Devin investigated whether salt reduces the boiling point of water. She poured 1 gallon of 18
2 answers:
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The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2M was neutralized by 0.01 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 0.125 mL
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is HCl (Stomach acid)
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 0.125 mL
Answer:
2,47 * (molecules)
Explanation:
Avogadro's number is a very important relationship: 1 mole = 6,02 * atoms, molecules, protons, etc.
To convert from moles to molecules, multiply the molar amount by Avogadro's number.
4,1 mol * 6,02 * ≈ 24,7 *
= 2,47 *
(molecules)
Therefore the third answer is correct