The solubility (in M) of O2 in the blood of a scuba diver at a depth of 100 feet is 2.32 × 10^-3 mol.
<h3>How to calculate the solubility?</h3>
It should be noted that the partial pressure of oxygen will be:
= Mole fraction × Total pressure
= 0.209 × 3
= 0.627
According to Henry's law, the solubility will be:
= 3.7 × 10^-2 × 0.627
= 2.32 × 10^-3 mol
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Check the picture, I balanced the ions with greater charges first, then I balanced the water last
Answer:
Iodide concentration: 0.02M I⁻
Persulfate concentration: 0.02M S₂O₈²⁻
Explanation:
In chemistry, a dilution is the process in which the concentration of solute is reduced by addition of more solvent.
Iodide (KI → I⁻), has a concentration of 0.20M. If you add 5mL resulting in a final volume of 50mL, the final concentration decreases in:
0.20M × (5mL /50mL) = <em>0.02M I⁻</em>
The persulfate ion (K₂S₂O₈ → S₂O₈²⁻) has a concentration of 0.10M in 10mL that result in a final volume of 50mL. That means final concentration is:
0.10M × (10mL /50mL) = <em>0.02M S₂O₈²⁻</em>
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Answer:
Eqv Pt pH = 8.73
Explanation:
HOAc + NaOH => NaOAc + H₂O
50ml(0.10M HOAc) + 50ml(0.10M NaOH) => 100ml(0.05M NaOAc) + H₂O
For neutralized system, 100ml of 0.05M NaOAc remains
NaOAc => Na⁺ + OAc⁻
Na⁺ + H₂O => No Rxn
OAc⁻ + H₂O => HOAc + OH⁻
C(i) 0.05M ----- 0M 0M
ΔC -x ----- +x +x
C(f) 0.05-x
≅ 0.05M ----- x x
Kb = Kw/Ka = [HOAc][OH⁻]/[OAc⁻] = 1 X 10⁻¹⁴/1.7 X 10⁻⁵ = (x)(x)/(0.05M)
=> x = [OH⁻] = SqrRt(0.05 x 10⁻¹⁴/1.7 x 10⁻⁵) = 5.42 x 10⁻⁶M
=> pOH = -log[OH⁻] = -log(5.42 x 10⁻⁶) = 5.27
pH + pOH = 14 => pH = 14 - pOH = 14 - 5.27 = 8.73 Eqv Pt pH