Insert the values from the question above for x, y, z, & n into KxFey(C204)z nH20 anod provide the correct coefficients to b
1 answer:
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Answer:
Phenols do not exhibit the same pka values as other alcohols;
They are generally more acidic.
Using the knowledge that hydrogen acidity is directly related to the stability of the anion formed, explain why phenol is more acidic than cyclohexane.
Explanation:
According to Bromsted=Lowry acid-base theory,
an acid is a substance that can release ions when dissolved in water.
So, acid is a proton donor.
If the conjugate base of an acid is more stable then, that acid is a strong acid.
In the case of phenol,
the phenoxide ion formed is stabilized by resonance.
The resonance in phenoxide ion is shown below:
Whereas in the case of cyclohexanol resonance is not possible.
So, cyclohexanol is a weak acid compared to phenol.
Answer:
2.387 mol/L
Explanation:
The reaction that takes place is:
- 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O
First we <u>calculate how many moles of each reagent were added</u>:
- HCl ⇒ 200.0 mL * 3.85 M = 203.85 mmol HCl
- Ba(OH)₂ ⇒ 100.0 mL * 4.6 M = 460 mmol Ba(OH)₂
460 mmol of Ba(OH)₂ would react completely with (2*460) 920 mmol of HCl. There are not as many mmoles of HCl so Ba(OH)₂ will remain in excess.
Now we <u>calculate how many moles of Ba(OH)₂ reacted</u>, by c<em>onverting the total number of HCl moles to Ba(OH)₂ moles</em>:
- 203.85 mmol HCl *
= 101.925 mmol Ba(OH)₂
This means the remaining Ba(OH)₂ is:
- 460 mmol - 101.925 mmol = 358.075 mmoles Ba(OH)₂
There are two OH⁻ moles per Ba(OH)₂ mol:
- OH⁻ moles = 2 * 358.075 = 716.15 mmol OH⁻
Finally we <u>divide the number of OH⁻ moles by the </u><u><em>total</em></u><u> volume</u> (100 mL + 200 mL):
- 716.15 mmol OH⁻ / 300.0 mL = 2.387 M
So the answer is 2.387 mol/L
<u>Answer:</u> 6.57 L of solution can be made.
<u>Explanation:</u>
Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:
.....(1)
Given values:
Molarity of LiBr = 3.5 M
Moles of LiBr = 23 moles
Putting values in equation 1, we get:
Hence, 6.57 L of solution can be made.