Answer: -
1.8
End point passed.
Explanation: -
Volume of HI solution = 47.0 mL = 0.047 L
Strength of HI solution = 0.47 M
Since HI is a strong acid, all of HI will dissociate to give H +.
[H+ ] =0.47 M x 0.047 L
= 0.02209 mol
Volume of KOH = 25.0 mL = 0.025 L
Strength of KOH = 0.25 M
Since KOH is a strong base, all of KOH will dissociate to give OH-.
[OH-] = 0.25 M x 0.025L
= 0.00625 mol
Since [H+] and [OH-] react to form water,
[H+] unreacted = 0.02209 – 0.00625 = 0.01584 mol
Using the formula
pH = - log [H+]
= - log 0.01584
= 1.8
As the strong acid HI is being titrated by strong base KOH, the pH at the end point should be 7.
The pH has already crossed that. Thus the titration end point has already passed
After titration, there are more moles of HI than KOH, implying excess HI (acid) is present. The remaining acid concentration is 0.2 M and consequently, the final pH of the solution is approximately 0.70.
In the case of the titration of a 47.0 mL of 0.47 M HI solution with 25.0 mL of 0.25 M KOH, we first need to understand that HI is a strong acid and KOH is a strong base. When we titrate a strong acid with a strong base, the equivalence point occurs at a pH of 7.0.
First, we calculate the moles of the acid and the base: moles of HI = 0.47 mol/L * 0.047 L = 0.02209 mol, and moles of KOH = 0.25 mol/L * 0.025 L = 0.00625 mol. Since there are more moles of HI than KOH, we will have extra HI left after the titration. Hence, it is a strong acid-strong base titration before the equivalence point i.e. when we have excess acid.
The remaining acid concentration is (0.02209 mol - 0.00625 mol) / (0.047 L + 0.025 L) = 0.2 M and pH of a strong acid is basically the negative logarithm of the acid's concentration. Therefore, the pH is -log[H+] = -log(0.2) = approx. 0.70.
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B. 1/1 H
C. 0/1 H
D. 2/1 H
Davisson and Germer is not the correct answer, I just finished the test. The answer has to be Maxwell I hope this helps someone else fo sho
The work of Davisson and Germer helped to explain light's ability to propagate through a vacuum. They designed and built a vacuum apparatus for the purpose of measuring the energies of electrons scattered from a metal surface. Electrons from a heated filament were accelerated by a voltage and allowed to strike the surface of nickel metal. They observed that at certain angles, there was a peak in the intensity of the scatters electron beam. The peak indicated wave behavior for the electrons.
c. molecular
b. elemental
d. radial
Answer:
c. molecular
Explanation:
b. The valence electrons stay in a single orbital.
c. The valence electrons break free of the crystals.
d. The valence electrons move closer to the nuclei.
Answer: THE answer is A
Explanation: took the exam