Answer: The information indicates more isotopes have an atomic mass of 1 than of 2 or 3
Explanation:
Average atomic mass is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]
We are given:
Three isotopes of Hydrogen, which are H-1, H-2 and H-3
Average atomic mass of hydrogen = 1.00794 amu
As, the average atomic mass of hydrogen is closer to the mass of H-1 isotope. This means that the relative abundance of this isotope is the highest as compared to the other two isotopes.
Percentage abundance of H-1 isotope = 99.98%
Percentage abundance of H-2 isotope & H-3 isotope = 0.02 %
Hence, the information indicates more isotopes have an atomic mass of 1 than of 2 or 3
Answer:
The given information indicates that:
3. more isotopes have an atomic mass of 1 than of 2 or 3.
Explanation:
Isotopes are defined as the elements differing based on the number of neutrons. Isotopes have same number of protons.
Average atomic mass is defined as the multiplication of fraction abundance with the mass of each isotope. It can be calculated by:
[tex]\text {Average\; Atomic\; Mass} &=& \sum^{2}_{i=1} (\text {Atomic mass of isotopes}})_{i} * (\text {Fractional Abundance})_{i}[/tex]
Since, the isotopes of Hydrogen are [tex]\text H_{1}, \text H_{2}, \text H_{3}[/tex]. The average of atomic mass of hydrogen is given as 1.00794 amu.
From the given information, it is seen that the average atomic mass of Hydrogen is closer to [tex]\text H_{1}[/tex]. Thus, the relative abundance of [tex]\text H_{1}[/tex] is relatively higher than other two isotopes of Hydrogen. The percentage of natural abundance of
Thus, the given data in the question indicated that isotopes having atomic mass of 1 will be more abundant than [tex]\text H_{2}, \text {and}\; \text H_{3}[/tex].
For Further Reference:
https://brainly.com/question/13460782?referrer=searchResults
