Ask question

Ask question

All categories

3 years ago

8

Describe how the size of sediment particles affects their movement during deflation.

2 answers:

Solnce55 [7]3 years ago

8 0

Answer:

Deflation refers to the process in which the wind blowing over the ground picks up small particles of sediments along with it. It has been seen that the size of the sediment particles influences their movement at the time of deflation.

When the wind blowing has more energy, it can move larger particles of the sediments along with it. Usually, the small, and fine particles can be picked up and blown along the air. The particles of medium size skip and bounce along the ground, and the large size particles roll and slide along the ground.

lutik1710 [3]3 years ago

7 0

The larger the piece the longer it will take to break down. This is because it has more mass that needs to be broken down. Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

You might be interested in

What layers are composed of metals? Choose all that apply.<br> Inner core<br> Outer core<br> Mantle

svlad2 [7]

Answer:

to be honest ask your teacher

3 0

1 year ago

Never been taught this. Need help.

N76 [4]

Definitely not D i would guess A or B

7 0

3 years ago

alina1380 [7]

Answer:

A

Explanation:

Mutualism is A symbiotic relationship between two species in which both partners benefit

8 0

3 years ago

Many scientists have been involved in the development of the theory of the structure of

Genrish500 [490]

Answer:

bohr descirbed the atomic structure and found that electrons travel in separate orbits around the nucleus. jj thomson created the plum pudding model.

7 0

2 years ago

A chemist fills a reaction vessel with 9.20 atm nitrogen monoxide (NO) gas, 9.15 atm chlorine (CI) gas, and 7.70 atm nitrosyl ch

ivanzaharov [21]

Answer:

The reactions free energy $\Delta G = -49.36 kJ$

Explanation:

From the question we are told that

The pressure of (NO) is $P_{NO} = 9.20 \ atm$

The pressure of (Cl) gas is $P_{Cl} = 9.15 \ atm$

The pressure of nitrosly chloride (NOCl) is $P_{(NOCl)} = 7.70 \ atm$

The reaction is

$2NO_{(g)} + Cl_2 (g)$ ⇆ $2 NOCl_{(g)}$

From the reaction we can mathematically evaluate the $\Delta G^o$ (Standard state free energy ) as

$\Delta G^o = 2 \Delta G^o _{NOCl} - \Delta G^o _{Cl_2} - 2 \Delta G^o _{NO}$

The Standard state free energy for NO is constant with a value

$\Delta G^o _{NO} = 86.55 kJ/mol$

The Standard state free energy for $Cl_2$ is constant with a value

$\Delta G^o _{Cl_2} = 0kJ/mol$

The Standard state free energy for $NOCl$ is constant with a value

$\Delta G^o _{NOCl} =66.1kJ/mol$

Now substituting this into the equation

$\Delta G^o = 2 * 66.1 - 0 - 2 * 87.6$

$= -43 kJ/mol$

The pressure constant is evaluated as

$Q = \frac{Pressure \ of \ product }{ Pressure \ of \ reactant }$

Substituting values

$Q = \frac{(7.7)^2 }{(9.2)^2 (9.15) } = \frac{59.29}{774.456}$

$= 0.0765$

The free energy for this reaction is evaluated as

$\Delta G = \Delta G^o + RT ln Q$

Where R is gas constant with a value of $R = 8.314 J / K \cdot mol$

T is temperature in K with a given value of $T = 25+273 = 298 K$

Substituting value

$\Delta G = -43 *10^{3} + 8.314 *298 * ln [0.0765]$

$= -43-6.36$

$\Delta G = -49.36 kJ$

4 0

3 years ago