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2 years ago

What is the pH of a solution that contains [H+]=3.18 x 10^-8 M?

Chemistry

1 answer:

Xelga [282]2 years ago

4 0

Answer:

7.50

Explanation:

pH=-log(H3O+)

- Hope that helps! Please let me know if you need further explanation.

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If oxygen is removed from a sample of air as iron nusts, what happens to the partial pressure of oxygen in the air?

trasher [3.6K]

Answer:

It increases because the less oxygen the more pressure there will be on that specific object

Explanation:

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2 years ago

Which of the following best compares the structure of carbohydrates and nucleic acids?

Serga [27]

I believe the best description comparing the structure of carbohydrates and nucleic acids would be A.

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3 years ago

Read 2 more answers

Why are the Flemish called Flemish

Dmitriy789 [7]

Explanation:

The flemish are an ethnic group which belong to the northern part of Belgium called Flanders. They also speak a language called flemish. The major population of Belgium are the flemish. The language they speak is a Low Franconian dialect cluster of the Dutch language.

There was a country called the flanders located in the lowlands of Europe. Any person from this country was considered as flemish previously.

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2 years ago

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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2 years ago

Question 10(Multiple Choice Worth 5 points)

jarptica [38.1K]

Answer:

O Generation of electricity

Explanation:

because the others dont make sense to me and tbh carbon is used everyway and everywhere

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1 year ago