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Harman [31]
3 years ago
7

A chemist heats a fixed amount of gas in a sealed glass container. Which law is the chemist most likely investigating?

Chemistry
2 answers:
PilotLPTM [1.2K]3 years ago
8 0

The law which this chemist is most likely investigating heating a fixed amount of gas in a sealed glass container is: Gay Lussac’s law.

Gay Lussac's law states that when the volume of an ideal amount of gas is held (kept) constant, the pressure of the gas is directly proportional to the absolute (Kelvin) temperature of the gas.

Mathematically, Gay Lussac's law is calculated by using the formula;

PT = K\\\\\frac{P_1}{T_1} = \frac{P_2}{T_2}

<u>Where:</u>

  • P_1 is the original pressure.
  • P_2 is the final pressure.
  • T_1 is the original temperature.
  • T_2 is the final temperature.

In conclusion, the chemist is most likely investigating how changes in temperature affect the pressure of the gas by heating a fixed amount in a sealed glass container in accordance with Gay Lussac's law.

Read more: brainly.com/question/24928274

galina1969 [7]3 years ago
4 0

Gases Unit Test Honors Chemistry A

1) B. The volume increases to twice its original value.

2) A. volume and temperature directly proportional

pressure and volume inversely proportional

pressure and temperature directly proportional

3) D. Gay-Lussac’s law, by seeing how changes in temperature affect the pressure of the gas

4) B. V1P1/T1=V2P2/T2

5) A. It is a straight line with a positive slope showing that an increase in temperature results in an increase in volume.

6) D. keeping the pressure constant and increasing the temperature

7) A. When temperature is held constant and volume increases, the pressure increases.

8) D. pressure, volume, temperature, number of moles

9) A. volume

10) A. 7.10 L/mol

11) A. 0.105 mol

12) B. 27 g/mol

13) B. the temperature increasing by a factor of 2

14) brainly.com/question/24915187

15) brainly.com/question/24544023

16)brainly.com/question/24544061

17) brainly.com/question/24598785

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Explanation:

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[1] ÷ [2]

\frac{E\times N_A}{E'\times N_A}=\frac{\frac{hc\trimes 6.022\times 10^{23} mol^{-1}}{800 \times 10^{-9} m}}{\frac{hc\times 6.022\times 10^{23} mol^{-1}}{400 \times 10^{-9} m}}

\frac{E}{E'}=\frac{1}{2}

E=\frac{1}{2}\times E'

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