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:
0.027 mole of NaOH.
Explanation:
We'll begin by obtaining the number of mole H2SeO4 in 45mL of 0.30M H2SeO4
This is illustrated below:
Molarity of H2SeO4 = 0.3M
Volume of solution = 45mL = 45/1000 = 0.045L
Mole of H2SeO4 =...?
Mole = Molarity x Volume
Mole of H2SeO4 = 0.3 x 0.045
Mole of H2SeO4 = 0.0135 mole
Next, the balanced equation for the reaction. This is given below:
H2SeO4 + 2NaOH –> Na2SeO4 + 2H2O
From the balanced equation above,
1 mole of H2SeO4 required 2 moles of NaOH.
Therefore, 0.0135 mole of H2SeO4 will require = 0.0135 x 2 = 0.027 mole of NaOH.
Therefore, 0.027 mole of NaOH is needed for the reaction.
Answer:
Weight % of NH₃ in the aqueous waste = 2.001 %
Explanation:
The chemical equation for the reaction
Moles of HCl = Molarity × Volume
= 0.1039 × 31.89 mL × 
= 0.0033 mole
Total mass of original sample = 25.888 g + 73.464 g
= 99.352 g
Total HCl taken for assay = 
= 0.1083 g
Moles of NH₃ = 
= 0.03047 moles
Mass of NH₃ = number of moles × molar mass
Mass of NH₃ = 0.03047 moles × 17 g
Mass of NH₃ = 0.51799
Weight % of NH₃ = 
%
Weight % of NH₃ in the aqueous waste = 2.001 %
Answer:
1. Percentage by weight = 0.5023 = 50.23 %
2. molar fraction =0.153
Explanation:
We know that
Molar mass of HClO4 = 100.46 g/mol
So the mass of 5 Moles= 5 x 100.46
Mass (m)= 5 x 100.46 = 502.3 g
Lets assume that aqueous solution of HClO4 and the density of solution is equal to density of water.
Given that concentration HClO4 is 5 M it means that it have 5 moles of HClO4 in 1000 ml.
We know that
Mass = density x volume
Mass of 1000 ml solution = 1 x 1000 =1000 ( density = 1 gm/ml)
m'=1000 g
1.
Percentage by weight = 502.3 /1000
Percentage by weight = 0.5023 = 50.23 %
2.
We know that
molar mass of water = 18 g/mol
mass of water in 1000 ml = 1000 - 502.3 g=497.9 g
So moles of water = 497.7 /18 mole
moles of water = 27.65 moles
So molar fraction = 5/(5+27.65)
molar fraction =0.153
24g 4Al+3O2=2Al2O3
27g x(g)
108 96
X=24g
