Principles of General, Organic, Biological Chemistry
2nd Edition
ISBN: 9780073511191
Author: Janice Gorzynski Smith Dr.
Publisher: McGraw-Hill Education
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Chapter 7.4, Problem 7.11P
Why does a soft drink become “flat” faster when it is left open at room temperature compared to when it is left open in the refrigerator?
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Chapter 7 Solutions
Principles of General, Organic, Biological Chemistry
Ch. 7.1 - Classify each substance as a heterogeneous...Ch. 7.1 - Classify each product as a solution, colloid, or...Ch. 7.2 - Consider the following diagrams for an aqueous...Ch. 7.2 - Prob. 7.4PCh. 7.2 - Prob. 7.5PCh. 7.2 - Prob. 7.6PCh. 7.2 - Prob. 7.7PCh. 7.2 - If a solution contains 125 mEq of Na+ per liter,...Ch. 7.3 - Prob. 7.9PCh. 7.3 - Prob. 7.10P
Ch. 7.4 - Why does a soft drink become flat faster when it...Ch. 7.4 - Predict the effect each change has on the...Ch. 7.5 - Prob. 7.13PCh. 7.5 - Prob. 7.14PCh. 7.5 - Prob. 7.15PCh. 7.5 - A drink sold in a health food store contains 0.50%...Ch. 7.5 - Prob. 7.17PCh. 7.5 - Prob. 7.18PCh. 7.5 - Prob. 7.19PCh. 7.6 - Prob. 7.20PCh. 7.6 - Prob. 7.21PCh. 7.6 - Prob. 7.22PCh. 7.6 - Prob. 7.23PCh. 7.6 - Prob. 7.24PCh. 7.7 - Prob. 7.25PCh. 7.7 - Prob. 7.26PCh. 7.7 - Prob. 7.27PCh. 7.8 - Which solution in each pair exerts the greater...Ch. 7.8 - Describe the process that occurs when a 1.0 M NaCl...Ch. 7.8 - Prob. 7.30PCh. 7 - Prob. 7.31UKCCh. 7 - Prob. 7.32UKCCh. 7 - Prob. 7.33UKCCh. 7 - Prob. 7.34UKCCh. 7 - Prob. 7.35UKCCh. 7 - Prob. 7.36UKCCh. 7 - Prob. 7.37UKCCh. 7 - Prob. 7.38UKCCh. 7 - Prob. 7.41UKCCh. 7 - Prob. 7.42UKCCh. 7 - Prob. 7.43APCh. 7 - Prob. 7.44APCh. 7 - Prob. 7.45APCh. 7 - Prob. 7.46APCh. 7 - Prob. 7.47APCh. 7 - Prob. 7.48APCh. 7 - Prob. 7.49APCh. 7 - Prob. 7.50APCh. 7 - Prob. 7.51APCh. 7 - Prob. 7.52APCh. 7 - Prob. 7.53APCh. 7 - Prob. 7.54APCh. 7 - Prob. 7.55APCh. 7 - Prob. 7.56APCh. 7 - Prob. 7.57APCh. 7 - Prob. 7.58APCh. 7 - Prob. 7.59APCh. 7 - Prob. 7.60APCh. 7 - Prob. 7.61APCh. 7 - Prob. 7.62APCh. 7 - Prob. 7.63APCh. 7 - Prob. 7.64APCh. 7 - Prob. 7.65APCh. 7 - Prob. 7.66APCh. 7 - Prob. 7.67APCh. 7 - Prob. 7.68APCh. 7 - Prob. 7.69APCh. 7 - Prob. 7.70APCh. 7 - Prob. 7.71APCh. 7 - Prob. 7.72APCh. 7 - Prob. 7.73APCh. 7 - Prob. 7.74APCh. 7 - Prob. 7.75APCh. 7 - Prob. 7.76APCh. 7 - Prob. 7.77APCh. 7 - Prob. 7.78APCh. 7 - Prob. 7.79APCh. 7 - Prob. 7.80APCh. 7 - Prob. 7.81APCh. 7 - Prob. 7.82APCh. 7 - Prob. 7.83APCh. 7 - Prob. 7.84APCh. 7 - Prob. 7.85APCh. 7 - Prob. 7.86APCh. 7 - Prob. 7.87APCh. 7 - Prob. 7.88APCh. 7 - Prob. 7.89APCh. 7 - Prob. 7.90APCh. 7 - If the concentration of glucose in the blood is 90...Ch. 7 - Prob. 7.92APCh. 7 - Prob. 7.93APCh. 7 - Prob. 7.94APCh. 7 - Prob. 7.95APCh. 7 - Prob. 7.96APCh. 7 - Prob. 7.97APCh. 7 - Prob. 7.98APCh. 7 - Prob. 7.99APCh. 7 - Prob. 7.100APCh. 7 - Prob. 7.101APCh. 7 - Prob. 7.102APCh. 7 - Prob. 7.103CPCh. 7 - Prob. 7.104CP
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- . What does it mean to say that a state of chemical or physical equilibrium is dynamic?arrow_forwardDescribe a nonchemical system that is not in equilibrium, and explain why equilibrium has not been achieved.arrow_forwardJulia did an experiment to study the solubility of two substances. She poured 100 mL of water at 20 °C into each of two beakers labeled A and B. She put 50 g of Substance A in the beaker labeled A and 50 g of Substance B in the beaker labeled B. The solution in both beakers was stirred for 1 minute. The amount of substance left undissolved in the beakers was weighed. The experiment was repeated for different temperatures of water and the observations were recorded as shown. Substance Mass of Undissolved Substance at Different Temperatures (gram) 20 °C 40 °C 60 °C 80 °C A 40 37 34 30 B 15 14 13 12 Part 1: Which substance has a higher solubility?Part 2: Explain your answer for Part 1.arrow_forward
