Assuming it's a perfect gas, we have PV=nRT hence if T goes down, V goes down up. The volume will decrease.
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 level of toxins in the fish's cell is equivalent to the level of toxins in the water. Therefore, in order to reduce the toxins further, we should replace the now contaminated water with clean water. After the level of toxins in the fish's cell stops reducing, we replace the water with clean water once again.
The final volume of the gas that was heated from -25.0 °C to standard temperature is 2.2L.
<h3>How to calculate volume?</h3>
The volume of a given gas can be calculated using the Charles law equation as follows:
V1/T1 = V2/T2
Where;
- V1 = initial volume
- V2 = final volume
- T1 = initial temperature
- T2 = final temperature
- V1 = 2L
- V2 = ?
- T1 = -25°C + 273 = 248K
- T2 = 273K
2/248 = V2/273
273 × 2 = 248V2
546 = 248V2
V2 = 546/248
V2 = 2.2L
Therefore, the final volume of the gas that was heated from -25.0 °C to standard temperature is 2.2L
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Answer:
mass CaI2 = 23.424 Kg
Explanation:
From the periodic table we obtain for CaI2:
⇒ molecular mass CaI2: 40.078 + ((2)(126.90)) = 293.878 g/mol
∴ mol CaI2 = (4.80 E25 units )×(mol/6.022 E23 units) = 79.708 mol CaI2
⇒ mass CaI2 = (79.708 mol CaI2)×(293.878 g/mol) = 23424.43 g
⇒ mass CaI2 = 23.424 Kg
