Answer: Dipole-dipole forces: electrostatic interactions of permanent dipoles in molecules; includes hydrogen bonding.
Explanation:
https://courses.lumenlearning.com/introchem/chapter/dipole-dipole-force/- More on the Topic not sure if this will help though?
<u>Answer:</u> The osmotic pressure is 54307.94 Torr.
<u>Explanation:</u>
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

where,
= osmotic pressure of the solution = ?
i = Van't hoff factor = 3
C = concentration of solute = 0.958 M
R = Gas constant = 
T = temperature of the solution = ![30^oC=[30+273]K=303K](https://tex.z-dn.net/?f=30%5EoC%3D%5B30%2B273%5DK%3D303K)
Putting values in above equation, we get:

Hence, the osmotic pressure is 54307.94 Torr.
Answer:
2) HClO3 is stronger because chlorine is more electronegative than iodine.
Explanation:
The more electronegative the element is the more strong or acidic it becomes.
Chlorine being more electronegative than Iodine makes it easier for it to pull the electron of hydrogen more strongly and hence has a higher tendency to release a H+ unit. Hence that makes it stronger.
Answer:
0.641 moles of ethane
Explanation:
Based on the equation:
C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)
We can determine ΔH of reaction using Hess's law. For this equation:
<em>Hess's law: ΔH products - ΔH reactants</em>
ΔH = {2ΔHCO2 + 3ΔHH2O} - {ΔHC2H6}
<em>Pure monoatomic substances have a ΔH = 0kJ/mol; ΔHO2 = 0kJ/mol</em>
<em />
ΔH = {2*-393.5kJ/mol + 3*-285.8kJ/mol} - {-84.7kJ/mol}
ΔH = -1559.7kJ/mol
That means when 1 mole of ethane is in combustion there are released 1559.7kJ of heat. To produce 1.00x10³kJ there are needed:
1.00x10³kJ * (1mole ethane / 1559.7kJ) =
<h3>0.641 moles of ethane</h3>
Answer:
The pH of a solution with a (H+] = 0.80 M is 
Explanation:
As we know
pH = -[log H+]
Substituting the value of H+ ion concentration in the above equation, we get -
pH = -log [0.80]
pH ![= -[-0.096] = 0.096 = 9.6 e^{-2}](https://tex.z-dn.net/?f=%3D%20-%5B-0.096%5D%20%3D%200.096%20%20%3D%209.6%20e%5E%7B-2%7D)