A reaction is known to follow first-order kinetics. What is the half-life (in seconds) if it takes (7.96x10^2) seconds for the concentration of the only reactant to drop from (3.460x10^0) M to (3.470x10^-1) M? Enter your answer in scientific notation with 3 sig figs. Do not include any units in your answer. Do not round any intermediate calculations. Note: Your answer is assumed to be reduced to the highest power possible.

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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A reaction is known to follow first-order kinetics. What is the half-life (in seconds) if
it takes (7.96x10^2) seconds for the concentration of the only reactant to drop from
(3.460x10^0) M to (3.470x10^-1) M?
Enter your answer in scientific notation with 3 sig figs. Do not include any units in
your answer.
Do not round any intermediate calculations.
Note: Your answer is assumed to be reduced to the highest power possible.
Transcribed Image Text:A reaction is known to follow first-order kinetics. What is the half-life (in seconds) if it takes (7.96x10^2) seconds for the concentration of the only reactant to drop from (3.460x10^0) M to (3.470x10^-1) M? Enter your answer in scientific notation with 3 sig figs. Do not include any units in your answer. Do not round any intermediate calculations. Note: Your answer is assumed to be reduced to the highest power possible.
A reaction is known to follow zero-order kinetics and has k = (2.90x10^-2) M1s-1.
What is the half-life (in seconds) when the concentration of the only reactant is
(7.670x10^-1) M?
Enter your answer in scientific notation with 3 sig figs. Do not include any units in
your answer.
Do not round any intermediate calculations.
Note: Your answer is assumed to be reduced to the highest power possible.
Transcribed Image Text:A reaction is known to follow zero-order kinetics and has k = (2.90x10^-2) M1s-1. What is the half-life (in seconds) when the concentration of the only reactant is (7.670x10^-1) M? Enter your answer in scientific notation with 3 sig figs. Do not include any units in your answer. Do not round any intermediate calculations. Note: Your answer is assumed to be reduced to the highest power possible.
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