Suppose a 250 mL flask is filled with 1.6 mol of NO and 0.40 mol of NO2. The following reaction becomes possible: NO3 (g) + NO (g) = 2NO2 (g) The equilibrium constant K for this reaction is 0.253 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 1.6 mol of NO and 0.40 mol of...
Suppose a 500. mL flask is filled with 1.4 mol of NO and 0.60 mol on NO2. The following reaction becomes possible: NO3(g) + NO(g) <--> 2NO2(g) The equilibrium constant K for this reaction is 0.162 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 0.50 mol of H, and 0.40 mol of HCl. The following reaction becomes possible: H2(g) + Cl2(g) = 2HCl (g) The equilibrium constant K for this reaction is 5.61 at the temperature of the flask. Calculate the equilibrium molarity of Cl. Round your answer to two decimal places. TOM xo?
Suppose a 250. ml flask is filled with 1.6 mol of Cl, and 1.3 mol of HCI. The following reaction becomes possible: H2(g) + Cl2(g) → 2HCI (g) The equilibrium constant K for this reaction is 0.967 at the temperature of the flask. Calculate the equilibrium molarity of C1. Round your answer to two decimal places. OM I ?
Suppose a 250. mL flask is filled with 2.0 mol of NO and 0.30 mol of NO . The following reaction becomes possible: NO(g) + NO(g) - 2NO() The equilibrium constant K for this reaction is 0.662 at the temperature of the flask. Calculate the equilibrium molarity of NO2. Round your answer to two decimal places. x ?
Suppose a 500. mL flask is filled with 0.40 mol of CO, 0.60 mol of NO and 0.70 mol of CO2. The following reaction becomes possible: NO2(g) +CONO( CO2() The equilibrium constant K for this reaction is 9.06 at the temperature of the flask. Calculate the equilibrium molarity of CO. Round your answer to two decimal places. Џи
Suppose a 500 ml flask is filled with 1.2 mol of NO, and 0.40 mol of NO the following reaction becomes possible: NO,() +NO(g) - 2NO(g) The equilibrium constant K for this reaction is 7.29 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 1.3 mol of NO, and 1.5 mol of NO. The following reaction becomes possible: NO2(g) +NO(g) + 2NO2(g) The equilibrium constant K for this reaction is 9.06 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. i M x 6 ?
Suppose a 250 ml flask is filled with 1.0 mol of NO and 0.70 mol of NO . The following reaction becomes possible: NO3(8) + NO(g) - 2NO(8) The equilibrium constant K for this reaction is 5.78 at the temperature of the flask. Calculate the equilibrium molarity of NO3. Round your answer to two decimal places. Ом 3 ? Expiat tion Check 20:30 MCGHE AB
Suppose a 250. ml flask is filled with 0.50 mol of CO, 0.60 mol of NO 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 3.29 at the temperature of the flask. Calculate the equilibrium molarity of CO. Round your answer to two decimal places. IM x 6 ?
= Objective Knowledge Check Question 12 Suppose a 250 ml flask is filled with 1.6 mol of Bry, 1.7 mol of Oct, and 1.0 mol of BrCl. The following reaction becomes possible: Br2(8) +OC1,() - BrOCI(g) +BrCl(g) The equilibrium constant K for this reaction is 2.54 at the temperature of the flask. Calculate the equilibrium molarity of Brz. Round your answer to two decimal places. OM X 5 ?