A gas with a vapor density greater than that of air, would be most effectively displaced out off a vessel by ventilation.
The two following principles determine the type of ventilation: Considering the impact of the contaminant's vapour density and either positive or negative pressure is applied.
Consider a vertical tank that is filled with methane gas. Methane would leak out if we opened the top hatch since its vapour density is far lower than that of air. A second opening could be built at the bottom to greatly increase the process' efficiency.
A faster atmospheric turnover would follow from air being pulled in via the bottom while the methane was vented out the top. The rate of natural ventilation will increase with the difference in vapour density. Numerous gases that require ventilation are either present in fairly low concentrations or have vapor densities close to one.
Answer:
4.34atm
Explanation:
The following data were obtained from the question:
P1 = 3.50 atm
T1 = 24°C = 24 +273 = 297K
T2 = 95°C = 368K
P2 =?
Using P1 /T1 = P2 /T2, we can obtain the new pressure as follows:
P1 /T1 = P2 /T2
3.5 / 297 = P2 / 368
Cross multiply to express in linear form:
297 x P2 = 3.5 x 368
Divide both side by the 297
P2 = (3.5 x 368) /297
P2 = 4.34atm
Therefore, the gas pressure in the container at 95°C will be 4.34atm
Answer:
<h3>5.80000000E+ 46...</h3>
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Answer:
PI = 6.04
In the attached figure are the missing hydrogen atoms
Explanation:
Given both pk, the isoelectric point will be equal to:
PI = (1/2)*(pka + pka2), where pka = 2.32 and pka2 = 9.76
PI = (1/2)*(2.32 + 9.76) = 6.04
