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:
A
Explanation:
Molecules of a gas are relatively more compressible than those of liquids and solids because they are relatively far apart without any intermolecular forces between them. However, at lower temperature and higher pressure, there is now a significant intermolecular interaction between the gas molecules and they are no longer relatively far apart. Hence they are more compressible than liquids and solids which already possess significant intermolecular interaction and thus a definite volume.
Elements of Group 1 and group 2 in the periodic
table contain elements so reactive that they are never found in the free state
<u>Explanation</u>:
The metals in group 1 of periodic table consisting of 'alkali metals' which include lithium, potassium, sodium, rubidium, Francium and caesium. They are highly reactive because they have low ionisation energy and larger radius. The group 2 metals consist of 'alkaline earth metals' which include calcium, strontium, barium, beryllium, radium and magnesium. These alkaline earth metal have +2 oxidation number, hence are highly reactive.
These both group metals are mostly reactive and so are never found in a free state. When they are exposed to air they would immediately react with oxygen. Hence, are stored in oils to avoid oxidation.
The question is incomplete, the complete question is;
Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should assume ideal van't Hoff factors where applicable. 0.100 m C6H12O6 0.100 m AlCl3 0.100 m NaCl 0.100 m MgCl2 They all have the same boiling point.
Answer:
AlCl3 0.100 m
Explanation:
Let us remember that the boiling point elevation is given by;
ΔTb = Kb m i
Where;
ΔTb = boiling point elevation
Kb = boiling point constant
m = molality of the solution
i = Van't Hoff factor
We can see from the question that all the solutions possess the same molality, ΔTb now depends on the value of the Van't Hoff factor which in turn depends on the number of particles in solution.
AlCl3 yields four particles in solution, hence ΔTb is highest for AlCl3 . The solution having the highest value of ΔTb also has the highest boiling point.
Answer:
magnesium has a greater charge, there will be greater attraction between delocalised electrons and the positively charged ion.
Explanation:
I did this once and got it right I hope you get it right too
