Given:
A compound with:
Number of carbon atoms = 9
Number of double bonds = 1
A double bond between 5th and 6th carbon
A propyl group (CH2CH2CH3) branching off the 3rd carbon from the left
Try to illustrate the given and observe the formation of the atoms. Now, follow the correct IUPAC naming system. The name of the compound is
4-propyl-1-hexene
Count from the right to the left, the double bond is between the 1st and 2nd carbon, thus, 1-hexene. The propyl branches out the 4th carbon from the right, thus 4-propyl.
Answer:
a=28600J; b=90.6 J/K; c=402 torr
Explanation:
(a) considering the data given
Vapour pressure P1 =0 at Temperature T1 = 42.43˚C,
Vapour pressure P2 = 273.15 at Temperature T2= 315.58 K)
Using the Clausius-Clapeyron Equation
ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)
In 760/140 = ΔH/8.314 J/mol/K × (1/315.58K -- 1/273.15K)
ΔH vap= +28.6 kJ/mol or 28600J
(b) using the Equation ΔG°=ΔH° - TΔS to solve forΔS.
Since ΔG at boiling point is zero,
ΔS =(ΔH°vap/Τb)
ΔS = 28600 J/315.58 K
= 90.6 J/K
(c) using ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)
ln P298 K/1 atm = 28600 J/8.314 J/mol/K × (1/298.15K - 1/315.58K)
P298 K = 0.529 atm
= 402 torr
Making toast is a physical change.
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We all know, Density = Mass/ volume.
When volume increases and the mass remains the same , the density will decrease considerably.
Answer:
The answer would be cation
Explanation:
