Answer:
[Ne]3s2
Explanation:
ahora tenemos que mirar cada una de las configuraciones electrónicas de cada átomo de cerca antes de tomar una decisión.
considerando la configuración electrónica más externa de cada una de las especies mostradas;
para la primera configuración, ns2 np6 corresponde a un gas noble.
para la segunda configuración ns2 np3 corresponde a un elemento no metálico del grupo 5.
para la tercera configuración, ns2 corresponde a un elemento metálico del grupo 2.
para la cuarta configuración, ns2 np4 corresponde a un elemento no metálico del grupo 6
The answer is an igneous rock.
Hope this helps!!
<span>1. Na
2. Ni
3. Na2S
4. Electrolytic Cell (Galvanic and Voltaic are synonyms, both are electric cells without an external power source.
5. 3
6. The first one
7. Na
8. Reducing agents are oxidized
hope this helps you
</span>
The intermolecular forces that are responsible for the dissolution of Ethylene glycol in water is hydrogen bonding dipole-dipole forces and dispersion forces.
Both ethylene glycol and water contains the pair of hydrogen and oxygen.
The hydrogen of one atom create a bond with the oxygen of other atom this results in the formation of intra molecular hydrogen bonding.
The electron are non uniformly distributed over the molecule or the atom which results in the fluctuation of the electron density in the atom.
So it creates are dispersion forces which is present all over the molecule this forces helps to increase the strength of the bond formed between the ethylene glycol and water because they have large masses.
Both ethylene glycol and water are polar molecules because of being polar they form dipole and the dipole of both the molecules interact with each other in order to form bond between the atoms which eventually results in the formation dissolution of ethylene glycol in water.
To know more about intermolecular forces, visit,
brainly.com/question/2193457
#SPJ4
Compete Question - which intermolecular forces are responsible for the dissolution of ethylene glycol? select all that apply hydrogen bonding, dipole-dipole, dispersion and Ion dipole interaction.
Answer:
Amount of pyridine required = 0.0316 M
Explanation:
pH of a buffer solution is calculated by using Henderson - Hasselbalch equation.
![pH=pK_a+log\frac{[Conjugate\ base]}{[weak\ acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2Blog%5Cfrac%7B%5BConjugate%5C%20base%5D%7D%7B%5Bweak%5C%20acid%5D%7D)
Pyridinium is a weak acid and in the presence of its conjugate base, it acts as buffer.
Henderson - Hasselbalch equation for pyridine/pyridinium buffer is as follows:
![pH=pK_a+log\frac{[Py]}{PyH^+]}](https://tex.z-dn.net/?f=pH%3DpK_a%2Blog%5Cfrac%7B%5BPy%5D%7D%7BPyH%5E%2B%5D%7D)
pH = 4.7
(Pyridinium)=0.100 M
Substitute the values in the formula
![pH=pK_a+log\frac{[Py]}{PyH^+]}\\4.7=5.2 log\frac{[Py]}{0.100}](https://tex.z-dn.net/?f=pH%3DpK_a%2Blog%5Cfrac%7B%5BPy%5D%7D%7BPyH%5E%2B%5D%7D%5C%5C4.7%3D5.2%20log%5Cfrac%7B%5BPy%5D%7D%7B0.100%7D)
![4.7-5.2=log\frac{[Py]}{0.100} \\-0.5=log\frac{[Py]}{0.100}\\\frac{[Py]}{0.100}=antilog -0.5\\\frac{[Py]}{0.100}=0.316](https://tex.z-dn.net/?f=4.7-5.2%3Dlog%5Cfrac%7B%5BPy%5D%7D%7B0.100%7D%20%5C%5C-0.5%3Dlog%5Cfrac%7B%5BPy%5D%7D%7B0.100%7D%5C%5C%5Cfrac%7B%5BPy%5D%7D%7B0.100%7D%3Dantilog%20-0.5%5C%5C%5Cfrac%7B%5BPy%5D%7D%7B0.100%7D%3D0.316)
![\frac{[Py]}{0.100} =0.316](https://tex.z-dn.net/?f=%5Cfrac%7B%5BPy%5D%7D%7B0.100%7D%20%3D0.316)
[Py]=0.0316\ M
Amount of pyridine required = 0.0316 M
