Respuesta :
This is how far away from the ammonia-dampened plug[tex]&x=51.7 \mathrm{~cm} f r 0 \mathrm{~m} \mathrm{NH}[/tex]
How can you figure out the diffusion ratio of NH3 and HCl?
- Diffusion and the effusion process are connected. Effusion is the process by which a confined gas escapes from its container through a very small opening.
- An effusion is shown by the fact that a helium-filled balloon eventually stops floating and sinks to the ground. This is caused by the helium gas evaporating through the balloon's microscopic pores.
- Both diffusion and effusion are influenced by the velocity at which different gas molecules move. Lower molar mass gases diffuse and effuse more quickly than higher molar mass gases.
Given,
molar mass of [tex]$\mathrm{HCl}=36.5$[/tex]
molar mass of [tex]$\mathrm{NH}_3=17$[/tex]
Now, Rate [tex]$\propto \frac{1}{\sqrt{\text { molar mass }}}$$[/tex]
[tex]\frac{\text { Rate of } \mathrm{NH}_3}{\text { Rate of } \mathrm{HCl}} &=\sqrt{\frac{36.5}{17}}=\sqrt{\frac{\text { molar mass of } \mathrm{HCl}}{\text { molar mass of } \mathrm{NH}_3}} \\[/tex]
[tex]&=\frac{6.04}{4.12}=\frac{1.47}{100}[/tex]
We know,
[tex]\frac{\text { Distance }}{\text { rate }}=\text { time }[/tex]
Let x = d distance travelled by[tex]$\mathrm{NH}_3$[/tex]
87 - x = distance travelled by [tex]$\mathrm{HCl}$[/tex]
[tex]&\frac{x}{1.47}=\frac{87-x}{1} \quad \text { (comparing both } \mathrm{NH}_3 \& \mathrm{HCl} \text { ) } \\[/tex]
[tex]&x=1.47(87-x) \\[/tex]
[tex]&2.47 x=127.89 \\[/tex]
[tex]&x=51.7 \mathrm{~cm} f r 0 \mathrm{~m} \mathrm{NH}[/tex] is
Hence, This is how far away from the ammonia-dampened plug[tex]&x=51.7 \mathrm{~cm} f r 0 \mathrm{~m} \mathrm{NH}[/tex]
To learn more about NH3, refer to:
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