An aqueous solution of Br2 is shaken with CCl4 (a liquid immiscible with H2O) at 25°C until?
474679
0.035=7/200 therefore for every 7 parts Br2 you must have 200 CCl4. (200/7)*50 = 1428.6mL
400328
Since both HCl and KCl are strong elecrolyes, both should remain totally dissassocitated in solution.  I dont think any HCl will react with KCl in this case.\n\nHowever, have you ever heard the rule "Always Add Acid"?\nWhenever you dilute an acid with water, the resulting solution gives off energy (dissolving HCl in water is an exothermic reaction).  This when you dilsute 1 M HCl to .5 M HCl, energy will be released and will cause a temperature change in the solution.\n\na)  The number of moles of HCl used in this experiement is constant. This value will be equal to the number of moles of HCl present in the initial 100 ml of 1 M solution.\n1 M = x moles / .1 L, thus you have .1 moles of HCL.\n\nb)  The heat of the reaction will be the heat given off when the HCl solution is diluted.\n\nc)  Once you know the heat energy given off by the reaction, you can calculate the change in temperatutre if you know the specific heat of the final solution.  I am not sure if you are suppose to estimate the final solution's specific heat as being equal to that of water (4.186 J/g C), but that would be my guess.\nQ = m*c*(delta T)\n\n\nEDIT:\n"davidchingwong13" is correct in his application of the common ion effect.  By dissolving additional Cl- ions in the solution by adding HCl, the equilibrium between solid and aqeuous KCl is disturbed.  However, KCl is a strong electrolyte, it readily dissolves and breaks in to ions in water.  I doubt the addion of so few moles of Cl- ions to an already unsaturated KCl solution willl effect it that much.\n\nHowever....dissolving KCl in water is a endothermic reaction, so if, indeed, the KCl did precipitate out, it would be exothermic and give off additional energy.
136403
You can calculate the freezing point depression of the solution is you know three things, firstly, the freezing point of the pure solvent, and secondly, the concentration of the solute, and thirdly, the freezing point depression constant of the solvent.\nIn this problem, you know both (One is given, two of them are assumed knowledge).\n\nThe freezing point of pure water is 0 degrees Celcius.\nThe concentration of the solute is 52 percent.  You need to convert this to a molaltiy concentration to use it (or fiind a different freezing point depression constant).\nLook up for your self the k_f value.\n\nthe freezing point depression = k_f * m,\nwhere k_f is the freezing point depression and m is the concentraion of the solute in terms of molality.\n\nmolality = moles of solute / kg is solvent
838309
b a solution, the salt will dissolve
923888
CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)
621890
CaCl2+ 2NaHCO3 + 2H2O------> 2NaCl + 2H2CO3 + Ca(OH)2 \n\n\nyou have to add water molecules on left hand side         \n                                                                                  \nwill this help you?
327703
The amount is diluted by a factor of 25.  So the resulting solution would have a concentration of 11.6M divided by 25 or \nor 0.464M.
423379
let me see if i can at least help you with part of your problem, solubility in chemistry is simple, basically the equation will always be:\n\n[Solubility equalibrium constant] = concentration of product divided by concentration of reagents (concentration of each reagents multiplying by each other).\n\nthe easiest way to understand this is to simplify things for you so as you know, a chemical equation always go like this:\n\nreagents <==> products\n\nJust know this from the above, the dynamic here is that when you have an excess of either the reagents or the product, the balance of the equation tend to shift in the opposite direction (ie excess concentration of product will cause more reagents to form and vice versa), in a chemical equation such as the one from above, the reaction can actually go either way, products or reagents can form.  What really matters is the equalibrium constant that I talked about in the first equation. This constant represent the concentration of the products and the reagents when they are in equailibrium or when they are balance, meaning that the rate at which products or reagent form stays the same and there are no change in the concentration of either the products or the reagents so the whole thing is static at the point.   Think about this, more concentration of products wouldn't form unless you have extra reagents to spare that can shift the balance.  When something is saturated it means that it can longer dissolve in a solvent (let's use water to simplify the explanation)so it just accumulates at the bottom, water molecules can only separate so many ions from each other (the reagents) and preventing them from forming a solid product since water itself has limited molecules (if you mix enough table salt in a glass of water, it'll get to the point where it'll accumulate at the bottom and won't dissolve).  If remember correctly, in a chemical equation, when there's different amount of reagents needed to form the product, there's a slight modification to the equation I gave you, for example:\n\n2 CaCO3 + 4 ZnO2 <----> 3 ZnCO3 + 3 CaO2\n\nthen  [equilibrium constant] = [Cao2]^3 x [ZnCO3]^3 divided by [CaCO3]^2 x [ZnO2]^2\n\nnote that you should balance the chemical equations first, I am just giving an example.\n\nhere's a summary of how to compare numbers, if you plug given concentrations of reagents and product into the equation and the number you get is larger than the real equilibrium constant, then the concentration of products is higher than it should be and the equation will shift towards forming more reagents (in the case of the solid product, it'll dissolve) and if there's more reagents, more solid product is gonna form.  Hope that helps. Remember the Chemistry textbook is still your best reference.
