How many mL of this solution must be diluted with water in order to make 1.00L of 0.100 moles Ba(OH)_2?
459258
start with computation of molarity of solution:\nM_BaOH = (51.24g/ MW BAOH)/ 1.2 L\n\nthen \nusing M1V1 = M2V2\nM_BaOH (x) = (0.1) (1L)\nsolve for x
1006885
I think you are leaving out the first step.\n\nRemember that 12.0 M means 12 moles of HCL per liter of aqueous solution.\n\nTo understand the equations so that you can use them correctly, first think that you will have to add some amount of water to some amount of the concentrated solution to make it the dilute solution.\n\nso, 12 mol/liter , assuming you had a liter of the solution, you keep the same number of moles and add water\n\nhow much water do you add to get to the weak, .6 mol/liter concentration\n\n12 mol/ (1+x) liters = .6 mol/liter\n\n12/(1+x)=.6\n12=.6(1+x)=.6+.6x\n11.4=.6x\n19 = x\n\nso if you had a liter of the solution at 12 M, you would have to add 19 liters of water and then you would have 20 liters of solution at the diluted .6 M\n\nbut you don't need that much of the final solution, so you use the same ratio of water addition, but only start with enough concentrated to end with 300 ml\n\nthe ratio is 19 to 1\n\nso 1/20 * 300 = 15   you need 15 ml of the conc. solution\n   19/20 * 300 = 285  you add 285 ml of water \n\nthe total is 300 ml of the .6 M solution\n\nnow look back at the equations and you will see where they came from and how to apply them properly
725559
Take, say, a standard 2M solution and dilute it by a factor of 20.\n\nThat is, add 19cm3 of water to 1cm3 of 2M HCl.
168718
The eq constant is 5.286E-16, so this means that 5.287E-16 = [H][OH]. (The [H2O] doesn't matter as it is a pure liquid.) The H and OH concentrations are the same (because of mole ratios), so solve this equation:\n\nx^2 = 5.287E-16\n\nx is the eq. concentration of H, and also the eq. concentration of OH.
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)
1125846
NaVO3, MW = 122 g/mol\n\n100 mM = 0.1 M, so 0.1 moles in 1 L.  Dissolve 12.2g in water, dilute to 1L.
777607
Since they are both strong acids and bases you can assume: \n\nMaVa = MbVb\n\n(25/1000) * (1.2) = Mols NaOH\n\nconvert Mols to grams and that's your answer
939677
Long answer:  Use the vol. and concentration of NaOH to figure out how many moles of NaOH you have which will = moles of OH (since NaOH has only one OH).  Formula is:\n\nMolarity = moles/liters (make sure the volume is in liters)\n\nIn a neutralization reaction the moles of OH = moles of H+\n\nOnce you have moles of H+, use that answer and the volume of HCl to calculate concentration of H+ which will be = to concentration of HCl since HCl has only one H+\n\nUse the same Molarity formula\n\nShort answer:  As long as the base (NaOH) and the acid (HCl) have only one OH and one H, the short form is:\n\nMb x Vb = Ma x Va\n\nwhere Mb = molarity of the base\nwhere Vb = volume of the base\nwhere Ma = molarity of the acid\nwhere Va = volume of the acid
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.
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
964051
First you must know the equation of the reaction, so you can establish the mole connections between the involved substances\n\n2HNO3(aq) + Mg(OH)2(aq) → Mg(NO3)2(aq) + 2H2O(l)\n\nNow, you have 60,0 mL of 0,100 M Mg(OH)2(aq)\nTherefore:\n0,100M*(60,0 mL)(1L/1000 mL)= 6,00*10^-3 moles of pure Mg(OH)2\nNow, 2 moles of HNO3(aq) react with 1 mol of Mg(OH)2(aq)\n\n6,00*10^-3 moles Mg(OH)2*(2 moles HNO3/1mol Mg(OH)2 )= 0,0120 moles of HNO3(aq)\nMolarity = moles solute/volume of solution\n0,200M= 0,0120 moles /?\n?= 2,40*10^-3 L of solution= \n2,40 mL of 0.200M HNO3(aq)  are needed to completely neutralize 60.0 ml of 0.100 M Mg(OH)2 solution
798989
1) Calculate the mass of acid in 1 liter:\nIn one liter we have 0.01 mol = 1.321 gram\n\n2) Calculate the dilution factor:\nWe have to dilute by a factor 5/1.321 = 3.785..  \n\n3) Calculate the volume of the 5g/litre solution we have to dilute to obtain 100 ml:\nAs the dilution factor was 3.785, we need 100/3.785 = 26.42 ml of the original solution.
862256
Going from 10^12 cells/mL to 10^10 is a 1:100 dilution.\nIf you want a final of 2mL, you need to add 20uL of your stock solution.\nSo do 20uL of stock (10^12)+ 1980uL of salt solution = 2000uL (10^10), which is the same as 2mL.
710157
ANSWER: 0.482 M\n\n\nEXPLANATION:\n\nM = mol solute / L soln\n\nTherefore, you need to find the mol of solute and L of soln to find M.  \n\nTo find mol solute:\n10800mg Na(1g Na/1000mg Na)(1mol Na/22.99g Na) = 0.4698mol Na\n\nTo find L of soln:\n1kg soln(1000g soln/1kg soln)(1ml soln/1.025g soln)(1L soln/1000ml soln) = 0.9756L\n\nUsing the two values found plug in to M = mol solute / L soln\n\n0.4698mol Na / 0.9756L soln = 0.482 M
575750
This can be done using the formula M1V1=M2V2.  For example, if you have 2M HCl and wanted 2L of .5M HCl, then putting that into this equation as 2(V1)=(.5)(2).  Solve accordingly.  This is just an example.  This formula can be used in a variety of ways that all.  But, to make it to the desired you, a volumetric flask is needed in which water is added until the etched line on the flask to equal the preset volume of the flask.
558772
the correct answer is 1.30.  the answer would be two if the concentration was 0.01M.  you just take the negative log of the concentration of hydronium ion.
773833
2NaOH + H2SO4  = Na2SO4 +2H2O                                      Sodium Hydroxide is 0.5moles.Two Moles neutralise One mole of Acid.   0.5Moles neutralise of Alkali neutralises 0.25Moles of Acid                                                                                           50mls of 1M SULPHURIC ACID = .05M THEREFORE MOLAR CONCENTRATION IS 5