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Lyrx [107]
3 years ago
15

What is the number of atoms in a mole of any element?

Chemistry
2 answers:
Elan Coil [88]3 years ago
7 0

Answer : The number of atoms in a mole of any element is, 6.022\times 10^{23} (Avogadro's number).

Explanation :

According to the mole concept, the one mole of an any element contains 6.022\times 10^{23} number of atoms.

For example : In the Cl_2 there are two chlorine element or molecule. So, in the Cl_2 the number of atoms will be, 2\times 6.022\times 10^{23}.

Hence, the number of atoms in a mole of any element is, 6.022\times 10^{23} (Avogadro's number).

ozzi3 years ago
4 0

6.022×10^23 should be correct. Are there any options to choose from?


<u>Avogadros number</u>

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You have to prepare some 2 M solutions, with 10 g of solute in each. What volume of solution will you prepare, for each solute b
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Answer:

             0.045 L  or 45 mL

Explanation:

Moles = Mass/M.Mass

Moles = 10 g / 109.94 g/mol

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Also,

Molarity = Moles / Vol in L

Or,

Vol in L = Moles / Molarity

Vol in L = 0.09 mol / 2 mol/L

Vol in L = 0.045 L

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2 years ago
the atomic number of an element x is 14 and its relative atomic mass is 28 write the electronic distribution in an atom of x​
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the electronic distribution in an atom of x is 14

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How are bacteria and humans different?
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4 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)+
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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^-

7 0
3 years ago
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