For some hypothetical metal, the equilibrium number of vacancies at 600°C is 1 × 1025 m-3. If the density and atomic weight of this metal are 7.40 g/cm3 and 85.5 g/mol, respectively, calculate the fraction of vacancies for this metal at 600°C. Enter your answer using scientific notation.

fraction of vacancies for this metal FV = 1.918*10⁻⁴

Explanation:

Given:

The number of vacancies per unit volume => ( Nv = 1*10²⁵ m⁻³ )

But we know that Avogrado's constant NA = 6.022*10²³ atoms/mol

Density of the material is given in g/cm3 we need to convert it to g/m³

Density of material ( p ) in g/m³ :

To convert we know that

1 g/cm³ = 1000000 g/m³ then

7.40 * ( 1000000 ) = 7.40*10⁶ g/m³

So, Density of material ( p ) in g/m³ = 7.40*10⁶ g/m³

Given Atomic mass = 85.5 g/mol

To Calculate the number of atomic sites per unit volume , we will use the below formula by substituting those values above

N = NA * p / A

N = ( 6.022*10²³ ) * ( 7.40*10⁶ ) / 85.5

N = 4.45*10³⁰ / 85.5

N = 5.212*10²⁸ atoms/m³

We can now Calculate the fraction of vacancies using the formula below;

Fv = Nv / N

Fv = 1*10²⁵ / 5.212*10²⁸

fraction of vacancies for this metal at 600c.= 1.918*10⁻⁴

adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation:

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