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.
<h2>Question:- </h2>
A solution has a pH of 5.4, the determination of [H+].
<h2>Given :- </h2>
- pH:- 5.4
- pH = - log[H+]
<h2>To find :- concentration of H+</h2>
<h2>Answer:- Antilog(-5.4) or 4× 10-⁶</h2>
<h2>Explanation:- </h2><h3>Formula:- pH = -log H+ </h3>
Take negative to other side
-pH = log H+
multiple Antilog on both side
(Antilog and log cancel each other )
Antilog (-pH) = [ H+ ]
New Formula :- Antilog (-pH) = [+H]
Now put the values of pH in new formula
Antilog (-5.4) = [+H]
we can write -5.4 as (-6+0.6) just to solve Antilog
Antilog ( -6+0.6 ) = [+H]
Antilog (-6) × Antilog (0.6) = [+H]

put the value in equation
![{10}^{ - 6} \times 4 = [H+] \\ 4 \times {10}^{ - 6} = [H+]](https://tex.z-dn.net/?f=%20%7B10%7D%5E%7B%20-%206%7D%20%20%20%5Ctimes%204%20%3D%20%5BH%2B%5D%20%5C%5C%204%20%5Ctimes%20%20%20%7B10%7D%5E%7B%20-%206%7D%20%20%3D%20%5BH%2B%5D)
Answer:
6 atm
Explanation:
Using the formula P1V1=P2V2
P1= Initial Pressure
V1= Initial Volume
P2= Final Pressure
V2= Final Volume
And knowing that at stp gas will always be at 1 atm
250L(P2) = 1500
P2= 6 atm
Answer:
The reactions free energy 
Explanation:
From the question we are told that
The pressure of (NO) is 
The pressure of (Cl) gas is 
The pressure of nitrosly chloride (NOCl) is 
The reaction is
⇆ 
From the reaction we can mathematically evaluate the
(Standard state free energy ) as

The Standard state free energy for NO is constant with a value

The Standard state free energy for
is constant with a value

The Standard state free energy for
is constant with a value

Now substituting this into the equation

The pressure constant is evaluated as

Substituting values


The free energy for this reaction is evaluated as

Where R is gas constant with a value of 
T is temperature in K with a given value of 
Substituting value
![\Delta G = -43 *10^{3} + 8.314 *298 * ln [0.0765]](https://tex.z-dn.net/?f=%5CDelta%20%20G%20%20%3D%20-43%20%2A10%5E%7B3%7D%20%2B%208.314%20%2A298%20%2A%20ln%20%5B0.0765%5D)

