There are 2.32 x 10^6 kg sulfuric acid in the rainfall.
Solution:
We can find the volume of the solution by the product of 1.00 in and 1800 miles2:
1800 miles2 * 2.59e+6 sq m / 1 sq mi = 4.662 x 10^9 sq m
1.00 in * 1 m / 39.3701 in = 0.0254 m
Volume = 4.662 x 10^9 m^2 * 0.0254 m
= 1.184 x 10^8 m^3 * 1000 L / 1 m3
= 1.184 x 10^11 Liters
We get the molarity of H2SO4 from the concentration of [H+] given by pH = 3.70:
[H+] = 10^-pH = 10^-3.7 = 0.000200 M
[H2SO4] = 0.000100 M
By multiplying the molarity of sulfuric acid by the volume of the solution, we can get the number of moles of sulfuric acid:
1.184 x 10^11 L * 0.000100 mol/L H2SO4 = 2.36 x 10^7 moles H2SO4
We can now calculate for the mass of sulfuric acid in the rainfall:
mass of H2SO4 = 2.36 x 10^7 moles * 98.079 g/mol
= 2.32 x 10^9 g * 1 kg / 1000 g
= 2.32 x 10^6 kg H2SO4
Answer: The major product of the reaction between Hydrobromic Acid and 2-bromo-1-butene in the presence of ether and acid is 2,2-dibromobutane.
Explanation:
The mechanism of the reaction is supported by the Markovnikov's rule which explains that in the addition reaction of alkenes by hydrogen-halogen compounds, the incoming halogen substituent goes to the more substituted Carbon. It can also be stated that incoming hydrogen atom goes to the Carbon with more Hydrogen atoms.
The only case when the reverse of Markovnikov's rule takes place is when Hydrogen peroxide is present and the addition reagent is HBr.
This case is not like that and it simply follows the Markovnikov's rule.
I'll add an attachment of the reaction to this now.
Answer:

Explanation:
Hello there!
In this case, since there is a 2:2 mole ratio between sodium peroxide and water according to the given reaction, it is possible to apply the following stoichiometric setup for the calculation of the required mass of water:

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Answer: a) The
of acetic acid at
is 
b) The percent dissociation for the solution is 
Explanation:

cM 0 0

So dissociation constant will be:

Give c= 0.10 M and
= ?
Also ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![[H^+]=1.35\times 10^{-3}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D1.35%5Ctimes%2010%5E%7B-3%7DM)
![[CH_3COO^-]=1.35\times 10^{-3}M](https://tex.z-dn.net/?f=%5BCH_3COO%5E-%5D%3D1.35%5Ctimes%2010%5E%7B-3%7DM)
![[CH_3COOH]=(0.10M-1.35\times 10^{-3}=0.09806M](https://tex.z-dn.net/?f=%5BCH_3COOH%5D%3D%280.10M-1.35%5Ctimes%2010%5E%7B-3%7D%3D0.09806M)
Putting in the values we get:


b) 


