Answer:
It is the third one
Explanation:
"Both atoms achieve a more stable configuration."
Answer:
ΔE = -2661 KJ/mole
ΔH = -2658 KJ/mole
Explanation:
ΔH = q - PΔV
ΔE = q + w
<u>First, to find ΔE:</u>
The reaction PRODUCES 2658 kJ of h (q), and does 3 kJ of work (w).
2658 kJ(q) + 3 kJ(w) = 2661 kJ, BUT the reaction <u><em>PRODUCES</em></u> heat, which means ΔE is negative.
ΔE = -2661 KJ/mole
<u>Second, to find ΔH:</u>
ΔH = q - PΔV
ΔH = 2658 kJ(q) - PΔV
Now, the question states that butane burns at a constant pressure; that just translates to the pressure of the reaction is equal to 0.
ΔH = 2658 KJ(q) - (0)ΔV
ΔH = 2658 KJ - 0
ΔH = 2658 kJ, BUT, like before, the reaction PRODUCES heat, which also mean ΔH is negative.
ΔH = -2658 KJ/mole
I hope this helped! Have a nice week.
The molar mass is 64.07g/mo
Answer:
New pH = 3.84
Explanation:
First of all we may think that if the buffer has pH 3.98 and we're adding H⁺, pH's buffer will be lower, as the [H⁺] is been increased.
Let's determine the moles of each compound:
0.23 M . 1.3L = 0.299 moles of NaBz
0.38 M . 1.3L = 0.494 moles of HBz
We add 0.058 of HCl, which is the same as 0.058 moles of H⁻
HCl → H⁺ + Cl⁻
As we add the moles of protons, these are going to react to the Bz⁻
In the buffer system we have these dissociations:
NaBz → Na⁺ + Bz⁻
HBz → H⁺ + Bz⁻
So, as we add protons, we have a new equilibrium:
Bz⁻ + H⁺ ⇄ HBz
In 0.299 0.058 0.494
Eq 0.241 - 0.552
Protons are substracted to benzoate, so the [HBz] is now higher than before. We calculate the new pH, with the Henderson Hasselbach equation
pH = pKa + log (Bz⁻/HBz)
pH = 4.20 + log (0.241 / 0.552) → 3.84
