Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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- Suppose a 250. mL flask is filled with 0.10 mol of N, and 1.8 mol of NO. The following reaction becomes possible: N2(g) + 0,(g) == 2NO (g) The equilibrium constant K for this reaction is 0.422 at the temperature of the flask. Calculate the equilibrium molarity of O2. Round your answer to two decimal places. IM ?arrow_forward3NO₂(g) N₂O5(g) + NO(g) K₂=1.0 × 10-11 If a 4.00 L container initially holds 0.13 mol of NO2, how many moles of N₂O5 will be present when this system reaches equilibrium? Enter your answer in scientific notation. moles of N₂O5 = i x 10 i molarrow_forwardNitrogen dioxide is one of the many oxides of nitrogen (often collectively called "NOx") that are of interest to atmospheric chemistry. It can react with itself to form another form of NOx, dinitrogen tetroxide. A chemical engineer studying this reaction fills a 125 L tank with 47. mol of nitrogen dioxide gas. When the mixture has come to equilibrium he determines that it contains 22. mol of nitrogen dioxide gas. The engineer then adds another 16. mol of nitrogen dioxide, and allows the mixture to come to equilibrium again. Calculate the moles of dinitrogen tetroxide after equilibrium is reached the second time. Round your answer to 2 significant digits. ||mol x10arrow_forward
- Suppose a 250. mL flask is filled with 0.60 mol of N₂ and 0.40 mol of O₂. The following reaction becomes possible: N₂(g) + O₂(g) → 2NO(g) The equilibrium constant K for this reaction is 9.69 at the temperature of the flask. Calculate the equilibrium molarity of N₂. Round your answer to two decimal places. Xarrow_forwardData Table Mg (s) MgO (s) Initial Temperature of HCI (aq) ("C) Final Temperature of solution ("C) Change in temperature ("C) 23.0 23.0 34.8 38.0 11.8 15.0 Mass of substance 0.13 0.97 (g) Concentration of HCI (aq) = 1.00mol/L Volume of HCI (ag) used in each trial = 0.050Larrow_forwardplease explain and give complete answerarrow_forward
- "Synthesis gas" is a mixture of carbon monoxide and water vapor. At high temperature synthesis gas will form carbon dioxide and hydrogen, and in fact this reaction is one of the ways hydrogen is made industrially. A chemical engineer studying this reaction fills a 125 L tank with 20. mol of carbon monoxide gas and 24. mol of water vapor. When the mixture has come to equilibrium he determines that it contains 7.0 mol of carbon monoxide gas, 11. mol of water vapor and 13. mol of carbon dioxide. The engineer then adds another 7.0 mol of carbon monoxide, and allows the mixture to come to equilibrium again. Calculate the moles of hydrogen after equilibrium is reached the second time. Round your answer to 2 significant digits. molarrow_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_forwardSuppose a 250. mL flask is filled with 1.7 mol of NO2, 0.80 mol of CO and 0.70 mol of CO2. The following reaction becomes possible: NO2(g) + CO(g) = NO(g) + CO2(g) The equilibrium constant K for this reaction is 2.90 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places.arrow_forward
- Suppose a 500. mL flask is filled with 1.5 mol of H₂ and 1.2 mol of Cl₂. The following reaction becomes possible: H₂(g) + Cl₂(g) → 2HCl(g) The equilibrium constant K for this reaction is 3.45 at the temperature of the flask. Calculate the equilibrium molarity of H₂. Round your answer to two decimal places. M Śarrow_forwardUse 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_forwardFor the equilibrium 3NO₂(g) K₂=1.0 x 10-11 N₂O5(g) + NO(g) If a 4.00 L container initially holds 0.25 mol of NO₂, how many moles of N₂O5 will be present when this system reaches equilibrium? Enter your answer in scientific notation. moles of N₂O5 = i x 10 Mi molarrow_forward
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