solve part b and c

Part A an Review | Constants | Periodic T Based on a Kc value of 0.220 and the initial concentrations given in the table, determine in which direction the net reaction will proceed to attain equilibrium. Part B Initial concentrations (M) Learning Goal: Based on a Ke value of 0.220 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively? Mixture [XY] [X] [Y] To determine equilibrium concentrations from initial conditions. Express the molar concentrations numerically. A 0.100 0 The reversible reaction View Available Hint(s) B 0.500 0.100 0.100 XY(aq) - X(aq) + Y(aq) has a reaction quotient Q- defined as AEd C 0.200 0.300 0.300 Lallyi = Drag the items into the appropriate bins. (xY]. [x]. [Y]= .261,.239,.239 M Because the reaction is reversible, both the forward and reverse reactions will occur simultaneously. The reaction View Available Hint(s) wil eventually reach equilibrium, at which point the Submit Previous Answers concentrations do not change, and Q is equal to a constant known as Ke- Reset Help X Incorrect; Try Again Calculating equilibrium concentrations when the net reaction proceeds in reverse Consider mixture C, which will cause the net reaction to proceed in reverse. + net Concentration (M) [XY] X + [Y] initial: 0.200 ).300 0.300 Figure change: -I -I equilibrium: 0.200 + z 0.300 - z 0.300 - I The change in concentration, a, is positive for the reactants because they are produced and negative for the products because they are consumed Reverse Forward K KO Part C Based on a Ke value of 0.220 and the data table given, what are the equilibrium concentrations of XY. X, and Y, respectively? Express the molar concentrations numerically Mixture Mixture A Mixture B View Available Hint(s) Reaction Squibbrium Reaction forms forms products reactants