Answer:
c. Compound 2 is more acidic because its conjugate base is more resonance stabilized
Explanation:
You haven't told us what the compounds are, so let's assume that the formula of Compound 1 is HCOCH₂OH and that of Compound 2 is CH₃COOH.
The conjugate base of 2 is CH₃COO⁻. It has two important resonance contributors, and the negative charge is evenly distributed between the two oxygen atoms.
CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺
The stabilization of the conjugate base pulls the position of equilibrium to the right, so the compound is more acidic than 1.
The original question is to find the pH and the pOH of 0.023 M of perchloric acid.
Answer:
pH = 1.638
pOH = 12.362
Explanation:
1- getting the pH:
pH can be calculated using the following rule:
pH = -log[H+]
Since the given solution is an acid, this means that [H+] is the same as the concentration of the solution.
This means that:
[H+] = 0.023
Substitute in the above equation to get the pH as follows:
pH = -log[0.023]
pH = 1.638
2- getting the pOH:
We know that:
pH + pOH = 14
We have calculated that pH = 1.638.
Substitute in the above equation to get the pOH as follows:
pOH + 1.638 = 14
pOH = 14 - 1.638
pOH = 12.362
Hope this helps :)
Answer:
∆H° rxn = - 93 kJ
Explanation:
Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.
We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.
N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)
1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)
∆H° rxn = ∑ H bonds broken - ∑ H bonds formed
∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]
∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ
be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .
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