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
Answer:
Robert
Explanation:
There is not more than one colour
Explanation:
Dehydrohalogenation reactions occurs as elimination reactions through the following mechanism:
Step 1: A strong base(usually KOH) removes a slightly acidic hydrogen proton from the alkyl halide.
Step 2: The electrons from the broken hydrogen‐carbon bond are attracted toward the slightly positive carbon (carbocation) atom attached to the chlorine atom. As these electrons approach the second carbon, the halogen atom breaks free.
However, elimination will be slower in the exit of Hydrogen atom at the C2 and C3 because of the steric hindrance by the methyl group.
Elimination of the hydrogen from the methyl group is easier.
Thus, the major product will A