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 1.4 mol of NO and 0.60 mol on...
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
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 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 1.6 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 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 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 0.20mol of NO3 and 1.9mol of NO . The following reaction becomes possible: +NO3 (g) + NO (g) ---> 2NO2 (g) The equilibrium constant K for this reaction is 5.74 at the temperature of the flask. Calculate the equilibrium molarity of NO3 . Round your answer to two decimal places.
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 0.70 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 ?