How much pure salt must be added to 25 grams of a 12% salt solution to make it a 20% solution?
831841
Let g = grams of salt added.\n\nThere are already 3 g of salt in there.  So there will be 3+g grams of salt when you are done, and there will be 25+g grams of total solution.\n\nWrite a proportion:  (3+g)/(25+g) = 20/100 and solve
239849
.7M x 20mL = .25M x (Volume of new IA solution)\nVol = (.7 x 20) / .25\nVol= 56mL
784999
.15 M solution of NaCl means .15 moles of NaCl (solute) in 1000g of water (solvent).\ntherefore, to make 250 ml (or g) solution [ as density of H2O is 1]\nmoles of NaCl required = .15/4 = .0375 moles \n& moles = given wt * molecular wt \nso, given wt (or required wt in this case)= moles/ molecular wt. = .0375/58.1 = 645.4388 * 10^-6 g
470567
To do this problem we need to do some convertions.  Convertion factors are simply ratios that allow us to change the units of a certial quantity.  For example: the molar concentration (molarity) is a convertion factor that is a ratio of the moles of dissolved substance to the volume of solution (moles/L). Using the concentration  we can convbert moles to liters or liters to moles depending on what we have to start with.  Another convertion factor that we will need is the molar mass of KOH.  This is the ratio of the mass of KOH to the moles of KOH (this is usually measeured in grams per mole (g/mol)).  We can find this number by looking up the molar masses of K, O, and H on a periodic table and adding them together. K =39.0983g/mol, O = 15.9994g/mol, and H = 1.00794g/mol. \nAdding these together we have a molar mass of 56.10564g/mol for KOH.\nSchematicaly this is the road map we will follow to find our answer:\n\n-------(moles/L)-------(g/mol)\nvolume <------> mole <-------> mass\n\nWhere volume, moles, and mass are actuall ammounts and the convertion factors between them are on the arrows.\n\nNow for your specific problem we will start with the given volume 750mL.  First we will convert it from mL to L then convert it to moles of KOH then finally to grams of KOH.\n\n750ml(L/1000mL)(15.0mol KOH/L)(56.1056g KOH/mol KOH)= 631.188g KOH\n\nTo three significant figures our answer is 631g KOH.  This is a big number but a 15.0 M solution is very concentrated so it is not surprizing.
880704
The pH of the salt depends on the strengths of the original acids and bases:\n\nAcidBaseSalt pH\nstrongstrongpH = 7\nweakstrongpH > 7\nstrongweakpH < 7\nweakweakdepends on which is stronger\n\nThat should be what you want.
1136961
Are you preparing a buffer soln with NaHCO3 and NaCl ? Then take equimolar amounts of both and dissolve them in 100 ml of water and then dilute them to the desired strength.
127092
It would depend on the fluid that you are weighing.  Water for example is 1 g/mL at 20 degree Celcius. Other fluids with different densities will convert differently.
607804
The easiest way to prepare an insoluble salt is to produce it from two aqueous solutions each containing an ion present in the salt!\n\nSo, you'll just mix two solutions, and the insoluble salt appears! Of course, if you intend to prepare a certain amount/ mass of the salt, you may need to carry out some calculations to determine the necessary concentrations/ volumes of the 2 initial solutions. \n\nYou'll also get another product in the reaction, but it should be soluble and will not contaminate the desired product (the insoluble salt).  \n\nYou may refer to this website to gain a better understanding...\nhttp://www.wpbschoolhouse.btinternet.co.uk/page01/AqueousChem/AqueousChem.htm#salts [Refer to point (4)]
646163
Presuming that the base is monobasic.\n\na) with 25 ml added, you are at the equivalence point.  The pH is 7.\nb) after adding 25.01 ml, you will have 0.01 ml of base that did not get reacted.  The molarity of the base in solution will then be 0.1 * (0.01/50.01) = 2.0 x 10^-5.  Your pOH is 4.70, pH would be 9.30\nc) Now, it's 0.1 ml of unreacted base.  Molarity is 0.1 * (0.1/50.1) = 2.0 x 10^-4.  Your pOH is 3.70, pH is 10.30.\nd) Now 1 ml of unreacted base.  Molarity is 0.1 * (1/51) = 2.0 x 10^-3.  pOH is 2.70, pH is 11.30.\ne) Now 25 ml of unreacted base.  Molarity is 0.1 * (25/75) = 3.33 x 10^-2.  pOH is 1.48, pH is 12.52.\n\nIn each case, you have to figure out the amount of unreacted titrant and recalculate the molarity as if you were simply diluting it, using McVc = MdVd, where the volume is now the total volume of the solution (the unreacted amount plus the 25 ml of acid and the 25 ml of base that have neutralized each other).
