The solubility of silver bromide is 6.0 x 10-7 mol L-1 at 18°C. Calculate?
482555
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
1167833
sounds like an AP question.  So you are trying to calculate Ksp for silver acetate. \n\nAgC2H3O2 <--> Ag+  + C2H3O2-\n\nSolubility will be darn low so from there set up your ice chart and go to town.
706351
KCl(aq) + AgNO3(aq)  --> KNO3(aq) + AgCl(s)\n\nNet ionic: Cl-(aq) + Ag+(aq)  --> AgCl(s)\n\nNa2SO4(aq) + Ba(NO3)2  --> 2NaNO3(aq) + BaSO4(s)\n\nnet ionic: SO4(-2)(aq) + Ba(+2)(aq)  --> BaSO4(s)
517995
67.  AgCl <---> Ag+ + Cl-  Ksp = [Ag+][Cl-] = 1.8 x 10^-10\n[Ag+] = 0.001/2 = 0.0005 M; [Cl-] = 0.0025 M; [Ag+][Cl-] = 1.25 x 10^-6, as this is greater than Ksp, a precipitate will form.\n\n68.  SrCrO4; Ksp = [Sr2+][CrO42-] = 3.5 x 10^-5; molar solubility = [Sr2+] = [CrO42-] = sqrt (3.5 x 10^-5) = 0.0059 M\n\n69.  Total volume = 3L  [Fe2+] = 0.150 * 1/3 = 0.050M; [OH-] = 0.0333 * 2 / 3 = 0.0222 M.  Ksp Fe(OH)2 = 4 x 10^-38.  (0.050) * (0.0222)^ 2 >>> 4 x 10^-38, so a precipitate will form.\n\n73.  Pb3(AsO4)2 <---> 3 Pb2+ + 2 AsO43-.  If solubility = x, then [Pb2+] = 3x, [AsO43-] = 2x.\n4.0 x 10^-36 = [3x]^3[2x]^2 = 108 x^5 ==> x = 3.26 x 10^-8
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.
211996
Well, I'll start you off.\n\nAbsorbances are additive, so you subtract the blank from the measurement (.214 - .009 = .205)\n\nSo the absorbance is .205. Now you use beer's law, which is:\n\nA = ebc\n\ne is molar absorptivity (in L/mol cm)\nb is path length (cm)\nc is concentration. \n\nYou use that equation to figure out the concentration of the complex.\n\nFrom there, you can figure out the concentration of iodine based on the amount of starch used. Unfortunately, you've not provided enough information to go beyond here.
227577
Molar mass is 61.97894 g/mol.
473551
Silver:   242 btu/hr-ft-F\nCopper:   230 btu/hr-ft-F\nAluminum: 125 btu/hr-ft-F
663838
I agree with Thinkstoomuch's Answers in most part.  \n\nSolid Cu, C will not be soluble in water.  However glucose and  CuBr2 will be soluble in water.  At STP solubility of glucose (C6H12O6) is about 90g/ml.  That eliminates choice 1 and 3.\n\nWhen a compound ionizes in soln it makes electrolyte.  Glucose will not ionize but CuBr2 will ionize to Cu++ and 2Br-, choice 4 can be eliminated.\n\nThe third clue is just there to confirm your answer and CuBr2 being an ionic compound will indeed have a relatively high melting point.  498°C to be exact.  \n\nClearly, Cupric Bromide should be your Answer.\n\nI hope this helps.
959805
HBr, -1\nHOBr, +1\nBr2, 0\nHBrO4,  +7\nBrF3  +3\n\nThere is nothing to explain.  There are rules, and you need to learn them and follow them.\n\nHere is a site with a good set of rules\nhttp://www.science.uwaterloo.ca/~cchieh/cact/c123/oxidstat.html
768648
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
809008
is this your school homework,  are there suppossed to be other questions before these because my computer shows them starting from No. 4.\nnway,\n4.  filter the solution while still hot because as the soltion cools crystallization occurs, and you will nloose your product on the fiolter paper.\n5. crystallization can be induced by reduction in temperature i.e put in an ice bath or you can use a technique called seeding where you put a few crystals of the expected substance in the solution, the other method is to increase the concentration of the solution by evaporating it, to saturation point, ensure you do not evaporate it dry.\n6.i do not know what you started with so i can not tell.\n7.slow cooling will definately give bigger crystalls.\n8.sodium chloride reacts with silver nitrate to form a white precipitate, actually silver nitrate solution is used to analyse NaCl in industrial setting.\n\nope you will find these information sufficient.
739042
2Al + 6HNO3 ----> 2Al(NO3)3(s) + 3H2(g)
958410
1.\nMn(s) + SnCl2(aq) --> MnCl2(aq) + Sn(s)\nCr(s) + 2HBr(aq) --> CrBr2(aq) + H2(g)\nCo(s) + CaBr2(aq) --> N.R\n2Al(s) + 3Ni(NO3)2(aq) --> 2Al(NO3)3(aq) + 3Ni(s)\n\n2.\nMg + 2AgNO3 --> Mg(NO3)2 + 2Ag\nmole of Ag = 3.580 / 108 = 0.033 mole\nmole of AgNO3 = 0.033 mole = 33 mmole\nmolarity of AgNO3 = 33 / 50 = 0.66 M\n\n3.\na.soluble\nb.soluble\nc.soluble\nd.insoluble\ne.soluble\nf.soluble\ng.insoluble in water, soluble in acid\nh.soluble\n\n4.\n3NaOH(aq) + Fe(ClO4)3(aq) --> Fe(OH)3(s) + 3NaClO4(aq)\n3Na+ + 3OH- + Fe3+ + 3ClO4- --> Fe(OH)3 + 3Na+ + 3ClO4-\nFe3+ + 3OH- --> Fe(OH)3\n(NH4)2SO4(aq) + Pb(NO3)2(aq) --> PbSO4(s) + 2NH4NO3(aq)\n2NH4+ + SO42- + Pb2+ + 2NO3- --> PbSO4 + 2NH4+ + 2NO3-\nPb2+ + SO42- --> PbSO4\n\nHope it helps\npeace\nvixklen
706398
CaCl2 + 2AgNO3 --> Ca(NO3)2 + 2AgCl. AgCl is the precipitate.\n(80g cacl2) x (1mol cacl2 / molar mass cacl2) x (2mol agcl / 1mol cacl2) x (molar mas agcl / 1mol agcl)= g agcl= 206.6 g AgCl
184409
1.75mol  /  2mol of C6H6 times specfic heat ( -98.0kJ)\n\n=  .875 * -98.0kJ = -85.75kJ
967201
Write the dissociation reaction:\n\nAg2Co3(s) <--> 2Ag+1(aq) + CO3-2(aq)\n\nSince this is a solubility equilibrium, the solubility product expression is just the equilibrium expression.\n\nKsp=[Ag2]^2[CO3]\n\nTo find the actual value, you'll need to look it up in a table.\nBTW, it's 8.1*10^-12