742488
C1V1=C2V2\n40%x100mL=25%xV2\n4000=25%xV2\n4000/25=V2   ((25%xV2)/25=V2)\n160=V2\nV2-V1=60\nshe should add 60 mL of water.
793017
Take 1.875 L of the .2 molar stock solution of NaBr and add .625 L of water to it and you get 2.5 L of a .15 molar solution of NaBr this just use the c1v1=c2v2 equation.\n\nNote:\nThis is the only way it could work. You do not want to add more NaBr because your concentration is already higher then desired.   If you just added 0.5L of water to the 2.0 L of .2 molar stock solution your molarity would still be to high at .16 meaning you have to dilute a smaller portion of your 2L .2 molar stock solution.
861717
2 HNO3(aq) + Sr(OH)2(aq) --> Sr(NO3)2 (aq) + 2 H2O\n\n2 H3PO4(aq) + 3 Ba(OH)2(aq) --> Ba3(PO4)2 (cr)* + 6 H2O\n\n* Barium Phosphate will precipiate out of an aqueous solution in this reaction
277253
MgCl + H2O = MgO + H2Cl
827978
PV = nRT\nwhere P must be in atmospheres, V must be in L, n is moles, R is a constant, and T must be in Kelvins.\nyou have to compensate for the water vapor.\n(see source, 25 degrees is 23.76 mmHg which is the same as torr.)\nP = 1 atm @ STP\nV = unknown\nto find n: (641-23.76/ 760) atm x .5 L = n x (R is constant: .08206) x (25 + 273 to be in Kelvins)\nn =  .0166 mol\nR = .08206\nT = 273 K @ STP\n\nsolve for V\n1 x V = .0166 x .08206 x 273\nV = .372 L = 372 mL
533415
[H+] is the concentration of hydronium ion in mol/L.  Assuming HBr completely dissociates in water, it would 2.0 mol/L in this example - one mole of HBr provides one mole of H+.  If the dissociation is not complete, then you'd have to use the equilibrium constant to calculate [H+].
558840
When you dissolve a salt in water, the freezing point of the resulting solution is lower than the freezing point of the pure solvent (water).\nThe amount of the freezing point depressure depends on two factors, the concentration of the solute (salt) and the freezing point depression constant of the solvent.\n\nThe formula to calculate the freezing point depression in therms of the change in the original freezing point (delta T) is,\n(delta T) = k_f * m\nwhere k_f is the freezing point depression constant and m is the concentration of the solute in terms of molarity.\n\nIt is mathmatically possible for the concentration of the solute to be so greate at to create a change in the freezing point of water of more than 32 degrees F.\n\nThe freezing point depression constant of water is -1.86 degrees C per molality.  So mathmatically, if one had a solute concentration of about 9.57 molality or greater, the freezing point of water would be below 0 degree F.
386784
Well first of all you have to specify if the H's you want to account for are either comming from HCl alone or both HCl and some H's comming from the dissociation of water.\n\nI suppose that only the number of moles of H's comming from HCl is really worth while working the question for, since the H's of the dissociated water does not change the pH of the 0.10M solution.\n\nTherefore:\n\n0.10moles are in 1000mL\n? moles are in 50mL\n\n50 x 0.1 / 1000 = 0.005moles\n\nTherefore we can say that 0.005moles of HCl in 50mL of water results in a concentration of 0.10M
853922
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)