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 viewsThe idea with that have two naturally occurring is that the percent abundances of those two must add up to give $100%$. In calculations, it is often easier to...
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 viewsWe know that: $x_(1)106.9059 + x_(2)108.9047 = 107.8682$ $(i)$ $x_(1) + x_(2) = 1$ $(ii)$ We assume that silver has only the 2 , which for a first approximation...
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 viewsThe mass of antimony's second naturally occurring is $"122.902 u"$ The thing to know when doing isotope abundance problems is that the abundances of the two must add up to...
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 viewsThe average isotopic mass is the weighted average of the mass of the individual . Because the quoted average, $63.6$ $"amu"$, is closer to $""^63A$ than $""^65A$, the $""^63A$...
1 Answers 1 views