According to Le cha telier's principle when we increase the pressure of a equilibrium reaction the reaction shift to the side where few moles of gas present.
Your reaction (Notice - I guess the reaction you written is wrong, but still I'm solving with your given prediction)
3NaI (g) + H2 (g) = 2NaHI (g)
Where access of sodium iodide is reacting with Hydrogen gas to form NaHI molecule.
Number of moles of gas on reaction side - 3+2 = 5 moles
Number of moles of gas on product side - 2 moles
<em>Conclusion- the reaction will shift to the right of the reaction</em>
Answer:
<h3>I think, answer is threshold energy.</h3>
OR
<h3>activation energy. </h3>
Explanation:
<h3>Hope it helps you....</h3>
<h2>Thank you..</h2>
Any buffer exists in this equilibrium
HA <=>

In a buffer, there is a large reservoir of both the undissociated acid (HA) and its conjugate base (

)
When a strong acid is added, it reacts with the large reservoir of the conjugate base (

) forming a salt and water. Since this large reservoir of the conjugate base is used, the ph does not alter drastically, but instead resist the pH change.
Density= mass/volume
= 100/25
density = 4g/ml