Suppose a 250. mL flask is filled with 1.0 mol of NO, and 1.5 mol of NO. The following reaction becomes possible: 3. NO3(g)+NO(g)-2NO,(g) The equilibrium constant K for this reaction is 6.65 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 0.70 mol of NO₂, 0.20 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 0.715 at the temperature of the flask. Calculate the equilibrium molarity of NO₂. Round your answer to two decimal places.

Suppose a 250. mL flask is filled with 0.60 mol of SO₂ and 1.0 mol of SO3. This reaction becomes possible: 2SO₂(g) + O₂(g) — 2SO3(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 O₂. You can leave out the M symbol for molarity. initial change equilibrium SO₂ 0₂ 0 X 0 SO₂ 0 010 X Ś

Suppose a 500. mL flask is filled with 1.5 mol of H₂ and 1.2 mol of Cl₂. The following reaction becomes possible: H₂(g) + Cl₂(g) → 2HCl(g) The equilibrium constant K for this reaction is 3.45 at the temperature of the flask. Calculate the equilibrium molarity of H₂. Round your answer to two decimal places. M Ś

Suppose a 500. mL flask is filled with 0.70 mol of CO, 1.7 mol of NO and 1.0 mol of CO,. The following reaction becomes possible: NO, (g) +CO(g)- NO(g) +CO, (g) The equilibrium constant K for this reaction is 5.97 at the temperature of the flask. Calculate the equilibrium molarity of CO. Round your answer to two decimal places. IM Continue MacBook Air

Suppose a 250. mL flask is filled with 0.30 mol of NO and 0.80 mol of NO2. The following reaction becomes possible: NO3(g)+NO(g)<->2NO2(g) The equilibrium constant K for this reaction is 0.717 at the temperature of the flask.  Calculate the equilibrium molarity of NO2. Round your answer to two decimal places.

Suppose a 500. mL flask is filled with 0.60mol of NO3 and 1.7mol of NO. The following reaction becomes possible: NO 3(g) + NO(g) 2NO2(g) The equilibrium constant for this reaction is 5.55 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 N₂ and 1.2 mol of NO. The following reaction becomes possible: N₂(g) + O₂(g) → 2NO(g) The equilibrium constant K for this reaction is 6.17 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places.

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O KINETICS AND EQUILIBRIUM Calculating equilibrium composition from an equilibrium constant Suppose a 250. mL flask is filled with 1.8 mol of NO3 and 1.1 mol of NO2. The following reaction becomes possible: NO₂(g) + NO(g) + 2NO₂(g) The equilibrium constant K for this reaction is 0.707 at the temperature of the flask. Calculate the equilibrium molarity of NO3. Round your answer to two decimal places. M × S 1/5