Answer:
there are approximately n ≈ 10²² moles
Explanation:
Since the radius of the earth is approximately R=6378 km= 6.378*10⁶ m , then the surface S of the earth would be
S= 4*π*R²
since the water covers 75% of the Earth's surface , the surface covered by water Sw is
Sw=0.75*S
the volume for a surface Sw and a depth D= 3 km = 3000 m ( approximating the volume through a rectangular shape) is
V=Sw*D
the mass of water under a volume V , assuming a density ρ= 1000 kg/m³ is
m=ρ*V
the number of moles n of water ( molecular weight M= 18 g/mole = 1.8*10⁻² kg/mole ) for a mass m is
n = m/M
then
n = m/M = ρ*V/M = ρ*Sw*D/M = 0.75*ρ*S*D/M = 3/4*ρ*4*π*R² *D/M = 3*π*ρ*R² *D/M
n=3*π*ρ*R² *D/M
replacing values
n=3*π*ρ*R² *D/M = 3*π*1000 kg/m³*(6.378*10⁶ m)² *3000 m /(1.8*10⁻² kg/mole) = 3*π*6.378*3/1.8 * 10²⁰ = 100.18 * 10²⁰ ≈ 10²² moles
n ≈ 10²² moles
HI
So, the formula for water is H2O
When you have the same amount of the reactants , hydrogen will be the limiting reactant.
Limiting reactant is the thing that runs out first.
Answer: The mass percentage of 
is 5.86%
Explanation:
To calculate the mass percentage of in the sample it is necessary to know the mass of the solute ( in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).
To calculate the mass of the solute, we must take the mass of the 
precipitate. We can establish a relation between the mass of and using the stoichiometry of the compounds:
Since for every mole of Tl in there are two moles of Tl in , we have:
Using the molar mass of we have:
Finally, we can use the mass percentage formula:
In comparison see it is very easy in goolge
