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Suppose a 250. mL flask is filled with 0.30 mol of H2 and 1.3 mol of...
Suppose a 250 ml flask is filled with 0.30 mol of I, and 1.5 mol of HI. The following reaction becomes possible: H2(g) +12(g)=2HI(g) The equilibrium constant K for this reaction is 0.532 at the temperature of the flask. Calculate the equilibrium molarity of 12. Round your answer to two decimal places. Пм x s ?
Suppose a 250. ml flask is filled with 0.10 mol of H and 0.30 mol of I. The following reaction becomes possible: H2(g) +12(g) - 2HI(g) The equilibrium constant K for this reaction is 5.61 at the temperature of the flask. Calculate the equilibrium molarity of H. Round your answer to two decimal places. OM x ?
Suppose a 250. ml flask is filled with 1.7 mol of H, and 0.30 mol of Cly. The following reaction becomes possible: H2(g) + Cl2(g) - 2HCI(g) The equilibrium constant K for this reaction is 6.15 at the temperature of the flask. Calculate the equilibrium molarity of Cl2. Round your answer to two decimal places. OM 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 1.5 mol of Cl, and 1.3 mol of HCl. The following reaction becomes possible: H2(g) + Cl2(g) 2HCl (8) The equilibrium constant K for this reaction is 0.560 at the temperature of the flask. Calculate the equilibrium molarity of H. Round your answer to two decimal places. OM * 5 ?
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 1.3 mol of H2 and 0.10 mol of HC1. The following reaction becomes possible: H2(g) + Cl2(g)-2HCl (g) The equilibrium constant K for this reaction is 3.03 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 0.50 mol of H2 and 1.9 mol of HI. The following reaction becomes possible: H2(g)+12g2HIg The equilibrium constant K for this reaction is 4.94 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 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 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?