Answer:
Question 1
C) polarizability
Question 2
C) London dispersion forces
Question 3:
D)Kr
Question 4:
E) strong enough to hold molecules relatively close together but not strong enough to keep molecules from moving past each other
Explanation:
It is known that equation for ideal gas is as follows.
PV = nRT
The given data is as follows.
Pressure, P = 1500 psia, Temperature, T =
= 104 + 460 = 564 R
Volume, V = 2.4 cubic ft, R = 10.73 
Also, we know that number of moles is equal to mass divided by molar mass of the gas.
n = 
m = 
=
= 9.54 lb
Hence, molecular weight of the gas is 9.54 lb.
- We will calculate the density as follows.
d = 
=
= 3.975 
- Now, calculate the specific gravity of the gas as follows.
Specific gravity relative to air =
= 
= 51.96
Explanation:
Let us assume that the value of
= 
Also at 1500 K,
= 

Relation between
and
is as follows.

Putting the given values into the above formula as follows.



Also, 
or, 
= 
= 
Thus, we can conclude that the value of
is
.
Answer:
See explaination
Explanation:
The electrons geometry shows the special distribution of the electrons around of the central atom of the molecule.
The molecular geometry shows the special distribution of the atoms that form the molecule.
Please kindly check attachment for further solution.
Answer:
Percent Composition
1. Find the molar mass of all the elements in the compound in grams per mole.
2. Find the molecular mass of the entire compound.
3. Divide the component's molar mass by the entire molecular mass.
4. You will now have a number between 0 and 1. Multiply it by 100% to get percent composition.