Reviewer For
Chemical Engineering
Licensure Examination
3rd Edition
Solutions Manual
Monroe H. De Guzman, Bsche
Batangas State University
,I. Physical and Chemical Principles
A. General Inorganic Chemistry
4. For Zn2+, Atomic Number = 30, Atomic Mass = 65.38 (~65),
Number Of Protons = 30; Neutrons = 65-30 = 35; Electrons = 30 – 2 = 28
E- + Proton + Neutron = 30 + 35 + 28 = 93
6. The Balmer Series Or Balmer Lines In Atomic Physics, Is The Designation Of One Of A Set Of Six
Different Named2 Series Describing The Spectral Line Emissions Of The Hydrogen Atom (Wikipedia, 2014).
𝑛
𝜆=( )(𝑅∞ /4); For M = 2 As N → ∞
𝑛2−𝑚2
Balmer Series = 1/𝜆 = ∞ (𝑅 /4) = 10973731.57 M/4 =
∞ ∞ 27434.3289 cm-1
7. Named After The American Physicist Frederick Sumner Brackett Who First Observed The
Spectral Lines In 1922 (Wikipedia,
1 1 2014).
)(𝑅 ∞ ); M = 4 As N → ∞ = 10973731.57 M/16 = 6858.5822 cm-1
Brackett Series = ( −
2 2
𝑚 𝑛
10. Minimum Wavelength Of Light For Work Functions (De Broglie
Wavelength) For Φ = 2.90 Ev (J/C Of Electron)
ℎ𝑐𝑜 (6.6261 𝑥 10−34)(299792458)
𝜆= = (2.90)(1.6022 𝑥 10−19)
= 427.5317 nm
𝜙𝑒
11. For Φ = 5.00 Ev
ℎ𝑐𝑜 (6.6261 𝑥 10−34)(299792458)
𝜆= = (5.00)(1.6022 𝑥 10−19)
= 247.9584 nm
𝜙𝑒
13. Work Function And Kinetic Energy
Rb Φ = 2.16 Ev; Λ = 350 Nm Of Light
Wavelength. Ephoton = Φ + Ek = Hv = Hco/Λ; Ek =
Hco/Λ – Φ
(6.6261 𝑥 10−34)(299792458)
EK = 350 𝑋 10 −9
− 2.16(1.6022 𝑥 10−19)
Ek = 2.1667 X 10-19 J
Ek = (Mev2)/2 = 689721.8665 M/S ~ 7.00 X 105 M/S
14. For N =2 And Ms = -1/2. For L=0 And L = 1, It Is Composed Of 3 Counter-Clockwise At P
Orbital And 1 At The S-Orbital = 4 Electrons
15. Cl = 35.45 Composed Of Cl-35 And Cl-37
35.45 = 37(X) + 35(1-X) Where X Is The Fraction Of
Cl-37 X= 0.225; Cl-35 Is 77.5 % In Abundance.
17. If L = 0, Ml Should Be = 0.
18. Outermost Shell Is At N = 4; Taking The Electrons At The Outermost Shell Of 4s2 And 4p3, 2 +
3=5.
31. T = 1600 °C; Br2(G) → 2br(G)
Nbr2 = 1.05 Moles ; Tv = 2 L; Α = 0.025 Br2(g) 2Br(g)
[Br2] = 1.05/2 = 0.525 M
0.525 0
Kc = {([Br]2)/Br
2}Eq = 1.3462 X M 10-3 0.525(1-α) 0.525x2α
Kp = Kc(Rt)Σv
Kp = 1.3462 X 10-3(0.08206 X 1873.15)2-1 = 0.2069 Atm 0.511875 M 0.02625 M
36. Vph2o° = 23.756 Torr (25°C); Msolute = 18.913 G; Msolvent = 36 G H2o; Vpsoln = 20.234
Torr Mw Solute = ?
, Vpsoln = Vph2o°Xsolvent
20.234 = 23.756(X); X = 0.8517 Therefore, The Mole Fraction Of The Solute Is
0.1483 36 G/ 18.02g/Mole = 1.9978 Mole Solvent; 0.8517 = 1.9978/(1.9978 +
Nsolute)
Nsolute = 0.3479 Moles
Mw = 18.913 G/ 0.3479 Moles = 54.3693 G/Mol ~ 56 G/Mol
37. 30% Wt Urea In Water.
