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3 years ago

Use the periodic table to determine the electron configuration for aluminum (al) and arsenic (as) in noble-gas notation.

Chemistry

2 answers:

stellarik [79]3 years ago

6 0

Answer: Electronic configuration of

1. Aluminium is $[Ne]3s^23p^1$

2. Arsenic is $[Ar]4s^23d^{10}4p^3$

Explanation: To write the electronic configuration of the elements in the noble gas notation, we first count the total number of electrons and then write the noble gas which lies before the same element.

For Aluminium:

Atomic number = 13

Total number of electrons = 13

Noble gas which lies before this element is Neon (Ne).

Neon has 10 electrons, rest 3 electrons are filled in 3s and 3p sub-shells.

Electronic configuration of Aluminium becomes $[Ne]3s^23p^1$

For Arsenic:

Atomic number = 33

Total number of electrons = 33

Noble gas which lies before this element is Argon (Ar).

Neon has 18 electrons, rest 15 electrons are filled in 4s , 3d and 4p sub-shells.

Electronic configuration of Arsenic becomes $[Ar]4s^23d^{10}4p^3$ .

inessss [21]3 years ago

5 0

Al: [Ne] 2s2 2p1

As: [Ar] 4s2 3d10 4p3

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Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79

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Answer : The mole fraction and partial pressure of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

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Now we have to calculate the mole fraction of $CH_4,CO_2$ and $He$ gases.

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267$

and,

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179$

and,

$\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554$

Thus, the mole fraction of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of $CH_4,CO_2$ and $He$ gases.

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 5.78 atm

$X_i$ = mole fraction of gas

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$p_{CH_4}=0.267\times 5.78atm=1.54atm$

and,

$p_{CO_2}=X_{CO_2}\times p_T$

$p_{CO_2}=0.179\times 5.78atm=1.03atm$

and,

$p_{He}=X_{He}\times p_T$

$p_{He}=0.554\times 5.78atm=3.20atm$

Thus, the partial pressure of $CH_4,CO_2$ and $He$ gases are, 1.54, 1.03 and 3.20 atm respectively.

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