Chemistry, 01.09.2019 15:50 ddarby

Which property of gases allows you to smell popcorn when you walk into a movie theater?

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Answers

Answer from: daty0355

Gasses contain highly energetic particles which tend to spread out to fill the room/ container they are in. Often when a smell rises, there is a high concentration of gas molecules in the rising smell. Here diffusion occurs where the molecules spread out randomly moving from an area of high concentration of gas molecules to areas of low gas molecules.

Answer from: saharB

The correct answer is diffusion.

Explanation:

Diffusion refers to the net movements of atoms or molecules from a place of high chemical potential or higher concentration to a place of low chemical potential or lower concentration. Diffusion takes place due to a gradient in chemical potential of the diffusing species.

One smells popcorn after walking into a movie theater is because of the phenomenon of diffusion. The smells travel in the air by diffusion. As molecules are always traveling, the tiny molecules from popcorn travel through the air all through the theater and enter into the noses.

Answer from: Quest

some of the qualities most frequently associated with the baroque are grandeur, sensuous richness, drama, vitality, movement, tension, emotional exuberance, and a tendency to blur distinctions between the various arts.

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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