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Delvig [45]
2 years ago
11

A rotameter calibration curve (flow rate versus float position) obtained using a liquid is mistakenly used to measure a gas flow

rate. Would you expect the gas rate determined in this manner to be too high or too low?
Chemistry
1 answer:
Veronika [31]2 years ago
5 0

Answer:

I would expect the gas rate determined in this manner to be too low

Explanation:

A Rotameter can be designed to respond to the sensitivity of density, velocity, to measure the flow rate of liquid or gas enclosed in a tube. Liquids are denser than gas, and since the gas rate to be determined needed to respond to the velocity head alone of the rotameter so as to bring the forces in the tube equilibrium. Knowing if there is no flow, then the float would remain at the bottom, so gas has to flow at a higher rate compared to the liquid so the float would be in a similar position making it easier to measure the flowrate. This leaves the gas rate to be determined too low.

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Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib
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<u>Answer:</u> The value of K_c for the final reaction is 7.16\times 10^{25}

<u>Explanation:</u>

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<u>Equation 1:</u>  H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1

<u>Equation 2:</u>  HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2

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S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c

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The value of equilibrium constant for net reaction is:

K_c=\frac{1}{K_1}\times \frac{1}{K_2}

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Putting values in above equation, we get:

K_c=\frac{1}{(9.57\times 10^{-8})}\times \frac{1}{(1.46\times 10^{-19})}=7.16\times 10^{25}

Hence, the value of K_c for the final reaction is 7.16\times 10^{25}

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