Chemistry, asked by rush4611, 16 hours ago

At what temperature will water boil under apressure of 787mm?The latent heat of vaporization is 536 k cal per gram.​

Answers

Answered by tejaswinibavana2003
0

Answer:

T1 =370.65K

Explanation:

T_(2)=1000^(@)C=373K, P_(2)=760mm(atb.pt.ofwater)DeltaH=540 "cal"//gT_(1)=? P_(1)=700mm DeltaH=540xx18 "cal"//moltherefore 2.303 "log"(760)/(700)=(540xx18)/(2)[(373-T_(1))/(T_(1)Xx373)]6.34xx10^(-3)T_(1)=373-T_(1)T_(1)=370.65K`.

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