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jeka57 [31]
3 years ago
11

List all orbitals from 1s through 5s according to increasing energy for multielectron atoms. Rank orbitals from smallest to larg

est energy.
1s,3p,4p,3s,3d,4s,2p,2s,5s.
Chemistry
1 answer:
Paha777 [63]3 years ago
3 0

Answer:

The given list of orbitals can be ranked as follow:

1s 2s 2p 3s 3p 4s 3d 4p 5s.

Explanation:

The given list of orbitals can be ranked as follow:

1s 2s 2p 3s 3p 4s 3d 4p 5s.

According to the Aufbau principal in ground state of elements electron first occupy the lower energy level then fill the higher energy levels. We know that there four subshells s, p, d and f. The maximum number of electrons in these subshells can be calculated by following formula:

2 (2l +1 )

and l = 0,1,2,3,....

maximum numbers of electrons in s subshell are,

l=0

2 ( 2(0) + 1)

2

so maximum electrons in s subshell are 2.

maximum numbers of electrons in p subshell are,

l = 1

2 ( 2(1) + 1)

2( 2 + 1

6

so maximum electrons in p subshell are 6.

maximum numbers of electrons in d subshell are,

l = 2

2 ( 2(2) + 1)

2( 4 + 1)

10

so maximum electrons in d subshell are 10.

maximum numbers of electrons in f subshell are,

l = 3

2 ( 2(3) + 1)

2( 6 + 1)

14

so maximum electrons in f subshell are 14.

Electron first fill 1s subshell then 2s subshell and in this way they goes to higher energy levels.

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Fe + O2 —(H2O)—> Fe2O3

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Q_{metal} = -6799\,J

Explanation:

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The formation of SO3 from SO2 and O2 is an intermediate step in the manufacture of sulfuric acid, and it is also responsible for
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Answer:

118.22 atm

Explanation:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)      

KP = 0.13 = \frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}

Where p(SO₃) is the partial pressure of SO₃, p(SO₂) is the partial pressure of SO₂ and p(O₂) is the partial pressure of O₂.

  • With 2.00 mol SO₂ and 2.00 mol O₂ if there was a 100% yield of SO₃, then 2 moles of SO₃ would be produced and 1.00 mol of O₂ would remain.
  • With a 71.0% yield, there are only 2*0.71 = 1.42 mol SO₃, the moles of SO₂ that didn't react would be 2 - 1.42 = 0.58; and the moles of O₂ that didn't react would be 2 - 1.42/2 = 1.29.

The total number of moles is 1.42 + 0.58 + 1.29 = 3.29. With that value we can calculate the molar fraction (X) of each component:

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  • XO₂ = 1.29/3.29 = 0.392
  • XSO₃ = 1.42/3.29 = 0.432

The partial pressure of each gas is equal to the total pressure (PT) multiplied by the molar fraction of each component.

  • p(SO₂) = 0.176 * PT
  • p(O₂) = 0.392 * PT
  • p(SO₃) = 0.432 * PT

Rewriting KP and solving for PT:

\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}=0.13\\\frac{(0.432*P_{T})^{2}}{(0.176*P_{T})^{2}(0.392*P_{T})} =0.13\\\frac{0.1866*P_{T}^{2}}{0.0121*P_{T}^{3}} =0.13\\\frac{15.369}{P_{T}}=0.13\\P_{T}=118.22 atm

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