Answer:
The ratio of HC2H3O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) to HC2H3O2(aq) in the flask after the addition of 1.0mL of NaOH(aq) is 15 : 19 .
Explanation:
HC2H3O2 is CH₃⁻ COOH, which is also known as Acetic acid.
IUPAC name of this compound is Ethanoic acid.
Acetic acid has a basicity of 1. so there is one acidic hydrogen is acetic acid.
Given that, equivalence point was reached when 20.0mL of NaOH is added.
let the normality of acetic acid is N₁ and that of NaOH is N₂.
volume of acetic acid is V₁ and that of NaOH is V₂.
Equivalence point occurs when, N₁ × V₁ = N₂ × V₂.
⇒ N₁ × V₁ = N₂ × 20.
after the addition of 5.0mL of NaOH(aq), remaining N₁ × V° = N₂ × (20 - 5).
= N₂ × 15.
after the addition of 1.0mL of NaOH(aq), remaining N₁ × Vˣ = N₂ × (20 - 1).
= N₂ × 19.
⇒ V° : Vˣ = 15 : 19 .
⇒
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Abca I think it’s right not sure
Answer:
Mass released = 8.6 g
Explanation:
Given data:
Initial number of moles nitrogen= 0.950 mol
Initial volume = 25.5 L
Final mass of nitrogen released = ?
Final volume = 17.3 L
Solution:
Formula:
V₁/n₁ = V₂/n₂
25.5 L / 0.950 mol = 17.3 L/n₂
n₂ = 17.3 L× 0.950 mol/25.5 L
n₂ = 16.435 L.mol /25.5 L
n₂ = 0.644 mol
Initial mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.950 mol × 28 g/mol
Mass = 26.6 g
Final mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.644 mol × 28 g/mol
Mass = 18.0 g
Mass released = initial mass - final mass
Mass released = 26.6 g - 18.0 g
Mass released = 8.6 g
