The rate at which a certain drug is eliminated by the body follows first-order kinetics, with a half life of 68 minutes. Suppose in a particular patient the concentration of this drug in the bloodstream immediately after injection is 0.37 ug/mL. What will the concentration be 272 minutes later? Round your answer to 2 significant digits. mL O
The rate at which a certain drug is eliminated by the body follows first-order kinetics, with a half life of 68 minutes. Suppose in a particular patient the concentration of this drug in the bloodstream immediately after injection is 0.37 ug/mL. What will the concentration be 272 minutes later? Round your answer to 2 significant digits. mL O
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
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![### Drug Elimination First-Order Kinetics Problem
The rate at which a certain drug is eliminated by the body follows first-order kinetics, with a half-life of 68 minutes.
**Problem Statement:**
Suppose in a particular patient the concentration of this drug in the bloodstream immediately after injection is 0.37 µg/mL. What will the concentration be 272 minutes later?
**Instructions:**
Round your answer to 2 significant digits.
**Text Box for Input:**
There is a text box where you need to input your answer in µg/mL.
**Buttons Available:**
1. **Checkmark (√)**: Likely used to submit the answer.
2. **Reset (↻)**: Likely used to clear the input and reset the problem.
3. **Question Mark (?)**: Likely provides help or hints related to solving the problem.
If you need to solve the problem, use the first-order decay formula:
\[ C(t) = C_0 \times \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \]
Where:
- \( C(t) \) is the concentration at time \( t \)
- \( C_0 \) is the initial concentration
- \( t_{1/2} \) is the half-life
- \( t \) is the time elapsed
Given:
- \( C_0 = 0.37 \, \mu g/mL \)
- \( t_{1/2} = 68 \, \text{minutes} \)
- \( t = 272 \, \text{minutes} \)
By applying the formula, you can calculate the concentration 272 minutes later.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F223155f5-176c-4d15-b176-e7c491fe06e3%2F8e292b57-9ac8-4ae3-a02d-37e12d655725%2Fyzukgwg_processed.png&w=3840&q=75)
Transcribed Image Text:### Drug Elimination First-Order Kinetics Problem
The rate at which a certain drug is eliminated by the body follows first-order kinetics, with a half-life of 68 minutes.
**Problem Statement:**
Suppose in a particular patient the concentration of this drug in the bloodstream immediately after injection is 0.37 µg/mL. What will the concentration be 272 minutes later?
**Instructions:**
Round your answer to 2 significant digits.
**Text Box for Input:**
There is a text box where you need to input your answer in µg/mL.
**Buttons Available:**
1. **Checkmark (√)**: Likely used to submit the answer.
2. **Reset (↻)**: Likely used to clear the input and reset the problem.
3. **Question Mark (?)**: Likely provides help or hints related to solving the problem.
If you need to solve the problem, use the first-order decay formula:
\[ C(t) = C_0 \times \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \]
Where:
- \( C(t) \) is the concentration at time \( t \)
- \( C_0 \) is the initial concentration
- \( t_{1/2} \) is the half-life
- \( t \) is the time elapsed
Given:
- \( C_0 = 0.37 \, \mu g/mL \)
- \( t_{1/2} = 68 \, \text{minutes} \)
- \( t = 272 \, \text{minutes} \)
By applying the formula, you can calculate the concentration 272 minutes later.
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