Lower flammable limit means the lowest concentration of a material that will propagate a flame.
What is hazardous atmosphere?
It is an atmosphere that may expose employees to risk of death, incapacitation, impairment of ability to self-rescue, injury, or acute illness from one or more of following causes
- Flammable gas, vapor, or mist in excess of 10 percent of lower flammable limit (LFL)
- Airborne combustible dust at concentration that meets or exceeds its LFL
What is lower flammable limit?
- It means the lowest concentration of a material that will propagate a flame.
- The LFL is usually expressed as percent by volume of material in air (or other oxidant)
- Atmospheres with concentration of flammable vapors at or above 10 percent of lower explosive limit (LEL) are considered hazardous when located in confined spaces.
- However, atmospheres with flammable vapors below 10 percent of LEL are not necessarily safe. Such atmospheres are too lean to burn
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The pH of the monoprotic weak acid is 2.79.
<h3>What are weak acids?</h3>
The weak acids are the acids that do not fully dissociate into ions in the solution. Strong acids fully dissociate into ions.
The chemical reaction is HA(aq) ⇄ A⁻(aq) + H⁺(aq).
c (monoprotic acid) = 0.33 M.
Ka = 1.2·10⁻⁶
[A⁻] = [H⁺] = x
[HA] = 0.33 M - x
Ka = [A⁻]·[H⁺] / [HA]
2. 6 × 10⁻⁶ = x² / (0.33 M - x)
Solve quadratic equation: [H⁺] = 0.000524 M.
pH = -log[H⁺]
pH = -log(0.000524 M)
pH = 2.79
Thus, the pH of the monoprotic weak acid is 2.79
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Answer:
Step 1;
q = w = -0.52571 kJ, ΔS = 0.876 J/K
Step 2
q = 0, w = ΔU = -7.5 kJ, ΔH = -5.00574 kJ
Explanation:
The given parameters are;
= 100 N·m
= 327 K
= 90 N·m
Step 1
For isothermal expansion, we have;
ΔU = ΔH = 0
w = n·R·T·ln(
/
) = 1 × 8.314 × 600.15 × ln(90/100) = -525.71
w ≈<em> -0.52571</em> kJ
At state 1, q = w = -0.52571 kJ
ΔS = -n·R·ln(
/
) = -1 × 8.314 × ln(90/100) ≈ 0.876
ΔS ≈ 0.876 J/K
Step 2
q = 0 for adiabatic process
ΔU = 25×(27 - 327) = -7,500
w = ΔU = <em>-7.5 kJ</em>
ΔH = ΔU + n·R·ΔT
ΔH = -7,500 + 8.3142 × 300 = -5,005.74
ΔH = ΔU = <em>-5.00574 kJ</em>