A balloon has a volume of 125mL with 10 pumps of gas. The balloon is reduced in volume to 88mL, how much gas is in the balloon?

What are examples of smaller bodies

Inga [223]

comets, astroids oort clouds which surround the kuiper belt

8 0

2 years ago

Of the following: H2(g); He(g); CO2(g); which would behave least like an ideal gas? Why?

Margarita [4]

Answer:

CO2(g)

Because CO2 is the larges molecule with specific geometric, therefore it is not likely to behave as an ideal gas.

4 0

2 years ago

The removal of silver tarnish from silverware using aluminum foil and a solution of electrolyte is an example of

eimsori [14]

Answer:

The answer is C. An electrochemical cell.

Explanation:

The aluminum ion react with the sulfide to form aluminum sulfide.

8 0

3 years ago

At what time of the day is most photosynthesis happens

Elodia [21]

I assume that it would be whenever the sun is at it's highest point because it is normally the hottest and covers the most surface area in a place. I'd say like 2pm maybe.

8 0

3 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%

4 0

3 years ago