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Tpy6a [65]
3 years ago
5

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

terial is 5.22 g/cm3, calculate its atomic packing factor. The atomic weights of cr
Chemistry
2 answers:
GarryVolchara [31]3 years ago
8 0

Answer:

The atomic packing factor is 0.76

Explanation:

The area is

A = 6r²√3 = 6(a/2)²√3 = 1.5a²√3

For HCP, a is equal to:

a = 2r

Replacing

A = 1.5 * (4.961x10⁻⁸)²√3 = 6.39x10⁻¹⁵ cm²

The cell volume is

V = A*c = 6.39x10⁻¹⁵ * 1.36x10⁻⁷ = 8.7x10⁻²² cm³

For n yields

n = (e * N * V)/(∑Ac + ∑An) = (5.22 * 6.022x10²³ * 8.7x10⁻²²)/(2 * 52 + (3 * 16) = 18 formula unit/unit cell

There are 18 unit of Cr₂O₃ or 36 ions of Cr₂O₃ The total volume is

V = 36 * (4π/3) * (rCr³) + 54 * (4π/3) * (rO³) = 36 * (4π/3) * (6.2x10⁻⁹)³ + 54 * (4π/3) * (1.4x10⁻⁸)³ = 6.57x10⁻²² cm³

The APF is equal to:

APF = 6.57x10⁻²²/8.7x10⁻²² = 0.76

IRINA_888 [86]3 years ago
6 0

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}

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

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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Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1
givi [52]

According to Hasselbach-Henderson equation:

pH=pK_{a}+log\frac{[A^{-}]}{[HA]}

Here, [A^{-}] is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is CH_{3}COOK and acid is CH_{3}COOH thus, Hasselbach-Henderson equation will be as follows:

pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{2}{1}=5.06

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{1}{3}=4.28

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{5}{1}=5.45

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{1}{1}=4.76

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{1}{10}=3.76

Therefore, pH of solution is 3.76.

4 0
3 years ago
An aqueous solution in a 55 gallon (208 l drum), characterized by minimal buffering capacity, received 4kg of phenol and 1.5 kg
lianna [129]

An aqueous solution in a 55 gallon (208 l drum), characterized by minimal buffering capacity, received 4kg of phenol and 1.5 kg of sodium phenate. What is the ph of the solution. The pka of phenol = 9.98. Mw of phenol and sodium phenate are 94 g/mol and 116 g/mol, respectively.


Volume of solution = 55 gallons = 208.2 L [ 1 gallon = 3.78 L]

moles of phenol = mass / molar mass = 4000 g / 94 = 42.55 moles

moles of sodium phenate = mass / molar mass = 1500 / 116 = 12.93 moles

pKa of phenol = 9.98

We know that the pH of buffer is calculated using Hendersen Hassalbalch's equation

pH = pKa + log [salt] / [acid]

volume is same for both the sodium phenate and phenol has we can directly take the moles of each in the formula

pH = 9.98 + log [12.93 / 42.55] = 9.46



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