You have to use the Henderson-Hasselbalch equation. Keep in mind that because the Pka is given the equation changes form slightly:
PH = Pka + log[acid/base]
Step 1 (Figure out the concentrations):
0.282 M of Acid (C6H5OOH) - 0.150 M = 0.132 M of acid
0.282 M of Base (C6HCOO) + 0.150 M = 0.432 M of bas3
Step 2 (Plug into equation):
PH = Pka + log[acid/base]
PH = 4.20 + log[0.132 M/0.432 M]
PH = 3.69
The formation of nitric acid from nitrogen, hydrogen, and oxygen can be written as,
N₂ + H₂ + 3O₂ --> 2HNO₃
The net enthalpy of formation of nitric acid is calculated by,
Hrxn = Hproduct - Hreactant
Since all the reactants are in their elemental forms, the simplified equation would be,
Hrxn = Hproduct
Substituting,
Hrxn = (-186.81 kJ/mol)(2 mols)
<em>Answer: -372.42 kJ</em>
Answer:
See explanation
Explanation:
The first step in this reaction is a unimolecular reaction. It involves the formation of the carbocation. This is so because tertiary alkyl halides only undergo substitution by SN1 mechanism due to sterric crowding.
The second step in the reaction is bi molecular. In this step, the carbocation now combines with the OH^- to yield the alcohol.
Net equation of the reaction is;
(CH3)3CBr + OH^- -------> (CH3)3COH + Br^-
The intermediate here is the carbocation, (CH3)3C^+
