Answer:
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Explanation:
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Answer:
![\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}](https://tex.z-dn.net/?f=%5Cmathbf%7Bs%20%3D%5Csqrt%20%5B3%5D%7B%5Cdfrac%7BK_%7Bsp%7D%7D%7B4%7D%7D%7D)
Less than the concentration of Pb2+(aq) in the solution in part ( a )
Explanation:
From the question:
A)
We assume that s to be the solubility of PbI₂.
The equation of the reaction is given as :
PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹
[Pb²⁺] = s
Then [I⁻] = 2s
![K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} = 4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BPb%24%5E%7B2%2B%7D%24%5D%5BI%24%5E%7B-%7D%24%5D%7D%5E%7B2%7D%20%3D%20s%5Ctimes%20%282s%29%5E%7B2%7D%20%3D%20%204s%5E%7B3%7D%5C%5Cs%5E%7B3%7D%20%3D%20%5Cdfrac%7BK_%7Bsp%7D%7D%7B4%7D%5C%5C%5C%5Cs%20%3D%5Cmathbf%7B%20%5Csqrt%20%5B3%5D%7B%5Cdfrac%7BK_%7Bsp%7D%7D%7B4%7D%7D%7D%5C%5C%5C%5C%5Ctext%7BThe%20mathematical%20expressionthat%20can%20be%20used%20to%20determine%20the%20value%20of%20%20%7D%5Cmathbf%7Bs%20%3D%5Csqrt%20%5B3%5D%7B%5Cdfrac%7BK_%7Bsp%7D%7D%7B4%7D%7D%7D)
B)
The Concentration of Pb²⁺ in water is calculated as :
![\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}](https://tex.z-dn.net/?f=%5Cmathbf%7Bs%20%3D%5Csqrt%20%5B3%5D%7B%5Cdfrac%7BK_%7Bsp%7D%7D%7B4%7D%7D%7D)
![\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}](https://tex.z-dn.net/?f=%5Cmathbf%7Bs%20%3D%5Csqrt%20%5B3%5D%7B%5Cdfrac%7B7%2A10%5E%7B-9%7D%7D%7B4%7D%7D%7D)
![\mathbf{s} =\sqrt[3]{1.75*10^{-9}}](https://tex.z-dn.net/?f=%5Cmathbf%7Bs%7D%20%3D%5Csqrt%5B3%5D%7B1.75%2A10%5E%7B-9%7D%7D)

The Concentration of Pb²⁺ in 1.0 mol·L⁻¹ NaI




The equilibrium constant:
![K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \ m/L](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5BPb%5E%7B2%2B%7D%7D%5D%5BI%5E-%5D%5E2%20%5C%5C%20%5C%5C%20K_%7Bsp%7D%20%3D%20s%2A%281.0%2A2s%29%5E2%20%3D7%2A1.0%5E%7B-9%7D%20%5C%5C%20%5C%5C%20s%20%3D%207%2A10%5E%7B-9%7D%20%5C%20%5C%20%20m%2FL)
It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.
Ammonium is NH₄⁺ and Carbonate is CO₃⁻² => Ammonium Carbonate is (NH₄)₂CO₃
Au, N, O ( give me brainliest please)
Since one atmosphere is 29.9213 inches of mercury, we can simply find this by taking 29.9213 and multiplying that by 1.24. You end up with 37.102412 inches of mercury. I hope this helps you out, and good luck on the rest of your homework.