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)
Answer:
Prokaryotic is the answer!
Explanation:
I know this because, the nucleolus is absent in the image.
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They give information about wheatger patterns which th use to pedict the wheather
Answer:
11.43g of Aluminum Hydroxide
Explanation:
Since we know that the sulfuric acid is the limiting reactant in this chemical reaction, we know that we are going to be left with excess aluminum hydroxide. So to find the amount of leftover aluminum hydroxide we are going to need to convert the given amount of sulfuric acid to the amount of aluminum hydroxide needed to react with the sulfuric acid.
![\frac{35g H_{2}SO_{4}}{1}*\frac{1 mole H_{2}SO_{4}}{98.079 g H_{2}SO_{4}} *\frac{2 moles Al(OH)_{3} }{3 moles H_{2}SO_{4}} * \frac{78.003 g Al(OH)_{3} }{1 mole Al(OH)_{3} } = 18.557 g Al(OH)_{3}](https://tex.z-dn.net/?f=%5Cfrac%7B35g%20H_%7B2%7DSO_%7B4%7D%7D%7B1%7D%2A%5Cfrac%7B1%20mole%20H_%7B2%7DSO_%7B4%7D%7D%7B98.079%20g%20H_%7B2%7DSO_%7B4%7D%7D%20%2A%5Cfrac%7B2%20moles%20Al%28OH%29_%7B3%7D%20%7D%7B3%20moles%20H_%7B2%7DSO_%7B4%7D%7D%20%2A%20%5Cfrac%7B78.003%20g%20Al%28OH%29_%7B3%7D%20%7D%7B1%20mole%20Al%28OH%29_%7B3%7D%20%7D%20%3D%2018.557%20g%20Al%28OH%29_%7B3%7D)
Once you do that, you need to subtract that number from the amount of aluminum hydroxide given to get the amount of left over aluminum hydroxide.
![30 g Al(OH)_{3} - 18.557 g Al(OH)_{3} = 11.43 g Al(OH)_{3}](https://tex.z-dn.net/?f=30%20g%20Al%28OH%29_%7B3%7D%20-%2018.557%20g%20Al%28OH%29_%7B3%7D%20%3D%2011.43%20g%20Al%28OH%29_%7B3%7D)
Hope this helps!
Answer:
Along period electronegativity and ionization energy increases.
Along group electronegativity and ionization energy decreases.
Explanation:
Along period:
As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. Thus the attraction of the atoms for valance electrons increases. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required, and electronegativity also increases.
Along group:
As we move from top to bottom in periodic table the atomic sizes increases.The electrons are added in next energy level in every next element. Thus the valance electrons farther away from the nucleus and hold of nucleus becomes weaker, because of weak nuclear attraction atomic radii increases and electronegativity and ionization energy decreases.