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:
75 mg
Explanation:
We can write the extraction formula as
x = m/[1 + (1/K)(Vaq/Vo)], where
x = mass extracted
m = total mass of solute
K = distribution coefficient
Vo = volume of organic layer
Vaq = volume of aqueous layer
Data:
m = 75 mg
K = 1.8
Vo = 0.90 mL
Vaq = 1.00 mL
Calculations:
For each extraction,
1 + (1/K)(Vaq/Vo) = 1 + (1/1.8)(1.00/0.90) = 1 + 0.62 = 1.62
x = m/1.62 = 0.618m
So, 61.8 % of the solute is extracted in each step.
In other words, 38.2 % of the solute remains.
Let r = the amount remaining after n extractions. Then
r = m(0.382)^n.
If n = 7,
r = 75(0.382)^7 = 75 × 0.001 18 = 0.088 mg
m = 75 - 0.088 = 75 mg
After seven extractions, 75 mg (99.999 %) of the solute will be extracted.
Answer:
[ S2- ] = 4.0 E-47 M
Explanation:
- PbS(s) → Pb2+ + S2-
- HgS(s) → Hg2+ + S2-
∴ Ksp PbS = 3.4 E-28 = [Pb2+]*[S2-]
∴ [Pb2+] = 0.181 M
∴ Ksp HgS = 4.0 E-53 = [Hg2+]*[S2-]
∴ [Hg2+] = 0.174 M
∴ Ksp PbS > Ksp HgS ⇒ precipitate first Hg2+:
∴ [ Hg2+ ] = 1.0 E-6 M
⇒ [S2-] = 4.0 E-53 / 1.0 E-6 = 4.0 E-47 M
There would be 79 electrons present in each atom of gold. I hope this helps :)