Answer:
rise in temperature and the effect it has on our global weather patterns
Answer:
red supergiants is the answer
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.
Caffeine has the following percent composition: carbon 49.48%, hydrogen 5.19%, oxygen 16.48% and nitrogen 28.85%. Its molecular weight is 194.19 g/mol.
The change in temperature had the greatest effect at changing the volume of the balloon.
<h3>What are the gas laws?</h3>
The gas laws are used to describe the parameters that has to do with gases.
Given that;
P1 = 98.5 kPa
T1 = 18oC or 291 K
V1 = 74.0 dm3
P2 = 7.0 kPa
V2 = ?
T2 = 18oC or 291 K
P1V1/T1 = P2V2/T2
P1V1T2 =P2V2T1
V2= P1V1T2/P2T1
V2 = 98.5 kPa * 74.0 dm3 * 291 K/ 7.0 kPa * 291 K
V2 = 1041.3 dm3
When;
V1 = 1041.3 dm3
T1 = 291 K
V2 = ?
T2 = 80oC or 353 K
V1/T1 = V2/T2
V1T2 = V2T1
V2 = V1T2/T1
V2 = 1041.3 dm3 * 353 K/291 K
V2 = 1263 dm3
The change in temperature had the greatest effect at changing the volume of the balloon.
Given that
V1 = 100 cm^3
T1 = 273 K
P1 = 1.01 * 10^5 Pa
V2 = ?
P2 = 3.00 x 10^-4 Pa
T2 = -180oC or 255 K
V2= P1V1T2/P2T1
V2 = 1.01 * 10^5 Pa * 100 cm^3 * 255 K / 3.00 x 10^-4 Pa * 273 K
V2 = 3.14 * 10^10 cm^3
Learn more about gas laws:brainly.com/question/12669509
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