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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Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 3.16 L of chlorine were consumed at STP.
Be sure your answer has the correct number of significant digits.
Answer: Thus volume of carbon tetrachloride that would be produced is 0.788 L
Explanation:
According to ideal gas equation:

P = pressure of gas = 1 atm (at STP)
V = Volume of gas = 3.16 L
n = number of moles = ?
R = gas constant =
T =temperature =



According to stoichiometry:
4 moles of chlorine produces = 1 mole of carbon tetrachloride
Thus 0.141 moles of methane produces =
moles of carbon tetrachloride
volume of carbon tetrachloride =
Thus volume of carbon tetrachloride that would be produced is 0.788 L
Answer:
The correct answer is pOH= 11
Explanation:
From the aqueous acid-base equilibrium we know that
pH + pOH = 14
If we know pH, we can calculate pOH as follows:
pOH = 14 - pH
In this problem, the solution has a pH of 3, so:
pOH = 14 - 3 = 11
Answer:
Not exactly But you can take the slope of the curved portion and the slope of the flatline.
It wont do you much good since your working for absorbance but if you ever see something like a temperature change you can use the slope(s) to find freezing points/melting
Explanation:
If you need to submit a slope you could use a best fit which is just point to point or you could break it up like i mentioned