HBr and HF are both monoprotic Arrhenius acids—that is, in aqueous solution, they dissociate and ionize to give hydrogen ions. A strong acid ionizes completely; a weak acid ionizes partially.
In this case, HBr, being a strong acid, would ionize completely in water to yield H+ and Br- ions. However, HF, being a weak acid, would ionize only to a limited extent: some of the HF molecules will ionize into H+ and F- ions, but most of the HF will remain undissociated.
pH is, by definition, a measurement of the concentration of hydrogen ions in solution (pH = -log[H+]). A higher concentration of hydrogen ions gives a lower pH, while a lower concentration of hydrogen ions gives a higher pH. At 25 °C, a pH of 7 indicates a neutral solution; a pH less than 7 indicates an acidic solution; and a pH greater than 7 indicates a basic solution.
If we have equal concentrations of HBr and HF, then the HBr solution will have a greater concentration of hydrogen ions in solution than the HF solution. Consequently, the pH of the HBr solution will be less than the pH of the HF solution.
Choice A is incorrect: Strong acids like HBr dissociate completely, not partially.
Choice B is incorrect: While the initial concentration of HBr and HF are the same, the H+ concentration in the HBr solution is greater. Since pH is a function of H+ concentration, the pH of the two solutions cannot be the same.
Choice C is correct: A greater H+ concentration gives a lower pH value. The HBr solution has the greater H+ concentration. Thus, the pH of the HBr solution would be less than that of the HF solution.
Choice D is incorrect for the reason why choice C is correct.
Answer: option B. 0.59
Explanation:Please see attachment for explanation
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
The maximum oxygen uptake is known as the VO2 max.
