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.
Product
Explanation:
Water in this reaction is called the product of the reaction.
The product is the new kind of matter formed as a result of a chemical change. They are produced from the combination of reactants.
2H₂ + O₂ → 2H₂O
Reactants Products
A chemical reaction involves the combination of chemical species to form a product. The type of reaction above is a called synthesis reaction.
learn more:
Chemical equation brainly.com/question/5247791
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CFCs and their associated compounds were developed in the early 1900s as a non-toxic, non-flammable solution to other more dangerous products such as ammonia.
Answer:
(FeSCN⁺²) = 0.11 mM
Explanation:
Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2
M (Fe(NO₃)₃ = 0.200 M
V (Fe(NO₃)₃ = 10.63 mL
n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol
M (KSCN) = 0.00200 M
V (KSCN) = 1.42 mL
n (KSCN) = 0.00200 * 1.42 = 0.00284 mmol
Total volume = V (Fe(NO₃)₃ + V (KSCN)
= 10.63 + 1.42
= 12.05 mL
Limiting reactant = KSCN
So,
FeSCN⁺² = 0.00284 mmol
M (FeSCN⁺²) = 0.00284/12.05
= 0.000236 M
Excess reactant = (Fe(NO₃)₃
n(Fe(NO₃)₃ = 2.126 mmol - 0.00284 mmol
=2.123 mmol
For standard 2:
n (FeSCN⁺²) = 0.000236 * 4.63
=0.00109
V(standard 2) = 4.63 + 5.17
= 9.8 mL
M (FeSCN⁺²) = 0.00109/9.8
= 0.000111 M = 0.11 mM
Therefore, (FeSCN⁺²) = 0.11 mM
