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c. Calculate the [H'] concentration at equilibrium at a temperature where its K, is equal to 4.5 x 10 (H)N03] [H20][HNO,] 0.0600-X x 0.0S00 %3D 4.5x10-5= x= J4.5 KI0-6X0.0500 0.0500 -x X=1,5 K10-3 2. 4.5x 10-5= 0,0500 d. Calculate the pH at equilibrium at this temperature. e. Calculate the percent dissociation of HNO, at equilibrium at this temperature. Also answer this: If ye were to dilute this solution the percent dissociation would [increase; decrease].

c. Calculate the [H'] concentration at equilibrium at a temperature where its K, is equal to 4.5 x 10 (H)N03] [H20][HNO,] 0.0600-X x 0.0S00 %3D 4.5x10-5= x= J4.5 KI0-6X0.0500 0.0500 -x X=1,5 K10-3 2. 4.5x 10-5= 0,0500 d. Calculate the pH at equilibrium at this temperature. e. Calculate the percent dissociation of HNO, at equilibrium at this temperature. Also answer this: If ye were to dilute this solution the percent dissociation would [increase; decrease].

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Can you please help me with d, e, f, and g?
NAME PER 2. Suppose a 0.0500M solution of HNO, was created Write the dissociation reaction for this compound. (use correct symbols) a. HNO3 (aq)+H20(1)→ H(aq) + NO5 (aq) b. Write the equilibrium expression for the acid dissociation constant, K for this acid: (H+][NO5] し、 (H20)(HNO,] c. Calculate the [H'] concentration at equilibrium at a temperature where its K, is equal to 4.5 x 10s しa (H+][NO3] 0.0600-X X 0.DS00 [H20][HND;] 4.5 x10-5= X= J4.5 x10-5X0.0500 0.0500 -X 4.5x 10-5= x2 0,0500 d. Calculate the pH at equilibrium at this temperature. Calculate the percent dissociation of HNO, at equilibrium at this temperature. Also answer this: If you were to dilute this solution the percent dissociation would [increase; decrease]. e. CENGAGE Learning
Transcribed Image Text:NAME PER 2. Suppose a 0.0500M solution of HNO, was created Write the dissociation reaction for this compound. (use correct symbols) a. HNO3 (aq)+H20(1)→ H(aq) + NO5 (aq) b. Write the equilibrium expression for the acid dissociation constant, K for this acid: (H+][NO5] し、 (H20)(HNO,] c. Calculate the [H'] concentration at equilibrium at a temperature where its K, is equal to 4.5 x 10s しa (H+][NO3] 0.0600-X X 0.DS00 [H20][HND;] 4.5 x10-5= X= J4.5 x10-5X0.0500 0.0500 -X 4.5x 10-5= x2 0,0500 d. Calculate the pH at equilibrium at this temperature. Calculate the percent dissociation of HNO, at equilibrium at this temperature. Also answer this: If you were to dilute this solution the percent dissociation would [increase; decrease]. e. CENGAGE Learning
AP® EDITON NAME PER f. In a separate beaker, 5.00 grams of KNO,(1) are dissolved in water to make a 500, mL solution. Calculate the initial molar concentration of NO, ions. ii. Calculate the K, for NO, at this temperature. Assume K, is still equal to 1014 at this temperature. ii. Calculate the pH of this solution. At 25°C, the K of HNO, = 4.0 x 104. If the overall process of the dissociation of the acid is exothermic, was the temperature originally suggested (in part c) higher or lower than 25°C. Justify your response.
Transcribed Image Text:AP® EDITON NAME PER f. In a separate beaker, 5.00 grams of KNO,(1) are dissolved in water to make a 500, mL solution. Calculate the initial molar concentration of NO, ions. ii. Calculate the K, for NO, at this temperature. Assume K, is still equal to 1014 at this temperature. ii. Calculate the pH of this solution. At 25°C, the K of HNO, = 4.0 x 104. If the overall process of the dissociation of the acid is exothermic, was the temperature originally suggested (in part c) higher or lower than 25°C. Justify your response.
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