Answer:
c. HF can participate in hydrogen bonding.
Explanation:
<u>The boiling points of substances often reflect the strength of the </u><u>intermolecular forces</u><u> operating among the molecules.</u>
If it takes more energy to separate molecules of HF than of the rest of the hydrogen halides because HF molecules are held together by stronger intermolecular forces, then the boiling point of HF will be higher than that of all the hydrogen halides.
A particularly strong type of intermolecular attraction is called the hydrogen bond, <em>which is a special type of dipole-dipole interaction between the hydrogen atom in a polar bond</em>, such as N-H, O-H, or F-H, and an electronegative O, N, or F atom.
Answer:
The correct answer is B.
The
is samller than
of the reaction . So,the reaction will shift towards the left i.e. towards the reactant side.
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.
For the given chemical reaction:

The expression for
is written as:
![Q=\frac{[PCl_3][Cl_2]}{[[PCl_5]^1}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5B%5BPCl_5%5D%5E1%7D)


Given :
= 0.0454
Thus as
, the reaction will shift towards the left i.e. towards the reactant side.
4.92 grams / 6 mL = .82 grams / mL
<span>(A) 0.82 g/mL</span>
Answer: 5.72 g mass of carbon dioxide was produced
Explanation:

Moles of C =
Mass of
reacted = 59.1 g - 17.5 g = 41.6 g
Moles of 
According to reaction 1 mole
produces 1 mole of
then 1.3 mole of
will produce
moles of 
Mass of
:
=Moles of
× molar mass of
= 1.3 ×44 g/mol =
=57.2 grams
5.72 g mass of carbon dioxide was produced