I think the answer is a and c
I
Henry's law constant for oxygen is 0,0013 mol/L·<span>atm. Air has 21,0% oxygen.
concentration of oxygen at 1 atm: 0,0013 mol/L</span>·atm · 0,21 · 1 atm = 0,000273 mol/l.
concentration of oxygen at 1 atm: 0,0013 mol/L·atm · 0,21 · 0,892 atm = 0,000243 mol/l.
difference in concentration: 0,000273 - 0,000243 = 0,00003 mol/L.
n(oxygen) = 0,00003 mol/L · 4,40 L = 0,000132 mol.
Answer:
option D is the correct answer of this question.
The neutralization equation is:
3 Ca(OH)₂ + 2 H₃PO₄ → Ca₃(PO₄)₂ + 6 H₂O
From this equation we can see that 3 moles of Ca(OH)₂ react with 2 moles of H₃PO₄
Numbers of mmol of Ca(OH)₂ = M x V = 0.04345 x 54.93 = 2.387 mmol
Number of mmol of H₃PO₄ = 2.387 x (2/3) = 1.591 mmol
Molarity of solution = n (mmol) / V(ml) = 1.591 / 25 = 0.0636 M
