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3 years ago

6

At 1.3 atm and 23 degrees Celsius, 12.5 grams of a gas occupies a volume of 8.2 liters. What is the molar mass (grams/mole) of t

he gas?

1 answer:

klemol [59]3 years ago

7 0

Answer:

0.44 moles Molar Mass = 28.496

Explanation:

1.3(8.2)=n(0.0821)(23+273)

PV=nRT

12.5/0.43865

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If you wanted to find a sample of fermium,which has an atomic mass of 100,where would i look???

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If you wanted to find a sample of fermium,which has an atomic mass of 100 I would look deep under the earth

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3 years ago

Chemical properties of an element are primarily dependent upon Group of answer choices the electron configuration of the element

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Answer:

The answer is "the electron configuration of the element".

Explanation:

Electronics are distributed in atomic and molecular orbit via electrons from an atom or a molecule.

It reflects a most frequent dependence on valence electrons in the outer.

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2 years ago

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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3 years ago

Plz help and plz tell me if u need any thing else

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Pick "A" as the answer.

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3 years ago

Which of the following statements explains how temperature affects the rate of a reaction ? (5 points ) When temperature decreas

harkovskaia [24]

Answer:

As the kinetic energy of the gaseous solute increases, its molecules have a greater tendency to escape the attraction of the solvent molecules and return to the gas phase. Therefore, the solubility of a gas decreases as the temperature increases.

Explanation:

As the kinetic energy of the gaseous solute increases, its molecules have a greater tendency to escape the attraction of the solvent molecules and return to the gas phase. Therefore, the solubility of a gas decreases as the temperature increases.

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2 years ago