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.
Answer: Esta tendencia es tan regular que el poder de combinación, o valencia, de un elemento se definió una vez como el número de átomos de hidrógeno unidos al elemento en su hidruro. El hidrógeno es el único elemento que forma compuestos en los que los electrones de valencia están en la capa n = 1.
Explanation:
¡Espero que esto ayude!
Answer:
In Fahrenheit it is 294.8°F and in Celsius it is 146°C.
Answer:
D (The last answer)
Explanation:
In a transverse wave, particles oscillate perpendicular to the direction of wave motion.
In a longitudinal wave, the oscillations of particles are parallel to the direction of propagation.
the answer is option D "people emphasized obtaining knowledge through scientific experiments" (on plato)
