Given:
m1 = 45.95263 amu
A1 = 72.5 %
m2 = 47.94795 amu
A2 = 16.4 %
m3 = 49.94479 amu
A3 = 11.1 %
use:
atomic mass = sum of (mass of isotope * abundance) / 100
= (m1*A1 + m2*A2 + m3*A3)/100
= (45.95263*72.5 + 47.94795*16.4 + 49.94479*11.1)/100
= 46.72299 amu
Answer: 46.72299 amu
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help Assignment Score: 3744/5200 L Give Up? EFeedback Resources Resume < Question 2 of 52 > OAttempt 2 On another planet, the isotopes of titanium have the given natural abundances. Abundance Mass (u) Isotope 46Ti 48TI 78.100% 45.95263 18.100% 47.94795 50Ti 3.800% 49.94479 What is the average atomic mass of titanium on that planet? average atomic mass 46.4 about us privacy policy terms of use contact us help careers
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