An aqueous solution of Br2 is shaken with CCl4 (a liquid immiscible with H2O) at 25°C until\nequilibrium is attained. The final ratio of the molarity of Br2 in the aqueous solution to the\nmolarity of Br2 in the CCl4 solution is 0.035. Find the volume of CCl4 that must be shaken\nwith 50 mL of an aqueous solution of Br2 in order to extract 95% of the Br2 from the aqueous\nlayer.
215438
0.035=7/200 therefore for every 7 parts Br2 you must have 200 CCl4. (200/7)*50 = 1428.6mL
5946
The instruction is not correct because the total volume yielded when you mix 100ml of concentrated solution and 900ml water will be somewhat less than 1000ml. The concentration will be slightly higher than 1.0M\n\nThe correct procedure is to take a 1l volumetric flask  and add the concentrated solution to about 800ml water, then fill up to the mark.
304492
5. KHCrO4 ----------------> K+   +   HCrO4-\nSince KHCrO4 dissociates complteley,\n the molarity of HCrO4- = molarity of KHCrO4\n the molarity of HCrO4- = 0.05M\n\nNow HCrO4 ------------------>    H+    +     (CrO4)2-\nCalculate the concentration of H+ from the pH value\n[H+] = 10^(-pH)   = 10^(-3.80)  = 1.58 * 10^(-4)\nIt is clear the only source of H+ is HCrO4-, therefore the concentration of (CrO4)2- is the same as concentration of H+\n[(CrO4)2-]   =   [H+]   = 1.58 * 10^-4\n\nKa = {H+} * [(CrO4)2-] / [HCrO4-]\nKa = [1.58 * 10^(-4)]^2/0.05\nKa = 5.02 * 10^ (-7)\n\n6.       KOH----------------> K+    +    OH-\n         HClO4 -------------> H+    +    ClO4-\n(a) \n    Moles of KOH = Molarity * Volume = 0.200M * 0.030L\n    Moles of KOH = 0.006 moles\n    From the equation above, \n    Moles of OH- = moles of KOH = 0.006 moles\n    \n    Moles of HClO4 = Molarity * Volume = 0.150M * 0.030L\n    Moles of HClO4 = 0.0045 moles\n    From the equation above,\n    Moles of H+ = moles of HClO4  = 0.0045 moles\n\n     H+    +    OH-    ------------>   H2O\nWhen this reaction is finished, 0.0015 moles of OH- are left\n[OH-]  =  moles of remaining OH- / Total Volume\n[OH-]  =  0.0015 moles/ 0.06L (30ml + 30ml)\n[OH-]  =  0.025M\npOH = -log [OH-]\npOH = -log (0.025) = 1.60\npH = 14 - pOH  = 14 - 1.60 = 12.40 \n\nDo the same for parts b, c, d, and e\n\n7. I will assume that \n   H2S --------------------------> 2H+    +    S2-\n   Ksp of HgS = 2.0 * 10^(-53)\n   Ksp of FeS = 6.3 * 10^(-18)\n\nKsp of HgS = [Hg2+] * [S2-]\n[Hg2+] = 0.01 M\n[S2-] = x\n2.0 * 10^(-53) = x*0.01\nx = 2.0 * 10^(-51)\nFrom the above equation, for every 1 mole of S2- there are 2 moles of H+\ntherefore,\n[H+] = 2 * 2.0*10^(-51) = 4*10^(-51)\npH = -log (4*10^(-51)) = 50.39\n\nKsp of FeS = [Fe2+] * [S2-]\n[Fe2+] = 0.01M\n[S2-] = x\n6.3* 10^(-18) = x* 0.01\nx = 6.3 * 10^(-18)\nFrom the above equation, for every 1 mole of S2- there are 2 moles of H+\ntherefore,\n[H+] = 2 * 6.3 * 10^(-18) = 1.26*10^(-17)\npH = -log (1.26*10^(-17)) = 16.90\n\nthe pH range is from 16.90-50.39\nThis is the best explanation I can gibe. By the way, pH level beyond 0 and 14 do exist, but, I'm not 100% about this answer.\nI fixed it, I think it is correct now.
