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.
There are a number of
ways to express concentration of a solution. This includes molarity. Molarity
is expressed as the number of moles of solute per volume of the solution. The
concentration of the solution is calculated as follows:
<span> </span><span>Molarity = 15.5 g NaOH (1 mol NaOH / 40 g NaOH) / .250 L
solution</span>
<span>Molarity = 1.55 M</span>
It would be CH2! you’re just simplifying C4H8, 4 can go into C4 1 time (so we just say C) and 4 can go into H8 2 times (H2)
Answer:
i odnt knwonwta whyour sayong
Explanation: