What is the atomic mass of $""^6Li$, if the atomic mass of lithium is 6.941 amu, and $""^7Li$ is 92.58% abundant and weighs 7.015 amu?

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...

It follows that whatever atomic mass is quoted on the Periodic Table, the most abundant isoptope of that element must make the greatest statistical contribution. Calcium has an atomic mass...

To find the average ($"A"_r$) of an element, given the percentage rarities of its , simply multiply each isotope's $"A"_r$ by its percentage that is expressed as a decimal...

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"$

$"Atomic mass of titanium"={0.785xx45.95263+0.123xx47.94795+0.092xx49.94479}*"amu"=??$ Is this the same as on this planet?

The idea here is that you need to use the mass of a single proton and the mass of a single neutron to calculate the mass of a lithium-6 nucleus,...

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 $"amu"$.

Since the given information is in moles, we need only look at the ratio between the chemicals: Six moles of Li will produce 2 moles of lithium nitride. So, the...