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2 years ago

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What is the edge length of a 369-g iron cube?(The density of iron is 7.86 g/cm3, and the volume of a cube is equal to the edge l

ength cubed.)

1 answer:

Galina-37 [17]2 years ago

6 0

Answer:

edge length = 3.61 cm

Explanation:

Givens

m = 369 gm

d = 7.86 gm / cm^3

v = ?

Formula

v = m / d

Solution

v = 369 / 7.86

v = 46.95 cm^3

============

That's not the answer. You must take the cube root of the volume

side = cube root (46.95)

side = 3.61 cm

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To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

$A=6R^{2}\sqrt{3}$

Here, R is radius and is related to a as follows:

$R=\frac{a}{2}$

Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

Since, $1 nm=10^{-7}cm$

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Now, volume can be calculated as follows:

$V=Area\times c$

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Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

Here, A is atomic mass of $Cr_{2}O_{3}$ is 151.99 g/mol.

Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

Converting them into cm:

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Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

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Volume of sphere will be sum of volume of total number of cations and anions thus,

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Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

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