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zvonat [6]
1 year ago
11

an atmosphere is considered hazardous if it contains a hazardous gas in excess of 10 percent of the hazardous material's:

Chemistry
1 answer:
svlad2 [7]1 year ago
6 0

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

Learn more about lower flammable limit at brainly.com/question/2456135

#SPJ4

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What is the free energy change if the ratio of the concentrations of the products to the concentrations of the reactants is 22.3
snow_lady [41]

The free energy change(Gibbs free energy-ΔG)=-8.698 kJ/mol

<h3>Further explanation</h3>

Given

Ratio of the concentrations of the products to the concentrations of the reactants is 22.3

Temperature = 37 C = 310 K

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Required

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Solution

Ratio of the concentration : equilbrium constant = K = 22.3

We can use Gibbs free energy :

ΔG = ΔG°+ RT ln K

R=8.314 .10⁻³ kJ/mol K

\tt \Delta G=-16.7~kJ/mol+8.314.10^{-3}\times 310\times ln~22.3\\\\\Delta G=-8.698~kJ/mol

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C(s) + S8(s) → CS2(l)
Vanyuwa [196]
The balanced chemical reaction is written as:

<span>4C(s) + S8(s) → 4CS2(l)

We are given the amount of carbon and sulfur to be used in the reaction. We need to determine first the limiting reactant to be able to solve this correctly.

</span>7.70 g C ( 1 mol / 12.01 g) =0.64 mol C
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The limiting reactant would be S8. We use this amount to calculate.

0.08 mol S8 ( 4 mol CS2 / 1 mol S8 ) ( 256.48 g / 1 mol ) = 78.8 g CS2
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