The concentration of
in the stack gas = 12 ppmv
That means 12 L of
is present per ![10^{6} L gas](https://tex.z-dn.net/?f=%2010%5E%7B6%7D%20L%20gas%20)
The given temperature is 273 K (0 C) and pressure is 1 atm. At these conditions, 1 mol of gas would occupy,
![PV = nRT](https://tex.z-dn.net/?f=%20PV%20%3D%20nRT%20)
![(1 atm) (V) = (1 mol)(0.08206\frac{L.atm}{mol.K}) (273 K)](https://tex.z-dn.net/?f=%20%281%20atm%29%20%28V%29%20%3D%20%281%20mol%29%280.08206%5Cfrac%7BL.atm%7D%7Bmol.K%7D%29%20%28273%20K%29%20)
V = 22.4 L
1 mol
occupies 22.4 L
Moles of
= ![12 L *\frac{1 mol}{22.4 L} = 0.5357 mol SO_{2}](https://tex.z-dn.net/?f=%2012%20L%20%2A%5Cfrac%7B1%20mol%7D%7B22.4%20L%7D%20%3D%200.5357%20mol%20SO_%7B2%7D%20%20%20%20)
Mass of
=
=
μg
Converting
:
= ![10^{3} m^{3}](https://tex.z-dn.net/?f=%2010%5E%7B3%7D%20%20%20m%5E%7B3%7D%20%20%20)
Calculating the concentration in μg/
:
![\frac{3.432 * 10^{7} microgram}{10^{3} L} = 3.432 * 10^{4} microgram/m^{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3.432%20%2A%2010%5E%7B7%7D%20microgram%7D%7B10%5E%7B3%7D%20L%7D%20%20%20%20%3D%203.432%20%2A%2010%5E%7B4%7D%20%20microgram%2Fm%5E%7B3%7D%20%20)
B.
its the only one with ozone in the reaction O₃
Answer:
The amount of glue used, the material and how much of it you're using every time you test each glue, the temperature of the room, etc.
Explanation:
There are many answers to this question but think about the obvious ones first. Obviously you'll need to use the same amount of glue every time you're testing each type. The material used is also important and how much of the material you're using because different types of glue can work better with different materials, as well as how much material you're using. The temperature of the room can affect how well each glue holds up because often glue dries easier/quicker with heat, so in your experiment to make it fair you need to make sure that each glue is tested in the same environment.
Answer:
![r=\frac{r(NO_{2})}{2} =\frac{r(O_{2})}{1/2}=\frac{r(N_{2}O_{5})}{1}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Br%28NO_%7B2%7D%29%7D%7B2%7D%20%3D%5Cfrac%7Br%28O_%7B2%7D%29%7D%7B1%2F2%7D%3D%5Cfrac%7Br%28N_%7B2%7DO_%7B5%7D%29%7D%7B1%7D)
Explanation:
Let us consider the reaction:
2 NO₂ + 1/2 O₂ ⇄ N₂O₅
The rate of formation of a substance is equal to the change in concentration of the product divided the change in time:
![r(N_{2}O_{5})=\frac{\Delta [N_{2}O_{5}] }{\Delta t}](https://tex.z-dn.net/?f=r%28N_%7B2%7DO_%7B5%7D%29%3D%5Cfrac%7B%5CDelta%20%5BN_%7B2%7DO_%7B5%7D%5D%20%7D%7B%5CDelta%20t%7D)
The rate of disappearance of a reactant is equal to to the change in concentration of the reactant divided the change in time, with a negative sign so that the rate is always a positive variable.
![r(NO_{2})=-\frac{\Delta[NO_{2}] }{\Delta t}](https://tex.z-dn.net/?f=r%28NO_%7B2%7D%29%3D-%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%20%7D%7B%5CDelta%20t%7D)
![r(O_{2})=-\frac{\Delta[O_{2}] }{\Delta t}](https://tex.z-dn.net/?f=r%28O_%7B2%7D%29%3D-%5Cfrac%7B%5CDelta%5BO_%7B2%7D%5D%20%7D%7B%5CDelta%20t%7D)
The rate of the reaction is equal to the rate of any substance divided its stoichiometric coefficient. In this way, we can relate these expressions:
![r=\frac{r(NO_{2})}{2} =\frac{r(O_{2})}{1/2}=\frac{r(N_{2}O_{5})}{1}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Br%28NO_%7B2%7D%29%7D%7B2%7D%20%3D%5Cfrac%7Br%28O_%7B2%7D%29%7D%7B1%2F2%7D%3D%5Cfrac%7Br%28N_%7B2%7DO_%7B5%7D%29%7D%7B1%7D)
Answer:
We will Multiply the number of moles with Avagadro's number
So it will be (2.04×1023)×6.022×10^23
<h3>or 2.04×1023 NA</h3>
NA means Avagadro's number