req2req3At 298K, the equilibrium constant for the following reaction is 5.90x10-²:H₂C₂O4(aq) + H₂0 H30+ (aq) + HC₂04 (aq)80DThe equilibrium constant for a second reaction is 6.40x10-5:HC₂04 (aq) + H₂0 H30+ (aq) + C₂04² (aq)Use this information to determine the equilibrium constant for the reaction:H₂C₂O4(aq) + 2H₂02H30+ (aq) + C₂04² (aq)KSubmit Answer$4RF4F%Use the References to access important values if needed for this question.5TalvCRetry Entire Group 9 more group attempts remainingCengage Learning Cengage Technical SupportTG^6MacBook AirYH&7F7UV B N* 008JF81M(9K▶▶O)<C0F10P4)DELPrevious Next>FIL{+Save and ExitF12

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req 2req At 298K, the equilibrium constant for the following reaction is 5.90x10-2: H₂C₂O4(aq) + H₂0H30+ (aq) + HC₂04 (aq) Use the References to access important values if needed for this question. The equilibrium constant for a second reaction is 6.40x10-5: HC₂04 (aq) + H₂0 H30+ (aq) + C₂04² (aq) Use this information to determine the equilibrium constant for the reaction: H₂C₂04(aq) + 2H₂O2H30+ (aq) + C₂04² (aq) K = Submit Answer Retry Entire Group 9 more group attempts remaining Previous

req 2req At 298K, the equilibrium constant for the following reaction is 5.90x10-2: H₂C₂O4(aq) + H₂0H30+ (aq) + HC₂04 (aq) Use the References to access important values if needed for this question. The equilibrium constant for a second reaction is 6.40x10-5: HC₂04 (aq) + H₂0 H30+ (aq) + C₂04² (aq) Use this information to determine the equilibrium constant for the reaction: H₂C₂04(aq) + 2H₂O2H30+ (aq) + C₂04² (aq) K = Submit Answer Retry Entire Group 9 more group attempts remaining Previous

Chemistry for Engineering Students

4th Edition

ISBN:9781337398909

Author:Lawrence S. Brown, Tom Holme

Publisher:Lawrence S. Brown, Tom Holme

Chapter12: Chemical Equilibrium

Section: Chapter Questions

Problem 12.66PAE

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The reaction just described is reversible. Deprotonation of the conjugate acid of an organic base by water provides another example of simultaneous making and breaking of sigma bonds. Thus, in the deprotonation of anilinium ion by water, the base is water, which has unshared electrons on the ________ atom. The acid is ________ ion. A pair of ________ electrons on the oxygen atom of water is pushed toward the ________ atom. Simultaneously, the pair of ________ electrons between the hydrogen and ________ atom of the anilinium ion is pushed toward the ________ atom. Thus, the oxygen- ________ sigma bond is made and a hydrogen- ________ sigma bond is broken. The nitrogen atom, which possessed a positive charge, is now ________, and the oxygen atom, which was neutral, now possesses a formal ________ charge.
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