Answer:
Ethanol is completely miscible due to <u><em>presence</em></u> of Hydrogen bonding.
Ethanethiol is partially miscible due to <u><em>absence</em></u> of Hydrogen Bonding.
Explanation:
The miscibility of liquids depend upon the intermolecular interactions between the two liquids. The stronger the intermolecular interactions the more miscible will be the liquids.
Among the two given examples, Ethanol is more miscible in water because it exhibits hydrogen bonding which is considered the strongest intermolecular interaction. Hydrogen bonding occurs when the hydrogen atom is bonded to more electronegative atoms like Fluorine, Oxygen and Nitrogen. In this way the hydrogen atom gets partial positive charge and the electronegative atom gets partial negative charge. Hence, these partial charges results in attracting the opposite charges on other surrounding atoms.
While, in case of Ethanethiol the hydrogen atom is not bonded to any high electronegative atom hence, there will be no hydrogen bonding and therefore, there will be less interactions between the neighbour atoms.
I believe it is 3 maybe
Hope I got it right
An atom that has 13 protons and 15 neutrons is isotope of Aluminium (answer C)
<u><em>Explanation</em></u>
- Isotope is a form of the same element with the equal number of protons but difference number of neutrons in their nuclei.
- In other words isotope has the same atomic number but different mass number.
- Atomic number of a element is determined by number of protons of an element.
- from the periodic table Aluminum in atomic number 13 therefore it has 13 protons <em>therefore an atom that has 13 protons and 15 neutrons is a isotope of Aluminium. </em>
Answer:
Don't post any question if isn't related to the topic or to your homework or assignment.
Explanation:
Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :
[butane]=1.0 M , [isobutane]=2.5 M
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?
Answer:
The equilibrium concentration of each gas:
[Butane] = 1.14 M
[isobutane] = 2.86 M
Explanation:
Butane ⇄ Isobutane
At equilibrium
1.0 M 2.5 M
After addition of 0.50 M of butane:
(1.0 + 0.50) M -
After equilibrium reestablishes:
(1.50-x)M (2.5+x)
The equilibrium expression will wriiten as:
![K_c=\frac{[Isobutane]}{[Butane]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BIsobutane%5D%7D%7B%5BButane%5D%7D)
x = 0.36 M
The equilibrium concentration of each gas:
[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M
[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M
