At a certain temperature, 0.4011 mol of N2 and 1.501 mol of H₂ are placed in a 4.50 L container. N2(g) + 3H2(g) 2 NH3(g) At equilibrium, 0.1401 mol of N₂ is present. Calculate the equilibrium constant, Kc. K₁ = Kc 5.225 Incorrect

Suppose a 500. mL flask is filled with 1.6 mol of N₂ and 0.40 mol of NO. The following reaction becomes possible: N₂(g) + O₂(g) 2NO(g) The equilibrium constant K for this reaction is 6.10 at the temperature of the flask. Calculate the equilibrium molarity of N₂. Round your answer to two decimal places. M X Ś

Suppose a 500. mL flask is filled with 1.3 mol of N₂ and 1.4 mol of O₂. The following reaction becomes possible: N₂(g) + O₂(g) → 2NO(g) The equilibrium constant K for this reaction is 7.02 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. M X S

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At a certain temperature, 0.840 mol SO3 is placed in a 4.50 L container. 2 SO₂ (g) 2 SO₂(g) + O₂(g) At equilibrium, 0.100 mol O₂ is present. Calculate K. Kc = 6.914 ×10−5 Incorrect

For the reaction 2H2(g)+O2(g)↔2H2O(g)2H2(g)+O2(g)↔2H2O(g),  K = 1083. Which species, H2, O2, or H2O, will be present in the largest amount at equilibrium?

Suppose a 250. mL flask is filled with 1.1 mol of NO3 and 2.0 mol of NO. The following reaction becomes possible: NO3(g) + NO(g) → 2NO,(g) The equilibrium constant K for this reaction is 8.88 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 0.30 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.

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At a certain temperature, 0.4611 mol of N, and 1.561 mol of H, are placed in a 5.00 L container. N₂(g) + 3H₂(g)2NH₂(g) At equilibrium, 0.1601 mol of N₂ is present. Calculate the equilibrium constant, Ke. Kc = 23.83 Incorrect