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:
Answer is Ca2+(aq)+S2-(aq)=>CaS(s)
Explanation:
I hope it's helpful!
Answer:
The answer is in the explanation.
Explanation:
The KHP is an acid used as standard in titrations to find concentration of bases as NaOH.
The reaction that explain this use is:
KHP + NaOH → KNaP + H2O
<em>where 1 mole of KHP reacts per mole of NaOH</em>
That means, at equivalence point of a titration in which titrant is NaOH, the moles of KHP = Moles of NaOH added
With the moles of KHP = Moles of NaOH and the volume used by titrant we can find the molar concentration of NaOH.
The moles of KHP are obtained from the volume and the concentration as follows:
Volume(L)*Concentration (Molarity,M) = moles of KHP
If the concentration is more or less than 0.100M, the moles will be higher or lower. For that reason, we need to know the concentration of KHP but is not necessary to be 0.100M.
Answer:
The equilibrium temperature of the coffee is 72.4 °C
Explanation:
Step 1: Data given
Mass of cream = 15.0 grams
Temperature of the cream = 10.0°C
Mass of the coffee = 150.0 grams
Temperature of the coffee = 78.6 °C
C = respective specific heat of the substances( same as water) = 4.184 J/g°C
Step 2: Calculate the equilibrium temperature
m(cream)*C*(T2-T1) = -m(coffee)*c*(T2-T1)
15.0 g* 4.184 J/g°C *(T2 - 10.0°C) = -150.0g *4.184 J/g°C*(T2-78.6°C)
62.76T2 - 627.6 = -627.6T2 + 49329.36
690.36T2 = 49956.96
T2 = 72.4 °C
The equilibrium temperature of the coffee is 72.4 °C
The natural environment or natural world encompasses all living and non-living things occurring naturally, meaning in this case not artificial.
