Answer:
The gas argon does not reach a state of vibrational excitation when infrared radiation strikes this gas.
Explanation:
The dry atmosphere is composed almost entirely of nitrogen (in a volumetric mixing ratio of 78.1%) and oxygen (20.9%), plus a series of oligogases such as argon (0.93%), helium and gases of greenhouse effect such as carbon dioxide (0.035%) and ozone. In addition, the atmosphere contains water vapor in very variable amounts (about 1%) and aerosols.
Greenhouse gases or greenhouse gases are the gaseous components of the atmosphere, both natural and anthropogenic, that absorb and emit radiation at certain wavelengths of the infrared radiation spectrum emitted by the Earth's surface, the atmosphere and clouds . In the Earth's atmosphere, the main greenhouse gases (GHG) are water vapor (H2O), carbon dioxide (CO2), nitrous oxide (N2O), methane (CH4) and ozone (O3 ). There is also in the atmosphere a series of greenhouse gases (GHG) created entirely by humans, such as halocarbons (compounds containing chlorine, bromine or fluorine and carbon, these compounds can act as potent greenhouse gases in the atmosphere and they are also one of the causes of the depletion of the ozone layer in the atmosphere) regulated by the Montreal Protocol. In addition to CO2, N2O and CH4, the Kyoto Protocol sets standards regarding sulfur hexafluoride (SF6), hydrofluorocarbons (HFCs) and perfluorocarbons (PFCs).
The difference between argon and greenhouse gases such as CO2 is that the individual atoms in the argon do not have free bonds and therefore do not vibrate. As a consequence, it does not reach a state of vibrational excitation when infrared radiation strikes this gas.
Answer:
0.924 g
Explanation:
The following data were obtained from the question:
Volume of CO2 at RTP = 0.50 dm³
Mass of CO2 =?
Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:
1 mole of gas = 24 dm³ at RTP
Thus,
1 mole of CO2 occupies 24 dm³ at RTP.
Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e
Xmol of CO2 = 0.5 /24
Xmol of CO2 = 0.021 mole
Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.
Finally, we shall determine the mass of CO2 as follow:
Mole of CO2 = 0.021 mole
Molar mass of CO2 = 12 + (2×16) = 13 + 32 = 44 g/mol
Mass of CO2 =?
Mole = mass /Molar mass
0.021 = mass of CO2 /44
Cross multiply
Mass of CO2 = 0.021 × 44
Mass of CO2 = 0.924 g.
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Answer
is: volume is 20 mL.<span>
c</span>₁(CH₃COOH) = 2,5 M.<span>
c</span>₂(CH₃COOH) = 0,5 M.<span>
V</span>₂(CH₃COOH) = 100 mL.<span>
V</span>₁(CH₃COOH) = ?<span>
c</span>₁(CH₃COOH) · V₁(CH₃COOH)
= c₂(CH₃COOH) · V₂(CH₃COOH).<span>
2,5 M · V</span>₁(CH₃COOH)
= 0,5 M · 100 mL.<span>
V</span>₁(CH₃COOH) = 0,5 M · 100 mL ÷ 2,5 M.<span>
V</span>₁(CH₃COOH) = 20 mL ÷ 1000 mL/L =0,02 L.
I believe the answer is
At the moment it is the best way of explaining our scientific knowledge.
The volume of a gas that occupies 9 L at a temperature of 325K is 12.46L.
<h3>How to calculate volume?</h3>
The volume of a given gas can be calculated using the following Charle's law equation:
V1/T1 = V2/T2
Where;
- T1 = initial temperature
- T2 = final temperature
- V1 = initial volume
- V2 = final volume
- V1 = 9L
- V2 = ?
- T1 = 325K
- T2 = 450K
9/325 = V2/450
325V2 = 4050
V2 = 4050/325
V2 = 12.46L
Therefore, the volume of a gas that occupies 9 L at a temperature of 325K is 12.46L.
Learn more about volume at: brainly.com/question/2817451
