Answer:
I think it would be the last answer
Explanation:
The answer to this question will be C
D is the correct answer......
hope it help, please thank me if it did.....
Given the percentage composition of HC as C → 81.82 % and H → 18.18 %
So the ratio of number if atoms of C and H in its molecule can will be:
C : H = 81.82 12 : 18.18 1 C : H = 6.82 : 18.18 = 6.82 6.82 : 18.18 6.82 = 1 : 2.66 ≈ 3 : 8
So the Empirical Formula of hydrocarbon is:
C 3 H 8
As the mass of one litre of hydrocarbon is same as that of C O 2 The molar mass of the HC will be same as that of C O 2 i.e 44 g mol
Now let Molecular formula of the HC be ( C 3 H 8 ) n
Using molar mass of C and H the molar mass of the HC from its molecular formula is:
( 3 × 12 + 8 × 1 ) n = 44 n So 44 n = 44 ⇒ n = 1
Hence the molecular formula of HC is C 3 H 8
Does that help?
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
