22.5 g of an unknown metal at 97.3 oC is placed in 85.0 g of water (s = 4.18 J/g-oC) at 7.6 oC. What is the sp
340464
mmmmmmm....u sure know how to turn me on
327609
If I've understood what you're asking then:\nA\nCd(s) + Ag2+ -> Cd2+ + Ag(s) \n\nCadmium occurs at Anode and the Silver would deposit on the Cathode\n\nB (I hate Manganese equations and the partial above is incorrect)\n\nMnO2 + 4H+ + 2AgBr -> Mn2+ 2Br- + 2Ag + 2H2O\n\nMnO2 @ Anode\nAgBr @ Cathode\n\nC\n\n(PtCl4)2- + 4Ag -> Pt + 4AgCl \nAg @ anode\n(PtCl4)2- @ Cathode
7018
58.93\nJust go to your Yahoo search page and type in 'Periodic Table'
436191
Uh, the rest of the question isn't appearing. But it's an equilibrium solubility problem. Write a reaction:\n\nMgF2 --> Mg+2 + 2F-\n\nKsp = 5.52e-9\n\nFill in the values on the ICE table (I don't know them, the numbers are missing), and set an equation up.\n\nKsp = [F-]^2 * [Mg+2]\n\nSolve with the numbers from the ICE table. Remember that those concentrations in the above equation are the equilibrated concentrations, not the initial.\n\n\nHope this helped.
104563
My answer is:\n\n.217N
159316
[Ca(NO3)2]=n/V  ---> n=[]*V= 0.390M * 0.044L=0.01716 mol\nn=m/M ---> m=n*M= 0.01716mol * 164 g/mol=2.81424 g
76548
mc(delta T) = -mc(delta T)\nTf = the equilibrium temperature (final temperature) and is the same for both metals\n\n(2.6kg)(.22kcal/kgC)(Tf-13C) = -(10.2kg)(.093kcal/kgC)(Tf-93C)\n\n.572(Tf)-7.436 = -.9486(Tf)+88.2198\n1.5206(Tf) = 95.6558\n\nTf = 62.91 C
1008978
well this is really a chemistry problem but the formula is q=mC(delta) T, meaning heat= mass times the specific heat capacity times the change in temperature, so to set this up we use, 1.68^10 j = m (4.18 j/g) (80 degrees) becuase 4.18 is C and 80 degrees is the amount u wish to raise it, so then u get 1.68^10 j = 334.4m, divide both sides by 334.4 and you get .53558=m so your mass is .536g of water
422875
U have not mentioned the gas gravity to determone deviation factor to have an exact answer...generally..taking z=1\npsia ( absolute value) will also be used \nPsc*Vsc/Tsc=PV/T\n\nso: Vsc/V=PTsc/PscT=53.1/T \n\nThis is the simplest correct answer.... make is easier but worse that the process is isothermal so\n\nVsc/V=102 --> 1cu ft at TVD= 102 SCF\n\nThe most exact solution can be provided if u want..with taking temperatire and deviation,composition in account
830588
If I understand your question, you start out with 2.15 g of AgNO3 and you isolate 1.16 g of AgCL.\n\n(It saves time if you include the M.Wt. of the materials.)\n\nAg = 107.87g/mole\nNO3- has 14.0067 + 3*15.9994 = 62.005 g/mole\nCL = 35.453 g/mole\nAgNO3 = 169.875 g/mole\nAgCL = 143.323 g/mole\n\nIn theory, for every mole of AgNO3 you should get 1 mole of AgCl\n\n(2.15 g AgNO3)*(1 mole AgNO3/169.875g) = 0.0126 mole AgNO3, and also the same number of moles of AgCL\n\n0.0126 mole AgCL * (143.323 g AgCl/mole AgCl) = 1.81 g AgCl in theory (rounded)\n\nYou isolated only 1.16 g, therefore\n\n  (1.16/1.81)*100% = 64 % yield.\n\nYou say you want % error, I don' t know if that is the correct term but here is what the error term is:\n\n [abs (1.16 - 1.81)/1.81]*100% = 35.9%\n\nplease check for errors
832296
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)
867590
Well, for a start your equation shows copper(II) sulphide, not sulphate, and your equation is wrong even then.  That makes a bit of a difference!\n\nLet us imagine that you've got the formula right, but the name wrong.\n\nThe equation is simply Cu + S ---> CuS\n\nThen 64g will give (64+32)g.\n\nThus 64g will give 96g.\n\nSo 64 x (200/96)g of Cu will give 200g of product.
