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zaharov [31]
3 years ago
5

What is electron configuration of oxygen in its excited state​

Chemistry
2 answers:
VLD [36.1K]3 years ago
7 0

Answer:

If we look at the ground state (electrons in the energetically lowest available orbital) of oxygen, the electron configuration is 1s^{2} 2s^{2} 2p^{4} . If the element were to become excited, the electron could occupy an infinite number of orbitals. However, in most texts, the example will be the next available one. So for oxygen, it might look like this:  1s^{2} 2s^{2} 2p^{3}3s^{1}  - where the valence electron now occupies the 3s orbital in an excited (i.e. not ground) state.

So, the electron configuration of oxygen in its excited state​ is 1s^{2} 2s^{2} 2p^{3}3s^{1}.

timofeeve [1]3 years ago
3 0

Answer:

1 {s}^{2} 2 {s}^{2} 2 {p}^{4}

OR

2 : 6

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Is it difficult or easy to obtain hydro power
Paha777 [63]

Answer:yes

Explanation:

cause yes

3 0
2 years ago
Where is the fulcrum of this lever? circle and label its location on the diagram above
Sladkaya [172]

Answer:

Explanation:

1) A fulcrum is a pivot point that plays a central role (not necessarily located at the center) in a lever. The fulcrum of the attached picture has been circled (in blue).

2) The object placed on this lever's measurement tray is balanced by placing it at the center of the tray. This is the standard way of placing objects on any balance.

6 0
3 years ago
Which is the balanced single displacement reaction between Aluminum (Al3+) and Copper Sulfate (Cu2+(SO4)2-)
Aliun [14]

Answer:

The balanced single displacement reaction between aluminum and copper sulphate is as follows.

2Al(s)+3CuSO_{4}(aq)\rightarrow Al_{2}(SO_{4})_{3}(aq)+3Cu(s)

Explanation:

The given ions are as follows.

Al^{3+}- Aluminium ion and copper sulfate ion.

Single displacement reaction occurs one element replaces another element.

Hence, the balanced chemical reaction is as follows.

2Al(s)+3CuSO_{4}(aq)\rightarrow Al_{2}(SO_{4})_{3}(aq)+3Cu(s)

<u>Oxidation half reaction:</u>

Al\rightarrow Al^{3+}+3e^{-}

Aluminium loses three electrons.

<u>Reduction half reaction:</u>

<u>Cu^{2+}+2e^{-}\rightarrow Cu(s)</u>

Copper gain three electrons.

4 0
3 years ago
Which is the correct number of moles of NO that is produced from 13.2 moles of oxygen
Oksanka [162]
<h3>Answer:</h3>

10.6 mol NO

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 4NH₃ + 5O₂ → 4NO + 6H₂O

[Given] 13.2 mol O₂

<u>Step 2: Identify Conversions</u>

[RxN] 5 mol O₂ → 4 mol NO

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                     \displaystyle 13.2 \ mol \ O_2(\frac{4 \ mol \ NO}{5 \ mol \ O_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 10.56 \ mol \ NO

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

10.56 mol NO ≈ 10.6 mol NO

5 0
2 years ago
A certain metal forms a soluble nitrate salt M(NO3)3. Suppose the left half cell of a galvanic cell apparatus is filled with a 3
DiKsa [7]

Answer:

1.The electrode on the right is positive

2. 0.058V

Explanation:

The above cell is a concentration cell.

A concentration cell is an electrolytic cell that is made of two half-cells with the same electrodes, but differs in concentrations of the solutions. A concentration cell functions by diluting the more concentrated solution and concentrating the more dilute solution, creating a voltage as the cell reaches an equilibrium thereby transferring the electrons from the cell with the lower concentration to the cell with the higher concentration.

In the above cell, electrons flow from the left electrode (less concentrated) to the right electrode (more concentrated). Therefore, the right electrode is the positive electrode (cathode).

Part 2: Please, see the attachment below for the calculations.

7 0
3 years ago
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