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3 years ago

How are the shape of sand dunes determined?using one word

Chemistry

1 answer:

SVEN [57.7K]3 years ago

8 0

"windweathered" because the use of wind weathering would bring the sand all around and it would be wind weathered

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Four gases were combined in a gas cylinder with these partial pressures: 3. 5 atm N2, 2. 8 atm O2, 0. 25 atm Ar, and 0. 15 atm H

aksik [14]

The total pressure of the gas has been 11.9 atm.

The partial pressure has been the pressure exerted by the individual gas in the mixture.

According to the Dalton's law of Partial pressure, the pressure of the gas has been the sum of the partial pressure of each gas in the mixture.

<h3 /><h3>Computation for the pressure of gas</h3>

The given gas has partial pressure of Nitrogen, $P_{N_2}=3.5\;\rm atm$

The partial pressure of Oxygen, $P_{O_2}=8\;\rm atm$

The partial pressure of Argon, $P_{Ar}=0.25\;\rm atm$

The partial pressure of Helium, $P_{He}=0.15\;\rm atm$

The total pressure, (P) of the gas has been given as:

$P=P_{N_2}\;+\;P_{O_2}\;+\;P_{Ar}\;+\;P_{He}\\
P=3.5\;+8\;+\;0.25\;+\;0.15\;\text{atm}\\
P=11.9\;\rm atm$

The total pressure of the gas has been 11.9 atm.

Learn more about the vapor pressure, here:

brainly.com/question/25356241

7 0

3 years ago

What is the energy required to change a spherical drop of water to five smaller spherical drops of equal size? At room temperatu

pav-90 [236]

This is a straightforward question related to the surface energy of the droplet. 

You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. 

With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. 

The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ 

The five smaller droplets need to have the same volume as the original. Therefore 

5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. 

Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. 

Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². 

The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ 
From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. 

Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets.

3 0

3 years ago

The diagram shows two molecules (A and B) with different

Anarel [89]

Answer: chromatography

Explanation:

5 0

3 years ago

Which of these equations is balanced?

stealth61 [152]

It looks like A is balanced

5 0

3 years ago

Silver sulfate (Ag2SO4) is slightly soluble in water and it partly dissolves at the equilibrium according to the following balan

statuscvo [17]

Answer:

a) S = 0.0152 mol/L

b) S' = 4.734 g/L

Explanation:

Ag2SO4 ↔ 2Ag+ + SO42-

S 2S S...............in the equilibrium

Ksp = 1.4 E-5 = [ Ag+ ]² * [ SO42-]

a) molar solubility:

⇒ Ksp = ( 2S) ² * S = 1.4 E-5

⇒ 4S² * S = 1.4 E-5

⇒ S = ∛ ( 1.4 E-5 / 4 )

⇒ S = 0.0152 mol/L

b) solubility ( S' ) in grams per liter:

∴ Mw Ag2SO4 = 311.799 g/mol

⇒ S' = 0.0152 mol/L * ( 311.799 g/mol )

⇒ S' = 4.734 g/L

3 0

3 years ago