Answer:
6,78 mL of 12,0 wt% H₂SO₄
Explanation:
The equilibrium in water is:
H₂O (l) ⇄ H⁺ (aq) + OH⁻ (aq)
The initial concentration of [H⁺] is 10⁻⁸ M and final desired concentration is [H⁺] = , thus,
Thus, you need to add:
[H⁺] = = 5,31x10⁻⁸ M
The total volume of the pool is:
9,00 m × 15,0 m ×2,50 m = 337,5 m³ ≡ 337500 L
Thus, moles of H⁺ you need to add are:
5,31x10⁻⁸ M × 337500 L = 1,792x10⁻² moles of H⁺
These moles comes from
H₂SO₄ → 2H⁺ +SO₄²⁻
Thus:
1,792x10⁻² moles of H⁺ × = 8,96x10⁻³ moles of H₂SO₄
These moles comes from:
8,96x10⁻³ moles of H₂SO₄ × × × =
6,78 mL of 12,0wt% H₂SO₄
I hope it helps!
When you doing a titration, you need to use an indicator to confirm whether the reaction is completed. When the indicator has the color change and will not change back in one minute, the reaction is finished and you don't need to add more.
Answer is: C₃H₃N₃O₃.
Chemical reaction: CₓHₓNₓOₓ + O₂ → aCO₂ + x/2H₂ + x/2N₂.
m(CₐHₓNₓ) = 5,214 g.
m(CO₂) = 5,34 g.
m(H₂) = 1,09 g.
m(N₂) = 1,70 g.
n(CO₂) = n(C) = 5,34 g ÷ 44 g/mol = 0,121 mol.
n(H₂O) = 1,09 g ÷18 g/mol = 0,06 mol.
n(H) = 2 · 0,0605 mol = 0,121 mol.
n(N₂) = 1,7 g ÷ 28 g/mol = 0,0607 mol.
n(N) = 0,0607 mol · 2 = 0,121 mol.
n(C) : n(H) : n(N) = 0,121 mol : 0,121 mol : 0,121 mol /: 0,121
n(C) : n(H) : n(N) = 1 : 1 : 1.
M(CHN) = 27 g/mol.
m(O₂) = 8,13 g - 5,214 g = 2,914 g.
n(O₂) = 2,914 g ÷ 32 g/mol = 0,09 mol.
n(CₓHₓNₓOₓ) = 5,214 g ÷ 129,1 g/mol = 0,0404 mol.
n(CₓHₓNₓOₓ) : n(CO₂) = 1 : 3.
C because it is and I know