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.
In ionic bonds, one atom gives one or more electrons to another atom so both can get closer to 8 valence electrons. Example: In potassium chloride (KCl), Potassium gives up one valence electron to chlorine, so that the outer shell of potassium has 8 valence electrons. This happens only between metals and nonmetals.
In covalent bonds, atoms share their electrons to reach 8 valence electrons. Example: In water (H2O), Oxygen shares one valence electron with one atom of hydrogen, and another valence electron with another atom of hydrogen. Oxygen now has 8 (4 unshared + 2 of its own + 1 from hydrogen + 1 from hydrogen), and each hydrogen has 2 valence electrons: one of its own and one from oxygen [ note that hydrogen only needs 2 valence electrons to be complete instead of 8].
In metallic bonds between metals, the valence electrons move much more freely than in other bonds. This free characteristic makes metals how they are: ductile, malleable, sectile, conductive, etc.
Answer:
Potassium (K) and Fluorine (F)
Explanation:
A salt is formed with a metal and non-metal element from the periodic table of elements. if you look at the online ptable.com the upper right hand corner of each element shows the valence electrons for each element, and how many are in each shell for that element. Fluorine (F) has 7 valence electrons in its outer most shell, which means there is room for 1 more electron since the second shell can hold a max of 8. Potassium(K) has 1 electron in its outer most shell- which means is can fill in the 1 space available that fluorine has in its outer most shell. Since Potassium(K) is a metal and Fluorine(F) is a non-metal they can form an ionic compound, salt.
Because you need to know what you are looking for before actually trying something so you can prevent any accidents by doing stuff at random
