Answer:
C) hydrogen bonding
Explanation:
All atoms and molecules have London Dispersion Forces between them, but they are usually overshadowed but the much stronger forces. In this scenario the major attractive force in HF molecules are hydrogen bonds. Hydrogen bonds are electrostatic forces of attraction found when Hydrogen is bonded to a more electronegative atom such as Oxygen, Chlorine and Fluorine.
 
        
             
        
        
        
A catalyst will ALWAYS increase the reaction rate so C
        
             
        
        
        
Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb
 
        
             
        
        
        
Physical isn't so chemical
        
                    
             
        
        
        
The correct answer is option 3. The IUPAC name is Iron(II) sulfide. It is the less stable amorphous form.  When this is powdered, it is pyrophoric or it ignites spontaneously in air. It readily reacts with hydrochloric acid producing hydrogen sulfide.