The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)
<u>Answer:</u> The volume of HCl neutralized is 1.25 mL
<u>Explanation:</u>
To calculate the volume of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of stomach acid which is HCl
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Hence, the volume of HCl neutralized is 1.25 mL
Answer:
Keq = 0.217
Explanation:
Let's determine the equilibrium reaction.
In gaseous state, water vapor can be decomposed to hydrogen and oxygen and this is a reversible reaction.
2H₂(g) + O₂(g) ⇄ 2H₂O (g) Keq
Let's make the expression for the equilibrium constant
Products / Reactants
We elevate the concentrations, to the stoichiometry coefficients.
Keq = [H₂O]² / [O₂] . [H₂]²
Keq = 0.250² / 0.8 . 0.6² = 0.217
The answer is A. Water
Bronsted-Lowry base compounds are those that can accept protons
Bronsted-Lowry Acid Compounds are those that can recieve one
Water / H2O is an Amphoteric compund which mean that its molecul can act as a Base and Acid compound, so the answer is A.
The answer in chemistry is mol/L and altho it is represented by the symbol M usually.
Niether, they are equal.
1,000 cm3 = 1000 mL
