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Semmy [17]
3 years ago
6

Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH(aq). Calculate the a

mount of Ga(s) that can be deposited from a Ga(III) solution using a current of 0.760 A that flows for 50.0 min.
Chemistry
1 answer:
UkoKoshka [18]3 years ago
5 0

Answer:

m_{Ga}=0.550gGa

Explanation:

Hello!

In this case, since the applied current for the 50.0 mins provides the following charge to the system:

q=0.760\frac{C}{s}*50.0min*\frac{60s}{1min}=2,280C

As 1 mole of electrons carries a charge of 1 faraday, or 96,485 coulombs, we can compute the moles of electrons involved during the reduction:

n_{e^-}=2,280C*\frac{1mole^-}{96,485C}=0.0236mole^-

Then the reduction of Ga³⁺ to Ga involves the transference of three electrons, we are able to compute the moles and therefore the mass of deposited gallium:

m_{Ga}=0.0236mole^-*\frac{1molGa}{3mole^-}*\frac{69.72gGa}{1molGa}  \\\\m_{Ga}=0.550gGa

Best regards!

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Rank in increasing order of effective nuclear charge:

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<u>1) Effective nuclear charge definitions</u>

  • While the total positive charge of the atom nucleus (Z) is equal to the number of protons, the electrons farther away from the nucleus experience an effective nuclear charge (Zeff) less than the total nuclear charge, due to the fact that electrons in between the nucleus and the outer electrons partially cancel the atraction from the nucleus.

  • Such effect on on a valence electron is estimated as the atomic number less the number of electrons closer to the nucleus than the electron whose effective nuclear charge is being determined: Zeff = Z - S.

<u><em>2) Z eff for a F valence electron:</em></u>

  • F's atomic number: Z = 9
  • Total number of electrons: 9 (same numer of protons)
  • Period: 17 (search in the periodic table or do the electron configuration)
  • Number of valence electrons:  7 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 9 - 7 = 2
  • Zeff = Z - S = 9 - 2 = 7

<u><em>3) Z eff for a Li valence eletron:</em></u>

  • Li's atomic number: Z = 3
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  • Period: 1 (search on the periodic table or do the electron configuration)
  • Number of valence electrons: 1 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 3 - 1 = 2
  • Z eff = Z - S = 3 - 2 = 1.

<em>4) Z eff for a Be valence eletron:</em>

  • Be's atomic number: Z = 4
  • Total number of electrons: 4 (same number of protons)
  • Period: 2 (search on the periodic table or do the electron configuration)
  • Number of valence electrons: 2 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 4 - 2 = 2
  • Z eff = Z - S = 4 - 2 = 2

<u><em>5) Z eff for a B valence eletron:</em></u>

  • B's atomic number: Z = 5
  • Total number of electrons: 5 (same number of protons)
  • Period: 13 (search on the periodic table or do the electron configuration)
  • Number of valence electrons: 3 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 5 - 3 = 2
  • Z eff = Z - S = 5 - 2 = 3

<u><em>6) Z eff for a N valence eletron:</em></u>

  • N's atomic number: Z = 7
  • Total number of electrons: 7 (same number of protons)
  • Period: 15 (search on the periodic table or do the electron configuration)
  • Number of valence electrons: 5 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 7 - 5 = 2
  • Z eff = Z - S = 7 - 2 = 5

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  • <u>Conclusion</u>: the order is Li < Be < B < N < F
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