Respuesta :
Answer:
7+9.86+.56=17.42, 100-17.42= 82.58%
Explanation:
Add them all together then minus by 100%
From all the naturally occurring strontium isotopes (Sr-84, Sr-86, Sr-87, and Sr-88), the atomic mass and percent abundance of Sr-88 is 78.071amu and 82.58%, respectively.
The average atomic mass of Strontium is given by:
[tex] A_{Sr}*100\% = A_{Sr-84}*\%_{Sr-84} + A_{Sr-86}*\%_{Sr-86} + A_{Sr-87}*\%_{Sr-87} + A_{Sr-88}*\%_{Sr-88} [/tex] (1)
Where:
%: is the percent abundance of each isotope
A: is the atomic mass of each isotope
[tex] 87.62*100\% = 83.913*0.56\% + 85.909*9.86\% + 86.909*7.00\% + A_{Sr-88}*\%_{Sr-88} [/tex]
To find the percent abundance of Sr-88 we need to rest all the values of percent abundance of the isotopes of Sr to 100.
[tex] 100\% = 0.56\% + 9.86\% + 7.00\% + \%_{Sr-88} [/tex]
[tex] \%_{Sr-88} = 100\% - (0.56\% + 9.86\% + 7.00\%) = 82.58\% [/tex]
Now, after changing the percent values of equation (1) to decimal ones, we have:
[tex] 87.62*1 = 83.913*0.0056 + 85.909*0.0986 + 86.909*0.007 + A_{Sr-88}*0.8258 [/tex]
Solving for [tex]A_{Sr-88}[/tex]:
[tex] A_{Sr-88} = \frac{87.62 - (83.913*0.0056 + 85.909*0.0986 + 86.909*0.007)}{0.8258} = 94.54 amu [/tex]
Therefore, the mass and percent abundance of Sr-88 is 78.071amu and 82.58%, respectively.
You can find another example of average atomic mass here: https://brainly.com/question/21536220?referrer=searchResults
I hope it helps you!