This problem is to use the Claussius-Clapeyron Equation, which is:
ln [p2 / p1] = ΔH/R [1/T2 - 1/T1]
Where p2 and p1 and vapor pressure at estates 2 and 1
ΔH is the enthalpy of vaporization
R is the universal constant of gases = 8.314 J / mol*K
T2 and T1 are the temperatures at the estates 2 and 1.
The normal boiling point => 1 atm (the pressure of the atmosphere at sea level) = 101,325 kPa
Then p2 = 101.325 kPa
T2 = ?
p1 = 54.0 kPa
T1 = 57.8 °C + 273.15K = 330.95 K
ΔH = 33.05 kJ/mol = 33,050 J/mol
=> ln [101.325/54.0] = [ (33,050 J/mol) / (8.314 J/mol*K) ] * [1/x - 1/330.95]
=> 0.629349 = 3975.22 [1/x - 1/330.95] = > 1/x = 0.000157 + 1/330.95 = 0.003179
=> x = 314.6 K => 314.6 - 273.15 = 41.5°C
Answer: 41.5 °C
Answer:
0.02 moles.
Explanation:
volume of H₂ gas at R.T.P = 480 cm³
Where
R.T.P = room temperature and pressure
molar volume of gas at = 24000 cm³
no. of moles of hydrogen = ?
Solution:
formula Used
no. of moles = volume of gas / molar volume
put values in above equation
no. of moles = 480 cm³ / 24000 cm³/mol
no. of moles = 0.02 mol
So,
no. of moles of hydrogen in 480 cm³ is 0.02 moles.
Answer:
A is the correct option
Explanation:
batteries have chemical energy and will convert to electricity and when reached to bulb, it emits light as electromagnetic rays
Answer:
On the outside or something like that
