Answer:
T = 4.062V
Explanation:
from PV = nRT => T = PV/RT
P = 1 atm
V = Final Volume
n = 3 moles
R = 0.08206 L·atm/mol·K
T = ?
T = 1 atm · V(Liters)/(3 moles)(0.08206L·atm/mol·K) = 4.062·V(final) Kelvin
A) LITHIUM is the least reactive alkali metal
B) FRANCIUM is the most reactive alkali metal
C) BERYLLIUM is the least reactive alkaline earth metal
D) RADIUM is the most reactive alkaline earth metal
E) TENNESSINE is the most reactive halogen
For E it might also be Astatine because Tennessine is a fairly new
<u>Answer:</u> The potential of electrode is -0.79 V
<u>Explanation:</u>
When zinc is dipped in zinc sulfate solution, the electrode formed is 
Reduction reaction follows: 
To calculate the potential of electrode, we use the equation given by Nernst equation:
![E_{(Zn^{2+}/Zn)}=E^o_{(Zn^{2+}/Zn)}-\frac{0.059}{n}\log \frac{[Zn]}{[Zn^{2+}]}](https://tex.z-dn.net/?f=E_%7B%28Zn%5E%7B2%2B%7D%2FZn%29%7D%3DE%5Eo_%7B%28Zn%5E%7B2%2B%7D%2FZn%29%7D-%5Cfrac%7B0.059%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BZn%5D%7D%7B%5BZn%5E%7B2%2B%7D%5D%7D)
where,
= electrode potential of the cell = ?V
= standard electrode potential of the cell = -0.76 V
n = number of electrons exchanged = 2
![[Zn]=1M](https://tex.z-dn.net/?f=%5BZn%5D%3D1M)
(concentration of pure solids are taken as 1)
![[Zn^{2+}]=0.1M](https://tex.z-dn.net/?f=%5BZn%5E%7B2%2B%7D%5D%3D0.1M)
Putting values in above equation, we get:
Hence, the potential of electrode is -0.79 V
The reaction between the reactants would be:
CH₃NH₂ + HCl ↔ CH₃NH₃⁺ + Cl⁻
Let the conjugate acid undergo hydrolysis. Then, apply the ICE approach.
CH₃NH₃⁺ + H₂O → H₃O⁺ + CH₃NH₂
I 0.11 0 0
C -x +x +x
E 0.11 - x x x
Ka = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
Since the given information is Kb, let's find Ka in terms of Kb.
Ka = Kw/Kb, where Kw = 10⁻¹⁴
So,
Ka = 10⁻¹⁴/5×10⁻⁴ = 2×10⁻¹¹ = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
2×10⁻¹¹ = [x][x]/[0.11-x]
Solving for x,
x = 1.483×10⁻⁶ = [H₃O⁺]
Since pH = -log[H₃O⁺],
pH = -log(1.483×10⁻⁶)
<em>pH = 5.83</em>
