What is the volume of 2.268mol of oxygen gas at 35○C and 3.0atm?
1 answer:
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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
<u>Answer:</u> The number of moles of weak acid is moles.
<u>Explanation:</u>
To calculate the moles of KOH, we use the equation:
We are given:
Volume of solution = 43.81 mL = 0.04381 L (Conversion factor: 1L = 1000 mL)
Molarity of the solution = 0.0969 moles/ L
Putting values in above equation, we get:
The chemical reaction of weak monoprotic acid and KOH follows the equation:
By Stoichiometry of the reaction:
1 mole of KOH reacts with 1 mole of weak monoprotic acid.
So, of KOH will react with =
of weak monoprotic acid.
Hence, the number of moles of weak acid is moles.