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alina1380 [7]
3 years ago
9

If 18.0 g of CO2 is produced in the reaction of C2H2 with O2 to form CO2 and

Chemistry
1 answer:
Tanya [424]3 years ago
5 0

Answer:

3.68 g

Explanation:

n_{CO2} = 18 ÷ 44 = 0.4

2C2H2 + 5O2 → 4CO2 + 2H2O

   2       :    5      :    4      :    2  

                             0.409                 (moles)

⇒ n_{H2O} = 0.409 × 2 ÷ 4 = 0.2045 moles

⇒ m_{H2O} = 0.2045 × 18 = 3.68 grams

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Assume that each atom is a sphere, and that the surface of each atom is in contact with its nearest neighbor. Determine the perc
tatiyna

Answer:

  • The percentage of unit cell volume that is occupied by atoms in a face- centered cubic lattice is 74.05%
  • The percentage of unit cell volume that is occupied by atoms in a body-centered cubic lattice is 68.03%  
  • The percentage of unit cell volume that is occupied by atoms in a diamond lattice is 34.01%

Explanation:

The percentage of unit cell volume = Volume of atoms/Volume of unit cell

Volume of sphere = \frac{4 }{3} \pi r^2

a) Percentage of unit cell volume occupied by atoms in face- centered cubic lattice:

let the side of each cube = a

Volume of unit cell = Volume of cube = a³

Radius of atoms = \frac{a\sqrt{2} }{4}

Volume of each atom = \frac{4 }{3} \pi (\frac{a\sqrt{2}}{4})^3 = \frac{\pi *a^3\sqrt{2}}{24}

Number of atoms/unit cell = 4

Total volume of the atoms = 4 X \frac{\pi *a^3\sqrt{2}}{24} = \frac{\pi *a^3\sqrt{2}}{6}

The percentage of unit cell volume = \frac{\frac{\pi *a^3\sqrt{2}}{6}}{a^3} =\frac{\pi *a^3\sqrt{2}}{6a^3} = \frac{\pi \sqrt{2}}{6} = 0.7405

= 0.7405 X 100% = 74.05%

b) Percentage of unit cell volume occupied by atoms in a body-centered cubic lattice

Radius of atoms = \frac{a\sqrt{3} }{4}

Volume of each atom =\frac{4 }{3} \pi (\frac{a\sqrt{3}}{4})^3 =\frac{\pi *a^3\sqrt{3}}{16}

Number of atoms/unit cell = 2

Total volume of the atoms = 2X \frac{\pi *a^3\sqrt{3}}{16} = \frac{\pi *a^3\sqrt{3}}{8}

The percentage of unit cell volume = \frac{\frac{\pi *a^3\sqrt{3}}{8}}{a^3} =\frac{\pi *a^3\sqrt{3}}{8a^3} = \frac{\pi \sqrt{3}}{8} = 0.6803

= 0.6803 X 100% = 68.03%

c) Percentage of unit cell volume occupied by atoms in a diamond lattice

Radius of atoms = \frac{a\sqrt{3} }{8}

Volume of each atom = \frac{4 }{3} \pi (\frac{a\sqrt{3}}{8})^3 = \frac{\pi *a^3\sqrt{3}}{128}

Number of atoms/unit cell = 8

Total volume of the atoms = 8X \frac{\pi *a^3\sqrt{3}}{128} = \frac{\pi *a^3\sqrt{3}}{16}

The percentage of unit cell volume = \frac{\frac{\pi *a^3\sqrt{3}}{16}}{a^3} =\frac{\pi *a^3\sqrt{3}}{16a^3} = \frac{\pi \sqrt{3}}{16} = 0.3401

= 0.3401  X 100% = 34.01%

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