Answer:
![[Pb^{2+}]=3.9 \times 10^{-2}M](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%3D3.9%20%5Ctimes%2010%5E%7B-2%7DM)
this is the concentration required to initiate precipitation
Explanation:
⇄
Precipitation starts when ionic product is greater than solubility product.
Ip>Ksp
Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.
This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.
![Ip=[Pb^{2}][2Cl^-]^2=Ksp](https://tex.z-dn.net/?f=Ip%3D%5BPb%5E%7B2%7D%5D%5B2Cl%5E-%5D%5E2%3DKsp)

lets solubility=S
![[Pb^{2+}] = S](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%20%3D%20S)
![[Cl^-]=2S](https://tex.z-dn.net/?f=%5BCl%5E-%5D%3D2S)
![Ksp=[Pb^{2+}]\times [Cl^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BPb%5E%7B2%2B%7D%5D%5Ctimes%20%5BCl%5E-%5D%5E2)


![S=\sqrt[3]{\frac{Ksp}{4} }](https://tex.z-dn.net/?f=S%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BKsp%7D%7B4%7D%20%7D)

this is the concentration required to initiate precipitation
This question provides us –
- Weight of
is = 47 g - Volume, V = 375 mL
__________________________________________
- Molar Mass of
–


<u>Using formula</u> –






- Henceforth, Molarity of the solution is = 1.7M
___________________________________________
Answer:)
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Answer:
1120 gm
Explanation:
6. Consider the reaction: CzHo (g) + 02 (8) - 4 CO2(g) + 6H2O (1)
(a) Balance the equation.
(b) How many grams of oxygen are required to react with 10 moles of ethane for a complete
combustion reaction?
FIRST, CORRECT THE EQUATION THEN BALANCE
2C2H6(G) + 7O2------------> 4CO2 + 6H2O
so for 10 moles of ethane, we need
7 X 5 = 35 MOLES O2
=35 MOLES O2
O2 HAS A MOLAR MASS OF 2X16 = 32 gm
35 MOLES OF O2 HAS A MASS OF 35 X 32 =1120 gm