Efficiency of the ram is calculated by dividing the work done by the work or energy that is exerted to do so. This is calculated through the equation,
S = ((Wout) / (Win)) x 100%
Substiuting the known values to the equation.
S = (153 J)/(185 J) x 100%
Efficiency = 82.7%.
They can both be derived from fossil fuels, biomass, or perhaps from carbon dioxide water.
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:
962.2 K
Explanation:
Ideal Gas Law Equation:
PV = nRT
Writing a Equation to find T2:
PV = nRT
(P1V)/(nT1) = R = (P2V)/(nT2)
(P1)/(T1) = R = (P2)/(T2) V and n cancel out since they are constant.
(P1)/(T1) = (P2)/(T2)
T2 = (P2)(T1)/(P1)
Substituting Into the Equation:
T2 = (283 kPa * 170 K)/(50 kPa)
T2 = 962.2 K
Explanation:
Transition metals occupy the d block of the periodic table. Their location is right between the group 2A and 3A. Copper (Cu) is in the first row of the d-block elements. The image attached illustrates the position of the transition metals relative to other elements in the periodic table.
