Moles= mass divided by molar mass
Molar mass= 12.01(4) + 1.01(10)
= 58.14g/mol
Moles=14.5g / 58.14g/mol
=0.249
Therefore there are approx 0.249 moles in a 14.5g sample of C4H10
Missing question:
Nitrogen: <span>2.0 L; </span>1.0 atm; 25°<span>C.
Oxygen: 3</span>.0 L; 2.0 atm; 25°C.
<span>When the valve between the two containers is opened, nitrogen gas moves from one container to another container and gases are mixed together, total volume of nitrogen is than:
V(nitrogen) = 2,0 L + 3,0 L = 5,0 L.</span>
Your answer would be a change in odor! Hope this helps! ;D
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.
I can’t see the picture for some reason
