Answer:A
Explanation: I think I am right
Answer:
Gold, silver and sodium.
Explanation:
Gold, silver and sodium are the name of the elements which does not match to their abbreviation. It is because those names are comes from other language. For example, the symbol of sodium is Na which comes from Natrium that is a Latin word. The symbol of gold is Au which comes from Aurum that is also a Latin word. The symbol of silver is Ag which comes from a Latin word i. e. Argentum.
Answer:
Phosphorus does not exist in the atmosphere. The phosphorus cycle is the series of processes that move phosphorus among Earth systems. phosphorus cycle help move phosphorus through the geosphere, the hydrosphere, and the biosphere. ... Rocks exposed at Earth's surface release phosphates when they weather.
Answer:
The smallest possible inside length of the tank is 
m.
Explanation:
As we know that
Thus, volume of 
liter tank is also equal to 
cubic meter
The volume of a cube is equal to 
, where, x is the length of the side of the cube
With the give condition,
Solving the above equation, we get -
The smallest possible inside length of the tank is m.
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.
