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BE 699% Sun 3:48 PM E Chrome File Edit View History Bookmarks People Window Help MasteringChemistry: Post Lecture Homework Chapte 4 Adaptive Follow-Up Secure https session.masteringchemistry.co 81350210 m/m Chem 132-02 Pos Lecture Homework Chapter 14 Ada Exercise 1 Enhanced with Feedback Resources Exercise 14.56 Enhanced with Feedback where k is the rate constant and t is the time required for the reaction to reach a final reactant concentration of Alt, 6.80x10 2 M,beg nning from an initial reactant concentration of A 0.140 M. Rearrange the equation and substitute known values to solve for the time: E DATETIME The decomposition of XY is second order in XY and has a rate constant of 6.82x10 3 M 1.s i at 3/09/17 at 11:59pm a certain temperature. 3/09/17 at 11:59pm You may want to reference (D pages 635-641) Section 14.4 while completing this problem. 3/0917 at 11:59pm Part E 3/0917 at 11:59pm If the initial concentration of XY is 0.050 M, what is the concentration of XY after 50.0 s 7 at 11:59pm Express your answer using two significant figures. 3/09/17 at 11:59pm 4/02/17 at 11:59pm 04/17 at 11:59pm 4/09/17 at 11:59pm XYli 4.1.102 4/11/17 at 11:59pm Submit My Answers Give U 4/23/17 at 11:59pm 4/23/17 at 11:59pm incorrect; Try Again; 4 attempts remaining 4/25/17 at 11:59pm 4/25/17 at 11:59pm Part F If the initial concentration of XY is 0.050 M, what is the concentration of XY after 600 s? Master Express your answer using two significant figures. with KNEWT https://session.masteringchemistry.com/myct/assignment?assignmentID 5191477Explanation / Answer
The reaction is 2XY = > products
The differential rate law is
r = -(1/2)dc/dt = k[XY]2
where k = 6.88*10-3M-1s-1
The integral rate law is :
1/[XY] = 1/[XY]o +kt
1/[XY] = 1/0.050 + 0.00682x55
= 20.00 + 0.38 = 20.38
[XY] = 1/20.38= 0.049 M
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