3 years ago

I need help please. No links or I will report your whole profile

Chemistry

1 answer:

vekshin13 years ago

4 0

Answer:

i can't see the image. is that just me? but i would love to help you!

Explanation:

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3 years ago

For some hypothetical metal, the equilibrium number of vacancies at 600°C is 1 × 1025 m-3. If the density and atomic weight of t

makvit [3.9K]

Answer:

$\frac{N_{v}}{N}=1.92*10^{-4}$

Explanation:

First of all we need to find the amount of atoms per volume (m³). We can do this using the density and the molar mass.

$7.40 \frac{g}{cm^{3}}*\frac{1mol}{85.5 g}*\frac{6.023*10^{23}atoms}{1mol}*\frac{1000000 cm^{3}}{1m^{3}}=5.21*10^{28}\frac{atoms}{m^{3}}$

Now, the fraction of vacancies is equal to the N(v)/N ratio.

N(v) is the number of vacancies $1*10^{25}m^{-3}$
N is the number of atoms per volume calculated above.

Therefore:

The fraction of vacancies at 600 °C will be:

$\frac{N_{v}}{N}=\frac{1*10^{25}}{5.21*10^{28}}$

$\frac{N_{v}}{N}=1.92*10^{-4}$

I hope it helps you!

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3 years ago