Ksp for Ag2SO4 = 1.2x10^-5\n\nno idea how to solve it - any sources to equations such as websites to help solve these kinds of problems will help
135330
It is not possible.  Observe the chemical equation:\nAg2SO4(s) + AgNO3(s) --> Ag2SO4(s) + AgNO3(s)\n\nIn order for a substance to dissolve, it must be in an aqueous solution.
38967
My answer is,\n \nNH4NO3(ag) + KOH(aq) -> NH4OH(aq) + KNO3(aq)
56977
Pay more attention in class! And assuming it is an ideal gas, use PV=nRT. Since there is not change in volume you can basically drop it from the equation and make P=nRT. If you like algebra you can leave it in but it will cancel out eventually. You could solve for 'n' first using initial P and T, and then plug and chug.
440498
A) Ag2O (s) >>> 2 Ag(s) + 1/2 O2 (g)\nB) Nickel (II) Carbonate >>>>>Nickel (II) Oxide + [Carbon Dioxide], NiCO3 -> NiO + CO2\nC) Ammonium Nitrate >>>>> Dinitrogen monoxide + Water,  \nC does not balance as is... but this does:\nNH3NO3 -> N2O +  1.5 H2O + 0.5 O2\n\n48) A) 2 HCl (aq) + Ca (OH)2 (aq) >>>> 2 H2O + CaCl2 (aq)\nB) 3 Ag2 SO4 (aq) + 2 AlCl3 (aq) >>>> 6 AgCl (s) + Al2(SO4)3 (aq) (silver chloride is a precipitate)\nC) H2C2O4 (aq) + 2 KOH (aq)>>>> 2 H2O + K2C2O4 (aq)\nD) CdBr2 (aq) + Na2S (Aq)>>> CdS(s) + 2 NaBr (aq) (Cadium sulfide is a precipitate)\n50 Write a Balanced equation for the complete combustion of each compound:\nA) Butene (C4H8): C4H8 + 6 O2 -> 4 CO2 + 4 H2O\nB) Octane (C8H18): C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O\nC) glycerol (C3H8O3): C3H8O3 + 3.5 O2 -> 3 CO2 + 4 H2O\nD) Acetone (C3H6O): C3H6O + 4 O2 -> 3 CO2 + 3 H2O\n52: Complete these:\nA) Al (s) + H2SO4 (aq)>>>> AlSO4(aq) + H2(g)\nB) 2 HCl (aq) + Ba (OH)2 (aq)>>>> BaCl2 (aq) + 2 H2O\nC) Au (s) + 3 HCl (aq) >>>>> AuCl3(aq) + 1.5 H2 (g)\n54) EACH IS BALANCED BUT "INCORRECT", so correct these:\nA) Cl-2 + NaI(aq) >>> NaCl2(aq) + I-\nB) NH3(aq) >>> N- + H+3\nC) Na (s) + O2(g) >>NaO2(s)\nD) 2Mg (s) + H2SO4(aq) >>Mg2SO4(aq) + H2(g)\nE) MgCl2(aq) + Ca(OH)2(s) >>> Mg(OH)2(s) + CaCl2(aq)\nF) H2 (g) + Cl2 (g) >> H2Cl2 (g)
415149
This is a Hess' Law problem.\n\nThe strategy I like to teach for these is to get the biggest thing you need first then get rid of the biggest thing you don't need next.  So we want to get the 2 N2O5 first then get rid of the resulting 4 HNO3.\n\nUse -2 times the first eq to get your 2 N2O5, -4 times the third to get rid of your 4 HNO3, finally -2 times the second to get rid of the water:\n\n2 N2O5 + 2H2O > 4 HNO3   -152.4 kJ\n4 HNO3              >  2 N2 + 6 O2 + 2 H2  +696.0 kJ\nO2 + 2H2           >   2 H2O  -572 kJ\n\nEverything cancels properly to get the desired equation, and delta H is -28.4 kJ.
