Answer:
Electron-pair geometry: tetrahedral
Molecular geometry: trigonal pyramidal
Hybridization: sp³
sp³ - 4 p
Explanation:
There is some info missing. I think this is the original question.
<em>For NBr₃, What are its electron-pair and molecular geometries? What is the hybridization of the nitrogen atom? What orbitals on N and Br overlap to form bonds between these elements?</em>
<em>The N-Br bonds are formed by the overlap of the ___ hybrid orbitals on nitrogen with ___ orbitals on Br.</em>
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Nitrogen is a central atom surrounded by 4 electron domains. According to VESPR, the corresponding electron-pair geometry is tetrahedral.
Of these 4 electron domains, 3 represent covalent bonds with Br and 1 lone pair. According to VESPR, the corresponding molecular geometry is trigonal pyramidal.
In the nitrogen atom, 1 s orbital and 3 p orbitals hybridize to form 4 sp³ orbitals for each of the electron domains.
The N-Br bonds are formed by the overlap of the sp³ hybrid orbitals on nitrogen with 4p orbitals on Br.
A crushed garlic will have a lot of flavor when placed in food due to the surface area that is in contact with the food. When we have a large piece of garlic, only the external part touches the food and its full capacity is not used. When we reduce the size of the year by crushing the internal parts that were not in contact with the food, now they will be, in addition, liquids are also released due to the pressure exerted on the garlic and these liquids mix more easily with the food and they give it more flavor. For better understanding we can see the following figure:
Simply to understand it, in the figure, there is a clove of whole garlic represented by the rectangle that will have a height of 3 and a width of 1, the units do not matter in this case. The area that is in contact will be equal to 8, but if we divide the garlic into three equal parts, it will have a contact area greater than 12. Therefore, the more we divide the garlic, the more area it will be in contact with the food and will give it more flavor.
Explanation:
1.
Given parameters:
Frequency of the radiation = 8.4 x 10¹⁴Hz
Unknown:
Energy of the wave = ?
Solution:
The energy of a wave is given by the expression below;
E = hf
E is the energy
h is the Planck's constant = 6.63 x 10⁻³⁴m²kg/s
f is the frequency
Now insert the parameters and solve;
E = 6.63 x 10⁻³⁴m²kg/s x 8.4 x 10¹⁴Hz
E = 5.57 x 10¹ x 10⁻²⁰J
E = 5.57 x 10⁻¹⁹J
2.
Given parameters:
Wavelength = 2.13 x 10⁻¹³m
Unknown:
Frequency of the wave = ?
Solution:
The frequency of a wave can be determined using the expression;
C = f∧
C is the speed of light = 3 x 10⁸m/s
f is the frequency
∧ is the wavelength
f = = = 1.41 x 10²¹hz
This type of formula is use in chemistry it has Iron ,Sodium, Hydroxyl and Chloride unit s in the equation.
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