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:
F=ma
Explanation:
F=m×a
according to that F÷m=a and also F ÷a=m
Answer:
pH = 12.20
Explanation:
Ca(OH)2 is a strong base, so it dissociates completely. A 0.08 M solution of Ca(OH)2 is 0.16 M OH-, since every mole of Ca(OH)2 has 2 OH-.
Calculate pOH using [OH-] = 0.16 M
pOH = -log(0.16) = 0.80
pH = 14 - pOH = 14 - 0.80 = 12.20
As you move from left to right across a period,the number of valence electrons<span>-increases. </span>
