Suppose a 250. mL flask is filled with 0.50 mol of N2 and 0.80 mol of O2. The following reaction becomes possible: N2(g) + O2(g) 2NO(g) The equilibrium constant K for this reaction is 5.98 at the temperature of the flask. Calculate the equilibrium molarity of O2. Round your answer to two decimal places. Ом

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 500. mL flask is filled with 1.2 mol of H, and 2.0 mol of HCl. The following reaction becomes possible: H,(2) + Cl,(g) = 2HCI(g) The equilibrium constant K for this reaction is 0.641 at the temperature of the flask. Calculate the equilibrium molarity of HCl. Round your answer to two decimal places. ?

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

Suppose a 250. mL flask is filled with 2.0 mol of Cl₂ and 1.2 mol of HCl. The following reaction becomes possible: H2(g) + Cl2(g) 2HCl (g) The equilibrium constant K for this reaction is 3.70 at the temperature of the flask. Calculate the equilibrium molarity of C12. Round your answer to two decimal places. Шм 000 Ar

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 250. mL flask is filled with 0.80 mol of O2 and 0.30 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 r to stand for the unknown change in the molarity of O,. You can leave out the M symbol for molarity.

Calculating equilibrium composition from an equilibrium constant Suppose a 250, mil. flask is filled with 1.5 mol of NO₂, 0.10 mol of CO and 2.0 mol of CO₂. The following reaction becomes possible: NO₂(g) +CO(g) NO(g) + CO₂(g) The equilibrium constant K for this reaction is 0.226 at the temperature of the flask. Calculate the equilibrium molarity of NO₂. Round your answer to two decimal places. OM Explanation CIU Check DEC 9 X 3 6 tv 8

Suppose a 250. mL flask is filled with 1.8 mol of O2 and 0.30 mol of NO. The following reaction becomes possible: N₂(g) + O₂(g) → 2NO(g) The equilibrium constant K for this reaction is 2.61 at the temperature of the flask. Calculate the equilibrium molarity of O2. Round your answer to two decimal places. M X S