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dlinn [17]
3 years ago
11

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

x 1017/cm3. The atomic weight and density for silver are 107.9 g/mol and 9.5 g/cm3, respectively
Chemistry
1 answer:
MAXImum [283]3 years ago
8 0

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}

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1 year ago
You placed a sample of a hydrate of calcium chloride (CaCl2) in a weighed test tube, and weighed the filled test tube.
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Answer:

1. 5g

2. 2.3g

3. 2.7g

4. 0.02mol

5. 0.13mol

6. 7moles

Explanation:

From the question, the following were obtained:

Mass of empty tube = 13.5g

Mass of empty tube + hydrated salt = 18.5g

Mass of tube + anhydrous salt = 16.2g

1. Mass of empty tube = 13.5g

Mass of empty tube + hydrated salt = 18.5g

Mass of hydrated salt = 18.5 — Mass of empty tube

Mass of hydrated salt = 18.5 — 13.5 = 5g

2. Let us calculate the mass of the anhydrous salt.

Mass of tube + anhydrous salt = 16.2g

Mass of empty tube = 13.5g

Mass of anhydrous salt = 16.2 — Mass of empty tube = 16.2 — 13.5

Mass of anhydrous salt = 2.7g

Now we can calculate the mass of the water evolved as follows:

Mass of water = Mass of hydrated salt — Mass of anhydrous

Mass of water = 5 — 2.7 = 2.3g

3. Mass of empty tube = 13.5g

Mass of anhydrous salt = 16.2 — Mass of empty tube = 16.2 — 13.5

Mass of anhydrous salt = 2.7g

4. MM of CaCl2 = 40 +(2x35.5) = 40 + 71 = 111g

Mass of CaCl2 = 2.7g

Number of mole = Mass /Molar Mass

Number of mole of CaCl2 = 2.7/111 = 0.02mol

5. MM of H2O = (2x1) +16 = 2 + 16 = 18g/mol

Mass of H2O = 2.3g

Number of mole = Mass /Molar Mass

Number of mole of H20 = 2.3/18 = 0.13mol

6. To get the mole of water in the molecular formula, we will find the ratio of the number of mole of anhydrous salt to water as shown below:

Mole anhydrous : mole of water ie

0.02 : 0.13 = 1 : 7

Therefore, the mole of water in the formula is 7 ie

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