Answer:
Explanation:
The lewis structure (indicating all the atoms and patterns provided as hint in the question) of glycine can be seen in the attachment below. While the chemical structure of glycine can be seen below
H
|
H₂N - C - C =O
| \
H OH
The structure (of glycine) above provides a "fair idea" of how the lewis structure will be.
Answer:
There is None
Explanation:
This is because it is a derived function dependent on other factors.
Answer:
An alkali metal present in period 2 have larger first ionization energy.
Explanation:
Ionization energy:
The amount of energy required to remove the electron from the atom is called ionization energy.
Trend along period:
As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.
Trend along group:
As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.
As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus. Thus alkali metal present in period 2 have larger ionization energy because of more nuclear attraction as compared to the alkali metal present in period 4.
The 18o-labeled methanol (CH3O*H) will appear in the products side at position b.
<h3>
Position of 18o-labeled methanol in the products</h3>
The 18O label will appear at position b in the product as indicated in the image.
This methoxy group in the product formed in position b comes from the 18O-labeled methanol (CH3OH).
While the oxygens at positions a and c in the product come from the unlabeled hemiacetal.
Thus, the 18o-labeled methanol (CH3O*H) will appear in the products side at position b.
Learn more about methanol here: brainly.com/question/17048792
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<u>Answer:</u>
<u>For A:</u> The equation is 
<u>For B:</u> The equation is 
<u>For C:</u> The equation is 
<u>Explanation:</u>
Alpha decay process is the process in which nucleus of an atom disintegrates into two particles. The first one which is the alpha particle consists of two protons and two neutrons. This is also known as helium nucleus. The second particle is the daughter nuclei which is the original nucleus minus the alpha particle released.

Beta decay process is defined as the process the neutrons get converted into an electron and a proton. The released electron is known as the beta particle. In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.

<u>For A:</u> Uranium-238 emits an alpha particle
The nuclear equation for this process follows:

<u>For B:</u> Plutonium-239 emits an alpha particle
The nuclear equation for this process follows:

<u>For C:</u> Thorium-239 emits a beta particle
The nuclear equation for this process follows:
