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Tpy6a [65]
3 years ago
15

An engineer turns the temperature of a gas down, holding the volume constant at 1,250 L. Initially, the temperature of the gas w

as 625. °C and had a pressure of 275. KPa. Determine the pressure of the gas when the temperature decreased to 240°C
Chemistry
1 answer:
Natalija [7]3 years ago
5 0

Answer:

625=275

240=x

625x=275×240

625x=66000

x=66000/625

x=105 .6

Explanation:

If 625°C gives us 275KPa then 240°C will give us the the sum of 275×240 divided by 625

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For the following reaction, 2HI(g) – H (9) +1,9 the equilibrium constant k, = 2.60X10-2 at 1110 K and 2.68x10-2 at 1140 K. Calcu
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Answer:

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\ln (\frac{K_1}{K_2})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

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\Delta H = ?

Putting values in above equation, we get:

\ln(\frac{2.68\times 10^{-2}}{2.60\times 10^{-2}})=\frac{\Delta H}{8.314J/mol.K}[\frac{1}{1110 K}-\frac{1}{1140 K}]

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Explanation:

<h2>I hope it's helpful for you</h2>
6 0
3 years ago
Read 2 more answers
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