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Sagot :
To solve this problem, we need to understand the relationship between the isotopes' masses and their respective natural abundances in order to determine the average atomic mass of Boron in nature.
Here's a step-by-step solution for the given problem:
1. Identify and list the given data:
- There are two isotopes of Boron.
- One of the isotopes is [tex]\(^{11}B\)[/tex] with a mass of 11.009 amu and a natural abundance of 80.1%.
- The other isotope is unknown, and we denote it as B[tex]\(_{x}\)[/tex].
2. Determine the abundance of the unknown isotope:
- Since the total natural abundance must add up to 100%, the abundance of B[tex]\(_{11}\)[/tex] and B[tex]\(_{x}\)[/tex] together should be 100%.
- The abundance of B[tex]\(_{11}\)[/tex] is given as 80.1%, so the abundance of B[tex]\(_{x}\)[/tex] is:
[tex]\[ \text{Abundance of B}_{x} = 100\% - 80.1\% = 19.9\% \][/tex]
3. Calculate the contribution to the average atomic mass from isotope [tex]\(^{11}B\)[/tex]:
- The formula to calculate an isotope's contribution to the average atomic mass is:
[tex]\[ \text{Contribution} = \text{mass} \times \left(\frac{\text{natural abundance}}{100}\right) \][/tex]
- For [tex]\(^{11}B\)[/tex]:
[tex]\[ \text{Contribution of }^{11}B = 11.009 \, \text{amu} \times \left(\frac{80.1}{100}\right) \][/tex]
[tex]\[ \text{Contribution of }^{11}B = 11.009 \times 0.801 = 8.818209 \, \text{amu} \][/tex]
At this juncture, we have computed the detailed numerical results for the [tex]\(^{11}B\)[/tex] isotope:
- Mass: 11.009 amu
- Natural Abundance: 80.1%
- Contribution to average atomic mass: 8.818209 amu
4. Analysis:
- The total average atomic mass of Boron is typically calculated by summing the contributions from all isotopes. However, without the mass of the other isotope, B[tex]\(_{x}\)[/tex], we can only determine the existing contribution from [tex]\(^{11}B\)[/tex].
Thus, the intermediate calculations yield the following:
- For Boron isotope [tex]\(^{11}B\)[/tex]:
- Mass: 11.009 amu
- Natural abundance: 80.1%
- Contribution to the average atomic mass: 8.818209 amu.
These values constitute our current understanding based on the provided isotope information.
Here's a step-by-step solution for the given problem:
1. Identify and list the given data:
- There are two isotopes of Boron.
- One of the isotopes is [tex]\(^{11}B\)[/tex] with a mass of 11.009 amu and a natural abundance of 80.1%.
- The other isotope is unknown, and we denote it as B[tex]\(_{x}\)[/tex].
2. Determine the abundance of the unknown isotope:
- Since the total natural abundance must add up to 100%, the abundance of B[tex]\(_{11}\)[/tex] and B[tex]\(_{x}\)[/tex] together should be 100%.
- The abundance of B[tex]\(_{11}\)[/tex] is given as 80.1%, so the abundance of B[tex]\(_{x}\)[/tex] is:
[tex]\[ \text{Abundance of B}_{x} = 100\% - 80.1\% = 19.9\% \][/tex]
3. Calculate the contribution to the average atomic mass from isotope [tex]\(^{11}B\)[/tex]:
- The formula to calculate an isotope's contribution to the average atomic mass is:
[tex]\[ \text{Contribution} = \text{mass} \times \left(\frac{\text{natural abundance}}{100}\right) \][/tex]
- For [tex]\(^{11}B\)[/tex]:
[tex]\[ \text{Contribution of }^{11}B = 11.009 \, \text{amu} \times \left(\frac{80.1}{100}\right) \][/tex]
[tex]\[ \text{Contribution of }^{11}B = 11.009 \times 0.801 = 8.818209 \, \text{amu} \][/tex]
At this juncture, we have computed the detailed numerical results for the [tex]\(^{11}B\)[/tex] isotope:
- Mass: 11.009 amu
- Natural Abundance: 80.1%
- Contribution to average atomic mass: 8.818209 amu
4. Analysis:
- The total average atomic mass of Boron is typically calculated by summing the contributions from all isotopes. However, without the mass of the other isotope, B[tex]\(_{x}\)[/tex], we can only determine the existing contribution from [tex]\(^{11}B\)[/tex].
Thus, the intermediate calculations yield the following:
- For Boron isotope [tex]\(^{11}B\)[/tex]:
- Mass: 11.009 amu
- Natural abundance: 80.1%
- Contribution to the average atomic mass: 8.818209 amu.
These values constitute our current understanding based on the provided isotope information.
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