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...
$""^63Cu$ 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...
Multiply 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"$
The average of an element is determined by taking the weighted average of the atomic masses of its naturally occurring . Now, weighted average simply means that each...
The 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...
The $"Average Atomic Mass"$ 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...
The is the weighted average of the individual isotopic masses: $(23.99xx78.99%+24.99xx10.00%+25.98xx11.01%)*g=24.31*g$ I have given you answer in $g$ which here is equivalent to $"amu"$.
$"Average atomic mass"={18.473_"X"xx28.812%+19.962_"Y"xx 28.757% +21.469_"Z" xx(100-28.812-28.757)%}*"amu"$ $-=20.15*"amu"$...if I have done my 'rithmetic right... How did I get the percentage abundance of $Z$? And it is likely that we got the...