The molar mass of CO2 is 44 grams per mole.
165 grams / 44 grams per mole of CO2 = 3.75 moles CO2
Using Avogadro’s law where 1 mole of substance equals
6.023 x 10^23 molecules
3.75 moles CO2 (6.023 x 10^23 molecules /mole) = 2.26 x 10^24 molecules CO2
One way of knowing that oxygen was the gas removed from the volume of air and not another is to know what the volume of air is made of first. When the composition of the volume of air is already identified, then next would be the process of separating these elements from each other and as to which is to be separated first. This would usually lead to knowing their masses, their boiling and freezing points, the temperatures at which they condense, and so on. This is to identify their differences to each other and use those differences to successfully separate those elements to each other.
Newton’s 2nd law? Maybe I’m not completely sure
<u>Answer:</u> The equilibrium concentration of bromine gas is 0.00135 M
<u>Explanation:</u>
We are given:
Initial concentration of chlorine gas = 0.0300 M
Initial concentration of bromine monochlorine = 0.0200 M
For the given chemical equation:
<u>Initial:</u> 0.02 0.03
<u>At eqllm:</u> 0.02-2x x 0.03+x
The expression of 
for above equation follows:
![K_c=\frac{[Br_2]\times [Cl_2]}{[BrCl]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BBr_2%5D%5Ctimes%20%5BCl_2%5D%7D%7B%5BBrCl%5D%5E2%7D)
We are given:
Putting values in above equation, we get:
Neglecting the value of x = -0.96 because, concentration cannot be negative
So, equilibrium concentration of bromine gas = x = 0.00135 M
Hence, the equilibrium concentration of bromine gas is 0.00135 M
We first assume that this gas is an ideal gas where it follows the ideal gas equation. The said equation is expressed as: PV = nRT. From this equation, we can predict the changes in the pressure, volume and temperature. If the volume and the temperature of this gas is doubled, then the pressure still stays the same.
