What is the Kp for this equation?
968320
here, you just use the ideal gas law, PV=nRT.  since molarity is equal to moles of solute per liters of solution, M=n/V, so P=MRT.\ntherefore,\nKp=Pco*Ph2o/(Ph2*Pco2)\n  =[CO]RT*[H2O]RT/ ([H2]RT*[CO2]RT)\nall the RT coefficients end up canceling out, so:\nKp=[CO][H2O]/[H2][CO2]\n  =Kc\nsince there were equal numbers of gas molecules on each side of the equation, Kp was equal to Kc.  as a general rule, Kp=Kc*(RT)^N, where N is the change in the number of gas molecules.
783365
Use the gibbs free energy equation G=H-TS.  If you yous the boiling point as T, the system is at equilibrium at that point and G=0.  Dont forget to change T into Kalvin.
872075
V = kT/P, where V = volume, T = temp, P = pressure and k is the constant of proportionality.\nYou are given, T = 250, P = 76.2 and V = 750.\nUsing this information, you can solve for k.\na) 750 = k(250)/76.2\nk = 750(76.2)/250\nk = 228.6\nb) Go back to the equation at the top and replace k with 228.6\nV = 228.6T/P\nc) You are given T = 300 and P = 73.06\nPlug these in to find V.\nV = 228.6(300)/73.06\nV = 938.68
1017358
K = a^2 * sqrt(3)/4\nK = 16^2 * sqrt(3)/4\nK = (256/4) * sqrt(3)\nK = 64sqrt(3)\n\nANS : 64sqrt(3)cm^2 or 108.8cm^2
182644
° K = ° C + 273.\n\nIn this case:\n\nK=5C+273\nK=278
504715
You need the formula:\ndelta Tf=iKfXmolality\nyour delta Tf is 1.96\nyour i is 1 (because it does not dissociate)\nyour Kf is 5.12\nmolality is moles of solute divided by kilogram solvent, which is .0250 kg\nYour final equation should read:\n1.96=(1)(5.12)(x/.0250)\nThen use algebra to find x.
509528
1.00 LBP = 0.0483158 KES \n\n1 LBP = 0.0483158 KES   1 KES = 20.6972 LBP
296701
well its an inverse variation so the equation is y = K/x\nput y=32 in and x=0.5 in and you get 32 = K/0.5 so K = 16\nthe constant of variation is 16.
445873
K.E is given by m*(v^2)/2\nMomentum p = mv\nSquare both sides you get p^2 = m^2*v^2\nNow divide both sides by 2m and you get p^2/2m = m*(v^2)/2\nClearly the right hand side of the equation represents K.E\nso the answer is K.E = p^2/2m
912617
P=P(0)*exp(-1.1e-4*H)\n\nP(0) is the pressure at sea level. Note it's temperature dependent and it assumes a stable Boltzmann type distribution.
887006
pH= -log([H30+])\n\n\njust put the #'s in the above eqn.\n\nfor pOH, its pOH= -log([OH])\nand for pH given [OH], its pH= 14 - (-log([OH])
141442
KE per mole = (3/2) RT where R = 8.3145 J/mole*K and T is temperature in kalvin, so:\n\nKE = (3/2) RTn where n is the number of moles of gas\n\nYou can solve PV=nRT for n and substitute to get KE = (3/2) PV where P is pressure and V is volume.
974937
please complete the equation
884595
For reaction aA <--> bB + cC, the Eq.constant K is defined as\n\nK = (B^b * C^c) / A^a\n\nSince in your equation a=b=c=1, and concentration for B=C, the equation simplifies to\n\nK = B^2 / A\n\nwith B=0.046 and A=0.454    (0.454=0.500-0.046)\n\nso K = 0.046^2 / 0.454 = 0.0047
217977
f(E) = 1/(exp{(E-Ef)/kT}+1)
527802
N(Nitrogen), P(Phosphorus), K(Potassium)
650541
pH = -log ([H3O+])\npOH = -log ([OH-])\n\npH + pOH = pKw = 14\npOH = 14 - pH\n\n[OH-] = 10^-(pOH)\n\nPlug in and solve.