Answer:
The volume of the gas at given temperature is 
Explanation:
 = initial pressure of gas = 1 atm
 = initial pressure of gas = 1 atm
 = initial temperature of gas =
 = initial temperature of gas =  
 = initial volume of gas =
 = initial volume of gas =  
( , 1 mL = 0.001 L)
 , 1 mL = 0.001 L)
 ..[1]
..[1]
 = final pressure of gas = 2.1 atm
 = final pressure of gas = 2.1 atm
 = final temperature of gas =
 = final temperature of gas =  
 = final volume of gas = ?
 = final volume of gas = ?
 ..[2]
..[2]
By dividing [1] and [2] we get combined gas equation :,
 

Now put all the given values in the above equation, we get:
 

The volume of the gas at given temperature is 
 
        
                    
             
        
        
        
Answer:
A. weak acid and its conjugate base
Explanation:
A buffer solution can be made with a weak acid and conjugate base or a weak base and conjugate acid.
This may help you:
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Buffers/Introduction_to_Buffers
 
        
             
        
        
        
Answer:
Explanation:
The balanced reaction would be:
H3PO4+3KOH==>K3PO4+3H2O
 
        
             
        
        
        
Answer:
 for the given reaction is -99.4 J/K
 for the given reaction is -99.4 J/K
Explanation:
Balanced reaction: 
![\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20S%5E%7B0%7D%28NH_%7B3%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28N_%7B2%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28H_%7B2%7D%29_%7Bg%7D%5D)
where  represents standard entropy.
 represents standard entropy.
Plug in all the standard entropy values from available literature in the above equation:
![\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20192.45%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20191.61%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20130.684%5Cfrac%7BJ%7D%7Bmol.K%7D%5D%3D-99.4J%2FK)
So,  for the given reaction is -99.4 J/K
 for the given reaction is -99.4 J/K
 
        
             
        
        
        
False because particles stop moving or study slow down.