Answer:
Proton is a sub atomic particle while oxygen is an atom ....
 
        
                    
             
        
        
        
Answer:
What do <u>YOU</u> think?
Explanation:
This question is asking for an opinion. The word should is included.
 
        
             
        
        
        
Answer:It is a picture with a boat sailing on the water with sound waves going down to an old and sunken ship but the waves then go back up to the boat in a different angle.
Explanation:
hope it works thx
 
        
             
        
        
        
"Hypotonic" is the one solution among the choices given in the question that would <span>most likely cause a plant placed in it to become firmer and more rigid. The correct option among all the options that are given in the question is the second option or option "B". I hope the answer has come to your great help.</span>
        
                    
             
        
        
        
<u>Answer:</u> The vapor pressure of the liquid is 0.293 atm
<u>Explanation:</u>
To calculate the vapor pressure of the liquid, we use the Clausius-Clayperon equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
 = initial pressure which is the pressure at normal boiling point = 1 atm
 = initial pressure which is the pressure at normal boiling point = 1 atm
 = pressure of the liquid = ?
 = pressure of the liquid = ?
 = Heat of vaporization = 28.9 kJ/mol = 28900 J/mol     (Conversion factor: 1 kJ = 1000 J)
 = Heat of vaporization = 28.9 kJ/mol = 28900 J/mol     (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
 = initial temperature = 341.88 K
 = initial temperature = 341.88 K
 = final temperature = 305.03 K
 = final temperature = 305.03 K
Putting values in above equation, we get:
![\ln(\frac{P_2}{1})=\frac{28900J/mol}{8.314J/mol.K}[\frac{1}{341.88}-\frac{1}{305.03}]\\\\\ln P_2=-1.228atm\\\\P_2=e^{-1.228}=0.293atm](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7B1%7D%29%3D%5Cfrac%7B28900J%2Fmol%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B341.88%7D-%5Cfrac%7B1%7D%7B305.03%7D%5D%5C%5C%5C%5C%5Cln%20P_2%3D-1.228atm%5C%5C%5C%5CP_2%3De%5E%7B-1.228%7D%3D0.293atm)
Hence, the vapor pressure of the liquid is 0.293 atm