Classius claperyon equation
In (P2/ P2) = ΔHvap/R) × (1/T2-1/T1)
T2 occurs at normal boiling when vapor pressure P2 = 1 atm.
P1 = 55.1 mmHg, P2 = 1 atm = 760mmHg
T1 = 35°c = 308.15k, T2 =
ΔHvap = 32.1kJ/mol = 32100 J/mol
In (760/55.1) = (-32100/ 8.314) × ( 1/T2 - 1/308.15)
The normal boiling point T2 = 390k = 117°c
Answer:
4.9 x 10^-19 J, 2.7 x 10^-19 J
Explanation:
first wavelength, λ1 = 410 nm = 410 x 10^-9 m
Second wavelength, λ2 = 750 nm = 750 x 10^-9 m
The relation between the energy and the wavelength is given by
E = h c / λ
Where, h is the Plank's constant and c be the velocity of light.
h = 6.63 x 10^-34 Js
c = 3 x 10^8 m/s
So, energy correspond to first wavelength
E1 = (6.63 x 10^-34 x 3 x 10^8) / (410 x 10^-9) = 4.85 x 10^-19 J
E1 = 4.9 x 10^-19 J
So, energy correspond to second wavelength
E2 = (6.63 x 10^-34 x 3 x 10^8) / (750 x 10^-9) = 2.652 x 10^-19 J
E2 = 2.7 x 10^-19 J
The answer is definitely C.) their molecules move at the same average speed
Answer:
B-flood
Explanation:
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2.4 secs I think
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