Asked by confused

Benzene has a heat of vaporization of 30.72kj/mol and a normal boiling point of 80.1 C . At what temperature does benzene boil when the external pressure is 445 ? This is how I am solving this problem, but when I enter the answer it says im wrong. Can someone help with the math. Thanks
ln(445/760)= (30720/8.314)*(1/80 - 1/T2)
Answered by DrBob222
Your problem is that you are entering T as degrees C. You MUST use Kelvin (and don't throw away the 0.1 either).
Answered by confused
The final answer is suppose to be in C. So therefore, would I solve the problem in Kelvin then convert back to degrees?
Answered by DrBob222
Cid you mean, "The final answer is <b>supposed</b> to be in C?" Yes. Convert 80.1 C to Kelvin by adding 273.16, then convert back to C for the final answer.
Answered by confused
yes i was meaning C. For some reason im still getting the wrong answer probably my math, im going to keep trying it...thanks for your help
Answered by DrBob222
I worked it out and came out with 62.9 degrees C but check my math.
Answered by Chole
3.50 x 10^6 Km how do you express this in scientific notation in expaneded form.
Answered by Rachel
I got 51.5?
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