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Artyom0805 [142]
4 years ago
14

What is the pressure exerted by a 0.500 mol sample of nitrogen in a 10.0 L container at 20°C?

Chemistry
1 answer:
lidiya [134]4 years ago
6 0
The answer would be D.120 kPa
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The quantum numbers for the last electron placed in three elements are listed below. Which of these is(are) NOT correct? Er (4 3
UNO [17]

Answer:

The three elements Erbium, Thallium and Osmium have incorrect quantum numbers for the last electron placed.

Explanation:

The 4 quantum numbers are (<em>n,l,ml,ms</em>):

  • <em>n</em> (Principal quantum number): it is the <u>number of the shell (level)</u> where the electron is placed.
  • <em>l </em>(Angular momentum quantum number or Secondary): it represents the <u>sublevel where the electron is</u> placed. There are 4 subleves: s, p d and f so secondary quantum number can take the number 0 (s), 1 (p), 2 (d) or 3 (f) depending on which sublevel the electron is placed.
  • <em>ml</em> (Magnetic quantum number):  it represents the <u>spatial orientation</u> of the electron <u>in respect of the sublevel the electron</u> is placed. For example: if the electron occupies the <em>s sublevel</em> the magnetic number will be <em>0</em>, if the electron occupies the <em>p sublevel</em> the magnetic number could be <em>-1,0,1</em>, if the electron occupies the <em>d sublevel</em> the magnetic number could be <em>-2,-1,0,1,2</em> and if the electron occupies the <em>f sublevel</em> the magnetic number could be <em>-3,-2,-1,0,1,2,3</em>. You can see this in the attachment related to the correct sublevel for the example.
  • <em>ms</em> (Spin quantum number): this number represents the possible rotation of the electron so it could be 1/2 (which is represented by an up arrow) or -1/2 (represented by an down arrow).

Let's analyze the last electron of each element. You can see the attachment for better understanding. The last electron it is represented with orange color.

- Erbium:

This element has 68 electrons so following the Moeller's Diagram to fill the the electronic configuration, we found that the last electron of Erbium it is in the <u>4th level </u>(shell), in the <u>f sublevel</u>. As Erbium has 12 electrons in the f sublevel, it is necessary to follow the Hund's rule (electrons must be placed singly in every sublevel before place a parallel electron) to placed correctly all of them. Finally, the last electron of Erbium stays in the middle of the sublevel and it is represented by a down arrow so the correct quantum numbers in the Erbium element are (4,3,1,-1/2).

- Thallium:

This element has 81 electrons and following the Moeller's Diagram, we found that it last electron it is in the <u>6th level</u>, in the <u>p sublevel</u>. As Thallium has 1 electron in the p sublevel, it is placed singly in the sublevel. So the last electron of Thallium it is represented by an up arrow so the correct quantum numbers in the Thallium element are (6,1,-1,1/2).

- Osmium:

Osmium has 76 electrons and following the steps  that we did with we the other elements, we noticed that its last electron it is in the <u>5th level</u>, in the <u>d sublevel</u>. Following the Hund's rule the last electron of Osmium has a magnetic quantum number of -2 and its spin quantum number is -1/2, so the quantum numbers in the Osmium element are (5,2,-2,-1/2).

<u>Note:</u>

- Remember that the <em>s sublevel</em> has place for 2 electrons, the <u>p sublevel</u> has place for 6 electrons, the <u>d sublevel</u> has place for 10 electrons and the<em> f sublevel</em> has place for 14 electrons.

3 0
3 years ago
Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing indi
AlexFokin [52]

Explanation:

The given cell reaction is as follows.

       In(s)| In^{3+}(aq) || Cd^{2+}(aq) | Cd(s)

Hence, reactions taking place at the cathode and anode are as follows.

At anode ; Oxidation-half reaction : In(s) \rightarrow In^{3+}(aq) + 3e^{-} ...... (1)

At cathode; Reduction-half reaction : Cd^{2+}(aq) + 2e^{-} \rightarrow Cd(s) ....... (2)

Hence, balance the half reactions by multiplying equation (1) by 2 and equation (2) by 3.

Therefore, net cell reaction is as follows.

      2In(s) \rightarrow 2In^{3+}(aq) + 6e^{-}

      3Cd^{2+}(aq) + 6e^{-} \rightarrow 3Cd(s)

Net reaction: 2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)

Thus, we can conclude that the overall cell reaction is as follows.

        2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)

4 0
3 years ago
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