1172866
this can be a problem to you if you don't know anything about simultaneous equation or if u don't know how to form simultaneous equation. another thing is u should at  least have an  idea about chemistry. look\n  \nm1v1+m2v2=MV.............1\nm1+m2=total volume after dilution=150Lt..............2\n now\n              take m1 is the initial conc of saline soln=32%=0.32\n                   ''  v1                      volume                 =?\n                 m2 is the conc of water =0\n                 v2 is the volume of water =?\n               M is the final conc of the soln = 20%=0.20\n                V is the final volume of the soln obtained after mixing=150Lt\n\nnow\n             0.32v1+0*v2=0.2*150......................1\n\n             v1+v2=150.................2\nfrom eqn 1\n                      0.32v1=30\n           so v1= 93.75Lt's\n   so the remaining amount is of water\n                v2=150-93.75=56.25lts\nso you need 56.25litres of water and 93.75litres of 32%saline to prepare the solution you need.\n\nif you have these solution just try to mix at that proportion then u will tell me.
745889
We can solve this using a system of equations.\n\nLet x be the volume of the 25% solution. Let y be the volume of the 60% solution.\n\nThe sums of the volumes must total 140 ml:\n\nx + y = 140.\n\nThe mass of CuS04 in the desired solution must equal the amounts added through the use of both solutions.\n\n30 percent of 140ml gives 42 ml of "pure" copper sulfate solution.\n\nThe 25% solution would contribute .25x of "pure" copper sulfate solution. The 60% solution would contribute .60y of "pure" copper sulfate solution, so:\n\n0.25x + 0.60y = 42.\n\nHere is the system:\n\nx + y = 140\n0.25x + 0.60y = 42\n\nThere are many different ways to solve the system. I'll use substitution here:\n\nx + y = 140, so y = 140 - x\n\n0.25x + 0.6(140 - x) = 42\n0.25x + 84 - 0.6x = 42\n-0.35x = -42\nx = 120\ny = 20\n\nSince x is the amount of 25% solution and y is the amount of 60% solution, we need 120 ml of 25% solution and 20 ml of 60% solution.\n\nThis appears to match your answer!
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.
302581
add 2 parts HCl 1 part H20 to dilute properly
371296
well seeing that the molarity of the two solution is a direct ratio of how many molecules there are in the solution, i think that the ration would be 0.15/0.20\n\nso assuming that the reation is 1 to 1 (i cant remember so if it isnt 1 to 1, then i'm wrong) then you would need 18.75 mL of the 0.200 M HCl
775430
25% citric acid will have a concentration of 250 g/l.\n\nDividing by 192.1 g/mol we will get:\n\n(250 g/l) / (192.1 g/mol) = 1.3 mol/l = 1.3 M.
445032
Click on "Add details," post the problem, and, as SMeghead said, someone might be able to help.\n\n(EDIT: You didn't provide too much information. From what you gave, about all we can say is that 12.5% of the solute - 1% from 8% - evaporated. If you need to know how many grams, milliliters, cubic centimeters, etc. evaporated, we'll need more information.)
637181
I dont know the exact calculation but this site might be able to help you.\n\nhttp://www.fseric.demon.co.uk/skills/salinity.htm