∆𝑇 = 𝐾𝑓𝑚𝑖
Since Urea Is Non-Electrolyte, I = 1;
M = (30 G Urea/60.07 G/Mol)/70 G
H2o/1000g/Kg M = 7.1345 Mol/Kg
𝐶
∆𝑇 = (7.3145 𝑚) (1.86° 𝑚 ) (1) = 13.2702 𝐶°; Tsoln = 0-13.2702 = -13.2702°C
38. M = 0.205 M With Respect To Urea
200 G Of Solution Is Diluted With 250 G Of Water
First Find The Number Of Moles Of Urea Present In A 200 G
Solution. 0.205/1000 G = (X/60.07)/(200 – X); X = 2.4329 G Of
Urea
M’ = (2.4329 G/60.07 G/Mole)/(200 G – 2.4329 G + 250
G)/1000g/Kg M’ = 0.09049 M
𝐶
𝑇𝑓 = 0℃ − (0.09049 𝑚) (1.86° 𝑚 ) (1) = -0.1683 °C
39. Π = 38 Mm Hg (At 273.15 K)
Δtb =?
Π = Mrt
M = (38mm Hg/760 Mm Hg/1 Atm)/(0.08206 Atm-L/Mol-K)(273.15 K)
M = 2.2307 X 10-3 M
∆𝑇𝑏 = 𝐾𝑏𝑚𝑖 𝑎𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝑡ℎ𝑎𝑡 𝑚 ~ 𝑀; = (0.512 𝐶° ) (2.2307 X 10−3𝑚)(1) = 0.001142 𝐶°
𝑚
40. Ra-226 ----- → He + 222 86rn (Α Emission Atomic Mass Decreased By 4, Atomic Number By 2)
41. Half-Life Is 29 Years. Fraction (X) Remained After 100 Years
= ? Since Half-Life Exhibits First Order Reactions,
K = Ln2/T1/2 = Ln2/29 = 0.0239 Years-1
Ln|Ao| - Ln|A| = Kt = Ln|1| - Ln|X| =
0.0239(100) X = 0.0916 Remains
42. Another First Order
Reaction: Ln|1| -
Ln|0.01| = 50(K) K =
0.092103 Years-1
T1/2 = Ln2/K = 7.5257 Years
43. Ra-226 Half-Life = 1600 Years
Α Emission; T= 10 Mins; 10 Mg Sample Of Ra-226. 1 Year = 365 Days.
Since It Still Follows First Order Reaction, K = Ln2/1600 = 4.3322 X 10-4
Years-1 For T = 10 Mins, 10/(60)(24)(365) = 1.9026 X 10-5 Years Thus,
Kt = 8.24233 X 10-9 Disintegration
For 10 Mg, 10 Mg/(226)(1000 Mg/G) = 4.4248 X 10-5 Moles = 2.6646 X 1019 Particles
(2.6646 X 1019 Particles)(8.24233 X 10-9 Disintegration) = 2.1963 X 1011 Disintegrations
44. Activity = Number Of Disintegrations Per
Second 1ci = 3.7x1010
Disintegration/Atom-Sec
K = Ln2/(1600 X 365 X 24 X 3600) = 1.3737 X 10-11 Sec-1
Kn = (1.3737x10-11)(2.6646 X 1019) = 366041979.3 Particles/S
Activity = 366041979.3 Dps/3.7 X 1010 Dps/Ci = 9.8930 Ci
, 46. First Order Reaction; K-40 Decayed To Ar-40 T1/2 = 1.27x109 Years. Age Of Rock = ?
Mar/Mk = 4.
K = Ln2/1.27x109 Years = 5.4579 X 10-10/Year
Ln|K+Ar| – Ln|K| = Kt = Ln|4k + K| - Ln|K| = T(5.4579 X 10-10/Year)
Cancel K As Ln|5| = T(5.4579 X 10-10/Year); T = 2.9488 X109 Years.
