The idea here is that each isotope will contribute to the average of the element proportionally to their respective abundance. Now, the key to this problem lies in how you...
1 Answers 1 views$""^63Cu$ has $69.2%$ abundance. $""^65Cu$ has $30.8%$ abundance. So, the weighted average is $62.93xx69.2%$ $+$ $64.93xx30.8%$ $=$ $63.55$ $"amu"$. If we look at , copper metal (a mixture of isotopes...
1 Answers 1 viewsMultiply the atomic mass of each isotope times its percent abundance in decimal form. Add them together. $"Average atomic mass of Br"$$=$$(78.92xx0.5069)+(80.92xx0.49331)="79.92 u"$
1 Answers 1 viewsThe is the WEIGHTED average of the individual ..........$""^35Cl$, and $""^37Cl$......... And thus the atomic mass for this asteroid............ $(34.97xx0.1385+36.97xx0.8614)*"amu"$ $=$ $36.69*"amu".$ reports that on Earth, the isotopic...
1 Answers 1 viewsSo, $10.600 (mass$ $units) = 70%xx10.000 + 30%xxM_2;$ where $M_2$ is the isoptopic mass of the other isotope. So solve for $M_2$! A priori would you expect $M_2$ to...
1 Answers 1 viewsThe weighted average is the sum of each individual nuclide, multiplied by its isotopic abundance. So, we simply do the arithmetic: $0.7577xx34.969+0.2423xx36.966 = ???$ amu Then look at to...
1 Answers 1 viewsThe $"Average Atomic Mass"$ of an element is defined as "the weighted average mass of all naturally-occurring (occasionally radioactive) of the element." (and hence the name "average") [1] Dividing the...
1 Answers 1 viewsAnd thus the $"weighted average"$ is: $(62.93xx69.2%+64.93xx30.8%)*"amu"$ $=$ $63.55$ $"amu"$
1 Answers 1 viewsBackground Info Most elements have different isotopes, or atoms with the same number of protons, but different numbers of neutrons. Hence, it is possible to have two atoms that are...
1 Answers 1 viewsThe idea here is that you need to use the of the two in the two to find their . Once you have their , you also have the ratio...
1 Answers 1 views