Answer:
Releases a lot of energy. When hydrogen a highly reactive element reacts with oxygen, it does so using immense force, when the bonds break high amounts of emery is released.
Explanation:
Explanation:
The given data is as follows.
= 0.483,
= 0.173 M,
= 0.433 M,
= 0.306 M,
= 9.0 atm
According to the ideal gas equation, PV = nRT
or, P =
Also, we know that
Density = 
So, P = MRT
and, M = 
= 
= 
= 0.368 mol/L
Now, we will calculate the cell potential as follows.
E = ![E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}](https://tex.z-dn.net/?f=E%5E%7Bo%7D%20-%20%5Cfrac%7B0.0591%7D%7Bn%7D%20log%20%5Cfrac%7B%5BCo%5E%7B2%2B%7D%5D%5E%7B2%7D%5BCl_%7B2%7D%5D%7D%7B%5BCo%5E%7B3%2B%7D%5D%5BCl%5E%7B-%7D%5D%5E%7B2%7D%7D)
= 
= 
= 
= 0.483 - 0.0185
= 0.4645 V
Thus, we can conclude that the cell potential of given cell at
is 0.4645 V.
Answer:
the flow of charged particles found in an atom
Explanation:
Answer: They belong to the same kingdom but different phyla.
Explanation: