Answer:
[tex] 55.85 \, \textsf{u} [/tex]
Explanation:
To calculate the average atomic mass ([tex]A_\textsf{avg}[/tex]) for a sample of iron containing two isotopes ([tex]^{55}\textsf{Fe}[/tex] and [tex]^{56}\textsf{Fe}[/tex]), we can use the following formula:
[tex] \Large\boxed{\boxed{A_\textsf{avg} = (x \cdot M_1) + ((1 - x) \cdot M_2)}} [/tex]
where:
Given that [tex] x = 0.15 [/tex] (15% for [tex]^{55}\textsf{Fe}[/tex]), [tex] M_1 = 55 \, \textsf{u} [/tex], and [tex] M_2 = 56 \, \textsf{u} [/tex], we can substitute these values into the formula:
[tex] A_\textsf{avg} = (0.15 \cdot 55) + ((1 - 0.15) \cdot 56) [/tex]
Now, calculate this:
[tex] A_\textsf{avg} = 8.25 + (0.85 \cdot 56) [/tex]
[tex] A_\textsf{avg} = 8.25 + 47.6 [/tex]
[tex] A_\textsf{avg} = 55.85 \, \textsf{u} [/tex]
So, the average atomic mass for this sample of iron is [tex] 55.85 \, \textsf{u} [/tex].