How do you find the molarity of this solution?
601420
Well, I'm not doing your homework for you, but Molarity is Moles per liter.  So that means you would have to change grams to moles, hopefully you know how to do that.
755177
1. Make sure that HCl is standardized.\n2. Write the balance equation:\n\nCa(OH)2 + 2HCl ---> CaCl2 + 2H2O\n\n- this means that 1 mole of Ca(OH)2 reacts for every 2 moles of HCl\n\n3. Get the number of moles of HCl reacted by multiplying the volume used to react Ca(OH)2 by the actual concentration of HCl (value from standardization). Divide the result by 2 - and this is now your moles Ca(OH)2.\n\n4. Get the molar solubility of Ca(OH)2 by dividing its number of moles by its volume. Unit is in gram per liter.
832115
N2H4 + H2O -> N2H5+ + OH-\n\nKb = c[N2H5+] * c[OH-] / c[N2H4]\n\nAssuming that c[N2H5+] = c[OH-] and that c[N2H4] >> c[OH-] (so that we can use the starting concentration for c[N2H4]), we get approximately\n\nc[OH-] = sqrt( Kb * c[N2H4])\n\nor logarithmized\n\npOH = 1/2 ( pKb + 10^-(c[N2H4])\n\npH= 14 - pOH\n\n\ncalculator time.
724819
To determine the concentration of a reagent.
850392
You have to use the methods for equilibrium and pH of a salt to calculate the concentration of salt that would produce that pH.\nI cannot find my notes from chemistry right now,\nhowever you need to know the definition of Ka(or Kb)\nthen set up the equation as the salt reacts.\n\nnh4+ + h20 <--> nh3 + h30+\n\npH = -log[H30+}, so your [h30+] = 10^-pH\nyou can then calculate the concentration of nh4 using molar ratios and such, and then you will have molarity.
971190
Molarity is defined as number of moles in 1 liter.\nYou have 0.01 mol and 10 liters\nM = 0.01 / 10 mol/liter - 0.001 mol/liter = 0.001 M.  That's it.  You did it.\n\nNow pH = - log[H+].\n\nIf you write the equation\nHCl ---> H+ + Cl-\nThis means that you have the same molarity for HCl and H(+).  For both of them the coefficient in the chemical equation is equal to 1.  For other problems the coefficients may be different so always look at the balanced equation.\n\nso M(HCl) = M(H+) = 0.001.\n\nPutting this value into the pH equation:\n\npH = - log 0.001 = 3\n\npH + pOH = 14.\nCalculate the pH according to the pH definition and then use the above equation to find the pOH.
44293
Go here to find out about  "MOLAR ENTHALPY"\n\nhttp://www.carolina.com/chemistry/activities/Determining_Molar_Enthalpy_of_Solution.pdf
839352
If tha atomic mass of M is A, then the molecular weight is MW=2*A+60\n\nThe concentration of the M2CO3 solution is C=mole/V and since mole=mass/MW, C=mass/(MW*V)\n\nThen you do a titration for which we assume complete reaction and not reaching only the first end-point, so the reaction is\n\nM2CO3 + 2HCl -> 2MCl + CO2 + H2O\n\nTherefore 2*mole M2CO3=mole HCl =>\n2*C1*V1= C2*V2=>\n2*V1*mass/(MW*V) = C2*V2 =>\nMW= 2*V1*mass/(C2*V2*V)=\n=2*0.025*7.46/(0.1*0.027*1)\n=138\n\nand MW=2*A+60=138=> 2A=78 => A=39\n\nSo M, is potassium (K)
861217
molarity = moles/L\nso just plug in what you have!  6.3M= mol/0.216 L   just multiple 6.3 by 0.216 and you will get the moles of the acid (i dnt have a calculator on me right now, but just do the calc out and theres ur answer! ) :)
832170
Molarity = moles of solute/liters of solution\n15.8 g of HF = 1.01 + 19 = 20.01 g/mol\n15.8 g HF / 20.01 g/mol HF = 0.790 mols of HF\n0.790 moles of HF / 0.200 L of solution = \n3.95M solution
1117999
First make a chemical equation, balance it, and find a mole ratio.  Use that ratio and the above information to solve the problem.
39876
To answer this question you must look at your given data. 13% of 1.1 = .143 so you have .143g NaCl for every ml solution. Molarity is moles solute/liters solution. You just calculate how many grams NaCl are in a liter and convert to moles, do the division and your done.
479476
You need to know concentrations to determine the amount of substance that precipitates out by using solubility products. In this case the precipitate will be CaSO4, so the solubility product will be given by:\n\nk=[Ca(+2)]*[SO4(-2)]\n\nWhere k is the equilibrium constant (probably a given), and [Ca(+2)] & [SO4(-2)] are the concentrations of Ca(+2) and SO4(-2) in the solution. Any amounts greater than the equilbrium concentration will precipitate out as a solid.
715558
OK, but you need to look up atomic weights or calculate molar mass.\n\n   Grams = Moles x molar mass  OR  moles x atomic mass\n\n    Moles = Grams/(molar mass) OR Grams/(atomic mass)
876235
The product of a neutrilization reaction is always water and a salt, so use that as a guide.  For part 2, use the units as a guide.  Hint M=molarity=mols/L
715055
First off, you need the concentration of the NaOH solution.  From that you can determine the number of moles of NaOH that were used.  This is equal to volume (V) time the concentration, in mol/L (C).\nn = V * C = 0.02260 L * ? mol/L = ? mol\nn = 0.02200 L * ? mol/L = ? mol\n\nThe reaction of KHP (C8H5O4K) and NaOH is this:\nHOOC-C6H4-COOK + NaOH -> NaOOC-C6H4-COOK + H2O\nSo, 1 mol of NaOH will react with 1 mol of KHP.\nThat means that the number of moles of NaOH you calculated above is also equal to the number of moles of KHP.\nThe molecular weight of KHP is 12*8 + 1*5 + 16*4 + 39 = 204 g/mol\nMultiply that by the number of moles: 204 g/mol * ? mol = g actual KHP\n\nYou can divide that number by the 0.911 and 0.950 g and then multiply by 100% to find the purity.\n\nHere’s an example if your NaOH solution was 0.1 M (M means ‘molar,’ or mol/L):\nn = 0.02260 L * 0.1M = 0.002260 mol of NaOH = 0.002260 mol of KHP\n204 g/mol * 0.002260 mol of KHP = 0.461 g pure KHP\n0.461 g pure KHP / 0.911 g impure KHP * 100% = 50.6 % KHP\n\nHope this helps!
802376
YOUR ANSWER IS CORRECT!  Here's how i did it:\n\nYou can find the # mol of NaOH used:\n\nMolarity = (# mol)/(volume solution)\n0.4 = (# mol)/(.075 L)\n\nSolving, you get 0.03 mol of NaOH.\n\nUsing the balanced equation:\n\n3 NaOH + H3PO4 ==> Na3PO4 + 3 H2O \n\nYou see that 3 mol NaOH = 1 mol H3PO4.  Thus, if we have 0.03 mol NaOH, that means we have 0.01 mol H3PO4.\n\nNow we can solve for the molarity:\n\nMolarity = (0.01 mol H3PO4)/(0.05 L) = 0.2 M