The 4 types of air masses are polar, tropical, continental and maritime. Their classification depends on their location where they are formed.
Answer:
pH is the number given to a substance of how powerful it is before 7 or beyond 7.
the solution is an acid.
Explanation:
Pls mark BRAINLIEST
E
θ
Cell
=
+
2.115
l
V
Cathode
Mg
2
+
/
Mg
Anode
Ni
2
+
/
Ni
Explanation:
Look up the reduction potential for each cell in question on a table of standard electrode potential like this one from Chemistry LibreTexts. [1]
Mg
2
+
(
a
q
)
+
2
l
e
−
→
Mg
(
s
)
−
E
θ
=
−
2.372
l
V
Ni
2
+
(
a
q
)
+
2
l
e
−
→
Ni
(
s
)
−
E
θ
=
−
0.257
l
V
The standard reduction potential
E
θ
resembles the electrode's strength as an oxidizing agent and equivalently its tendency to get reduced. The reduction potential of a Platinum-Hydrogen Electrode under standard conditions (
298
l
K
,
1.00
l
kPa
) is defined as
0
l
V
for reference. [2]
A cell with a high reduction potential indicates a strong oxidizing agent- vice versa for a cell with low reduction potentials.
Two half cells connected with an external circuit and a salt bridge make a galvanic cell; the half-cell with the higher
E
θ
and thus higher likelihood to be reduced will experience reduction and act as the cathode, whereas the half-cell with a lower
E
θ
will experience oxidation and act the anode.
E
θ
(
Ni
2
+
/
Ni
)
>
E
θ
(
Mg
2
+
/
Mg
)
Therefore in this galvanic cell, the
Ni
2
+
/
Ni
half-cell will experience reduction and act as the cathode and the
Mg
2
+
/
Mg
the anode.
The standard cell potential of a galvanic cell equals the standard reduction potential of the cathode minus that of the anode. That is:
E
θ
cell
=
E
θ
(
Cathode
)
−
E
θ
(
Anode
)
E
θ
cell
=
−
0.257
−
(
−
2.372
)
E
θ
cell
=
+
2.115
Indicating that connecting the two cells will generate a potential difference of
+
2.115
l
V
across the two cells.
Answer:
The concentration of acetic acid is 8.36 M
Explanation:
Step 1: Data given
Volume of acetic acid = 1.00 mL = 0.001 L
Volume of NaOH = 32.40 mL = 0.03240 L
Molarity of NaOH = 0.258 M
Step 2: The balanced equation
CH3COOH + NaOH → CH3COONa + H2O
Step 3: Calculate the concentration of the acetic acid
b*Ca*Va = a*Cb*Vb
⇒with b = the coefficient of NaOH = 1
⇒with Ca = the concentration of CH3COOH = TO BE DETERMINED
⇒with Va = the volume of CH3COOH = 1.00 mL = 0.001L
⇒with a = the coefficient of CH3COOH = 1
⇒with Cb = the concentration of NaOH = 0.258 M
⇒with Vb = the volume of NaOH = 32.40 mL = 0.03240 L
Ca * 0.001 L = 0.258 * 0.03240
Ca = 8.36 M
The concentration of acetic acid is 8.36 M
