How many   mL   of   12.00 M   hydrochloric acid   must be diluted with water...?
961959
0,220M/L*2,20L= 0,484 moles of HCl\n12,00M = 0,484 moles/?\n? = 0,0403 L or 40,3 mL of 12,00 M HCl solution + 2,16 L of pure water, so you can get you 2,20 L of 0,220 M HCl solution
313620
M(conc.base) x vol.(conc.base) = M(dil. base) x vol.(dil. base)\n12 x vol. = 2.5 x 350\n vol = 72.9 ml \n\nIn a graduated cylinder, take 72.9 ml of concentrated NaOH. And in a standard flask, take 350 - 72.9 = 277.1 ml of water. To this flask, add the concentrated NaOH (72.9ml) from the graduated cylinder and shake the flask for uniform concentration. Now you've prepared 350 ml of diluted NaOH solution.
202477
I'm not sure if I understand your question.\n\nIf you are wanting to make a 0.92 M NaCl solution out of 0.39 M NaCl, well you are going to have to work hard.  Basically, you would be boiling the water and having to measure the mass of the flask.  This is a very tedious process.  Bottom line, you cannot use the dilution formula to make it happen.\n\nNow, if you are wanting to make a 0.39 M NaCl solution from a 0.92 M NaCl solution, then you use:\n\nM1V1 = M2V2\n\n0.92 * V1 = 0.39 * (.1)\n\nremember .1 L is 100 ml.\n\nSolve for V1\n\nV1 = 0.42 L\nV1 = 42 mL\n\nSo, take 42 ml of the 0.92 M NaCl and dilute it up to 100 ml.\n\nHope it helps!
422227
In your example, the concentration of the solution weight/weight would be (presuming the 100 ml is water, thus 100g) would be 10.7% (12 g solute/112 g total solution).
808705
Molarity of the solution = Number of moles of citric acid /1000 ml of solution\n\n= (250/192.1) moles of citric acid per 1000 ml of solution\n\n= 1.3014 M
186466
A quantity of 22% acid solution is being diluted with some 8% acid solution. The final solution must be 12% acid & contain 101 mL. How much of each solution must be added to accomplish this?- write and solve a system of equations that describes this.\n\nx = amount of the 22% soln\ny = amount of the 8% soln\nx + y = amount of the 12% soln = 101 ml\n\n.22x + .08y = (x + y)*(.12)\n.22x + .08y = .12x + .12y\n.10x - .04y = 0\n\nx + y = 101\nx = 101 - y\n\n.10(101 - y) - .04y = 0\n10.1 - .10y - .04y = 0\n10.1 - .14y = 0\n10.1 = .14y\n(10.1)/(.14) = y\n72.143 ml = y\n\nx + y = 101 ml\nx = 101 - y\nx = 101 - 72.143\nx = 28.857 ml
869667
Suppose your initial mixture (of acid & water) has a volume of x liters. So the volume of acid is 0.1x and the volume of water is 0.9x. When you add this mixture to 5L of acid, the percentage of water in the final solution is:\n\n[water/(acid + water) ]  = 0.4 (or 40%, as the question says)\n\nYou can substitute the values of water and acid in terms of the variable x (also include 5L of the acid), solve the eqn for x. \n\nOnce you have solved for x, you will know how much water you added (multiply x with 0.9). \n\nHope this helps!
124546
concentrated sulfuric acid is 36N, so if you dilute it by 17.9 times(56mls to 1L) then it would be 36 divided by 18(rounded off) or 2N.
2508
Weak acids and organic acids in water.\n\nBut remember the solution depends on what the solvent is.  In water, the seven strong acids essentially completely ionize...i.e. can be considered completely ionized.
295517
First you need to know what a mole is. A mole of HCl is it's formula mass, which is 36.5g. So, if you put 36.5g of HCl into one litre of water, you would have 1.0 molar solution. For 0.1 molar you use 3.65g HCl.
54175
If you have .15 Liters of a 6 Molar solution, that means you have .9 mol of HCl.  (A Molar = 1mol/Liter)\n\nTo make .9 mols back into a ratio of .1mol/1L or 1mol/10L, you need to reach a sum of 9 Liters.  Since you have .15 already, you need an additional 8.85 Liters.
1115475
To make 100mL: \n\n0.6% from a 10% solution is 6ml of the 10% diluted in 94ml of H2O.  \n\n0.6% from a 25% solution is 2.4ml of the 25% diluted in 97.6ml of H2O.\n\nThe math?\n\n0.6%*100mL / 10% = 6ml\n\n0.6%*100mL / 25% = 2.4ml
710905
i hope i rember this from 10th grade correctly:\n\nhousehold bleach is 12% so its 0.12 M right?\n\nM( molarity) = mol/L \n\n.12 = mol/4\n\nmol = .48,  \n\nnow you get the molecular mass of NaOCl:\n\n Na: 1 x 22.99\n O: 1 x 16\n Cl: 1 x 35.45\n--------------------\n            74.44 g\n\n74.44 grams of NaOCl in a mole of that\n\nsince it was .48 mol, you times that to the molecular mass:\n\n.48 x 74.44 = 35.7312g of NaOCl in 4 liters of household bleach,\n\n------------------------------------------------------------------------------\n\npool chlorine: same procedure:\n\n30% = 0.3 M\n\n0.3 = mol/ 4 \n\nmol = 1.2 \n\n74.44 x 1.2 = 89.328 g of NaOCl in 4l of pool Chlorine,\n----------------------------------------------------------------------------\nI hope i rembered correctly, i did this last year though. This would be true if we assume that it is only an NaOCl solution.
564420
.1ml of 100:1 dilution had 280\n1.0 ml of dilution has 2800\n1.0 ml of undiluted has 280,000
516306
Until the concentration of salt is large enough to form a saturated solution, ie. one in which no more solute will dissolve, the volume of the solvent (water) will not change.  20g of salt is not nearly enough to form a saturated solution, so the water level will remain at 500ml.
478026
1400 mL of NaOH needed...\n\nneed to make sure the number of moles 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\n\nthanks redsuns :D
1102382
You can dissolve it in water. Upto 7.4 g dissolves in 100 mL of water.  This solution will be acidic.