1167951
HgCl2 - 0.48 mol / 100 mol water at 25 degrees C\nHg2Cl2 - very insoluble in water
70269
Cu+S+O*4+H*2+O\n63.55+32.07+16*4+1.008*2+16= 177.64 g/mol\nThat x (usually a dot) means that there is a water molecule attached to the CuSO4.  The water molecule's mass should be included in the molar mass calculation.
802579
"1g/18g of H2O=.0556g/mol\n.57g/(63.55*32.07*16*4) -->> .57/ 130435.104= 4.3699x10^-6" \n\nemm, I dont really understand what u'r trying to do here, molar mass of CuSO4 should be calculate this way:\n\nM[CuSO4] = M[Cu] + M [S] + 4*M[O], not M[Cu]* M [S]*4*M[O].\n\nANY molar mass has the unit of g mol^-1, not g. If you divide 1g by 18g, u will have a UNITLESS number.\n\n"So now we divide both of them by teh lowest mole which for this equation is .0556g, right? But is the CuSO4 right? (The Mole)" \n \nit should be 0.0556 mol, not 0556g. No, it's the mole of water lost.\n\nThis is what u should do. \n\nfind out the MOLAR MASS [unit g/mol] of CuSO4 and H2O [CuSO4 should be around 160 g/mol, H2O should be around 18 g/mol]\n\nthen, find out the MOLE of CuSO4 and H2O in the compound by dividing the mass of CuSO4/ H2O by their molar mass.\n\nThen divide the mole of H2O in the compound by the mole of CuSO4 in the compound, find out ur ratio.\n\nthe ratio turns out to be 15.6, a very strange number in my point of view [not an integer, and it's a little bit large for this compound]
75209
I'll answer it, but you'd best pick me for the best answer if I'm doing all this work. ;)\n\nConvert C to K:\n\n178.0C = 451.15K\n25C = 298.15K\n30.6 = 303.75K\n\nTotal energy of water initially: Ei\n\nEi = 185.0g * 298.15K * 4.184J/g*K = 2.308E5 J\n\nTotal energy of water, final: Ef\n\nEf = 185.0g * 303.75g * 4.184J/g*K = 2.315E5 J\n\nEf - Ei = 700J\n\nLet Q be the specific heat of the alloy\n\nQ * 36.0g * 451.15K = Q * 36.0g * 303.75K + 700J\n\nQ*{(36.0g * 451.15K)-(36.0g * 303.75K)} = 700J\nQ = 700J/{(36.0g * 451.15K)-(36.0g * 303.75K)} \nQ = 700J/(16241g*K - 10935g*K)\nQ = 700J/3506g*K\nQ = 0.200 J/g*K
959337
The molal freezing point depression constant is 1.86 degree/molal\n\nWe observe a 5.5 degree depression in freezing point therefore its concentration is (5.5 degree*1 molal/1.86 degree) = 2.96 molal\n\nWe assume complete ionization of each substance therefore for x moles of NaNO3 there is 2x moles of particles, and for y moles of Mg(NO3)2 there are 3y moles of particles.\n\nLets scale things up to a liter of water, so instead of 6.6 g of solid we use 6.6 g * 20 = 132 g\n\nM.Wt. of NaNO3 is  84.99 g\nM.Wt of Mg(NO3)2 is 148.3 g (both are rounded)\n\n2x + 3y = 2.96\n\n x(84.99g/mole) + y(148.3g/mole) = 132\n\nTo solve the system, multiply row 1 by -42.495 and add to row 2\n\n-127.485 y  + 148.3 y= 132 -125.785\n   20.815 y = 6.215 ----------------> y = 0.299 moles, plug this into equation 1 to solve for x to get x = 1.032 moles\n\nremember we had scaled this up by 20, therefore we will have to scale it back down.\n\n0.299 moles Mg(NO3)2* 148.3 g/mole *(1/20) = 2.217 g\n\n1.032 moles NaNO3 *(85.99g/mole)*(1/20) = 4.437 g\n\nSum to double check our answer 2.217 + 4.437 = 6.654 g, which is close enough to 6.6 since we expect rounding errors.\n\nPlease check for computational errors.