Suppose a 500.mL flask is filled with 1.1mol of NO2 1.6mol of CO and 0.90mol of NO. The following reaction becomes possible:
NO2(g) + CO(g) NO(g) +CO2(g)
The equilibrium constant K for this reaction is 0.211 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 1.1mol of NO2 1.6mol of CO and 0.90mol of...
Suppose a 500. mL flask is filled with 0.70 mol of NO2, 2.0 mol of NO and 0.90 mol of CO2. The following reaction becomes possible: NO2(e)+Co(g)NO(g)+Co,(g) The equilibrium constant K for this reaction is 0.331 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 0.20 mol of NO2, 1.7 mol of NO and 0.80 mol of CO2. The following reaction becomes possible: NO2(g) +CO(g) = NO(g) + CO2(g) The equilibrium constant K for this reaction is 0.457 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. MM x 6 ?
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 1.5 mol of CO, 1.2 mol of NO and 1.0 mol of CO,. The following reaction becomes possible: NO2(g) +CO(g) = NO(g) + CO2(g) The equilibrium constant K for this reaction is 6.78 at the temperature of the flask. Calculate the equilibrium molarity of NO2. Round your answer to two decimal places. Пм x 6 ?
Suppose a 500. ml flask is filled with 0.90 mol of NO,, 0.10 mol of CO and 0.50 mol of NO. The following reaction becomes possible: NO2(g) +CO(g) = NO(g) + CO2(g) The equilibrium constant K for this reaction is 0.172 at the temperature of the flask. Calculate the equilibrium molarity of CO. Round your answer to two decimal places. IM xs ?
Suppose a 250.mL flask is filled with 1.0mol of NO2 , 0.40mol of CO and 0.10mol of CO2 . The following reaction becomes possible: +NO2gCOg +NOgCO2g The equilibrium constant K for this reaction is 7.16 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 0.10mol of NO and 1.2mol of NO2. The following reaction becomes possible: +NO3gNOg 2NO2g The equilibrium constant K for this reaction is 0.541 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 0.50 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 ?
Suppose a 500 ml flask is filled with 0.50 mol of CO, 1.7 mol of H,0 and 1.1 mol of H. The following reaction becomes possible: CO(g) +H,0(8) -CO2() +H,() The equilibrium constant K for this reaction is 0.244 at the temperature of the flask. Calculate the equilibrium molarity of 1,0. Round your answer to two decimal places.
Suppose a 500. mL flask is filled with 1.5 mol of CO, 1.8 mol of H,O and 0.60 mol of CO,. The following reaction becomes possible: CO(g) +H2O(g) + CO2(g)+H2(g) The equilibrium constant K for this reaction is 3.75 at the temperature of the flask. Calculate the equilibrium molarity of 1,0. Round your answer to two decimal places. IM | xs ?