Lithium has the lowest. if fluorine is the highest then lithium is the lowest. i hope this helps you out!
Answer:
option 2 is correct answer. its nitrogen.
Answer:
2.2 moles of Fe will be produced
Explanation:
Step 1: Data given
Number of moles of hydrogen gas = 3.3 moles
Number of moles of iron oxide = 1.5 moles
Step 2: The balanced equation
3H2 + Fe2O3 → 2Fe + 3H2O
Step 3: Calculate the limiting reactant
For 3 moles H2 we need 1 mol Fe2O3 to produce 2 moles Fe and 3 moles H2O
Hydrogen gas is the limiting reactant. It will completely be consumed (3.3 moles). Fe2O3 is in excess. There will react 3.3 / 3 = 1.1 moles
There will remain 1.5 - 1.1 = 0.4 moles Fe2O3
Step 4: Calculate moles Fe
For 3 moles H2 we need 1 mol Fe2O3 to produce 2 moles Fe and 3 moles H2O
For 3.3 moles H2 we'll have 2/3 * 3.3 = 2.2 moles Fe
2.2 moles of Fe will be produced
<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
= ?
![[Ni^{2+}_{diluted}]](https://tex.z-dn.net/?f=%5BNi%5E%7B2%2B%7D_%7Bdiluted%7D%5D)
= 
![[Ni^{2+}_{concentrated}]](https://tex.z-dn.net/?f=%5BNi%5E%7B2%2B%7D_%7Bconcentrated%7D%5D)
= 1.0 M
Putting values in above equation, we get:
Hence, the cell potential of the cell is +0.118 V