294312
This is as good a place to start as any -->\n\nhttp://www-istp.gsfc.nasa.gov/Education/bh2_2.html
343722
sorry i am no good at math, but this may help:\nhttp://www.fourmilab.ch/earthview/pacalc.html
449239
look at the periodic table.  H is on top group 1, and has a +1 for it's ox. number normally.  S and O are in the same family and to make them have an octet they want 2 electrons so they are -2, though S can vary.  you probably are trying to find that out.\n\nthen it's all about setting up the algebra expression.\n\nH(2)+S+O(4) = -1\n\n1(2)+S+-2(4)=-1\n\nsolve and you have your answer\n\nif you're still having problems with this stuff go see your teacher and they'll help
162127
CuSO4·5H2O(s) → CuSO4(s) + 5H2O(g)
370512
I have forgotten all the chemical formulas & whatnot... but I'm going to do this assuming that I'm right for the following:\n\n(note that if i get any of these wrong you'll have to make appropriate changes, this is just to show you HOW to do this)\nCu = Copper\nS = Sulfur\nCuS2 = Copper Sulfide\n\nSo first step is to balance the equation\n\nCu + 2S -> CuS2\n\nso then you can set up a problem like this:\n\n0.90 mol Cu * 1 mol CuS2  = # mol copper sulfide yielded\n(this is because there was 1 mole of Copper in 1 mole of CuS2.. so if CuS2 is not the correct formula change the "1 mol CuS2" accordinginly)\n\n\nThen add up the # of grams in 1 mol of copper sulfide\n\ndo that by taking the # of grams in 1 mol of copper + the number of grams in 2 mol of sulfur (again, this is just assuming that copper sulfide IS CuS2).  Once you've gotten that figure, multiply it by the number found from the earlier problem (# mol copper sulfide yielded).  This gives you the amount of grams.  To change this number to kilograms, divide it by 1000.
404410
F=-kx and F=ma\n\nfind mass from period     T = 2pi(m/k)^(1/2)\n\nhmmmm,  you seem to be missing k\n\nfor fixed ratio m/k the period of 0.51s and amplitude 4 cm is not enough to determine the mass (and therefore acceleration).
497090
If s is the solubility of Ca3(AsO4)2 in mole/L then\n\n... ... .. .. Ca3(AsO4)2 <=> 3Ca(+2) + 2AsO4(-3)\nDissolve .. ...  s\nProduce .. .. .. .. .. . .. . . ... .. .. .3s .. .. .. . 2s\n\nKsp=[Ca+2]^3 [AsO4-3]^2= (3s)^3 *(2s)^2= 27s^3 * 4s^2 =>\n\nKsp =108s^5\n\nThe solubility of Ca3(AsO4)2 is \n0.034/398.078 =8.54*10^-5 M\n\nThus Ksp =108*(8.54*10^-5)^5 =4.9*10^-19
510144
use the formula\nA=2pir(h+r)\n=2pi*7*21\n=294pi
447896
I think you mean\n\n(sin^4 t - cos^4 t )/ sin 2t where t is for O\n= (sin ^2 t+ cos^2t)(sin^2 - cos ^ 2 t) as (a+b)(a-b) = a^2 - b^2\n= 1. (- cos 2t)/sin 2t  using sin^2t + cos^2 t = 1 and cos^2 t- sin^2 t = cos 2t\n= - cot 2t
348314
use the molarity x volume = molarity x volume equation, but beware that the H2SO4 shells out 2 protons, not 1.
451992
Surface Area = 4*PI*r^2\n\nr = SQRT(SA/4/PI)
425593
You are very close. If K = -0.032 is accurate enough, then use it.\n\npA = Ae^(19 * -0.032)\np = e^(19 * -0.032) ~ 0.544\n\nIf you leave K = ln(.5)/22 then\np ~ 0.5496