Copper-64 is used in the form of copper acetate to study brain tumors. It decays by a first-order process and has a half-life of 12.8 hours. If you start with 33.0 mg of 64Cu-labeled copper acetate, what mass will remain after 3 days and 16 hours?
Copper-64 is used in the form of copper acetate to study brain tumors. It decays by a first-order process and has a half-life of 12.8 hours. If you start with 33.0 mg of 64Cu-labeled copper acetate, what mass will remain after 3 days and 16 hours?
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
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![Copper-64 is used in the form of copper acetate to study brain tumors. It decays by a first-order process and has a half-life of
12.8 hours. If you start with 33.0 mg of 64Cu-labeled copper acetate, what mass will remain after 3 days and 16 hours?
Note: 1 day = 24 hr
EQUATIONS AND CONSTANTS
E cell = E°cel – (0.0592/n) log Q @ 25 °C
R=8.314 J/mol K
- =-nFE°sell
F= 9.65 x 10 J/V•mol e
AG°rn =-(RT) In K
E°cell = (RT/nF) In K
For electrolysis:
In (No/N) = kt
1A=1 C/s
t12 0.693/k
F= 9.65 x 104 C/mol e
O 0.765 mg
O 0.375 mg
0.0938 mg
O 0.281 mg
O 0.188 mg
DELL
F1
F3
F4
F6
F7
F8
F9
F10
F11
@
23
$4
1
2
4.
6
8
R
T
Y
K
The Periodic Tablo of th](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6da80f40-d950-402d-8192-6a2924bc947d%2F16f229e0-e1e9-46df-9827-cb005ba1ac74%2F5f0cfrg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Copper-64 is used in the form of copper acetate to study brain tumors. It decays by a first-order process and has a half-life of
12.8 hours. If you start with 33.0 mg of 64Cu-labeled copper acetate, what mass will remain after 3 days and 16 hours?
Note: 1 day = 24 hr
EQUATIONS AND CONSTANTS
E cell = E°cel – (0.0592/n) log Q @ 25 °C
R=8.314 J/mol K
- =-nFE°sell
F= 9.65 x 10 J/V•mol e
AG°rn =-(RT) In K
E°cell = (RT/nF) In K
For electrolysis:
In (No/N) = kt
1A=1 C/s
t12 0.693/k
F= 9.65 x 104 C/mol e
O 0.765 mg
O 0.375 mg
0.0938 mg
O 0.281 mg
O 0.188 mg
DELL
F1
F3
F4
F6
F7
F8
F9
F10
F11
@
23
$4
1
2
4.
6
8
R
T
Y
K
The Periodic Tablo of th
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