Question

Suppose a mixture contains two isotopes of fictitious element "A" shown below. 45 A 45 A and 46 A 46 A In the mixture 33% are 45 A 45 A and 67% are 46 A 46 A . What is the relative atomic mass for element "A"? Show your work below.

Answer 1

Suppose a mixture contains two isotopes of fictitious element "A" shown below. 45A and 46A. In the mixture 33% are 45A and 67% are 46A. What is the relative atomic mass for element "A"? Show your work below.

The element A, with 2 isotopes, 45A (33%) and 46A (67%), has a relative atomic mass of 45.67 u.

The fictitious element A has 2 isotopes. 45A with an atomic mass of 45 u and an abundance of 33%, and 46 A with an atomic mass of 46 u and an abundance of 67%. The relative atomic mass (am) for element A is a weighted average that takes into account the mass of each isotope ($m_i$) and its abundance ($ab_i$).

$am = \frac{\Sigma m_i \times ab_i }{100} = \frac{45u \times 33 + 46u \times 67 }{100} = 45.67u$

The element A, with 2 isotopes, 45A (33%) and 46A (67%), has a relative atomic mass of 45.67 u.

You can learn more about relative atomic mass here: brainly.com/question/21690748