The atmosphere provides C02 for us, it protects us from the Sun's UV rays, it protects us from many objects in space that could crash into earth without it, and it holds the moisture in the air.
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:
The correct matching of the air mass and the letters in the word bank are given as follows;
1. Warm and humid ↔ D
2. Extremely cold and dry ↔ B
3. Cold and dry ↔ A
4. Cold and humid ↔ C
5. Warm and dry ↔ E
Where;
A Represents continental polar
B Represents Artic
C Represents Maritime Polar
D Represents Maritime Tropical
E Represents Continental Tropical
Explanation:
A. A continental polar is one that can be described as a cold and dry climate as the region is located at a further away from the oceanic water bodies that add humidity to the climate
B. The regions of the Artic and the Antarctic have very limited amount of precipitation every year because the air is very cold as well as dry
C. A polar climate is a cold climate region, while a maritime climate is humid.
Therefore, the maritime polar climate combines both cold and humid conditions
D. A warm and humid region has high rainfall and humidity, as such the maritime climate which are humid and the tropical climate, which are warm, combine to give a warm and humid climate
E. The continental tropical climate can be described as warm and dry, compared to the continental water bodies, due to the location being distant from and therefore, the absence of high moisture containing wind that comes from the oceans.
Answer:
<h3>The answer is 3.0 g/mL</h3>
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 15 g
volume = 5 mL
We have
We have the final answer as
<h3>3.0 g/mL</h3>
Hope this helps you
Answer: The correct option is 1.
Explanation: Endothermic reactions are the reactions in which heat is provided to break down the reactant molecules.
In option 1:
The stronger intermolecular forces between the particles in solid molecule are broken down to convert into gaseous form. Hence, some energy in the form of heat is provided to move them far apart. Therefore, it is considered as an endothermic reaction.
In option 2, 3 and 4:
All the other processes involves the formation of bonds and thus there is no need to provide any energy.
