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.
There is 118 elements<span> in the periodic table</span>
Salt, silt, and clay particles
Answer:
A. Energy is transferred to different forms
.
Explanation:
Hello!
In this case, we need to consider the law of conservation of mass and energy which states that mass and energy cannot be neither created nor destroyed, just modified; it means we can rule out B. and C. so far.
Moreover, since D. is actually true for combustion reactions because they are used to provide energy in industrial operations, this is not the concern here because a combustion reaction is not considered.
Therefore the correct option is A. Energy is transferred to different forms as the energy provided by Rose is transferred to the pendulum system
.
Best regards!
