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Calculate the equilibrium number of vacancies per cubic meter for copper at 1000K. The energy for vacancy formation is 0.9eV/ato

nexus9112 [7]

Answer:

Therefore the equilibrium number of vacancies per unit cubic meter =2.34×10²⁴ vacancies/ mole

Explanation:

The equilibrium number of of vacancies is denoted by $N_v$ .

It is depends on

total no. of atomic number(N)
energy required for vacancy
Boltzmann's constant (k)= 8.62×10⁻⁵ev K⁻¹
temperature (T).

$N_v=Ne^{-\frac{Q_v}{kT} }$

To find equilibrium number of of vacancies we have find N.

$N=\frac{N_A\ \rho}{A_{cu}}$

Here ρ= 8.45 g/cm³ =8.45 ×10⁶m³

$N_A$ = Avogadro Number = 6.023×10²³

$A_{Cu}$ = 63.5 g/mole

$N=\frac{6.023\times 10^{23}\times 8.45\times 10^{6}}{63.5}$

$=8.01\times 10^{28$ g/mole

Here $Q_v$ =0.9 ev/atom , T= 1000k

Therefore the equilibrium number of vacancies per unit cubic meter,

$N_v=( 8.01\times 10^{28}) e^{-(\frac{0.9}{8.62\times10^{-5}\times 1000})$

=2.34×10²⁴ vacancies/ mole

3 0

3 years ago

In the symbol 3p4

kompoz [17]

Answer:

a. the 3 represents the principal energy level

Explanation:

3 is the principal energy level. The p is the sublevel. 4 is the possible occupying electron.

4 0

3 years ago

Which of the following elements would you expect to be most reactive, based

ki77a [65]

Answer:

C. Lithium

Explanation:

This image describes the trend in reactivity. Group 1 metals are always the most reactive so yeah :)

8 0

3 years ago

A gaseous compound is 30.4 % N and 69.6% OF. A 5.25 g sample of the gas occupies a volume of 1.00 L and exerts a pressure of 958

Harman [31]

Answer:

The molecular formula = N2O4

Explanation:

Step 1: Data given

A gaseous compound is 30.4 % N and 69.6%

Mass of the compound = 5.25 grams

Volume of the gas = 1.00 L

Pressure of the gas = 958 mmHg = 1.26 atm

Temperature of the gas = -4 °C = 273 -4°C = 269 Kelvin

Molar mass of N = 14 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate mass of N

Mass of Nitrogen = 5.25 grams * 0.304 = 1.596 grams

Step 3: Calculate mass of O

Mass of Oxygen = 5.25 grams * 0.696 = 3.654 grams

Step 4: Calculate number of moles N

Number of moles N = Mass of N/ Molar mass of N

Moles of N = 1.596 grams / 14g/mol

Moles of N = 0.114 moles N

Step 5: Calculate moles of O

Moles O = 3.654 grams / 16 g/mol

Moles 0 = 0.2884 moles

Step 6: Calculate empirical formule

We calculate the empirical formule by dividing number of moles by the smallest number of mol

N : 0.114 / 0.114 = 1

O: 0.2284 / 0.114 = 2

Empirical formule = NO2

Step 7: Calculate number of moles of 5.25 g sample via gas law:

p*V = nRT

⇒ with p = the pressure = 1.26 atm

⇒ with v = 1.00 L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/ K*mol

⇒ with T = the temperature = 269 K

number of moles n = (p*V)/(R*T)

n = (1.26*1L)/(0.08206*269)

n = 0.057 mol

Step 8: Calculate molar mass of the compound

This means 5.25 grams of the gas = 0.057 moles

So 1 mol of the compound has a molar mass of: 5.25 / 0.057 = 92.11 g/mol

Step 9: Calculate molar mass of the empirical formula NO2

N = 14 g/mol

O = 16 g/mol

NO2 = 14 + 16 + 16 = 46 g/mol

The empirical formule NO2 has a molar mass of 46 g/mol

Step 10: Calculate molecular formula

92.11 / 46 = 2

This means the empirical formula should be multiplied by 2

2*(NO2) = N2O4

The molecular formula = N2O4

8 0

3 years ago

Chemistry question please help!!

LiRa [457]

Answer: It’s the first one

8 0

3 years ago