The number of moles of b2o3 that will be formed is determined as 4 moles.
<h3>
Limiting reagent</h3>
The limiting reagent is the reactant that will be completely used up.
4 b + 3O₂ → 2b₂O₃
from the equation above;
4 b ------------> 2 b₂O₃
2b ------------> b₂O₃
2 : 1
3O₂ -------------> 2b₂O₃
3 : 2
b is the limiting reagent, thus, the amount of b2o3 to be formed is calculated as;
4 b ------------> 2 moles of b2o3
8 moles -------> ?
= (8 x 2)/4
= 4 moles
Thus, the number of moles of b2o3 that will be formed is determined as 4 moles.
Learn more about limiting reactants here: brainly.com/question/14222359
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1 mol of formic acid is correct. There are two oxygen atoms in formic acid, compared to just one for the other molecules.
One mole of methane has a mass of 16 g
Answer: 0.18 V
Explanation:-
Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.
=-0.40V[/tex]
=-0.24V[/tex]
Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.
Where both 
are standard reduction potentials.
![E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Cd^{2+}/Cd]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BNi%5E%7B2%2B%7D%2FNi%5D%7D-%20E%5E0_%7B%5BCd%5E%7B2%2B%7D%2FCd%5D%7D)
Using Nernst equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cd^{2+}]}{[Ni^{2+]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BCd%5E%7B2%2B%7D%5D%7D%7B%5BNi%5E%7B2%2B%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential = 0.16 V
![E_{cell}=0.16-\frac{0.0592}{2}\log \frac{[0.10]}{[0.5]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D0.16-%5Cfrac%7B0.0592%7D%7B2%7D%5Clog%20%5Cfrac%7B%5B0.10%5D%7D%7B%5B0.5%5D%7D)
Thus the potential of the following electrochemical cell is 0.18 V.
