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2 years ago

8

Metallic gold crystallizes in a face-centered cubic lattice, with one au atom per lattice point. if the edge length of the unit

cell is found to be 408 pm, what is the metallic radius of au in pm?

1 answer:

vaieri [72.5K]2 years ago

6 0

The metallic radius (r) of au is 144.2pm

A face-centered cubic unit cell structure is made up of atoms arranged in a cube with six additional whole atoms placed in the centre of each cube face and a fraction of an atom at each of the cube's four corners. Eight additional unit cells share the atoms at the cube's corner.

Atoms in an FCC crystal are in contact along the unit cell's diagonal. 4r = √2 a, where r is the radius of an atom and a is the edge length of a unit cell.

Putting the values in the above expression,we get

r = √2×408 / 4

r = $\frac{1.414*408}{4}$

r = $\frac{576.9}{4}$

r = 144.2 pm

Therefore, The metallic radius (r) of au is 144.2pm

Learn more about metallic radius (r) here;

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Answer:

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Explanation:

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En este problema usaremos la ley de Boyle.

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The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th

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According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

$E_i = E_f$

$0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh$

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What's is meant by relative motion?

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3 years ago

Read 2 more answers

Assume that the radius ????r of a sphere is expanding at a rate of 70 cm/min.70 cm/min. The volume of a sphere is ????=43???????

rodikova [14]

Answer:

the rate of change in volume with time is 280πr² cm³/min

Explanation:

Data provided in the question:

Radius of the sphere as 'r'

$\frac{d\textup{r}}{\textup{dt}}$ = 70 cm/min

Volume of the sphere, V = $\frac{\textup{4}}{\textup{3}}\pi r^3$

Surface area of the sphere as 4πr²

Now,

Rate of change in volume with time, $\frac{d\textup{V}}{\textup{dt}}$

= $\frac{d(\frac{\textup{4}}{\textup{3}}\pi r^3)}{dt}$

= $3\times\frac{\textup{4}}{\textup{3}}\pi r^2}\times\frac{dr}{dt}$

Substituting the value of $\frac{dr}{dt}$

= $3\times\frac{\textup{4}}{\textup{3}}\pi r^2}\times70$

= 280πr² cm³/min

Hence, the rate of change in volume with time is 280πr² cm³/min

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