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16 Nov 2019
s = (Ksp/4) The expression of the Ksp of Ca(OH)2 is: Ksp = [Ca2 + ][OH ] Ksp = [Ca2 + ]2[OH ] Ksp = [Ca2 + ][OH ] [Ca2 + ]2[OH ]2 [Ca2 + ][OH-]2/[Ca(OH)2] Calculate the solubility of Ag2CrO4 (Ksp = 1.2 Times 10 ). 4.5 Times 10-7 M 1.8 Times 10 M 7.8 Times 10 M 6.7 Times 10-5 M 8.1 Times 10-6 M Calculate the solubility of BaF2 (Ksp = 1.7 Times 10-6) in g/L 7.5 Times 10 g/L 8.9 Times 10-4 g/L 4.5 Times 10-3 g/L 1.3 g/L 6.8 g/L Delta G degree for reaction Calculate Delta G for the reaction when the respective partial N2O, N2, and O2 are 3.0 Times 10-2 atm, 4.0 Times 10-3 atm, and 1.15 atm. -218100 J -208400 J -33000 J -110000 J -180300 J Delta G degree for the reaction What is K for this reaction at 25 degree at 25 degree C and 1 atm. 4.8 Times 10-25 1.00 Times 10 8.51 Times 10 1.44 Times Based on the value of K in one can conclude that: Reactants and products are equally favored at equilibrium Products are favored at equilibrium Reactants are favored at equilibrium The reaction is exergonic A lot of work can be obtained from this reaction at equilibrium The equilibrium constant of the reaction? 50.1 kJ -50.1 kJ -43.9 kJ
s = (Ksp/4) The expression of the Ksp of Ca(OH)2 is: Ksp = [Ca2 + ][OH ] Ksp = [Ca2 + ]2[OH ] Ksp = [Ca2 + ][OH ] [Ca2 + ]2[OH ]2 [Ca2 + ][OH-]2/[Ca(OH)2] Calculate the solubility of Ag2CrO4 (Ksp = 1.2 Times 10 ). 4.5 Times 10-7 M 1.8 Times 10 M 7.8 Times 10 M 6.7 Times 10-5 M 8.1 Times 10-6 M Calculate the solubility of BaF2 (Ksp = 1.7 Times 10-6) in g/L 7.5 Times 10 g/L 8.9 Times 10-4 g/L 4.5 Times 10-3 g/L 1.3 g/L 6.8 g/L Delta G degree for reaction Calculate Delta G for the reaction when the respective partial N2O, N2, and O2 are 3.0 Times 10-2 atm, 4.0 Times 10-3 atm, and 1.15 atm. -218100 J -208400 J -33000 J -110000 J -180300 J Delta G degree for the reaction What is K for this reaction at 25 degree at 25 degree C and 1 atm. 4.8 Times 10-25 1.00 Times 10 8.51 Times 10 1.44 Times Based on the value of K in one can conclude that: Reactants and products are equally favored at equilibrium Products are favored at equilibrium Reactants are favored at equilibrium The reaction is exergonic A lot of work can be obtained from this reaction at equilibrium The equilibrium constant of the reaction? 50.1 kJ -50.1 kJ -43.9 kJ
Beverley SmithLv2
19 Jan 2019