Answer is: pH of aniline is 9.13.<span>
Chemical reaction: C</span>₆H₅NH₂(aq)+
H₂O(l) ⇌ C₆H₅NH₃⁺(aq) + OH⁻(aq).
pKb(C₆H₅NH₂) = 9.40.
Kb(C₆H₅NH₂) = 10∧(-9.4) = 4·10⁻¹⁰.
c₀(C₆H₅NH₂) = 0.45 M.
c(C₆H₅NH₃⁺) = c(OH⁻) = x.
c(C₆H₅NH₂) = 0.45 M - x.
Kb = c(C₆H₅NH₃⁺) · c(OH⁻) / c(C₆H₅NH₂).
4·10⁻¹⁰ = x² / (0.45 M - x).
Solve quadratic equation: x = c(OH⁻) = 0.0000134 M.
pOH = -log(0.0000134 M.) = 4.87.
pH = 14 - 4.87 = 9.13.
Answer:
12.7mol Na.
Explanation:
Hello there!
In this case, according to the concept of mole, which stands for the amount of substance, we can recall the concept of Avogadro's number whereby we understand that one mole of any substance contains 6.022x10²³ particles, for the given atoms of sodium, we can calculate the moles as shown below:

Thus, by performing the division we obtain:

Regards!
Answer:
B) 16 g
Explanation:
First we <u>convert 4 moles of O₂ into moles of H₂</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 4 mol O₂ *
= 8 mol H₂
Finally we <u>convert 8 moles of H₂ into grams</u>, using <em>its molar mass</em>:
- 8 mol H₂ * 2 g/mol = 16 g
Thus, the correct answer is option B).
Answer:
The charged carbon atom of a carbocation has a complete octet of valence shell electrons
Explanation:
A charged carbon atom of a carbocation has a valence shell that is not filled, <u>that's why it acts as an electrophile (or a Lewis base)</u>. This unfilled valence shell is also the reason of the nucleophilic attack that takes place during the second step of a SN1 reaction.