Answer:
The molar solubility of lead bromide at 298K is 0.010 mol/L.
Explanation:
In order to solve this problem, we need to use the Nernst Equaiton:
![E = E^{o} - \frac{0.0591}{n} log\frac{[ox]}{[red]}](https://tex.z-dn.net/?f=E%20%3D%20E%5E%7Bo%7D%20-%20%5Cfrac%7B0.0591%7D%7Bn%7D%20log%5Cfrac%7B%5Box%5D%7D%7B%5Bred%5D%7D)
E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.
At equilibrium, E = 0, therefore:
![E^{o} = \frac{0.0591}{n} log \frac{[ox]}{[red]} \\\\log \frac{[ox]}{[red]} = \frac{nE^{o} }{0.0591} \\\\log[red] = log[ox] - \frac{nE^{o} }{0.0591}\\\\[red] = 10^{ log[ox] - \frac{nE^{o} }{0.0591}} \\\\[red] = 10^{ log0.733 - \frac{2x5.45x10^{-2} }{0.0591}}\\\\](https://tex.z-dn.net/?f=E%5E%7Bo%7D%20%20%3D%20%5Cfrac%7B0.0591%7D%7Bn%7D%20log%20%5Cfrac%7B%5Box%5D%7D%7B%5Bred%5D%7D%20%5C%5C%5C%5Clog%20%5Cfrac%7B%5Box%5D%7D%7B%5Bred%5D%7D%20%3D%20%5Cfrac%7BnE%5E%7Bo%7D%20%7D%7B0.0591%7D%20%5C%5C%5C%5Clog%5Bred%5D%20%3D%20%20log%5Box%5D%20-%20%20%5Cfrac%7BnE%5E%7Bo%7D%20%7D%7B0.0591%7D%5C%5C%5C%5C%5Bred%5D%20%3D%2010%5E%7B%20log%5Box%5D%20-%20%20%5Cfrac%7BnE%5E%7Bo%7D%20%7D%7B0.0591%7D%7D%20%5C%5C%5C%5C%5Bred%5D%20%3D%2010%5E%7B%20log0.733%20-%20%20%5Cfrac%7B2x5.45x10%5E%7B-2%7D%20%20%7D%7B0.0591%7D%7D%5C%5C%5C%5C)
[red] = 0.010 M
The reduction will happen in the anode, therefore, the concentration of the reduced specie is equivalent to the molar solubility of lead bromide.
Cover with metal .Any elements other than hydrozen and helium concentration shows an absorbance
Answer:
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Answer: rise earlier and set later
Explanation: just trust me hope this helps
The correct answer its:
Inorganic
Hope it helps
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