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Grace [21]
2 years ago
14

For the n = 3 electron shell, which of the following quantum numbers are valid? Check all that apply.

Chemistry
2 answers:
stealth61 [152]2 years ago
7 0

Answer:

The valid quantum numbers are l=0, l=-2 and l= 2.

Explanation:

Given that,

n = 3 electron shell

Suppose, the valid quantum numbers are

l = 3

m = 3

l = 0

m = –2

l = –1

m = 2

We know that,

The value of n = 3

Principle quantum number :

Then the principal quantum number is 3. Which is shows the M shell.

So, n = 3

Azimuthal quantum number :

The azimuthal quantum number is l.

l=0,1,2

Magnetic quantum number :

The magnetic quantum number is

m=-2,-1,0,1,2

Hence, The valid quantum numbers are l=0, l=-2 and l= 2.

Ymorist [56]2 years ago
3 0

Answer: I = 0  m = -2 m = 2

Explanation: Edge.

the correct answers to the second part of this are:

n = 1, I = 0, m = 0

n = 3, I = 0, m = 0

n = 5, I = 4, m = -3

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The chlorination of methane occurs in a number of steps that results in the formation of chloromethane and hydrogen chloride. Th
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Answer:

Total pressure = 0,806 atm

Partial pressure of CH₄: 0,037 atm

Partial pressure of Cl₂: 0,396 atm

Partial pressure of CH₃Cl: 0,125 atm

Partial pressure of HCl: 0,125 atm

Partial pressure of Cl⁻: 0,125 atm

Explanation:

For the reaction:

2CH₄(g)+3Cl₂(g)⟶2CH₃Cl(g)+2HCl(g)+2Cl⁻(g)

295 mL≡ 0,295L of methane at STP are:

n = PV/RT

P = 1 atm; V = 0,295L; R = 0,082atmL/molK; T = 273K.

moles of methane: 0,0132 moles

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moles of chlorine at STP are: ≡ 0,0324 moles

For a complete reaction of 0,0132 moles of CH₄:

0,0132 mol CH₄× \frac{3molCl_{2}}{2 molCH_{4}} = <em>0,0198 moles</em>

The reaction reaches 77%, moles of Cl₂ that react are: 0,0198×77% = 0,0153 mol

As you have 0,0324 moles of Cl₂, moles that will not react are:

0,0324 - 0,0153 = <em>0,0171 mol Cl₂</em>

As the reaction reaches 77% completion, moles of CH₄ that react are:

0,0132×77% =<em> 0,0102 moles of CH₄ And the moles that don't react are </em><em>0,00300 mol</em>

Thus, moles of each compound are:

0,0102 moles of CH₄×\frac{3Cl_{2}}{2 molCH_{4}}= <em>0,0153 mol  + 0,0171 mol = 0,0324 mol Cl₂</em>

0,0102 moles of CH₄×\frac{2CH_{3}Cl}{2 molCH_{4}}= <em>0,0102 mol CH₄</em>

0,0102 moles of CH₄×\frac{2HCl}{2 molCH_{4}}= <em>0,0102 mol HCl</em>

0,0102 moles of CH₄×\frac{2Cl^{-}}{2 molCH_{4}}= <em>0,0102 mol Cl⁻</em>

Total pressure using:

P = nRT/V

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Partial pressure of CH₃Cl: 0,125 atm

Partial pressure of HCl: 0,125 atm

Partial pressure of Cl⁻: 0,125 atm

<em>-To obtain partial pressure you change the moles for each compound-</em>

<em />

I hope it helps!

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