Answer:
Explanation:
To calculate the cell potential we use the relation:
Eº cell = Eº oxidation + Eº reduction
Now in order to determine which of the species is going to be oxidized, we have to remember that the more the value of the reduction potential is negative, the greater its tendency to be oxidized is. In electrochemistry we use the values of the reductions potential in the tables for simplicity because the only thing we need to do is change the sign of the reduction potential for the oxized species .
So the species that is going to be oxidized is the Aluminium, and therefore:
Eº cell = -( -1.66 V ) + 0.340 V = 5.06 V
Equally valid is to write the equation as:
Eº cell = Eº reduction for the reduced species - Eº reduction for the oxidized species
These two expressions are equivalent, choose the one you fell more comfortable but be careful with the signs.
Answer:
it is a infectiousr bacterial disease characterized by the growth of nodules(tubercles) in tissues especially the lungs
<u>Answer:</u> The initial pH of the HCl solution is 3
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is HCl
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
1 mole of HCl produces 1 mole of 
ions and 1 mole of 
ions
To calculate the pH of the solution, we use the equation:
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
We are given:
![[H^+]=0.001M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.001M)
Putting values in above equation, we get:
Hence, the initial pH of the HCl solution is 3
Solution:
At the equivalence point, moles NaOH = moles benzoic acid
HA + NaOH ==> NaA + H2O where HA is benzoic acid
At the equivalence point, all the benzoic acid ==> sodium benzoate
A^- + H2O ==> HA + OH- (again, A^- is the benzoate anion and HA is the weak acid benzoic acid)
Kb for benzoate = 1x10^-14/4.5x10^-4 = 2.22x10^-11
Kb = 2.22x10^-11 = [HA][OH-][A^-] = (x)(x)/0.150
x^2 = 3.33x10^-12
x = 1.8x10^-6 = [OH-]
pOH = -log [OH-] = 5.74
pH = 14 - pOH = 8.26
Metalloid is the closest because noble gases are obviously out, nonmetals are terrible conductors, and metals are not malleable
