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:
C. The reaction can be broken down and performed in steps
Explanation:
Hess's Law of Constant Heat Summation states that irrespective of the number of steps followed in a reaction, the total enthalpy change for the reaction is the sum of all enthalpy changes corresponding to all the steps in the overall reaction. The implication of this law is that the change of enthalpy in a chemical reaction is independent of the pathway between the initial and final states of the system.
To obtain MgO safely without exposing magnesium to flame, the reaction sequence shown in the image attached may be carried out. Since the enthalpy of the overall reaction is independent of the pathway between the initial and final states of the system, the sum of the enthalpy of each step yields the enthalpy of formation of MgO.
Answer: 6.2 grams of the sodium acetate can dissolve in 5 milliliters of water. if 124 grams of the sodium acetate dissolves in 100 milliliters of water, then 6.2 grams of the sodium acetate can dissolve in 5 milliliters of water.
We determine the percent by mass of water in the compound by dividing the mass of water by the total mass. The total mass of Na2SO4.10H2O is equal to 322 g. The mass of water is 180 g.
percent by mass of water = (180 / 322)*(100 %) = 55.9%
The medium provides an opposing force to slow down the wave.
