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irga5000 [103]
2 years ago
8

The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.500 atm? (∆Hvap = 28.5

kJ/mol)
Chemistry
1 answer:
NNADVOKAT [17]2 years ago
7 0

For a normal boiling point of a liquid is 282 °C, the temperature is mathematically given as

T2=181.55°C\

<h3> What temperature (in °C) would the vapor pressure be 0.500 atm? </h3>

Generally, the equation for the gas  is mathematically given as

ln(p1/p2)=dHvap/R(1/T2-1/T1)

Therefore

ln(1/0.26)=23500/8.214(1/T2-1/555)

T2=181.55^C

In conclusion

T2=181.55°C

Read more about Temperature

brainly.com/question/13439286

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I hope this works.
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