Suppose a 500. mL flask is filled with 0.40 mol of CO, 0.60 mol of NO...
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 0.70 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 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 500. mL flask is filled with 0.70 mol of H2 and 0.60 mol of HI. The following reaction becomes possible: -2H1(g) The equilibrium constant K for this reaction is 2.38 at the temperature of the flask. Calculate the equilibrium molarity of H2. Round your answer to two decimal places.
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 ?
where did the 1.4, 1.6, 0.60 come from? QUESTION Suppose a 500 ml flask is filled with 0.70 mol of NO,, 0.80 mol of CO and 0.30 mol of NO. The following reaction becomes possible NO2(g) + CO(g) - NO(g) + CO2(g) The equilibrium constant K for this reaction is 0.986 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. initial change equilibrium [NO2][co] [no] [CO2] | 1.4 1.6 0.600...
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 1.3 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 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 1.7 mol of CO, 0.40 mol of H.O and 1.9 mol of CO,The following reaction becomes possible: CO(e)+H,0(B) - CO (8)+H.) The equilibrium constant K for this reaction is 0.305 at the temperature of the flask. Calculate the equilibrium molarity of CO. Round your answer to two decimal places