First, we need to get the value of Ka:
when Ka = Kw / Kb
we have Kb = 1.8 x 10^-5
and Kw = 3.99 x 10^-16 so, by substitution:
Ka = (3.99 x 10^-16) / (1.8 x 10^-5) = 2.2 x 10^-11
by using the ICE table :
NH4+ + H2O →NH3 + H+
intial 0.013 0 0
change -X +X +X
Equ (0.013-X) X X
when Ka = [NH3][H+] / [NH4+]
by substitution:
2.2 x 10^-11 = X^2 / (0.013 - X) by solving this equation for X
∴X = 5.35 x 10^-7
∴[H+] = X = 5.35 x 10^-7
∴PH = - ㏒[H+]
= -㏒(5.35 x 10^-7)
= 6.27
The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O
We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:
How to determine the mass of C
- Mass of CO₂ = 9.78 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 9.78
Mass of C = 2.67 g
How to determine the mass of H
- Mass of H₂O = 20.99 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 4
Mass of H = 0.44 g
How to determine the mass of O
- Mass of compound = 4.30 g
- Mass of C = 2.67 g
- Mass of H = 0.44 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 4.30 – (2.67 + 0.44)
Mass of O = 1.19 g
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as follow:
- C = 2.67 g
- H = 0.44 g
- O = 1.19 g
- Empirical formula =?
Divide by their molar mass
C = 2.67 / 12 = 0.2225
H = 0.44 / 1 = 0.44
O = 1.19 / 16 = 0.074
Divide by the smallest
C = 0.2225 / 0.074 = 3
H = 0.44 / 0.0744 = 6
O = 0.074 / 0.074 = 1
Thus, the empirical formula of the compound is C₃H₆O
Learn more about empirical formula:
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Answer:
The average rate is 2.84 X 10⁻³ Ms⁻¹
Explanation:
Average rate = -0.5*Δ[HBr]/Δt
given;
[HBr]₁ = 0.590 M
[HBr]₂ = 0.465 M
Δ[HBr] = [HBr]₂ - [HBr]₁ = 0.465 M - 0.590 M = -0.125 M
Δt Change in time = 22.0 s
Average rate = -0.5*Δ[HBr]/Δt
Average rate = - 0.5(-0.125)/22
Average rate = 0.00284 Ms⁻¹ = 2.84 X 10⁻³ Ms⁻¹
Therefore, the average rate is 2.84 X 10⁻³ Ms⁻¹
1) Answer is: molar mas of ammonia is 17.031 g/mol.
M(NH₃) = Ar(N) + 3 · Ar(H) · g/mol.
M(NH₃) = 14.007 + 3 · 1.008 · g/mol.
M(NH₃) = 17.031 g/mol.
2) Answer is: molar mas of lead(II) chloride is 278.106 g/mol.
M(PbCl₂) = Ar(Pb) + 2 · Ar(Cl) · g/mol.
M(PbCl₂) = 207.2 + 2 · 35.453 · g/mol.
M(PbCl₂) = 278.106 g/mol.
3) Answer is: molar mas of acetic acid is 60.052 g/mol.
M(CH₃COOH) = 2 · Ar(C) + 2 · Ar(O) + 4 · Ar(H) · g/mol.
M(CH₃COOH) = 2 · 12.0107 + 2 · 15.9994 + 4 · 1.008 · g/mol.
M(CH₃COOH) = 60.052 g/mol.
Well the color for battery acid would be purple due to the acidity in the battery acid but to lower the levels add some gypsy stone and it can lower it
