Answer:
δ N2(g) = 1.1825 g/L
Explanation:
- δ ≡ m/v
- Mw N2(g) = 28.0134 g/mol
ideal gas:
∴ P = (837 torr)×( atm/760 torr) = 1.1013 atm
∴ T = 45.0 °C + 273.15 = 318.15 K
∴ R = 0.082 atm.L/K.mol
⇒ n/V = P/R.T
⇒ n/V = (1.1013 atm) / ((0.082 atm.L/K.mol)(318.15 k))
⇒ n/V = 0.0422 mol/L
⇒ δ N2(g) = (0.042 mol/L)×(28.0134 g/mol) = 1.1825 g/L 
 
        
             
        
        
        
The ionization equation is:
HF ⇄ H(+) + F(-)
The ionization constant is Ka = [H(+)] * [H(-)] / [HF]
=> [H(+)] * [F(-)] = Ka * [HF]
Given that Ka < 1
[H(+)] * [F(-)] < [HF]
Which is [HF] >  [H(+)] * [F(-)] the option a. fo the list of choices. 
        
                    
             
        
        
        
Explanation:
It is known that  value of acetic acid is 4.74. And, relation between pH and
 value of acetic acid is 4.74. And, relation between pH and  is as follows.
 is as follows.
                     pH = pK_{a} + log ![\frac{[CH_{3}COOH]}{[CH_{3}COONa]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOOH%5D%7D%7B%5BCH_%7B3%7DCOONa%5D%7D)
                           = 4.74 + log 
So, number of moles of NaOH = Volume × Molarity
                                                    = 71.0 ml × 0.760 M
                                                     = 0.05396 mol
Also, moles of   = moles of
 = moles of 
                                           = Molarity × Volume
                                           = 1.00 M × 1.00 L
                                           = 1.00 mol
Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.
             
Initial :    1.00 mol                                  1.00 mol
NaoH addition:               0.05396 mol
Equilibrium : (1 - 0.05396 mol)    0           (1.00 + 0.05396 mol)
                     = 0.94604 mol                       = 1.05396 mol
As, pH = pK_{a} + log ![\frac{[CH_{3}COONa]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOONa%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)
                = 4.74 +  log 
                = 4.69
Therefore, change in pH will be calculated as follows.
                          pH = 4.74 - 4.69
                                = 0.05 
Thus, we can conclude that change in pH of the given solution is 0.05.
 
        
             
        
        
        
Refer to attachment for your answer