3. Let's try out the atomic mass equation. Suppose you have a mixture of two nitrogen isotopes. For every 1 Nitrogen-15 isotopes, there are 3 Nitrogen-14 isotopes. Determine the abundance of each isotope in this mixture. Predict the average atomic mass of this mixture. Hint: This is NOT the mixture of nitrogen found in nature. You can check your answer by clicking "My Mixture" under isc Next, create the mixture described in the problem. 14.50 amu

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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The relative ratio, or percent, of each isotope found in nature is the abundance number used to calculate the average mass.
Here's the general equation for average atomic mass:
(Abundance A)(Mass A) +
(Abundance B)(Mass B) = Atomic Mass
For example, here's how the average mass of of boron was calculated.
(Persant Boron-11 11 =10.81 amu
Percent Boron-10
10 +
11)11%3D10.81 amu
100
100
(.20)10 + (.80)11 = 10.8 amu
%3D
3. Let's try out the atomic mass equation.
Suppose you have a mixture of two nitrogen isotopes.
For every 1 Nitrogen-15 isotopes, there are 3 Nitrogen-14 isotopes.
Determine the abundance of each isotope in this mixture.
Predict the average atomic mass of this mixture.
Hint: This is NOT the mixture of nitrogen found in nature. You can check your answer by clicking "My Mixture" under isotope mixture
Next, create the mixture described in the problem.
14.50 amu
14.00 amu
14.25 amu
14.75 amu
Transcribed Image Text:The relative ratio, or percent, of each isotope found in nature is the abundance number used to calculate the average mass. Here's the general equation for average atomic mass: (Abundance A)(Mass A) + (Abundance B)(Mass B) = Atomic Mass For example, here's how the average mass of of boron was calculated. (Persant Boron-11 11 =10.81 amu Percent Boron-10 10 + 11)11%3D10.81 amu 100 100 (.20)10 + (.80)11 = 10.8 amu %3D 3. Let's try out the atomic mass equation. Suppose you have a mixture of two nitrogen isotopes. For every 1 Nitrogen-15 isotopes, there are 3 Nitrogen-14 isotopes. Determine the abundance of each isotope in this mixture. Predict the average atomic mass of this mixture. Hint: This is NOT the mixture of nitrogen found in nature. You can check your answer by clicking "My Mixture" under isotope mixture Next, create the mixture described in the problem. 14.50 amu 14.00 amu 14.25 amu 14.75 amu
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