393267
Actually it is not a solubility problem, but a buffer.\n\nLet's use the Henderson-Hasselbalch equation for simplicity\n\npH=pKa+log[CH3COONa]/[CH3COOH]\n\nIf V is the volume of the solution in L then\n[CH3COOH]= mole/V= 0.010/V\n[CH3COONa]= mole/V =x/V\n\nso substitute in the equation and you get\n\n5.37=4.74+ log((x/V)/(0.01/V))=>\n0.63=log(x/0.01) =>\nx/0.01= 10^0.63 =>\nx =0.01*10^0.63= 0.0427 mole =42.7 mmole or since MW=82\nm= 0.0427*82= 3.50 g
456556
These are the right ones:\n\n2K(s) + 2H2O(l) ---> 2KOH(aq) + H2(g)\nH2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l)\nCa(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)
36059
balance equation:\n \nCaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2CO3(aq)\nH2CO3(aq)->  H2O(l) + CO2(g)\n\nconbining the two equations,\n\n  CaCO3(s) + 2HCl(ag) -> CaCl2 (aq) + H2O(l) + CO2(g)
410812
Since the density is 1.0 g/mL, one liter of solution would be equal to 1000 g of solution and would contain  30 g of H2O2.  The molar mass of H202 is 34 g/mol so 30 g = 0.88 mol.  So you have 0.88 mol solute per 1.0 L solution which is the molarity.\n\nMolality is moles solute per 1000 g solvent.  You have 0.88 mol solute per 1000g solution.  That 1000 g of solution consists of 30 g of H2O2 so the amount of water must be 970 g.  So you have 0.88 mol solute per 970 g solvent = 0.91 mole / 1000 g solvent.\n\nMole fraction:  970 g water = 53.89 mole.  So the mole fraction is \n0.88 mole/ (53.89 + 0.88) mol = 0.016
348314
use the molarity x volume = molarity x volume equation, but beware that the H2SO4 shells out 2 protons, not 1.
341792
Na2CO3.xH2O + 2HCl --> 2NaCl + CO2 + (x+1)H2O ---(1)\nNa2CO3 + 2HCl --> 2NaCl + CO2 + H2O         --- (2)\n\nMass of Na2CO3.xH2O is 2.835g\nMean titration result (HCl) was 22.03cm3\n\n25.0cm3 of Na2CO3 solution was neutralised by 22.03cm3 of HCl\n\ncalculating no of moles of HCl,\nno of moles\n= (concentration) x (volume)\n= (concentration) x 0.2203 dm3\n= a [this is a variable]\n\nno of moles of Na2CO3.xH2O = half no of moles of HCl (according to the second chemical equation)\nso, no of moles of Na2CO3 = 0.5a\n\nI assume that the 25.0cm3 of Na2CO3 solution that you used for titration was measured from 250cm3 of solution. The latter had been made by dissolving solid Na2CO3.xH2O in water, then adding water until the total volume was 250cm3.\n\nSo since in 25cm3 there are 0.5a moles, in 250cm3 there'd be 10 times the no of moles, which is 5a.\n\nThus, the no of moles of Na2CO3 is 5a.\n\nYou can calculate the mass of Na2CO3:\nmass of Na2CO3\n= (no of moles) x (molar mass)\n= 5a x (2x23 + 12 + 3x16)\n= 5a x 106\n= b grams\n\nYou had 2.835g of hydrated Na2CO3 in the first place, so if you subtract the mass of Na2CO3 from the total mass, the remaining mass is the mass of water.\nMass of water = (2.835 - b) g\n\nSo then, you can find the no of moles of water:\nno of moles of water\n= (mass of water) / (molar mass)\n= (2.835 - b) / (2x1 + 16)\n= (2.835 - b) / 18\n= x\n\nAnd there you have it! :D You just have to substitute the concentration of the acid and everything will fall into place.
