Question 28 In the hydrolysis reaction C₂H₂Cl + H₂OC₂H₂OH + 2 HCI, the general rate law is rate= k[C, H,CI" [H₂O]". Assu

Business, Finance, Economics, Accounting, Operations Management, Computer Science, Electrical Engineering, Mechanical Engineering, Civil Engineering, Chemical Engineering, Algebra, Precalculus, Statistics and Probabilty, Advanced Math, Physics, Chemistry, Biology, Nursing, Psychology, Certifications, Tests, Prep, and more.
Post Reply
answerhappygod
Site Admin
Posts: 899603
Joined: Mon Aug 02, 2021 8:13 am

Question 28 In the hydrolysis reaction C₂H₂Cl + H₂OC₂H₂OH + 2 HCI, the general rate law is rate= k[C, H,CI" [H₂O]". Assu

Post by answerhappygod »

Question 28 In The Hydrolysis Reaction C H Cl H Oc H Oh 2 Hci The General Rate Law Is Rate K C H Ci H O Assu 1
Question 28 In The Hydrolysis Reaction C H Cl H Oc H Oh 2 Hci The General Rate Law Is Rate K C H Ci H O Assu 1 (31.87 KiB) Viewed 64 times
Question 28 In The Hydrolysis Reaction C H Cl H Oc H Oh 2 Hci The General Rate Law Is Rate K C H Ci H O Assu 2
Question 28 In The Hydrolysis Reaction C H Cl H Oc H Oh 2 Hci The General Rate Law Is Rate K C H Ci H O Assu 2 (18.92 KiB) Viewed 64 times
Question 28 In the hydrolysis reaction C₂H₂Cl + H₂OC₂H₂OH + 2 HCI, the general rate law is rate= k[C, H,CI" [H₂O]". Assuming that the water concentration is effectively constant, determine the order x from the following data No answer text provided. 02 Time (sec) CO 01 0.0 50.0 100.0 150.0 200.0 300.0 400.0 500.0 800.0 10000.0 KHCM 0.1000 0.0905 0.0820 0.0741 0.0671 0.0549 0.0448 0.0368 By using data in the Excel file 'Excel Spreadsheet Kinetics Day 2. What is x? 0.0200 0.0000
0 For the hydrolysis reaction C₂H₂Cl + H₂O +C₂H₂OH+ 2 HCl, for the previous questions, what is the magnitude of the rate constant (not the units) Question 30 4 pts include the units) How many seconds would it take for the [CH₂Clto decay to 0.0500M, given that we started with 0.10M? (Give the answer to 2 sig figs) (don't Question 31
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!
Post Reply