Suppose a 500. mL flask is filled with 0.30 mol of CO, 0.10 mol of NO and 1.9 mol of CO2. The following reaction becomes possible: NO2(g) + CO (g) NO(g) + CO2(g) = The equilibrium constant K for this reaction is 7.45 at the temperature of the flask. Calculate the equilibrium molarity of No. Round your answer to two decimal places.

Suppose a 500. mL flask is filled with 0.60 mol of CO, 0.30 mol of NO and 1.7 mol of CO,. The following reaction becomes possible: NO,(g) +cO(g) - NO(g) +CO, (g) The equilibrium constant K for this reaction is 0.953 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. ?

Suppose a 250. mL flask is filled with 1.0 mol of NO₂, 0.20 mol of NO and 1.7 mol of CO₂. The following reaction becomes possible: NO₂(g) +CO(g) NO(g) +CO, (g) The equilibrium constant K for this reaction is 5.54 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. M Images X

Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 75.0 L tank with 4.2 mol of ammonia gas, and when the mixture has come to equilibrium measures the amount of nitrogen gas to be 0.63 mol. Calculate the concentration equilibrium constant for the decomposition of ammonia at the final temperature of the mixture. Round your answer to 2 significant digits. K = 0 C

Suppose a 250. mL flask is filled with 2.0 mol of NO and 1.0 mol of NO₂. The following reaction becomes possible: NO₂(g)+NO(g) — 2NO₂(g) The equilibrium constant K for this reaction is 2.73 at the temperature of the flask. Calculate the equilibrium molarity of NO3. Round your answer to two decimal places. M S

Suppose a 250. mL flask is filled with 0.70 mol of NO,, 0.40 mol of CO and 2.0 mol of CO ,. The following reaction becomes possible: NO,(3) + CO (g) - No(g)+ CO,(g) The equilibrium constant K for this reaction is 1.04 at the temperature of the flask. Calculate the equilibrium molarity of CO. Round your answer to two decimal places.

Suppose a 500. mL flask is filled with 1.8 mol of CO, 1.9 mol of H,0 and 1.6 mol of H,. The following reaction becomes possible: co (g) +H,0(g)– CO,(g) +H2 (g) The equilibrium constant K for this reaction is 0.275 at the temperature of the flask. Calculate the equilibrium molarity of CO. Round your answer to two decimal places.

Suppose a 500. mL flask is filled with 1.9 mol of O, and 1.8 mol of SO2. This reaction becomes possible: 2s0,(g) +0,(g) – 2so,(g) Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of O,. You can leave out the M symbol for molarity. So, O2 initial 口 D. change equilibrium

Suppose a 250. mL flask is filled with 1.2 mol of NO and 0.90 mol of NO 2. The following reaction becomes possible: NO3(g) + NO(g) = 2NO2(g) The equilibrium constant K for this reaction is 0.661 at the temperature of the flask. Calculate the equilibrium molarity of NO 3. Round your answer to two decimal places. Ом ☑

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