Answer:
21.5mL of a 0.100M HCl are required
Explanation:
The sodium phenoxide reacts with HCl to produce phenol and NaCl in a 1:1 reaction.
To solve this question we need to find the moles of sodium phenoxide. These moles = Moles of HCl required to reach equivalence point and, with the concentration, we can find the needed volume as follows:
<em>Mass NaC6H5O:</em>
1.000g * 25% = 0.250g NaC6H5O
<em>Moles NaC6H5O -116.09g/mol-</em>
0.250g NaC6H5O * (1mol/116.09g) = 2.154x10⁻³ moles = Moles of HCl required
<em>Volume 0.100M HCl:</em>
2.154x10⁻³ moles HCl * (1L/0.100mol) = 0.0215L =
<h3>21.5mL of a 0.100M HCl are required</h3>
Answer:
t = 1862 s
Explanation:
To do this, we need first to determine the theorical detention time, which can be determined with the following expression:
t₀ = ∀/Q (1)
Where:
t₀: detention time
∀: Volume of the fluid in the reactor
Q: Flow rate in the reactor
With this time, we must use the following expression to determine the time that the workers will take to vent the tank:
C = C₀ e^(-t/t₀) (2)
From here, we must solve for time t, and the expression will be:
t = ln(C₀/C) * t₀ (3)
Now that we know the expression to use, let's solve for t. Using (1) to determine the detention time, ∀ is 1900 m³, and Q is 2.35 m³/s so:
t₀ = 1900 / 2.35 = 808.51 s
Now, let's solve for the time t. C will be 0.0015 mg/L (or 1.5 mg/m³ cause in 1 m³ we have 1000 L) and C₀ 15 mg/m³:
t = ln(15/1.5) * 808.51
<h2>
t = 1861.66 s or simply 1862 s</h2><h2>
</h2>
Hope this helps
A substance with at least one hydrogen atom that can dissociate to form an anion and an H+ ion (a proton) in aqueous solution, thereby foming an acidic solution.
A substance that produces one or more hydroxide ions (OH−) and a cation when dissolved in aqueous solution, thereby forming a basic solution.
According to Hasselbalch equation pH = pKa at the equal point. When pH equals the pKa value of the indicator, the base and acid forms of indicator are present in the ratio of 1:1
(9.3 + 10.5) / 2 =9.9
pKa = 9.9
Using the equation pH = pKa at equal point we get
Ka = 10^- pKa
Ka = 10^- 9.9
Ka = 1.26 x 10^-10
Ka = 1 x 10^-10
Answer:
The student did not lose any organic product in the aqueous layer
Explanation:
From the question;
After addition of hydrochloric acid, a student removes the lower aqueous layer and extracts the aqueous phase with diethyl ether (0.5 mL).
The only possible way the student could have lose the product was if there are some basic group e.g 
in the compound which have a tendency of forming 
salt.
Also these compound could be regenerated from aqueous layer by neutralizing the aqueous layer with basic solution , for example; by using sodium hydroxide (NaOH) in which the extraction is then followed by the usage of Diethyl ether.
