Nitration of aniline?
2 answers:
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Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.
There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6
Answer : The number of iron atoms present in each red blood cell are,
Explanation :
First we have to calculate the moles of iron.
Now we have to calculate the number of iron atoms.
As, 1 mole of iron contains number of iron atoms
So, 0.0519 mole of iron contains number of iron atoms
Now we have to calculate the number of iron atoms are present in each red blood cell.
Number of iron atoms are present in each red blood cell =
Number of iron atoms are present in each red blood cell =
Number of iron atoms are present in each red blood cell =
Therefore, the number of iron atoms present in each red blood cell are,