I believe it’s Non polar molecule
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:
Molarity = 0.24 M
Explanation:
Given data:
Number of moles of Mg(NO₃)₂ = 0.181 mol
Volume of solution = 0.750 L
Molarity of solution = ?
Solution:
Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.
Formula:
Molarity = number of moles of solute / L of solution
by putting values,
Molarity = 0.181 mol / 0.750 L
Molarity = 0.24 M
Answer:
V = 6.17 L
Explanation:
Given data:
Volume = ?
Number of moles = 0.382 mol
Pressure = 1.50 atm
Temperature = 295 k
R = 0.0821 L. atm. /mol. k
Solution:
According to ideal gas equation:
PV= nRT
V = nRT/P
V = 0.382 mol × 0.0821 L. atm. /mol. k ×295 k / 1.50 atm
V = 9.252 L. atm. / 1.50 atm
V = 6.17 L
