We know that 99.985% of all hydrogen atoms are hydrogen-1. Only 0.015% are hydrogen-2 atom What is the approximate atomic mass of hydrogen?

The question asked you to determine the mass of the hydrate, even though you could calculate the mass of anhydrous salt. And this is simply the intial mass of the...

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 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 molar mass of an element is its atomic weight (relative atomic mass) on in g/mol. Molar masses $"Cl":$$"35.45 g/mol"$ $"Al":$$"26.98 g/mol"$ The molar mass is the mass of one...

The quoted is the weighted average of the individual isotopic masses of $""^7Li$ and $""^6Li$. $6.941$ $"amu"$ $=$ $7.015*"amu"xx92.58%+("mass of "^6Li" isotope")xx7.42%$ And thus, $"6.495 amu"+("mass of "^6Li" isotope")xx7.42%=6.941*"amu"$ And...

The 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 $chi$.... $chi=(117.3-0.6543xx119.8)/(0.3457)=112.6*"amu"$

The answer is B) The must be exactly twice the for each atom. The number of protons is the atomic number of an element. The mass number is the sum...

Because the hydrogen atom has only one electron, so there are no electron repulsions to complicate the orbital energies. It is these electron repulsions that give rise to the different...

We know that $1*mol$ contains $6.022xx10^23$ individual sodium atoms, and has a mass of $22.99*g$. And thus for $1.$ there is a $2*g$ mass of $Na$. For $2.$ there is...