This question is incomplete, the complete question is;
Tonksite is a solid at 300.00K. At 300.00 K its enthalpy of sublimation is 66.00 kJ/mol. The sublimation pressure at 300.00 K is 5.00 × 10⁻⁴ atm
Calculate the sublimation pressure of the solid at the melting point of 400.00 K assuming that the enthalpy of sublimation is not a function of temperature.
Answer: the sublimation pressure of the solid at the melting point is 0.3727 atm
Explanation:
Given that;
T1 = 300 K
T2 = 400 K
H_sub = 66 kJ/mol = 66000 J/mol
P1 = 5.00 × 10⁻⁴ atm
p2 = ?
now using the expression
log( p2 / 5.00 × 10⁻⁴ ) = (H_sub / R × 2.303 ) (( T2 - T1) / T1T2)
now we substitute of given values into the expression
log(p2/p1) = (66000 / 8.314 × 2.303 ) (( 400 - 300) / 300 × 400 )
p2 = 0.3727 atm
therefore the sublimation pressure of the solid at the melting point is 0.3727 atm
How are you going to experiment means how will you show your project in a real life situation like a penny being cleaned with different acids.
Answer:
The reference point is the ground.
Explanation:
•Moves slow that you don't notice hope this helps you :) god loves you :)
