Suppose a 250. mL flask is filled with 0.70 mol of H, and 0.50 mol of HCI. The following reaction becomes possible: H, (g) +Cl, (g) = 2HC1(g) The equilibrium constant K for this reaction is 5.99 at the temperature of the flask. Calculate the equilibrium molarity of HCl. Round your answer to two decimal places.

Suppose a 500. mL flask is filled with 0.70 mol of Cl, and 0.90 mol of HCl. The following reaction becomes possible: H, (g)+ Cl, (g) = 2HC1 (g) The equilibrium constant K for this reaction is 8.22 at the temperature of the flask. Calculate the equilibrium molarity of Cl,. Round your answer to two decimal places. OM

Suppose a 500. mL flask is filled with 1.9 mol of H, and 1.1 mol of NH,. This reaction becomes possible: N, (g) +3H,(g) =2NH, (g) Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of N,. You can leave out the M symbol for molarity. olo Ar N2 H, NH, initial change equilibrium 미

MCPBA

Suppose a 250. mL flask is filled with 1.6 mol of NO,, 0.20 mol of CO and 0.40 mol of CO,. The following reaction becomes possible: NO,(2) +CO(g) – NO(g) +CO,(g) The equilibrium constant K for this reaction is 8.48 at the temperature of the flask. Calculate the equilibrium molarity of CO. Round your answer to two decimal places.

Suppose a 250. mL flask is filled with 1.0 mol of N, and 2.0 mol of NH,. This reaction becomes possible: N2 (g) + 3H,(g) – 2NH, (g) 2NH3 (g) Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of N,. You can leave out the M symbol for molarity. N2 H, NH, initial ? change equilibrium

Suppose a 500. mL flask is filled with 0.70 mol of Br,, 1.8 mol of BrOCl and 1.1 mol of BrCl. The following reaction becomes possible: Br, (g) + OCl,(g) =BROC1 (g)+BrCl (g) The equilibrium constant K for this reaction is 0.244 at the temperature of the flask. Calculate the equilibrium molarity of BrOCI. Round your answer to two decimal places. ?

Suppose a 500. mL flask is filled with 0.90 mol of NO, and 1.8 mol of NO,. The following reaction becomes possible: NO, (g) +NO(g) - 2NO, (g) The equilibrium constant K for this reaction is 5.58 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. ©2022 McGraw Hill LLC. All Rights Reserved. Terms 43,203 FEB 19 PP ESS MacBook Air

Suppose a 250. mL flask is filled with 0.10 mol of CO, 0.30 mol of H,O and 1.8 mol of H,. The following reaction becomes possible: co(g) + H,0(g) =CO,{g)+H2(g) The equilibrium constant K for this reaction is 0.296 at the temperature of the flask. Calculate the equilibrium molarity of H,0. Round your answer to two decimal places.

Suppose a  250.mL  flask is filled with  1.9mol  of  Br2 ,  1.4mol  of  OCl2  and  0.30mol  of  BrCl . The following reaction becomes possible: +Br2gOCl2g    +BrOClgBrClg The equilibrium constant  K  for this reaction is  2.78  at the temperature of the flask. Calculate the equilibrium molarity of  BrOCl . Round your answer to two decimal places.