Suppose a 500.mL flask is filled with 0.80 mol of CO, 0.90 mol of CO2 and 0.20mol of H2. The following reaction becomes possible: COg+H2Og--->CO2g+H2g The equilibrium constant K for this reaction is 6.70at the temperature of theflask.Calculate the equilibrium molarity of H2O. Round your answer to two decimal places.
Suppose a 500.mL flask is filled with 0.80 mol of CO, 0.90 mol of CO2 and...
Suppose a 500. ml flask is filled with 0.90 mol of NO, and 0.80 mol of NO. The following reaction becomes possible: NO2(g) + NO(g) + 2NO, (g) The equilibrium constant K for this reaction is 8.41 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. IM x 3 ?
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 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 ?
Suppose a 250. mL flask is filled with 0.50 mol of H20, 1.0 mol of Co2 and 1.5 mol of H2. The following reaction becomes possible cog)+H20(g)Co()+H28) The equilibrium constant K for this reaction is 4.10 at the temperature of the flask Calculate the equilibrium molarity of H20. Round your answer to two decimal places.
Suppose a 500. ml flask is filled with 1.2 mol of Cl, and 0.80 mol of HCl. The following reaction becomes possible: H2(g) + Cl2(g) = 2HCl (g) The equilibrium constant K for this reaction is 0.419 at the temperature of the flask. Calculate the equilibrium molarity of Cl. Round your answer to two decimal places. xs ?
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 OC1,, 0.20 mol of BroCl and 1.4 mol of BrCl. The following reaction becomes possible: Br2(g) +OC12(g) =BroCl(g) +BrCl(g) The equilibrium constant K for this reaction is 0.798 at the temperature of the flask. Calculate the equilibrium molarity of Br2. Round your answer to two decimal places. IM xs ?