Answer:
emf generated by cell is 2.32 V
Explanation:
Oxidation: 
Reduction: 
---------------------------------------------------------------------------------
Overall: 
Nernst equation for this cell reaction at 
-
![E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE_%7Bcell%7D%5E%7B0%7D-%5Cfrac%7B0.059%7D%7Bn%7Dlog%7B%5BAl%5E%7B3%2B%7D%5D%5E%7B2%7D%5BI%5E%7B-%7D%5D%5E%7B6%7D%7D)
where n is number of electrons exchanged during cell reaction, 
is standard cell emf , 
is cell emf , ![[Al^{3+}]](https://tex.z-dn.net/?f=%5BAl%5E%7B3%2B%7D%5D)
is concentration of 
and ![[Cl^{-}]](https://tex.z-dn.net/?f=%5BCl%5E%7B-%7D%5D)
is concentration of 
Plug in all the given values in the above equation -
![E_{cell}=2.20-\frac{0.059}{6}log[(4.5\times 10^{-3})^{2}\times (0.15)^{6}]V](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D2.20-%5Cfrac%7B0.059%7D%7B6%7Dlog%5B%284.5%5Ctimes%2010%5E%7B-3%7D%29%5E%7B2%7D%5Ctimes%20%280.15%29%5E%7B6%7D%5DV)
So, 
The answer is B. is the energy source of stars.
Fission is the type of nuclear energy simulated on Earth, as it is the one used to generate electricity. Fusion, on the other hand, is much more complicated to achieve because it requires extremely hot temperatures compared to fission. Fusion involves the combination of two hydrogen atoms to make helium, which releases a lot of energy. Stars such as the sun, exhibit fusion with its very hot temperature and abundant source of hydrogen.
Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
