Suppose a 500. mL flask is filled with 1.2 mol of N, and 0.10 mol of O2. The following reaction becomes possible: N2(g) + O2(g) ~ 2NO(g) The equilibrium constant K for this reaction is 9.60 at the temperature of the flask. Calculate the equilibrium molarity of O2. Round your answer to two decimal places.

Suppose a 500. mL flask is filled with 1.0 mol of CO, 1.8 mol of CO2 and 0.90 mol of H₂. The following reaction becomes possible: CO(g) + H₂O(g) → CO₂(g) + H₂(g) The equilibrium constant K for this reaction is 0.932 at the temperature of the flask. Calculate the equilibrium molarity of CO2. Round your answer to two decimal places. ☐M X 5

Consider the equilibrium system described by the chemical reaction below. A mixture of gas containing only N2 and H2 is reacted in a vessel at high temperature. At equilibrium, the 5.0 M H2, 8.0 M N2, and 4.0 M NH3 are present. Determine the initial concentrations of H2 and N2 that were present in the vessel. = N2(g) +3 H2(g) 2 NH3(g) 1 Based on the given values, fill in the ICE table to determine concentrations of all reactants and products. Initial (M) Change (M) Equilibrium (M) N2(g) + 3 H2(g) 2 NH3(g) RESET 0 5.0 8.0 4.0 -4.0 10.0 11.0 -2.0 2.0 -5.0 6.0 -6.0 MAR 22 F2 80 F3 F4 1775 % F5 ག|: MacBook Air 6 < 1 F6 27 & tvill AQ বর F7 E R T Y U Ꮴ $ 54 #3 43 S D F G H DII N&S F8 DD F9 D F10 I' 8 * ∞ ) ) 9 0 0 P J K L

Suppose a 500. mL flask is filled with 1.9 mol of N, and 0.20 mol of NO. The following reaction becomes possible: N₂(g) + O₂(g) → 2NO(g) The equilibrium constant K for this reaction is 3.27 at the temperature of the flask. Calculate the equilibrium molarity of N₂. Round your answer to two decimal places.

Suppose a 500. mL flask is filled with 1.3 mol of NO and 0.90 mol of NO2. The following reaction becomes possible: NO3(g)+NO(g) — 2NO₂(g) The equilibrium constant K for this reaction is 0.720 at the temperature of the flask. Calculate the equilibrium molarity of NO₂. Round your answer to two decimal places. M X Ś

Suppose a 250. mL flask is filled with 1.9 mol of Br,, 1.8 mol of BrOCl and 1.6 mol of BrCl. The following reaction becomes possible: Br, (g) + OC1, (g) = BIOC1(g) + BrC1(g) The equilibrium constant K for this reaction is 4.15 at the temperature of the flask. Calculate the equilibrium molarity of Br,. Round your answer to two decimal places. 0.53 M ?

Suppose a 500. mL flask is filled with 0.90 mol of NO, 1.6 mol of NO and 1.4 mol of CO,. The following reaction becomes possible: NO,(g) + CO (g) = NO (g)+ CO,(g) The equilibrium constant K for this reaction is 3.12 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. IM

Suppose a 500. mL flask is filled with 1.1 mol of NO2, 1.7 mol of NO and 0.90 mol of CO2. The following reaction becomes possible: NO, (g)+CO(g) NO(g) +CO,(s) The equilibrium constant K for this reaction is 0.292 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. м B C ol

Suppose a 500. mL flask is filled with 1.7 mol of OCl2, 0.70 mol of BrOCI and 1.1 mol of BrCl. The following reaction becomes possible: Br2(g) +OCl2(g) BrOCI(g) + BrCl (g) The equilibrium constant K for this reaction is 3.84 at the temperature of the flask. Calculate the equilibrium molarity of OCl2. Round your answer to two decimal places. ☐ M ☑

Suppose a 500. mL flask is filled with 1.0 mol of Br₂, 1.8 mol of OC12 and 0.40 mol of BrOC1. The following reaction becomes possible: Br₂(g) + OC1₂(g) → BrOC1 (g) + BrC1 (g) The equilibrium constant K for this reaction is 1.06 at the temperature of the flask. Calculate the equilibrium molarity of BrOC1. Round your answer to two decimal places. M X