Answer:
Moles of NO₂ = 0.158
Explanation:
SO 2 ( g ) + NO 2 ( g ) ⇄ SO 3 ( g ) + NO ( g )
According to the law of mass equation
= ![\frac{[SO_{3} ][NO]}{[SO_{2}][NO_{2} ]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BSO_%7B3%7D%20%5D%5BNO%5D%7D%7B%5BSO_%7B2%7D%5D%5BNO_%7B2%7D%20%20%5D%7D)
⇒ 3.10 =
At equilibrium [SO₃] = [NO]
⇒ [NO₂] = 
⇒ [NO₂] = 0.158
So. number of moles of NO₂ at equilibrium added = 0.158
<u>Answer:</u> The moles of oxygen and carbon dioxide in air is
and
respectively
<u>Explanation:</u>
To calculate the number of moles, we use the equation:

Given mass of atmosphere = 
Average molar mass of atmosphere = 28.96 g/mol
Putting values in above equation, we get:

We know that:
Percent of oxygen in air = 21 %
Percent of carbon dioxide in air = 0.0415 %
Moles of oxygen in air = 
Moles of carbon dioxide in air = 
Hence, the moles of oxygen and carbon dioxide in air is
and
respectively
0.114 mol/l
The equilibrium equation will be:
Kc = ([Br2][Cl2])/[BrCl]^2
The square factor for BrCl is due to the 2 coefficient on that side of the equation.
Now solve for BrCl, substitute the known values and calculate.
Kc = ([Br2][Cl2])/[BrCl]^2
[BrCl]^2 * Kc = ([Br2][Cl2])
[BrCl]^2 = ([Br2][Cl2])/Kc
[BrCl] = sqrt(([Br2][Cl2])/Kc)
[BrCl] = sqrt(0.043 mol/l * 0.043 mol/l / 0.142)
[BrCl] = sqrt(0.001849 mol^2/l^2 / 0.142)
[BrCl] = sqrt(0.013021127 mol^2/l^2)
[BrCl] = 0.114110152 mol/l
Rounding to 3 significant figures gives 0.114 mol/l
Photosynthesis is the ability for a plant to turn light into energy, thus feeding it