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ch4aika [34]
3 years ago
7

What property of a solution remains the same after it is diluted?

Chemistry
2 answers:
sineoko [7]3 years ago
7 0
The property of a solution which remain the same after it is diluted is THE AMOUNT OF SOLUTE.
Dilution is the process of adding solvent to a solution. This action decreases the concentration of the solute in the solution. When a solution is diluted, the amount of solvent increase, the total volume increases, the concentration decreases, but the amount of solute stays the same.

 






Mashutka [201]3 years ago
4 0
The amount of moles of solute remains constant, although the concentration decreases because the volume is increasing
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What is the ground-state electron configuration of a neutral atom of manganese?.
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Explanation:

elctronic configuration of manganese

Mn=1s²2s²2p⁶3s²3p⁶4s²3d⁵

ground state

Mn=Ar3d⁵4s²

note that Ar is argon

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Study the solutions in the glasses. Put the solutions in order from concentrated to dilute.
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A 10,0-L cylinder of gas is stored at room temperature (20.0°C) and a pressure of 1800 psi. If the gas is
makvit [3.9K]

Considering the Charles' law, the gas would have a temperature of -109.2 C.

<h3>Charles' law</h3>

Finally, Charles' law establishes the relationship between the volume and temperature of a gas sample at constant pressure. This law says that the volume is directly proportional to the temperature of the gas. That is, if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.

Charles' law is expressed mathematically as:

\frac{V}{T} =k

If you want to study two different states, an initial state 1 and a final state 2, the following is true:

\frac{V1}{T1} =\frac{V2}{T2}

<h3>Temperature of the gas in this case</h3>

In this case, you know:

  • P1= 1800 psi
  • V1= 10 L
  • T1= 20 C= 293 K (being 0 C= 273 K)
  • P2= 1800 psi
  • V2= 6 L
  • T2= ?

You can see that the pressure remains constant, so you can apply Charles's law.

Replacing in the Charles's law:

\frac{10 L}{293 K} =\frac{6 L}{T2}

Solving:

\frac{10 L}{293 K} T2=6 L

T2=\frac{6 L}{\frac{10 L}{293 K} }

<u><em>T2=163.8 K= -109.2 C</em></u>

The gas would have a temperature of -109.2 C.

Learn more about Charles's law:

brainly.com/question/4147359?referrer=searchResults

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I’m not completely sure
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