The electronic transition that will produce the lowest frequency is an electron falling from the 3rd to the 2nd energy level.
The question is incomplete, the complete question is;
As electrons fall from high energy orbitals to lower orbitals, energy is released in the form of electromagnetic radiation. The farther the electron falls, the more energy is released. Which of the following electronic transitions would produce a wave with the lowest frequency?
an electron falling from the 6th to the 2nd energy level
an electron falling from the 5th to the 2nd energy level
an electron falling from the 3rd to the 2nd energy level
an electron jumping from the 1st to the 2nd energy level
According to Bohr's theory, energy is absorbed or emitted when an electron moves from one energy level to another. This energy often occurs as visible light of known frequency and wavelength.
The magnitude of frequency of light depends on the difference in energy between the two energy levels. If the difference between the energy levels is high, the frequency of light is also high and vice versa.
The transition from 3rd to the 2nd energy level represents a low frequency transition because the energy levels are close together.
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Answer:
Diphosphorus pentoxide
Carbon dichloride
BCl3
N2H4
Explanation:
These are all covalent compounds. To name covalent compounds, you add prefixes to the beginning of their names depending on what the subscript is of each element. The prefixes are:
1: Mono
2: Di
3: Tri
4: Tetra
5: Penta
6: Hexa
7: Hepta
8: Octa
9: Nona
10: Deca
For example, since the first one is Phopsphorus with a 2 next to it, you add the prefix Di to it.
If the first element in the compound only has one, meaning no number next to it, you do not say mono. This is why we just say "Carbon" for the second one instead of "Monocarbon."
Finally, you always have to end the second element in the compound with "ide." So, "chlorine" becomes "chloride," "oxygen" becomes "oxide," and so on.
The volume (in mL) of 0.242 M NaOH solution needed for the titration reaction is 39.44 mL
<h3>Balanced equation </h3>
CH₃CH₂COOH + NaOH —> CH₃CH₂COONa + H₂O
From the balanced equation above,
- The mole ratio of the acid, CH₃CH₂COOH (nA) = 1
- The mole ratio of the base, NaOH (nB) = 1
<h3>How to determine the volume of NaOH</h3>
- Volume of acid, CH₃CH₂COOH (Va) = 46.79 mL
- Molarity of acid, CH₃CH₂COOH (Ma) = 0.204 M
- Molarity of base, NaOH (Mb) = 0.242 M
- Volume of base, KOH (Vb) =?
MaVa / MbVb = nA / nB
(0.204 × 46.79) / (0.242 × Vb) = 1
Cross multiply
0.242 × Vb = 0.204 × 46.79
Divide both side by 0.242
Vb = (0.204 × 46.79) / 0.242
Vb = 39.44 mL
Thus, the volume of NaOH needed for the reaction is 39.44 mL
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