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:
B. An element whose particles are molecules.
Explanation:
Answer:
A. 3.7 x 10⁻³
Explanation:
In the beginning of the reaction there are 0.200 moles of reactant. After 25 minutes, remain 0.108 moles. That means the moles that wer descomposed are:
0.200 moles - 0.108 moles = 0.092 moles of reactant were descomposed.
That descomposition occurs in 25 minutes. The average rate of descomposition in moles / minute are:
0.092 moles Methyl isonitrile / 25 minutes = 3.7x10⁻³ mol/min.
Right option is:
<h3>A. 3.7 x 10⁻³</h3>
Answer: A, 1.3*1020, 1326
A. 1326
B. 2960.9
C. 9804
D. 8559.6
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