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Anit [1.1K]
3 years ago
15

Luca wants to identify and quantity the chemicals causing air pollution in a city, but he is using a manual air sampler to colle

t air sample.
Why should he update his current sampler to an electronic one?
Chemistry
1 answer:
xenn [34]3 years ago
5 0

Answer

Manual samplers are prone to effects of temperature, speed of wind and air concentrations.

Explanation

Manual samplers face several challenges that can act as drawbacks to obtaining accurate results. They are subjected to effects of sampling duration where long sampling times are needed to obtain adequate mass for detection. Manual samplers face challenges when measuring non-volatile species because particles are observed into the adsorption medium at a slower rate of diffusion.

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Consider the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g). If 0.02 moles of PCl5, 0.04 moles of PCl3, and 0.08 moles of Cl2 are combined
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Answer:

The reaction quotient (Q) before the reaction is 0.32

Explanation:

Being the reaction:

aA + bB ⇔ cC + dD

Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b}  }

where Q is the so-called reaction quotient and the concentrations expressed in it are not those of the equilibrium but those of the different reagents and products at a certain instant of the reaction.

The concentration will be calculated by:

Concentration=\frac{number of moles of solute}{Volume}

You know  the reaction:

PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g).

So:

Q=\frac{[PCl_{3} ] *[Cl_{2} ] }{[PCl_{5} ]}

The concentrations are:

  • [PCl₃]=\frac{0.04 moles}{0.5 L} =0.08 \frac{moles}{L}
  • [Cl₂]=\frac{0.08 moles}{0.5 L} =0.16 \frac{moles}{L}
  • [PCl₅]=\frac{0.02 moles}{0.5 L} =0.04 \frac{moles}{L}

Replacing:

Q=\frac{0.08*0.16}{0.04}

Solving:

Q= 0.32

<u><em>The reaction quotient (Q) before the reaction is 0.32</em></u>

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