449502
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!!!!!
219638
I'm guessing that your question pertains to the series of standard solutions you would prepare using the stock solution and volumetric flasks/pipette.  \n\nYes, you can use the C1V1=C2V2 formula, however there's a more practical and faster method that can be arrived at through common sense.  That is the concept of dilution factor.  If you have a stock solution of .5M iron for instance and you wanted to make a .05M iron solution, just dilute the next solution by a factor of 10%.  One way this can be achieved is by taking 10 mL of this stock solution using a volumetric pipette and adding it to a 100 mL volumetric flask.  To lessen the sampling error, it's better to take a larger sample volume, but usually this becomes tedious since lab equipment can be limited.\n\nSo take the original x molarity stock solution.  And if your final desired concentration is y, then the dilution factor is\n\ny/x \n\nThis percentage, y/x, represents the volume percentage in the final solution, that is multiply y/x by the particular To Deliver volume of the volumetric flask, to determine how much volume of the stock solution you need.......place this solution into the volumetric flask with a volumetric pipette, and then dilute to the mark with the standard solvent.
394802
To solve part a, you need to know Kb, the equilibrium constant for the reaction:\n\nNH3 (aq) + H2O -> OH- + NH4+\n\nKb = [NH4+]*[OH-]/[NH3]\n\nFrom the first source, pKb = -log(Kb) = 4.75, so Kb = 10^-4.75, and we have that\n\n10^-4.75 = [NH4+]*[OH-]/[NH3]\n\nWe are told that [NH3] = 0.5 M. so plugging this in and solving for [NH4+], we have:\n\n[NH4+] = 10^-4.75 * 0.5/[OH-]\n\n\nWe also know that for aqueous solutions at room temperature,\n\n[H+]*[OH-] = 10^-14\n\nor\n\n[OH-] = (10^-14)/[H+]\n\npH is defined as:\n\npH = -log[H+], and we are told that the pH of the solution is 8.95, so:\n-8.95 = log[H+]\n[H+] = 10^-8.95\n\n[OH-] = 10^-14/10^-8.95 = 10^-5.05\n\nPlugging this into the equation for [NH4+], we get:\n\n[NH4+] = 10^-4.75 * 0.5 / 10^-5.05 = 0.5*10^0.3 = 0.998 M ~= 1.0 M\n\n----\n\nFor part B, we again have that:\n\n(eq 1) [NH4+]*[OH-]/[NH3] = Kb\n\n(eq 2) [OH-]*[H+] = 10^-14\n\nWe are told that 4 gm of NaOH is added to 1 liter of the previous solution.  The molecular mass of NaOH is 40 gm/mol, so (4 gm)/(1 liter)/(40 gm/mol) = 0.1 M.  The only source of Na+ in the solution is the NaOH that's added, and each mole of NaOH gives 1 mole of Na+ ions:\n\n(eq 3) [Na+] = 0.1\n\nIn the previous solution, there were 0.5 mol/liter of NH3 and 1.0 mol/liter NH4+ ions.  The total concentration of ammonia species is 0.5 + 1.0 = 1.5 M because increasing the concentration of NH4+ ions requires that the concentration of NH3 be decreased by an equivalent amount:\n\n(eq 4) [NH3] + [NH4+] = 1.5\n\nFinally, the solution must be electrically neutral.  That means the charges on the positive ions must be balanced by an equal number of negative charges:\n\n(eq 5)  [Na+] + [H+] + [NH4+] = [OH-]\n\nThis is 5 equations in 5 unknowns (the concentrations of H+, OH-, NH4+, NH3, and Na+), so we can solve for all the unknowns.