<u>Answer:</u> The cell potential of the cell is +0.118 V
<u>Explanation:</u>
The half reactions for the cell is:
<u>Oxidation half reaction (anode):</u> 
<u>Reduction half reaction (cathode):</u> 
In this case, the cathode and anode both are same. So,
will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BNi%5E%7B2%2B%7D_%7Bdiluted%7D%5D%7D%7B%5BNi%5E%7B2%2B%7D_%7Bconcentrated%7D%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 2
= ?
= 
= 1.0 M
Putting values in above equation, we get:


Hence, the cell potential of the cell is +0.118 V
In order to follow the law of conservation of mass, each side must contain an equal amount of each atom
Answer: https://van.physics.illinois.edu/qa/listing.php?id=1508&t=why-is-air-a-mixture
Explanation: The best reason I can think of for why we believe that air is a mixture is that freezing air slowly yields different liquids at different temperatures. Liquid nitrogen has a different boiling point than liquid oxygen. They also freeze at different temperatures. If air were only 1 compound, then air in its entirety would have a single boiling point and a single freezing point.
Jason (w. mike)
air is not a mixture because of scientists freezing it and finding different liquids, it is a mixture because the compounds that make up air e.g. oxygen (o2), Carbon dioxide (co2) and the most important Nitrogen which is an element and makes up 78.09% of air are not chemically bound in the way that compounds are because they can be separated easily and there has been no change in state to any of the compounds or elements in air
Explanation:
Relation between
and
is as follows.
![K_{p} = K_{c} [RT]^{\Delta n}](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20K_%7Bc%7D%20%5BRT%5D%5E%7B%5CDelta%20n%7D)
Given, temperature =
= (830 + 273) K = 1103 K
R = 8.314 J/mol K
= 1 - 0 = 1
Now, putting the given values into the above formula as follows.
![K_{p} = K_{c} [RT]^{\Delta n}](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20K_%7Bc%7D%20%5BRT%5D%5E%7B%5CDelta%20n%7D)
0.5 = 

ICE table for the given reaction will be as follows.

Initial: c - -
Equilibrium: (c - x) x x
= ![[CO_{2}]](https://tex.z-dn.net/?f=%5BCO_%7B2%7D%5D)
= x
Hence, same amount of CaO is produced.
Moles of CaO = 
Mass of CaO = 
= 0.310 g
Thus, we can conclude that 0.310 g of CaO produced when equilibrium is established.
Answer:
C. 1.17 grams
Explanation:
- The molarity is the no. of moles of solute in a 1.0 L of the solution.
<em>M = (mass/molar mass)solute x (1000/ V)</em>
M = 0.1 M, mass = ??? g, molar mass of NaCl = 58.44 g/mol, V = 200.0 mL.
∴ mass of NaCl = (M)(molar mass)(V)/1000 = (0.1 M)(58.44 g/mol)(200.0 mL)/1000 = 1.168 g ≅ 1.17 g.