Round to 4 significant figures. 0.007062
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
Answer:
a. alkyne
b. alkane
c. alkyne
d. alkene
Explanation:
The general formula for each class of compound is given below
Alkane:
Alkene:
Alkyne: (assuming single multiple bonds)
Now let us classify according to the above formulas:
a. It has two hydrogen atoms less than the two times of carbon atoms hence, it's alkyne
b. It has two hydrogen atoms more than the two times of carbon atoms hence, it's alkane
c. It has two hydrogen atoms less than the two times of carbon atoms hence, it's alkyne
d. It has hydrogen atoms two times of carbon atoms hence, it's alkene