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
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 
= 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