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aleksley [76]
3 years ago
5

Which electron configuration represents a violation of the pauli exclusion principle?

Chemistry
2 answers:
Katarina [22]3 years ago
7 0
<span>The electron configuration that represents a violation of the pauli exclusion principle is:

</span><span>1s: ↑↓ 
2s: ↑↑ 
2p: ↑</span>
The Pauli exclusion principle  refers to the quantum mechanical rule which expresses that at least two indistinguishable fermions (the particles with half-integer spin) can't involve a similar quantum state inside a quantum framework all the while.
Flauer [41]3 years ago
5 0

Answer:

Explanation:

The electron configuration that represents a violation of the pauli exclusion principle is:

1s: ↑↓ 

2s: ↑↑ 

2p: ↑

The Pauli exclusion principle  refers to the quantum mechanical rule which expresses that at least two indistinguishable fermions (the particles with half-integer spin) can't involve a similar quantum state inside a quantum framework all the while.

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Which of the following species contains manganese with the highest oxidation number?
ioda

In NaMnO₄, Mn has the highest oxidation number.

The question is incomplete, the complete question is;

Which of the following species contains manganese with the highest oxidation number?

A) Mn

B) MnF₂

C) Mn₃(PO₄)₂

D) MnCl₄

E) NaMnO₄

In order to ascertain the specie that contains manganese with the highest oxidation number, we must calculate the oxidation number of manganese in each of the species one after the other.

1) For Mn, the oxidation number of Mn is zero because the atom is uncombined.

2) For MnF₂;

Mn has an oxidation number of +2

3) For Mn₃(PO₄)₂

Mn has an oxidation number of +2

4) For MnCl₄

Mn has an oxidation number of +4

5) For NaMnO₄

Mn has an oxidation number of +7

Hence in NaMnO₄, Mn has the highest oxidation number.

Learn more: brainly.com/question/10079361

7 0
2 years ago
how many grams of oxygen gas are needed to produce 10.0 grams of carbon dioxide according to the balanced equation of CH4
Alinara [238K]
Molar mass of :

O2 = 16 * 2 = 32 g/mol

CO2 = 12 + 16 * 2 = 44 g/mol

<span>Balanced chemical equation :
</span>
1 CH4 + 2 O2 = 1 CO2 + 2 H2O
               ↓              ↓
             2 moles     1 mole

2* 32 g O2 ----------> 1* 44 g CO2
     x g O2 ------------> 10.0 g CO2

44 x = 2 * 32*10.0

44 x = 640

x =  \frac{640}{44}

x = 14.54 g of O2





7 0
2 years ago
Iron (III) oxide and hydrogen react to form iron and water, like this: Fe 03(s)+3H9)2Fe(s)+3HO) At a certain temperature, a chem
belka [17]

The question is incomplete, here is the complete question:

Iron (III) oxide and hydrogen react to form iron and water, like this:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

At a certain temperature, a chemist finds that a 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.

Compound             Amount

  Fe₂O₃                     3.95 g

     H₂                        4.77 g

     Fe                        4.38 g

    H₂O                      2.00 g

Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

<u>Answer:</u> The value of equilibrium constant for given equation is 1.0\times 10^{-4}

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M

  • <u>For water:</u>

Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M

For the given chemical equation:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

The expression of equilibrium constant for above equation follows:

K_{eq}=\frac{[H_2O]^3}{[H_2]^3}

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}

Hence, the value of equilibrium constant for given equation is 1.0\times 10^{-4}

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Answer: 43.125g

Explanation:4Na+ O2=2Na2O

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