<span>A small proportion of helium in the crust is helium that was trapped in the Earth when the Earth formed and has not yet escaped. Most helium on Earth, however, forms as a result of alpha decay of uranium and thorium. The emitted alpha particles, once they grab a couple of stray electrons, become helium atoms and can accumulate in gas reservoirs along with things such as methane.</span>
Yes your choices of B C D are right
Answer:A mole is an arbitrary number of molecules in a single unit - refer to avogadro's number. Essentially, 1 mole is 6.022x10^23 molecules for ALL molecules or atoms, however one must remember that not all atoms/molecules are the same size, this is where mass comes into play. When you measure out 2 grams of carbon powder, there will be a lot more molecules present than if you weighed out 2 grams of thorium powder; this is because carbon is much smaller - kind of like a car filled with clowns, one given car can hold a lot of small clowns but only a few big ones; so the same volume is occupied but the amount of substance (clowns) varies on their own size. The arbitrary mass (relative to the hydrogen atom) for a molecule is the sum of its atomic components' atomic masses; e. g. C2H6's will have 2x12.00 (carbon) + 6x1.01 (hydrogen) = ~30 grams / mole.
Explanation:
Explanation:
will dissociate into ions as follows.
Hence, 
for this reaction will be as follows.
![K_{sp} = [Pb^{2+}][Br^{-}]^{2}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BPb%5E%7B2%2B%7D%5D%5BBr%5E%7B-%7D%5D%5E%7B2%7D)
We take x as the molar solubility of when we dissolve x moles of solution per liter.
Hence, ionic molarities in the saturated solution will be as follows.
![[Pb^{2+}]](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D)
= ![[Pb^{2+}]_{o}](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D_%7Bo%7D)
+ x
![[Br^{-}]^{2}](https://tex.z-dn.net/?f=%5BBr%5E%7B-%7D%5D%5E%7B2%7D)
= ![[Br^{-}]_{o}](https://tex.z-dn.net/?f=%5BBr%5E%7B-%7D%5D_%7Bo%7D)
+ 2x
So, equilibrium solubility expression will be as follows.
= ![([Pb^{2+}]_{o} + x)([Br^{-}]_{o} + 2x)^{2}](https://tex.z-dn.net/?f=%28%5BPb%5E%7B2%2B%7D%5D_%7Bo%7D%20%2B%20x%29%28%5BBr%5E%7B-%7D%5D_%7Bo%7D%20%2B%202x%29%5E%7B2%7D)
Each sodium bromide molecule is giving one bromide ion to the solution. Therefore, one solution contains = 0.10 and there will be no lead ions. So, = 0
So, ![[Br^{-}]_{o} + 2x](https://tex.z-dn.net/?f=%5BBr%5E%7B-%7D%5D_%7Bo%7D%20%2B%202x)
will approximately equals to .
Hence, ![K_{sp} = x[Br^{-}]^{2}_{o}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20x%5BBr%5E%7B-%7D%5D%5E%7B2%7D_%7Bo%7D)
x = 
M
Thus, we can conclude that molar solubility of is M.
