Answer: B: Symbols of Elements
Explanation:
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:
Reaction 1: Kc increases
Reaction 2: Kc decreases
Reaction 3: The is no change
Explanation:
Let us consider the following reactions:
Reaction 1: A ⇌ 2B ΔH° = 20.0 kJ/mol
Reaction 2: A + B ⇌ C ΔH° = −5.4 kJ/mol
Reaction 3: 2A⇌ B ΔH° = 0.0 kJ/mol
To predict what will happen when the temperature is raised we need to take into account Le Chatelier Principle: when a system at equilibrium suffers a perturbation, it will shift its equilibrium to counteract such perturbation. This means that <em>if the temperature is raised (perturbation), the system will react to lower the temperature.</em>
Reaction 1 is endothermic (ΔH° > 0). If the temperature is raised the system will favor the forward reaction to absorb heat and lower the temperature, thus increasing the value of Kc.
Reaction 2 is exothermic (ΔH° < 0). If the temperature is raised the system will favor the reverse reaction to absorb heat and lower the temperature, thus decreasing the value of Kc.
Reaction 3 is not endothermic nor exothermic (ΔH° = 0) so an increase in the temperature will have no effect on the equilibrium.
Explanation:
high energy to Low energy
=the electron gains energy (K.E)
Melting point- the temperature at which a substance has changed from a solid to a liquid
freezing- the temperature at which a substance chanes from liquid to a solid
boiling point- the temperature at which a substance changes from a liquid to a gas phase