[ H₃O⁺] = 10 ^ - pH
[ H₃O⁺ ] = 10 ^ - 7.30
[ H₃O⁺ ] = 5.011 x 10⁻⁸ M
hope this helps!
Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.
See attached figure for the structure.
Vanillin have 3 functional groups:
1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen
2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring
3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons
Now for the hybridization we have:
The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.
The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a <u>sp³</u> hybridization because they are involved only in single bonds.
8A+2B——> 6C
since you multiply by a factor of 2 you do that to each letter
4*2=8
1*2=2
3*2=6
Answer : The cell emf for this cell is 0.118 V
Solution :
The half-cell reaction is:

In this case, the cathode and anode both are same. So,
is equal to zero.
Now we have to calculate the cell emf.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}{diluted}]}{[Cl^{-}{concentrated}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BCl%5E%7B-%7D%7Bdiluted%7D%5D%7D%7B%5BCl%5E%7B-%7D%7Bconcentrated%7D%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 1
= ?
= 0.0222 M
= 2.22 M
Now put all the given values in the above equation, we get:


Therefore, the cell emf for this cell is 0.118 V
Answer : The molarity of solution is, 1.00 M
Explanation : Given,
Moles of
= 0.500 mol
Volume of solution = 0.500 L
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :

Now put all the given values in this formula, we get:

Therefore, the molarity of solution is, 1.00 M