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faust18 [17]
3 years ago
8

Metals are really good thermal conductors for which of the following reasons? Select all that apply. Metals are really good ther

mal conductors for which of the following reasons?
Select all that apply.

Metals can absorb more heat flow than other elements before damage occurs.
Particles of metallic structure begin to vibrate faster as they absorb energy.
Metals have a larger number of freely moving electrons.
Metals have high densities.
Metals have high melting points.
Chemistry
2 answers:
prisoha [69]3 years ago
8 0

Explanation:

Metals are the substances that hold greater number of free electrons. Also, metals as utensil, cookware etc have molecules closer to each other.

Therefore, heat transfer becomes fast from one molecule to another. As a result, due to increase in degree of compactness, that is, density makes metals good conductor of heat.

Thus, we can conclude that metals are really good thermal conductors for the following reasons:

  • Particles of metallic structure begin to vibrate faster as they absorb energy.
  • Metals have a larger number of freely moving electrons.
  • Metals have high densities.
Natasha_Volkova [10]3 years ago
7 0

B. Particles of metallic structure begin to vibrate faster as they absorb.

C. Metals have a larger number of freely moving electrons.

D. Metals have high densities.

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​In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.

The average pea weighs between 0.1 and 0.36 grams.

If we take the lower value (0.1 g/pea), the number of peas in 454 g is:

454 g \times \frac{1pea}{0.1 g} = 4540pea

If we take the higher value (0.36 g/pea), the number of peas in 454 g is:

454 g \times \frac{1pea}{0.36 g} \approx  1261pea

​In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.

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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)+
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<u>Answer:</u> The spontaneous cell reaction having smallest E^o is I_2+Cu\rightarrow Cu^{2+}+2I^-

<u>Explanation:</u>

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:

  • <u>Case 1:</u>

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}

  • <u>Case 2:</u>

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}

  • <u>Case 3:</u>

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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