It Emeterio near the North Pole of the worlddddddd
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
Add proton + neutrons = Mass 37
Proton is same as your atomic mass= 17
If you go on the periodic table look up 17 and it will give u the letter.
Particles below the surface of a liquid
We need to know the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure.
The relationship is: As air pressure in an area increases, the density of the gas particles in that area increases.
For any gaseous substance, density of gas is directly proportional to pressure of gas.
This can be explained from idial gas edquation:
PV=nRT
PV=
RT [where, w= mass of substance, M=molar mass of substance]
PM=
RT
PM=dRT [where, d=density of thesubstance]
So, for a particular gaseous substance (whose molar mass is known), at particular temperature, pressure is directly related to density of gaseous substance.
Therefore, as air pressure in an area increases, the density of the gas particles in that area increases.
