Answer:
hypochlorite ion
Explanation:
The hypochlorous acid, HClO, is a weak acid with Ka = 1.36x10⁻³, when this acid is in solution with its conjugate base, ClO⁻ (From sodium hypochlorite, NaClO) a buffer is produced. When a strong acid as HCl is added, the reaction that occurs is:
HCl + ClO⁻ → HClO + Cl⁻.
Where more hypochlorous acid is produced.
That means, the HCl reacts with the hypochlorite ion present in solution
Answer:
Explanation:
L
=
1.10
L
of solution
Explanation:
The Molarity
M
is calculated by the equation comparing moles of solute to liters of solution
M
=
m
o
l
L
For this question we are given the Molarity 0.88M
We are told the solute is a 25.2 gram sample of LiF, Lithium Fluoride
We can convert the mass of LiF to moles by dividing by the molar mass of LiF
Li = 6.94
F = 19.0
LiF = 25.94 g/mole
25.2
g
r
a
m
s
x
1
m
o
l
25.94
g
r
a
m
s
=
0.97
moles
Now we can take the the molarity and the moles and calculate the Liters of solution
M
=
m
o
l
L
M
L
=
m
o
l
L
=
m
o
l
M
L
=
0.97
m
o
l
0.88
M
L
=
1.10
L
of solution i just did look at my papaer
Answer:
-1.05 V
Explanation:
A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.
Oxidation half equation;
Sn(s) ------> Sn^2+(aq) + 2e
Reduction half equation:
Mn^2+(aq) + 2e ----> Mn(s)
Cell voltage= E°cathode - E°anode
E°cathode= -1.19V
E°anode= -0.14 V
Cell voltage= -1.19 V - (-0.14V)
Cell voltage= -1.05 V
