3.16 X 10^-11 M is the [OH-] concentration when H3O+ = 1.40 *10^-4 M.
Explanation:
data given:
H30+= 1.40 X 10^-4 M\
Henderson Hasslebalch equation to calculate pH=
pH = -log10(H30+)
putting the values in the equation:
pH = -log 10(1.40 X 10^-4 M)
pH = 3.85
pH + pOH =14
pOH = 14 - 3.85
pOH = 10.15
The OH- concentration from the pOH by the equation:
pOH = -log10[OH-]
10.5= -log10[OH-]
[OH-] = 10^-10.5
[OH-] = 3.16 X 10^-11 is the concentration of OH ions when hydronium ion concentration is 1.40 *10^-4 M.
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
Mass and volume are two units used to measure objects. Mass- the amount of matter an object contains and volume- how much space it takes up
