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3 years ago

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Chemistry

1 answer:

Ad libitum [116K]3 years ago

7 0

Answer:

[Kr] 5s²

Explanation:

From the question given above, the following data were obtained:

Atomic number of strontium (Sr) = 38

Electronic configuration =?

Next, we shall determine the electronic configuration of the noble gas element before strontium (Sr).

The noble gas element before strontium (Sr) is krypton (Kr). Thus, the electronic configuration of krypton (Kr) is given below:

Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶

Finally, we shall determine the electronic configuration of strontium (Sr). This can be obtained as follow:

Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶

Sr (38) =>?

Sr (38) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶ 5s²

But

Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶

Therefore,

Sr (38) => [Kr] 5s²

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Answer:

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Calculate the energy (in J/atom) for vacancy formation in silver, given that the equilibrium number of vacancies at 800 C is 3.6

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Answer:

the energy vacancies for formation in silver is $\mathbf{Q_v = 3.069*10^{-4} \ J/atom}$

Explanation:

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the equilibrium number of vacancies at 800 °C

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Density of silver = 9.5 g/cm³

Let's first determine the number of atoms in silver

Let silver be represented by N

SO;

$N = \dfrac{N_A* \rho _{Ag}}{A_{Ag}}$

where ;

$N_A =$ avogadro's number = $6.023*10^{23} \ atoms/mol$

$\rho _{Ag}$ = Density of silver = 9.5 g/cm³

$A_{Ag}$ = Atomic weight of sliver = 107.9 g/mol

$N = \dfrac{(6.023*10^{23} \ atoms/mol)*( 9.5 \ g/cm^3)}{(107.9 \ g/mol)}$

N = 5.30 × 10²⁸ atoms/m³

However;

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$N_v = N \ e^{^{-\dfrac{Q_v}{KT}}$

From above; Considering the natural logarithm on both sides; we have:

$In \ N_v =In N - \dfrac{Q_v}{KT}$

Making $Q_v$ the subject of the formula; we have:

${Q_v = - {KT} In( \dfrac{ \ N_v }{ N})$

where;

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$Q _v =-( 8.62*10^{-5} \ eV/atom.K * 1073 \ K) \ In( \dfrac{3.6*10^{17}}{5.3 0*10^{28}})$

$\mathbf{Q_v = 2.38 \ eV/atom}$

Where;

1 eV = 1.602176565 × 10⁻¹⁹ J

Then

$Q_v = (2.38 \ * 1.602176565 * 10^{-19} ) J/atom }$

$\mathbf{Q_v = 3.069*10^{-4} \ J/atom}$

Thus, the energy vacancies for formation in silver is $\mathbf{Q_v = 3.069*10^{-4} \ J/atom}$

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