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:
[COF₂] = 0.346M
Explanation:
For the reaction:
2COF₂(g) ⇌ CO₂(g) + CF₄(g)
Kc = 5.70 is defined as:
Kc = [CO₂] [CF₄] / [COF₂]² = 5.70 <em>(1)</em>
Equilibrium concentrations of each compound after addition of 2.00M COF₂ will be:
[COF₂] : 2.00M - 2x
[CO₂] : x
[CF₄] : x
Replacing in (1):
5.70 = [X] [X] / [2-2x]²
22.8 - 45.6x + 22.8x² = x²
0 = -21.8x² + 45.6x - 22.8
Solving for x:
X = 1.265 <em>-False answer, will produce negative concentrations-</em>
<em>X = 0.827.</em>
Replaing, molar concentration of COF₂ is:
[COF₂] : 2.00M - 2×0.827 = <em>0.346M</em>
I hope it helps!
The given molarity of sodium hydroxide solution = 2.0 M
The required concentration of sodium hydroxide is 65 mL of 0.6 M NaOH
Converting 65 mL to L:
Calculating the moles of NaOH in the final solution:
Finding out the volume of 2.0 M solution taken to prepare the final solution:
Therefore, 19.5 mL of 2.0 M NaOH solution and make it up to 65 mL to prepare 0.6 M NaOH solution.
