87.71 amu (I am assuming degrees of significance here...) In order to determine the average of an element, we take the weighted average of all of the of that element....
1 Answers 1 viewsThe 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 views$M_r=(sum(M_ia))/a$, where: $M_r$ = relative attomic mass ($g$ $mol^-1$) $M_i$ = mass of each isotope ($g$ $mol^-1$) $a$ = abundance, either given as a percent or amount of $g$...
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 viewsLet the percentage of the isotope $""^71Ga$ be $=x%$ Then, The percentage of the isotope $""^69Ga$ is $=(100-x)%$ We write the mass balance equation $100*69.72=x*71+(100-x)*69$ $6972=71x+6900-69x$ $2x=6972-6900=72$ $x=36$ Therefore, The...
1 Answers 1 views