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    Exam 13: Chemical Kinetics
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    The Rate Law Predicted by the Following Two-Step Mechanism Is\(\begin{array}{lr} \mathrm{A} \rightarrow \mathrm{C}+\mathrm{B} & \text { slow } \\ \mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{E} & \text { fast } \end{array}\)
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The Rate Law Predicted by the Following Two-Step Mechanism Is

Question 50

Question 50

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The rate law predicted by the following two-step mechanism is rate = k[A][B]. A→C+B slow A+B→C+E fast \begin{array}{lr}\mathrm{A} \rightarrow \mathrm{C}+\mathrm{B} & \text { slow } \\\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{E} & \text { fast }\end{array}A→C+BA+B→C+E​ slow  fast ​

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