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2 years ago

What are 5 of the critical events in evolution that occurred in the Paleozoic era?!!

Chemistry

1 answer:

soldi70 [24.7K]2 years ago

5 0

Answer:

Cambrian, Ordovician, Silurian, Devonian, Carboniferous, and Permian

Explanation:

These all occurred in the paleozoic Era I hope you do great good luck <3

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You are given the drawing of 2 waves. Notice, wave A is taller and Wave B is thinner. Since wave A is taller, wave

nekit [7.7K]

Wave "B" is thinner. Is this the whole question?

4 0

3 years ago

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Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

Answer: The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

Explanation:

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

Case 1:

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 2:

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 3:

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

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3 years ago

An atom has 13 protons, 13 neutrons, and 13 electrons. Another atom has 13 protons, 14 neutrons, and 13 electrons. Analyze and e

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They are alike bc they both have 13 protons and neutrons

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3 years ago

The conductance of the electric current through the electrolytic solution increases with

zheka24 [161]

The conductance of the electric current through the electrolytic solution increases with increase in concentration.

<h3>What is electrolytic solution?</h3>

Electrolytic solutions are solutions that are capable of conducting an electric current due to presence of ions.

The current flowing in an Electrolytic solutions is calculated as;

Q = It

I = Q/t

where;

Q is charges

Increase in the concentration of the charges, increases the amount of charges in the solution and hence the conductance of the solutions will increase as well.

Learn more about Electrolytic solutions here: brainly.com/question/14654936

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2 years ago

How are babies born, explain . x'D

DaniilM [7]

After the mother goes into labor the mother pushes out the baby

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3 years ago

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