Answer is: <span>percent ionization of aqueous acetic acid solution is 0.341%.
c(CH</span>₃COOH) = 1.55 M.
Ka(CH₃COOH) = 1.8·10⁻⁵.
Chemical reaction: CH₃COOH ⇄ CH₃COO⁻ + H⁺.
[CH₃COO⁻] = [H⁺] = x.
[CH₃COOH] = 1.55 M - x.
Ka = [CH₃COO⁻] · [H⁺] / [CH₃COOH].
1.8·10⁻⁵ = x² / (1.55 M - x).
Solve quadratic equation: x = 0.0053 M.
percent of ionization:
α = 0,0053 M ÷ 1.55 M · 100% = 0,341%
Answer:
Explanation:
The expression for Bohr energy of the hydrogen-like species is:
Where, Z is the atomic number of the element
For, , Z = 2
For lowest energy, n = 1
Thus, Applying the values as:-
Answer:
(a) 0.2518 g; (b) 6
Explanation:
The Law of Multiple Proportions states that when two elements A and B combine to form two or more compounds, the masses of B that combine with a given mass of A are in the ratios of small whole numbers.
That is, if one compound has a ratio r₁ and the other has a ratio r₂, the ratio of the ratios r is in small whole numbers.
(a) Methane and ethane
In methane, the mass ratio of H:C is r₁ = 0.3357:1.00.
In ethane, the mass ratio of H:C is r₂
The ratio of the ratios r = ⁴/₃
r = r₁/r₂
⁴/₃ = 0.3357/r₂
(⁴/₃)r₂ = 0.3357
r₂ = 0.3357 × ¾ = 0.2518:1
Ethane contains 0.2518 g of H for every 1 g of C.
(b) Xenon fluorides
In XeF₂, the mass ratio of F:Xe is r₁ = 0.2894:1
In XeFₙ, the mass ratio of F:Xe is r₂ = 0.8682:1
r = r₂/r₁ = 0.8682/0.2894 = 3.000:1 ≈ 3:1
XeFₙ contains three times as many F atoms as XeF₂.
n = 6
Answer: 112 dm3
Explanation:
At STP, 1 mole occupied 22.4dm3
Hence
5mole will occupy 22.4×5= 112dm3