Answer:
b) Ion-dipole
Explanation:
Intermolecular forces are the forces of attraction or repulsion between molecules, they are significantly weaker than intramolecular forces like covalent or ionic bonds.
- <em>Hydrogen bonds</em> happen between a partially positively charged hydrogen and another partially negatively charged, it's a type of dipole-dipole interaction, one of the strongest among intermolecular forces.
- <em>Ion-dipole</em> involves an ion and polar molecule, its strength is proportional to the charge of the ion. It's stronger than hydrogen bonds because the ion and the polar molecule align so positive and negative charges are next to another allowing maximum attraction.
- <em>Dipole-dipole </em>is an interaction between two molecules that have permanent dipoles, aligning to increase attraction.
- <em>Ion-dipole</em> induced usually happens when a non-polar molecule interacts with an ion causing the molecule to be temporary partially charged. It's a weaker interaction.
- <em>Dipole- Induced Dipole</em>, like ion-dipole induced this interaction causes one of the two involved molecules to be temporary partially charged.
Considering this information we can conclude that Ion-Dipole interaction is the strongest force among intermolecular forces.
I hope this information is useful to you!
Answer:
20.(45)L or about 20.4545L
Explanation:
PV = nRT
Where:
P - pressure
V - volume
n - number of particle moles
R - a constant
T - temperature in K
We can assume the P and n (and definitely R) stay the same, so we infer that
Answer:
Water moves from the ground or oceans into the atmosphere through a process called evaporation. It's a process that happens on a molecular level when the molecules of water are really energized and rise into the air. Now you've got water in the air and water on land. Organisms all over the Earth need water to survive.
Explanation:
Answer:
Explanation:
Hello there!
In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:
![HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}](https://tex.z-dn.net/?f=HA%5Crightleftharpoons%20H%5E%2B%2BA%5E-%5C%5C%5C%5CKa%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:
![Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7Bx%5E2%7D%7B%5BHA%5D_0-x%7D%20%3D10%5E%7B-4.74%7D%3D1.82x10%5E%7B-5%7D)
Thus, it is possible to find x given the pH as shown below:
So that we can calculate the initial concentration of the acid:
![\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\](https://tex.z-dn.net/?f=%5Cfrac%7B%281.82x10%5E%7B-5%7D%29%5E2%7D%7B%5BHA%5D_0-1.82x10%5E%7B-5%7D%7D%20%3D1.82x10%5E%7B-5%7D%5C%5C%5C%5C%5Cfrac%7B1.82x10%5E%7B-5%7D%7D%7B%5BHA%5D_0-1.82x10%5E%7B-5%7D%7D%20%3D1%5C%5C%5C%5C)
![[HA]_0=3.64x10^{-5}M](https://tex.z-dn.net/?f=%5BHA%5D_0%3D3.64x10%5E%7B-5%7DM)
Therefore, the percent dissociation turns out to be:
![\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%](https://tex.z-dn.net/?f=%5C%25%20diss%3D%5Cfrac%7Bx%7D%7B%5BHA%5D_0%7D%2A100%5C%25%20%5C%5C%5C%5C%5C%25%20diss%3D%5Cfrac%7B1.82x10%5E%7B-5%7DM%7D%7B3.64x10%5E%7B-5%7DM%7D%2A100%5C%25%20%5C%5C%5C%5C%5C%25%20diss%20%3D%2050%5C%25)
Best regards!