- Consider the following reaction where K, = 9.52×10-² at 350 K. CHĄ(g) + CCl4(g) 2CH2Cl(g) A reaction mixture was found to contain 5.22x102 moles of CH4(g), 3.96×102 moles of CCL(g) and 1.21x102 moles of CH,Ch(g), in a 1.00 liter container. Is the reaction at equilibrium? If not, what direction must run in order to reach equilibrium? The reaction quotient, Qc. equals | The reaction A. must run in the forward direction to reach equilibrium. B. must run in the reverse direction to reach equilibrium. C. is at equilibrium.arrow_forwardConsider the reaction N₂(g) + O₂(g) = 2 NO(g) that has an equilibrium constant, Kc, fof 4.10 × 10−4 at 1700 °C. What percentage of O₂ will react to form NO if 0.635 mol N₂ and 0.635 mol O₂ are added to a 0.617 L container and allowed to come to equilbrium at 1700 °C? 2 percentage of 0₂: %arrow_forwardDetermine the equilibrium constant for this reaction given the following information. CO (g) + 2 H2S (g) CS2 (g) + H2O (g) + H2 (g) K =? = 5.49 · 102 CH. (g) + H20 (g) → CO (g) + 3 H2 (g) K CHA g) + 2 H2S (g) CS2 (g) + 4 H2 (g) K = 2.70 . 105 2.03 10-3 4.92 102 3.03 103 O 1.48 · 10° « Previous MacBook Air 80 888 DII F3 F4 F5 F6 F7 F8 F9 23 $ & 3 4 6. 7 8. E R Y U LIarrow_forward
- The equilibrium constant, K, for the following reaction is 8.93 at 739 K. 2NH3(g) N2(g) + 3H2(g) In order to Calculate Kc at this temperature for the following reaction: 1/2N2(g) + 3/2H2(g) NH3(g) The original Kc value will need to be multiplied by 1/2. True Falsearrow_forwardThe equilibrium constant is given for the following reaction. Determine the value of the missing equilibrium constant. N2 (g) + O2 (g) ⇋ 2 NO (g) Kc = 4.2 × 10-31 2 NO (g) ⇋ N2 (g) + O2 (g) Kc = ?arrow_forwardSuppose a 250. mL. flask is filled with 0.10 mol of NO₂, 1.1 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 0.306 at the temperature of the flask. Calculate the equilibrium molarity of CO. Round your answer to two decimal places. Ом M Xarrow_forward
- Consider the following reactions: CO (g) + 2 H2S (g) D CS2 (g) + H2O (g) + H2 (g) K1 = 1.3 x 105 ½ CH4 (g) + H2S (g) D ½ CS2 (g) + 2 H2 (g) K2 = 180 1. Find the equilibrium constant for: 2 CO (g) + 6 H2 (g) D 2 CH4 (g) + 2 H2O (g) 2. What molecules in the steps are reaction intermediates? 3. If 0.500 atm of CO, H2, CH4, and H2O are initially placed in the same container, then given your answer in part a., determine which direction the reaction will shift in order to establish equilibrium.arrow_forwardThe equilibrium constant is given for two of the reactions below. Determine the value of the missing equilibrium constant. NO(g) +½ Br2(g) = NOBr(g) K = 5.3 2 NO(g) = N2(g) + O2(g) Kc = 2.1 ×1030 N2(g) + Br2(g) + O2(g) = 2 NOBr(g) Kc = = ? Select one: O 1.3 x 10-20 O 0.12 O 1.3 x 10-29 O 2.8 x 104arrow_forward7) Based on the information below, determine the value of the unknown equilibrium constant, K.. 2 CO2 (g) + H20 (g) → 2 02 (g) + CH2CO (g) CH4 (g) + 2 02 (g) <→ CO2 (g) + 2 H20 (g) K. = 6.1 x 108 K. = 1.2 x 1014 %3D CH4 (g) + CO2 (g) → CH,CO (g) + H2O (g) K = ?arrow_forward
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