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3 years ago

What are the different phases of matter and their characteristics?

Chemistry

2 answers:

Nutka1998 [239]3 years ago

5 0

Solids, liquids, gases and plasma. The fifth state is the man-made Bose-Einstein condensates

Novay_Z [31]3 years ago

3 0

Answer:

The five phases of matter. There are four natural states of matter: Solids, liquids, gases and plasma. The fifth state is the man-made Bose-Einstein condensates. In a solid, particles are packed tightly together so they don't move much.

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How you could use a microscope to determine if a simple was either a prokaryote or a eukaryote

Black_prince [1.1K]

You could check if their cells had a nucleus or not. If it did it would be a eukaryote, if it did not then it would be a prokaryote.

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3 years ago

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When the temperature of a 3. 0-l sample of a gas is dropped from 200°c to 100°c, what will be the final volume of the gas sample

ipn [44]

P1V1T1=P2V2T2 Add 273 to convert degrees Celsius to Kelvin:

∴200×25/298=250×V2/273, ∴V2=200×25×273/298×250, ∴V2=18.32L

<h3>Where is the volume equation?</h3>

The basic formula for volume is length, breadth, and height, as opposed to length, width, and height for the area of a rectangular shape. The calculation is unaffected by how you refer to the various dimensions; for instance, you can use "depth" instead of "height."

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1 year ago

Does the Caribbean have high or low viscosity lava?

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3 years ago

What happens to energy from the sun that is neither reflected nor absorbed by the atmosphere

Valentin [98]

Energy from the sun that is neither reflected nor absorbed by the atmosphere passes through the atmosphere to the surface. The ozone layer absorbes most of the ultraviolet radiation, water vapor, and carbon dioxide absorbs infared radiation, clouds, dust, and other gases also absorb energy.

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3 years ago

1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.

Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice ) = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

$\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}$

$\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}$

$\Delta \ S = {(1*37.7)In \dfrac{263}{273}$

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0

3 years ago