prepare the solution.\nmole NaOH: 1 L * 0.5 mol/L = 0.5 mol\n\n2. Get the weight NaOH you need by multiplying the number of moles to the molecular mass of NaOH.\nweight NaOH: 0.5 mol * 40g/mol = 20g\n\n3. To prepare, weigh 20 grams of NaOH pellets. Dilute this with water to 1000 mL in a volumetric flask.
354896
1400 mL of NaOH needed...\n\nneed to make sure the concentrations are the same in order for them to completely neutralize\n\nM = Molarity... number of moles of solute in one liter of solution\n\nif you have 3 M for the sodium hydroxide you would need twice that amount in order to neutralize the 6 M of hydrochloric acid\n\nyou would need 1400 mL since the number of moles of the sodium hydroxide is exactly half of that of the hydrochloric acid.  With knowing that the number of moles NaOH is half, you would need double of what you have for the HCL...  700x2 is 1400 mL
343931
mix 50 % alcohol and 50 % distilled water. Apply with a soft cloth.
727216
use beers law to figure out the concentration and then take that to mole ratio
497859
A)\n\nConvert HCl into millimoles. .2M * 1 mL = .2 mmol. Now, you've diluted it to a volume of 5 L. So you convert back, keeping in mind that you need to divide by mL for the units to cancel:\n\n.2mmol/5000 mL = 4*10^-5 H30+\n\nKnowing that Kw = [H30] * [OH-], and that Kw = 1*10^-14, you can figure out OH- and pOH. (pOH is the -log of OH-)\n\nThe pH of the HCl soln is 4.39, based on -log (4*10^-5)\n\nB)\n\nSodium Oxide, like most alkali metal oxides, reacts with water to create sodium hydroxide an \n\nNa2O + H20 --> 2NaOH \n\nSo it's going to be .2 mol of OH- created, for every .1 mol created. So it's .2 mol NaOH/ 1L = .2M NaOH.\n\nThe OH- is .2M\nThe H+ is figured out using Kw = H+ * OH- (see above)\nThe pOH is 0.70\nThe pH, since pH + pOH = 14 at 25 degrees, is 13.3
320061
Someone else has asked this same question before, and it really isn't that hard . . .\n\n60 ml of 0.250M HCl plus 60 ml of 0.125M NaOH will result in 120 ml of solution total.  All of the NaOH will have been neutralized by the HCl, leaving an excess of HCl, so the solution will be acidic -- the pH will be much less than 7.\n\n60 ml * (1.0 liter/1000 ml) * (0.250 mole/liter HCl) = .015 mole HCl\n\n60 ml * (1.0 liter/1000 ml) * (0.125 mole/liter NaOH) = .0075 moles NaOH\n\nThat means you will have 0.015 - 0.075 = 0.075 moles of HCl left over after the neutralization.\n\nI leave it for you to calculate the pH of a solution of 0.075 moles of HCl in 120 ml of solution.  Remember that pH = -log(base 10)[H+]
747027
It'd be basic speaking qualitatively. NH3 is a base so any amount in water would make the solution basic.