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Oksi-84 [34.3K]
3 years ago
12

I need help with these question THAK YOU!!!! what is Hund's rule?

Chemistry
2 answers:
dezoksy [38]3 years ago
7 0
Every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied and all electrons in singly occupied orbitals have the same spin
Nitella [24]3 years ago
3 0

Answer:

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Explanation:

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Chlorine forms a number of oxides with the following oxidation numbers: 1, 3, 4, 6, and 7. These compounds are stable and neutra
SashulF [63]

Answer:

Cl_2O, Cl_2O_3, ClO_2, ClO_3, Cl_2O_7

Explanation:

Empirical formula of the compound is the simplest ratio of elements present in the compound.

Empirical formula of compounds of chlorine with oxygen is as follows:

Compounds in which oxidation state of Cl is +1

Cl_2O

Compounds in which oxidation state of Cl is +3

Cl_2O_3

Compounds in which oxidation state of Cl is +4

ClO_2

Compounds in which oxidation state of Cl is +6

ClO_3

Compounds in which oxidation state of Cl is +7

Cl_2O_7

8 0
3 years ago
What happens to the individual molecules of a liquid as they gain enough kinetic energy to escape the surface of the liquid?
Bumek [7]
Im pretty sure They freeze
7 0
3 years ago
Read 2 more answers
What is the value of for this aqueous reaction at 298 K? <br><br>A+B↽⇀C+D ΔG°=12.86 kJ/mol<br><br>K=
agasfer [191]

Answer:

The equilbrium constant is 179.6

Explanation:

To solve this question we can use the equation:

ΔG = -RTlnK

<em>Where ΔG is Gibbs free energy = 12.86kJ/mol</em>

<em>R is gas constant = 8.314x10⁻³kJ/molK</em>

<em>T is absolute temperature = 298K</em>

<em>And K is equilibrium constant.</em>

Replacing:

12.86kJ/mol = -8.314x10⁻³kJ/molK*298K lnK

5.19 = lnK

e^5.19 = K

179.6 = K

<h3>The equilbrium constant is 179.6</h3>

8 0
3 years ago
What material is least likely to be recognized as a mixture by looking under a microscope
I am Lyosha [343]
A homogenous mixture is uniform and thus hard to recognize as a mixture. An example is water.
6 0
4 years ago
Solid sodium iodide is slowly added to a solution that is 0.0050 M Pb 2+ and 0.0050 M Ag +. [K sp (PbI 2) = 1.4 × 10 –8; K sp (A
UkoKoshka [18]

Answer:

[Ag⁺] = 5.0x10⁻¹⁴M

Explanation:

The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:

Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸

Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷

The PbI₂ <em>just begin to precipitate when the product  [Pb²⁺] [I⁻]² = 1.4x10⁻⁸</em>

<em />

As the initial [Pb²⁺] = 0.0050M:

[Pb²⁺] [I⁻]² = 1.4x10⁻⁸

[0.0050] [I⁻]² = 1.4x10⁻⁸

[I⁻]² = 1.4x10⁻⁸ / 0.0050

[I⁻]² = 2.8x10⁻⁶

<h3>[I⁻] = 1.67x10⁻³</h3><h3 />

So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:

[Ag⁺] [I⁻] = 8.3x10⁻¹⁷

[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷

<h3>[Ag⁺] = 5.0x10⁻¹⁴M</h3>

6 0
3 years ago
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