Antimony has two stable isotopes. The first isotope 121Sb has a mass of 120.9038 amu and a natural
abundance of 57.2%. Calculate the percent abundance for the other stable isotope of antimony.

Respuesta :

The percent abundance for the other stable isotope of antimony is 121Sb = 57.2% and 123Sb = 42.8%.

Any of two or more species of atoms of a chemical element with the equal atomic quantity and nearly identical chemical conduct however with differing atomic mass or mass number and exceptional bodily isotope.

Given,

121.76amu is the average mass of all antimony isotopes.

x+y=1⇒y=1−x(both abundances add up to 100%),

=120.904x+122.904y=121.76

Using alzebra,

= 120.904x+122.904⋅(1−x)=121.76

∴x=0.572,

x+y=1

∴y≈0.428

changing into percentage = 0.428 *100 = 42.8%

Hence, 121Sb has a relative abundance of approximately 57.2%, and 123Sb has a relative abundance of approximately 42.8%.

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