There is an error in the first sentence of the question; the right format is:
Suppose a 500.mL flask is filled with 1.9mol of NO3 and 1.6mol of NO2.
It should be NO2 and not NO.
Answer:
The equilibrium molarity of NO = 0.21695 m
Explanation:
Given that :
the volume = 500 mL = 0.500 m
number of moles of 
number of moles of 
Then we can calculate for their respectively concentrations as :
![[NO_3] = \frac{number \ of \ moles}{volume}](https://tex.z-dn.net/?f=%5BNO_3%5D%20%3D%20%5Cfrac%7Bnumber%20%5C%20of%20%5C%20moles%7D%7Bvolume%7D)
![[NO_3] = \frac{1.9}{0.500}](https://tex.z-dn.net/?f=%5BNO_3%5D%20%3D%20%5Cfrac%7B1.9%7D%7B0.500%7D)
![[NO_3] = 3.8 \ M](https://tex.z-dn.net/?f=%5BNO_3%5D%20%3D%203.8%20%5C%20M)
![[NO_2] = \frac{number \ of \ moles}{volume}](https://tex.z-dn.net/?f=%5BNO_2%5D%20%3D%20%5Cfrac%7Bnumber%20%5C%20of%20%5C%20moles%7D%7Bvolume%7D)
![[NO_2] = \frac{}{} \frac{1.6}{0.500}](https://tex.z-dn.net/?f=%5BNO_2%5D%20%3D%20%5Cfrac%7B%7D%7B%7D%20%5Cfrac%7B1.6%7D%7B0.500%7D)
![[NO_2] = 3.2 \ M](https://tex.z-dn.net/?f=%5BNO_2%5D%20%3D%203.2%20%5C%20M)
The chemical reaction can be written as:

The ICE table is as follows;

Initial 3.8 - 3.2
Change +x x -2x
Equilibrium 3.8+x +x 3.2 - 2x
![K_c=\frac{[NO_2]^2}{[NO_3][NO]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BNO_3%5D%5BNO%5D%7D)



Using quadratic formula;

= 
= 0.21695 OR -3.7283
Going by the positive value;
x = 0.21695
![[NO_3] = 3.8 +x = 3.8 + 0.21695](https://tex.z-dn.net/?f=%5BNO_3%5D%20%3D%203.8%20%2Bx%20%20%3D%203.8%20%2B%200.21695)
= 4.01695 m
[NO] = x = 0.21695 m
![[NO_2] = 3.2 +x = 3.2 + 0.21695](https://tex.z-dn.net/?f=%5BNO_2%5D%20%3D%203.2%20%2Bx%20%20%3D%203.2%20%2B%200.21695)
= 3.41695 m
Assume that your electrical power company gets its energy from a hydroelectric dam. Outline all of the energy changes that occurred from the falling water that turned the generators at the dam to the hot air that was produced from your hair dryer this morning.
By using this formula of vapor pressure:
Pv(solu)= n Pv(water)
when we have Pv(solu)=231.16 torr & Pv(water)= 233.7 torr
from this formula, we can get n (mole fraction of water) by substitution:
231.16 = n * 233.7
∴ n(mole fraction of water) = 0.99
so mole fraction of solution = 1 - 0.99 = 0.01
when no.of moles of water = mass weight / molar weight
= 365g / 18g/mol = 20 moles
Total moles in solution = moles of water / mole fraction of water
= 20 / 0.99 =20.2
no. of moles of the solution= total moles in solution- moles of water
= 20.2 - 20 = 0.2 moles
when we assumed the mass weight of the solution = 16 g (missing in your question should be given)
∴ molar mass = mass weight of solute / no. of moles of solute
= 16 g / 0.2 mol = 80 g/mol
I want to say addition. But I have a tendency to be wrong
Answer:
Precautions
Explanation:(Grammar Check)