The atomic mass is the weighted average of the atomic masses of each isotope. In a weighted average, we multiply each value by a number representing its relative importance. In...
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 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 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 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 viewsNow I hope you have been exposed to the concept of weighted average; If not here is a video Now lets get our data; Lets call our element...
1 Answers 1 viewsAnd thus the $"weighted average"$ is: $(62.93xx69.2%+64.93xx30.8%)*"amu"$ $=$ $63.55$ $"amu"$
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 viewsThe weighted average of the is equal to $117.3*"amu"$. And thus.........with units of $"amu"$ $[65.43%xx119.8+34.57%xxchi] =117.3$ Where $chi-="mass of the other isotope"$.... And so we solve for $chi$.... $chi=(117.3-0.6543xx119.8)/(0.3457)=112.6*"amu"$
1 Answers 1 views