\n\nCombine eq 1 and eq 4 to get:\n\n[NH4+] = 1.5*Kb/([OH-] + Kb)\n\nand rewrite eq 2 as:\n\n[H+] = 10^-14/[OH-]\n\nUse these two equations to eliminate [H+] and [NH4+] in eq 5:\n\n[Na+] + 10^-14/[OH-] + 1.5*Kb/([OH-] + Kb) - [OH-] = 0\n\nMultiply through by [OH-] and again by ([OH-] + Kb) to clear the denominators, and we get:\n\n10^-14 + ([Na]*Kb + 1.5*Kb + 10^-14)*[OH-] + ([Na] - Kb)*[OH-]^2 - [OH-]^3 = 0\n\nPlugging in numerical values for Kb and [Na+] we have:\n\n10^-14 + (2.845*10^-5)*[OH-] + 0.1*[OH-]^2 - [OH-]^3 = 0\n\nThe first two terms are very small, and as an approximation to make it easier to solve this 3-rd order polynomial, we can ignore them.  We then have:\n\n0.1*[OH-]^2 - [OH-]^3 = 0\n\n0.1 M = [OH-], \n\nwhich is simply equal to the concentration of OH- ions added as NaOH.  That is, the presence of the ammonia species makes no difference to the pH, which we calculate as:\n\n[H+] = 10^-14/[OH-] = 10^-14/.1 = 10^-13\npH = -log(10^-13) = 13 \n\nYou can check to see if dropping the terms from the 3rd-order polynomial was justified by plugging the value for [OH-] value back into the equation and seeing if the equation is true:\n\n10^-14 + 2.845*10^-6 + 0.001 - 0.001 = 0\n\n2.845*10^-6 ~= 0\n\n\nIf one does the math, or uses numerical methods to solve the full third-order equation, one finds that the "correct" concentration of OH- is:\n\n[OH-] = 0.1003 M, which is well within typical experimental errors.
425847
From the reaction you had heat produced\n\nQ=m*Cp*ΔT\n\nfor water Cp= 1 calorie/(gram*Celsius)\n\nYou have a total of 50+50 =100mL and since d=1.00 g/mL you have 100 g\n\nso Q =100*1*(31.43-25.32) = 611 cal\n\nyou mixed mole HCl= C*V= 1*0.05= 0.05 with\nmole NaOH =C*V =1.1*0.05 =0.055 which means that, since the stoichiometry of the reaction is 1:1, HCl is the limiting reagent\n\n0.05 mole HCl give 611 cal\n1 mole HCl gives x\nx=611/0.05= 12220 cal/mole =12.22 Kcal/mole
498915
Firstly, please forgive me for not having a periodic table at hand. :)\n\nSecondly, please forgive yourself as even the most advanced scientists in the world sometimes forget how to do this calculation.\n\nThirdly, it is such a simple calculation, that it is no wonder that it is so easily forgotten.\n\n1.5% solution (mass/volume) means 1.5 g of KBr in 100 ml of solution\n\nIt's that simple. :)\n\nEnjoy your science. :)\n\nSincerely,\n\nCurtis
422184
Yes, you are correct.\nEveryone else who has posted so far is mistaken.\n\nGiven:\nM = mol of solute / liters of solution.\n\nFirst Step:\n\nSolving for x (liters of solution), you get:\n.40 / .096 = x\nx = 4.167 L\n\nSecond step.\nL => mL\n4.167 / 1000 = 4,167 ml or app. 4000 mL\n\nYour answer is approx. correct because you were asked for mL of water, which you found, not L of solution, which everyone else seemed to think.  You may lose points for not solving to the correct amount of significant figures though.
366895
Br+H20--->HBr+O\n\nNi(II)O+H20--->Ni+H202\n\nKO2+H2O--->KOH + H2\n\nHope that helps.
219453
AgBr-------->Ag+ + Br-\n(a) Ksp=[Ag][Br] solubility=s=[Ag]=[Br]=Ksp^0.5\ns=(6*10^-7)^0.5\n(b) [Br]=0.01 (because all of kBr was dissuciated)\ns=Ksp/[Br]=(6*10^-7)/0.01 mol/lit