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WARRIOR [948]
3 years ago
8

How many electrons in an atom can share the quantum numbers n=3, l =2 and m subscript l = 2? (1 point)

Chemistry
1 answer:
Rama09 [41]3 years ago
8 0

Answer : The number of electrons in an atom is, 2 electrons

Explanation :

As we know that there are four quantum numbers which are principle quantum number, azimuthal quantum number, magnetic quantum number and spin magnetic quantum number.

The Principle quantum number, n = 3. That means the electron present in the third energy level.

The Azimuthal quantum number, l = 2. That means the electron present in the d sub-shell.

The magnetic quantum number, m_l=2. That means it shows the 2 electrons present in one of the five 3d orbitals.

Hence, the number of electrons in an atom is, 2 electrons



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VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell =  0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

a)

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

 The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

b)

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

c)

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

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uysha [10]

Explanation:

Steps followed to practice laboratory safety during the experiment are as follows.

  • Used tongs or a test tube holder to hold materials over the Bunsen burner flame.
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When we heat a test tube over bunsen flame then the tube gets hot and when we hold it with bare hands then out hands will burn. Therefore, it is advised to hold test tube with the help of tongs or a holder so that our hands did not burn.

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It is also advised that we should not smell the gases produced but gases move freely from one place to another in a laboratory or any where else.

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Answer:

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Explanation:

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