Answer:
36.55 J
Explanation:
PE = Potential energy
KE = Kinetic energy
TE = Total energy
The following data were obtained from the question:
Position >> PE >>>>> KE >>>>>> TE
1 >>>>>>>> 72.26 >> 27.74 >>>> 100
2 >>>>>>>> 63.45 >> x >>>>>>>> 100
3 >>>>>>>> 58.09 >> 41.91 >>>>> 100
The kinetic energy of the pendulum at position 2 can be obtained as follow:
From the table above, at position 2,
Potential energy (PE) = 63.45 J
Kinetic energy (KE) = unknown = x
Total energy (TE) = 100 J
TE = PE + KE
100 = 63.45 + x
Collect like terms
100 – 63.45 = x
x = 36.55 J
Thus, the kinetic energy of the pendulum at position 2 is 36.55 J.
Answer:
The answer to your question is 1.11 M
Explanation:
Data
volume 1 = 287 ml
concentration 1 = 1.6 M
volume 2= 412 ml
concentration 2 = ?
Formula
Volume 1 x concentration 1 = Volume 2 x concentration 2
Solve for concentration 2
concentration 2 = (volume 1 x concentration 1) / volume 2
Substitution
concentration 2 = (287 x 1.6) / 412
Simplification
concentration 2 = 459.2 / 412
Result
concentration 2 = 1.11 M
Answer:
a) Kb = 10^-9
b) pH = 3.02
Explanation:
a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:
[NaA] and [A-] = 0.05/0.6 = 0.083 M
Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9
b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:
[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M
pKb = 10^-9
Ka = 10^-5
HA = H+ + A-
Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])
[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0
Clearing [H+]:
[H+] = 0.00095 M
pH = -log([H+]) = -log(0.00095) = 3.02
The answer is A
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