The only way to emit IR photons is if there are empty electronic energy levels really close to the ground state (lowest energy level). True or False
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Answer:
False. The statement is not entirely accurate.
Explanation:
Infrared (IR) photons can be emitted in various ways, not solely dependent on the presence of empty electronic energy levels close to the ground state. IR photons are typically associated with changes in molecular vibrational or rotational energy levels rather than electronic energy levels.
IR spectroscopy, for example, focuses on the interaction of IR radiation with molecules. When molecules undergo vibrational or rotational transitions, they can emit or absorb IR photons. These transitions are related to changes in the kinetic and potential energy of atoms within the molecule as they vibrate or rotate.
Electronic transitions, on the other hand, are responsible for phenomena like ultraviolet (UV) or visible light absorption and emission. In these cases, electrons move between electronic energy levels, which are typically at higher energy levels than the ground state. So, the emission of IR photons is not directly tied to electronic energy levels close to the ground state.
In summary, IR photon emission is primarily associated with vibrational and rotational transitions in molecules, and it is not contingent on the presence of empty electronic energy levels close to the ground state.
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Relative and average atomic mass both describe properties of an element related to its different isotopes. Out of these two Relative atomic mas is more accurate. Therefore, the amount of nitrogen in the second sample is 23.3g.
What is mass?
Mass defines the quantity of a substance. It is measured in gram or kilogram. Average mass is the mass of atoms of an element that are isotopes. It can be calculated by multiplying mass of a isotope to natural abundance of that isotope.
Average atomic mass = (mass of first isotope× percent abundance of first isotope)+(mass of second isotope× percent abundance of second isotope)
9 g of hydrogen - 42 g of nitrogen
5 g of hydrogen - x g of nitrogen
9x= 42 ×5
x=23.3g
Therefore, the amount of nitrogen in the second sample is 23.3g.
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5 g of hydrogen - x g of nitrogen
The mass of nitrogen in the second sample is 23.33 g.
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3 equations. 1: upper N upper O gas plus upper O subscript 3 gas right arrow upper N upper O subscript 2 gas plus upper O subscript 2 gas. Delta H subscript 1 equals 198.9 kilojoules. 2: StartFraction 3 over 2 EndFraction upper O gas right arrow upper O subscript 3 gas. Delta H subscript 2 equals 142.3 kilojoules. 3: upper O gas right arrow one half upper O subscript 2 gas. Delta H subscript 3 equals negative 247.5 kilojoules.
What is DeltaHrxn for the reaction shown below?
NO(g)+O(g)->NO2(g)
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Answer:
-304.1
I had this question and that’s what I got
Answer:
-867.7
Explanation:
yeah
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Answer the correct answer out of the four is option c spectroscopy
Explanation- The interaction between electromagnetic radiation and matter studied. This study is commonly known as spectroscopy. It can also be named as study of absorption spectra or emission spectra.
The former spectrum is formed when energy is absorbed from Photons or light energy by electrons while the latter spectrum is formed due to a wavelength of light that is released when electrons jump from higher to lower level.
Answer:
D spectroscopy
Explanation:
Answers
Answers
The concentration of ions in the solution is 0.0063 M
What is pH?
pH is a quantitative measure of the acidity or basicity of aqueous or other liquid solutions.
pH or pOH is the measure of acidity or alkalinity of a solution.
pOH is calculated by taking negative logarithm of hydroxide ion concentration.
pH = -log []
pH + pOH = 14
11.8 + pOH = 14
pOH = 2.2
Putting in the values:
2.2 =-log []
[] =0.0063 M
Hence, the concentration of ions in the solution is 0.0063 M
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Answer:
[OH⁻] = 6.31 × 10⁻³ M
SOlution:
Step 1: Calculate pOH as,
pH + pOH = 14
As, pH = 11.8
So,
11.8 + pOH = 14
Or,
pOH = 14 - 11.8
pOH = 2.2
Step 2: Calculate [OH⁻],
pOH is related to [OH⁻] as,
pOH = - log [OH⁻]
Putting value of pOH,
2.2 = -log [OH⁻]
Solving for [OH⁻],
[OH⁻] = 10⁻²'² ∴ 10 = Antilog
[OH⁻] = 6.31 × 10⁻³ M