@X@user.full_name@X@ @GMU: What is the solubility in mol/L of CH3Br at a partial pressure of 289.4 torr, if the solubility at a partial pressure of 394.0 torr is 0.08244 mol/L?

@X@user.full_name@X@ @GMU: What is Kp for the reaction below, if Kc = 3.76e-5 at 170.9°C?4A(s) + 4B(g) → 2C(g)

@X@user.full_name@X@ @GMU: NaBr has a face-centered cubic unit cell in which the Br- anions occupy corners and face centers, while the cations fit into the hole between adjacent anions. What is the radius of Na+ if the ionic radius of Br- is 195.0 pm and the density of NaBr is 3.200 g/cm3?

@X@user.full_name@X@ @GMU: Initially there were equilibrium concentrations of 0.3867 M A, 0.2230 M B, and 0.1026 M C. Which change will cause the equilibrium to shift from right to left if the reaction is exothermic? Use the reaction below.5A(aq) → 2 B(aq) +2 C(aq) K = 0.661

@X@user.full_name@X@ @GMU: By how much would the rate of reaction (see below) change if the concentration of CO increased by a factor of 2.413. The rate law of this reaction is Rate = k[NO2]2.NO2(g) + CO(g) → CO2(g) + NO(g)

@X@user.full_name@X@ @GMU: What is the pOH of a 0.1668 M H2S? See reaction belowH2S + H2O → HS- + H3O+ Ka = 9.100×10-8

@X@user.full_name@X@ @GMU: What is the rate of consumption of A (M/s), if the rate of consumption of B is 0.374 M/s? Use the reaction below:4A + 2B + 5C → products

@X@user.full_name@X@ @GMU: Which reaction below would result in a net decrease in entropy?

John Schreifels  @GMU: What is the change in mass of 15N when it is formed from protons, electrons and neutrons if the atom has a mass of 15.00010897 amu? The masses of the proton, electron and neutron are 1.00728 amu, 0.000549 amu and 1.00867 amu, respectively.

@X@user.full_name@X@ @GMU: What is the temperature at which the reaction below becomes spontaneous, if the standard entropies of reactants and products at 25oC are So (C(gr)) = 5.74 J/mol*K, So(H2(g)) = 130.68 J/mol*K, and So (C5H12(g)) = 349.0000J/mol*K and the heat of formation of C5H12(g)) is -146.4400 kJ/mol?C5H12(g) → 5C(gr) + 6H2(g)