The homologous structures and the analogous structures of different species.
Answer:
x = 33.52 amu
Explanation:
It is given that,
Isotope A has a mass of 34 amu and an abundance of 52%, isotope B has a mass of 33 amu and an abundance of 48%.
Let x is the average atomic mass of this element. It can be calculated as follows :
So, the average atomic mass of this element is 33.52 amu.
The correct answer is D. Mixtures can be easily separated solutions cannot.