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 abundance of $""_8^16"O"$ is 99.762 %, and the abundance of $""_8^18"O"$ is 0.201 %. Assume you have 100 000 atoms of O. Then you have 37 atoms of $""_8^17"O"$...
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 viewsAnd thus the $"weighted average"$ is: $(62.93xx69.2%+64.93xx30.8%)*"amu"$ $=$ $63.55$ $"amu"$
1 Answers 1 viewsUsing my (not very precise) periodic table, I get the mass of Ag as 107.87. This must be a combination of these two . So the % of the other...
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 viewsIn each 100 $mol$ of Sb you have: $57.21$ mol of $ _ ^(121)Sb$ and $100-57.21=42.79 mol$ of $ _ ^(123)Sb$ Assuming the mass of the isotopes is integer (they...
1 Answers 1 views