___Li+ __Br, -

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

MAXImum [283]

Answer:

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

Explanation:

Given that:

the equilibrium number of vacancies at 800 °C

i.e T = 800°C is 3.6 x 10¹⁷ cm3

Atomic weight of sliver = 107.9 g/mol

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;

The equation for equilibrium number of vacancies can be represented by the equation:

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

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;

K = Boltzmann constant = 8.62 × 10⁻⁵ eV/atom .K

Temperature T = 800 °C = (800+ 273) K = 1073 K

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

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

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

8 0

3 years ago

Pressure is defined as _______.1

Yakvenalex [24]

Answer:the perpendicular force per unit area

pressure, in the physical sciences, the perpendicular force per unit area, or the stress at a point within a confined fluid.

Explanation:

mark me brainliest!!

7 0

2 years ago

In IR spectroscopy, we normally talk about "frequencies" when in reality we are referring to wavenumbers. What is the mathematic

Svetach [21]

Answer:

Here's what I get.

Explanation:

(b) Wavenumber and wavelength

The wavenumber is the distance over which a cycle repeats, that is, it is the number of waves in a unit distance.

$\bar \nu = \dfrac{1}{\lambda}$

Thus, if λ = 3 µm,

$\bar \nu = \dfrac{1}{3 \times 10^{-6} \text{ m}}= 3.3 \times 10^{5}\text{ m}^{-1} = \textbf{3300 cm}^{-1}$

(a) Wavenumber and frequency

Since

λ = c/f and 1/λ = f/c

the relation between wavenumber and frequency is

$\bar \nu = \mathbf{\dfrac{f}{c}}$

Thus, if f = 90 THz

$\bar \nu = \dfrac{90 \times 10^{12} \text{ s}^{-1}}{3 \times 10^{8} \text{ m$\cdot$ s}^{-1}}= 3 \times 10^{5} \text{ m}^{-1} = \textbf{3000 cm}^{-1}$

(c) Units

(i) Frequency

The units are s⁻¹ or Hz.

(ii) Wavelength

The SI base unit is metres, but infrared wavelengths are usually measured in micrometres (roughly 2.5 µm to 20 µm).

(iii) Wavenumber

The SI base unit is m⁻¹, but infrared wavenumbers are usually measured in cm⁻¹ (roughly 4000 cm⁻¹ to 500 cm⁻¹).

8 0

3 years ago

Jai besoi daide

JulijaS [17]

Answer:

i dont know what you are saying

Explanation:

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4 0

2 years ago

a) One kmol of N2O4 dissociates at 25℃, 1 atm to form an equilibrium ideal gas mixture of N2O4 and NO2 in which the amount of N2

horsena [70]

Answer:

neq N2O4 = 0.9795 mol.....P = 0.5 atm; T = 25°C

Explanation:

ni change eq.

N2O4 1 1 - x 0.8154.....P = 1 atm; T = 25°C

NO2 0 0 + x x

∴ x = neq = Peq.V / R.T.....ideal gas mix

if P = 0.5 atm, T = 25°C; assuming: V = 1 L

⇒ x = neq = ((0.5 atm)(1 L))/((0.082 atm.L/K.mol)(298 K))

⇒ x = neq = 0.0205 mol

⇒ neq N2O4 = 1 - x = 1 - 0.0205 = 0.9795 mol

3 0

3 years ago

___Li+ __Br, - _LiBr,

_Li+ Br, - _LiBr,