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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Answer:
- What is the AGⓇ of this reaction? 0.
- Which will be favoured - the forward reaction, the reverse reaction, or neither? Neither.
- What effect does the presence of the enzyme aspartate transaminase have on the Key value when compared with its value in the absence of enzyme? It does not affect the value of Keq.
- If one of the products of reaction 1, oxaloacetate, is removed by converting it to citrate as follows: Reaction 2: oxaloacetate + acetyl-CoA citrate + COASH will the key for Reaction l be changed? No, the Keq does not change.
Explanation:
1. To calculate the delta G of a reaction given the K, we use the following equation:
ΔG°= -RT ln K.
Which gives us 0 when K is 1.
2.None of the reactions is favoured. Given that the K equals 1, the system will try to keep the concentration of both products and reagents the same.
3. A catalyst is a substance that, when added, provides a different and faster mechanism through which a reaction takes place. This only means that the speed at which the equilibrium is attained is reduced, but the enzyme does nothing to alter the difference in energy (ΔG°) of the start and end points of the reaction, which ultimately gives us the value of Keq.
4. The addition of a side reaction does not change the value of Keq for the main reaction. They are both separate ways of making oxaloacetate disappear. While the Keq does not change, keep in mind that the end concentrations will not be the same, for any set of starting concentrations of your substances.
Answer:
helium hydrogen
Explanation:
lithium beryllium bottom carbon
The problem is incomplete. However, there can only be two probable questions for this problem. First, you can be asked the individual partial pressures of each gas. Second, you can be asked the volume occupied by each gas. I can answer both cases for you.
1.
Let's assume ideal gas.
Pressure for N₂: 2 bar*0.4 = 0.8 bar
Pressure for CO₂: 2 bar*0.5 = 1 bar
Pressure for CH₄: 2 bar*0.1 = 0.2 bar
2. For the volume, let's find the total volume first.
V = nRT/P = (1 mol)(8.314 J/mol-K)(30 +273 K)/(2 bar*10⁵ Pa/1 bar)
V = 0.0126 m³
Hence,
Volume for N₂: 0.0126 bar*0.4 = 0.00504 m³
Volume for CO₂: 0.0126*0.5 = 0.0063 m³
Volume for CH₄: 0.0126*0.1 = 0.00126 m³
Ink can be separated into black and other color pigments. This can be done on filter paper by dotting the marker just above the edge and adding ethyl alcohol, which drags the pigments separately across the paper.
