<h3><u>Answer;</u></h3>
Dipole-dipole and hydrogen bonding
<h3><u>Explanation;</u></h3>
- <em><u>A solution of water and ethanol contains the dipole-dipole forces and hydrogen bonds as the intermolecular forces between molecules.</u></em>
- <em><u>Hydrogen bonding is a type of interactions between molecules that occurs when a partially negative atom such as oxygen end of one of the molecules is attracted to a partially positive hydrogen end of another molecule.</u></em>
- <em><u>Dipole-dipole forces</u></em> results from the unsymmetrical distribution of electrons, thus the polarity does not balance, thus resulting to a dipole attraction between molecules.
NH₃:
N = 8*10²²
NA = 6.02*10²³
n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol
O₂:
N=7*10²²
NA = 6.02*10²³
n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol
4NH₃ <span>+ 3O</span>₂ ⇒<span> 2N</span>₂<span> + 6H</span>₂<span>O
</span>4mol : 3mol : 2mol
0.133mol : 0.116mol : 0,0665mol
limiting reactant
N₂:
n = 0.0665mol
M = 28g/mol
m = n*M = 0.0665mol*28g/mol = <u>1,862g</u>
By decreasing n we can increase presure because decrease in n will shift equilibrium to either forward or reverse direction
Water vapor has more energy because it is a gas :)
