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an unknown element x has the following isotopes: 52x (90.00% abundant, …

Question

an unknown element x has the following isotopes:
52x (90.00% abundant, atomic mass = 52.04 amu),
40x (8.00% abundant, atomic mass = 48.99 amu), and
30x (2.00% abundant, atomic mass = 50.09 amu).
what is the average atomic mass of x in amu?

Explanation:

Step1: <Formula for average atomic mass>

The formula for average atomic mass \( M = \sum_{i}(m_i\times a_i) \), where \( m_i \) is the atomic mass of isotope \( i \), and \( a_i \) is the abundance of isotope \( i \) (in decimal form).

Step2: <Convert abundances to decimals>

For \( ^{52}X \): \( a_1=0.90 \), for \( ^{40}X \): \( a_2 = 0.08 \), for \( ^{50}X \): \( a_3=0.02 \).

Step3: <Substitute values into formula>

\( M=(52.04\times0.90)+(48.99\times0.08)+(50.09\times0.02) \)
\( M = 46.836+3.9192 + 1.0018\)

Step4: <Calculate the sum>

\( M=46.836+3.9192+1.0018 = 51.757\)

Answer:

\( 51.76 \) amu