Place the following compounds in order of increasing vapor pressure. rubbing alcohol [ (ch3)2choh) ]; propane, diethyl ether (ch3ch2-o-ch2ch3) a. diethyl ether < propane < rubbing alcohol b. diethyl ether < rubbing alcohol < propane c. rubbing alcohol < propane < diethyl ether d. propane < diethyl ether < rubbing alcohol e. propane < rubbing alcohol < diethyl ether f. rubbing alcohol < diethyl ether < propane

Vapor pressure is the pressure which is exerted by the vapors of a liquid in equilibrium with the liquid. Vapor pressure depends on the boiling point; lower the boiling point higher will be the vapor pressure.

Based on the given compounds:

rubbing alcohol is polar and has the strongest intermolecular force of attraction followed by relatively polar ether and finally the non-polar propane.

The increasing order of boiling point will be:

propane < diethyl ether < rubbing alcohol

Thus, the increasing order of vapor pressure will be:

rubbing alcohol < diethyl ether < propane

Ans : F

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%