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Luden [163]
2 years ago
10

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.)
Chemistry
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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The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma
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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

The value of c is 1.360 nm or 1.360\times 10^{-7} cm

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:

1 pm=10^{-10}cm

Thus,

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

and,

r_{O^{2-}}=1.4\times 10^{-8}cm

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

V_{S}=V_{Cr^{3+}}+V_{O^{2-}}

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

V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )

Putting the values,

V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}

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packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758

Thus, atomic packing factor is 0.758.

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