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Aleks04 [339]
3 years ago
8

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

Chemistry
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.
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as the elements period 3 are considered in order of increasing atomic number, the number of principal energy levels in each succ
Diano4ka-milaya [45]

Answer:

stay the same.

Explanation:  Period 3 consists of the full 1s, 2s, and 2p electron orbitals, plus the 3s and 3p valence orbitals, which are filled with a total of 8 more electrons as we move from left (Na) to the far right (Ar):

Na:  1s2 2s2 2p6 3s1

Ar:    s2 2s2 2p6 3s2 3p6

As we move from left to right, and ignoring the already-filled 1s, 2s, and 2p orbitals, the period three starting and ending elements have the following:

Na: 3s1

Ar: 3s2, 3p6

All the new electrons electrons filled the third energy level (3s and 3p).  So the energy level does not change, just the orbitals.

5 0
3 years ago
In a simple covalent compound, which statement is TRUE
Savatey [412]

Answer:

C:to the right side of the periodic table, and it is given

the suffix -ide.

5 0
2 years ago
It takes 495.0 kJ of energy to remove 1 mole of electron from an atom on the surface of sodium metal. How much energy does it ta
Zigmanuir [339]

Answer:

\lambda=241.9\ nm

Explanation:

The work function of the sodium= 495.0 kJ/mol

It means that  

1 mole of electrons can be removed by applying of 495.0 kJ of energy.

Also,  

1 mole = 6.023\times 10^{23}\ electrons

So,  

6.023\times 10^{23} electrons can be removed by applying of 495.0 kJ of energy.

1 electron can be removed by applying of \frac {495.0}{6.023\times 10^{23}}\ kJ of energy.

Energy required = 82.18\times 10^{-23}\ kJ

Also,  

1 kJ = 1000 J

So,  

Energy required = 82.18\times 10^{-20}\ J

Also, E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,  

79.78\times 10^{-20}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}

\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{82.18\times 10^{-20}}

\lambda=\frac{10^{-26}\times \:19.878}{10^{-20}\times \:82.18}

\lambda=\frac{19.878}{10^6\times \:82.18}

\lambda=2.4188\times 10^{-7}\ m

Also,  

1 m = 10⁻⁹ nm

So,  

\lambda=241.9\ nm

6 0
3 years ago
That's just the tip of the iceberg" is a popular expression you may have heard. It means that what you can see is only a small p
puteri [66]
Answer:

B 1.23 g/cc

Explanation:
For something to float on seawater, the density must be less than 1.03 g/mL. If the object sinks, the density is greater than 1.03 g/mL.

Let’s examine the answer choices. Keep in mind, the ice berg is mostly below the water level.

A. 0.88 g/cc
This is less than 1.03 g/cc, which would result in floating.

B. 1.23 g/cc
This is the best answer choice. The iceberg is mostly beneath the water, but some of it is exposed. The density is greater than 1.03 g/mL, but not so much greater that it would immediately sink.

C. 0.23 g/cc
This is less than 1.03 g/cc, which would produce floating.

D. 4.14 g/cc
This is much greater than 1.03 g/cc and the result would be sinking.
5 0
3 years ago
Why is there not a constant molar volume for solids and
iragen [17]

Answer:

all of the above

Explanation:

i got the answer right on cK-12

7 0
2 years ago
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