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sineoko [7]
3 years ago
9

What contains elements with similar properties in the periodic table?

Chemistry
2 answers:
evablogger [386]3 years ago
8 0
<h2>Answer:</h2>

<u>A) A column</u> contains elements with properties which are similar in periodic table.

<h2>Explanation:</h2>

The vertical section (column) in the periodic table has the similar type of chemical behavior so they are also called as groups or families. The electronic configuration gives the chemical description of the compound.

The chemical compounds in these families have the number of valence electron which is same in the outer cell hence, showing the same chemical properties. Halogens and Nobel gases are the groups of periodic table.

Nataly [62]3 years ago
6 0
A) a column

example: earth alkaline metals
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Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant
antoniya [11.8K]

Answer: The value of E^{o} for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

          AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)

Its solubility product will be as follows.

       K_{sp} = [Ag^{+}][Cl^{-}]

Cell reaction for this equation is as follows.

     Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)

Reduction half-reaction: Ag^{+} + 1e^{-} \rightarrow Ag(s),  E^{o}_{Ag^{+}/Ag} = 0.799 V

Oxidation half-reaction: Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-},   E^{o}_{AgCl/Ag} = ?

Cell reaction: Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)

So, for this cell reaction the number of moles of electrons transferred are n = 1.

    Solubility product, K_{sp} = [Ag^{+}][Cl^{-}]

                                               = 1.77 \times 10^{-10}

Therefore, according to the Nernst equation

           E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

At equilibrium, E_{cell} = 0.00 V

Putting the given values into the above formula as follows.

         E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

        0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}    

       E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}

                  = 0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}

                  = 0.577 V

Hence, we will calculate the standard cell potential as follows.

           E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}

       0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}

       0.577 V = 0.799 V - E^{o}_{AgCl/Ag}

       E^{o}_{AgCl/Ag} = 0.222 V

Thus, we can conclude that value of E^{o} for the half-cell reaction is 0.222 V.

3 0
3 years ago
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