Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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- Suppose a 500. mL flask is filled with 1.5 mol of SO, and 0.20 mol of SO2. This reaction becomes possible: 2 (3) 2S0, (g) + 0,(g) =2SO3(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 0,. You can leave out the M symbol for molarity. 2 so, 02 so, initial ? change equilibrium oloarrow_forwardSuppose a 250. mL flask is filled with 0.70 mol of H2 and 1.1 mol of 12. The following reaction becomes possible: H2(g) +I2(g) — 2HI(g) =0 The equilibrium constant K for this reaction is 8.37 at the temperature of the flask. Calculate the equilibrium molarity of 12. Round your answer to two decimal places. Омarrow_forwardSuppose a 250. mL flask is filled with 0.60 mol of SO₂ and 1.0 mol of SO3. This reaction becomes possible: 2SO₂(g) + O₂(g) — 2SO3(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 O₂. You can leave out the M symbol for molarity. initial change equilibrium SO₂ 0₂ 0 X 0 SO₂ 0 010 X Śarrow_forward
- Suppose a 500. mL flask is filled with 0.30 mol of H, and 1.4 mol of HI. This reaction becomes possible: H,(2) +I,(g) = 2HI(s) 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 I,. You can leave out the M symbol for molarity. H, HI initial change equilibrium Continue Submi O 2021 McGraw-Hill Education. All Rights Reserved. Terms of Use Privaarrow_forward2. A reaction is represented by this equation: A(aq) + 2B(aq) = 2C(aq) Ke = 1 × 10³ (a) Write the mathematical expression for the equilibrium constant. (b) Using concentrations S1 M, make up two sets of concentrations that describe a mixture of A, B, and C at equilibrium.arrow_forwardSuppose a 500. mL flask is filled with 0.80 mol of I, and 1.0 mol of HI. This reaction becomes possible: H, (g) +1,(g) = 2HI(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 I,. You can leave out the M symbol for molarity. H, HI initial change equilibriumarrow_forward
- Use this information to answer Questions 3, 4, and 5: The equilibrium constant (K) of the reaction below is K = 6.0 x 10-2, with initial concentrations as follows: [H2] = 1.0 x 102 M, [N2] = 4.0 M, and [NH3] = 1.0 x 10-4 M. N2(g) + 3H2(g) = 2NH3(g) Ton: N. x10-2 ent (O Pncca waythe chemical system would 5 the vara BI U = E T O Word(s) ------ T T; O Word(s) 5. If the concentration of the product NH3 was increased from 1.0x 10-4 M to 5.6 x 10-3 M, calculate the reaction quotient (Q) and determine which way the chemical system would shift by comparing the value of Q to K. B I U E T O Word(s)arrow_forwardSuppose a 250. mL flask is filled with 1.6 mol of I, and 1.3 mol of HI. This reaction becomes possible: H,(g) +I,(g) = -2HI(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 H,. You can leave out the M symbol for molarity. H, I, HI initial change equilibrium olo Xarrow_forwardSuppose a 500. mL flask is filled with 0.10 mol of O, and 1.6 mol of NO. The following reaction becomes possible: N2(g) + 0,(g) =2NO (g) The equilibrium constant K for this reaction is 0.679 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. | Marrow_forward
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