Answer: option B and option C
Explanation:
Answer:
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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:
2nd
Explanation:
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Answer:
7.5 L
Explanation:
At constant temperature and number of moles, Using Boyle's law
Given ,
V₁ = 3.00 L
V₂ = ?
P₁ = 36.74 psi = 2.5 atm (Conversion factor, 1 psi = 0.068046 atm)
P₂ = 1 atm (Atmospheric pressure as it comes to surface)
Using above equation as:
