Sedimentary rocks are formed from deposits of pre-existing rocks or pieces of once-living organism that accumulate on the Earth's surface. If sediment is buried deeply, it becomes compacted and cemented, forming sedimentary rock
hope this helps
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.
Learn more: brainly.com/question/10675485
Answer:
Since different isotopes of an element have different numbers of neutrons (but always the same number of protons) they have different mass numbers. Nitrogen-14 and nitrogen-15 are both stable isotopes of nitrogen. However, the other 5 isotopes are all unstable.
Answer:
D. oxygen atoms have twice as many protons as chlorine atoms
The number of sigma and pi bonds are,
Sigma Bonds =
16 Pi Bonds =
3Explanation: Every first bond formed between two atoms is sigma. Pi bond is formed when already a sigma bond is there. While in case of Alkyne (triple Bond) there is one sigma and one pi bond already present, so the third bond is formed by second side-to-side overlap of orbitals, hence, a second pi bond is formed.
Below all black bonds are sigma bonds, while in alkene there is one pi bond and in alkyne there are two pi bonds.