49. For C-13 (13.003355), What Is The Binding Energy In Mev?
Using Ebind = Δmco2, For C-13 With Number Of Protons = 12 And Neutron = 1;
12(1.007276466 Amu) = 12.08731759 (From (Mp)(Na)(1000))
1(1.008664915 Amu) = 1.008664915 (From (Mn)(Na)(1000))
M’ = Proton + Neutron = 13.09598251
Ebind = Δmco2 = (13.003355 - 13.09598251)(299792458)2 = -8.3249 X 1015 Amu M2/S2
= (-8.3249 X 1015 Amu M2/S2)/(6.022x1023 Amu/G)(1000g/Kg) = -1.3824 X10-11 J
Ebind/Nucleon = 1.3824 X10-11 J/13 = (1.0634 X 10-12 J/Nucleon)
= (1.0634 X 10-12 J/Nucleon)/(1.602176 X10-19 J/Ev) = 6.6372 Mev ~ 7 Mev
50. Since It Also Follows The Fo Kinetics, 1g Of Sr = 0.0111 Moles And 0.953 G = 0.01058888 Mol
Activity = ?
Ln|0.0111111| - Ln|0.01058888888| = K(2x365x24x3600); K = 7.632604 X 10-10 Dps
Initial Particle Concentration = (0.01111 X 6.02214179 X 1023) = 6.6913 X 1021 Particles
Kn = (7.632604 X 10-10 Dps)( 6.6913 X 1021 Particles) = 5.107182022 X1012 P-Dps
Activity = 5.107182022 X1012 P-Dps/3.7x1010 Dps/Ci = 138.0319 Ci ~ 140 Ci
Chemical Engineering
Licensure Examination
3rd Edition
Solutions Manual
Monroe H. De Guzman, Bsche
Batangas State University
,I. Physical and Chemical Principles
A. General Inorganic Chemistry
4. For Zn2+, Atomic Number = 30, Atomic Mass = 65.38 (~65),
Number Of Protons = 30; Neutrons = 65-30 = 35; Electrons = 30 – 2 = 28
E- + Proton + Neutron = 30 + 35 + 28 = 93
6. The Balmer Series Or Balmer Lines In Atomic Physics, Is The Designation Of One Of A Set Of Six
Different Named2 Series Describing The Spectral Line Emissions Of The Hydrogen Atom (Wikipedia, 2014).
𝑛
𝜆=( )(𝑅∞ /4); For M = 2 As N → ∞
𝑛2−𝑚2
Balmer Series = 1/𝜆 = ∞ (𝑅 /4) = 10973731.57 M/4 =
∞ ∞ 27434.3289 cm-1
7. Named After The American Physicist Frederick Sumner Brackett Who First Observed The
Spectral Lines In 1922 (Wikipedia,
1 1 2014).
)(𝑅 ∞ ); M = 4 As N → ∞ = 10973731.57 M/16 = 6858.5822 cm-1
Brackett Series = ( −
2 2
𝑚 𝑛
10. Minimum Wavelength Of Light For Work Functions (De Broglie
Wavelength) For Φ = 2.90 Ev (J/C Of Electron)
ℎ𝑐𝑜 (6.6261 𝑥 10−34)(299792458)
𝜆= = (2.90)(1.6022 𝑥 10−19)
= 427.5317 nm
𝜙𝑒
11. For Φ = 5.00 Ev
ℎ𝑐𝑜 (6.6261 𝑥 10−34)(299792458)
𝜆= = (5.00)(1.6022 𝑥 10−19)
= 247.9584 nm
𝜙𝑒
13. Work Function And Kinetic Energy
Rb Φ = 2.16 Ev; Λ = 350 Nm Of Light
Wavelength. Ephoton = Φ + Ek = Hv = Hco/Λ; Ek =
Hco/Λ – Φ
(6.6261 𝑥 10−34)(299792458)
EK = 350 𝑋 10 −9
− 2.16(1.6022 𝑥 10−19)
Ek = 2.1667 X 10-19 J
Ek = (Mev2)/2 = 689721.8665 M/S ~ 7.00 X 105 M/S
14. For N =2 And Ms = -1/2. For L=0 And L = 1, It Is Composed Of 3 Counter-Clockwise At P
Orbital And 1 At The S-Orbital = 4 Electrons
15. Cl = 35.45 Composed Of Cl-35 And Cl-37
35.45 = 37(X) + 35(1-X) Where X Is The Fraction Of
Cl-37 X= 0.225; Cl-35 Is 77.5 % In Abundance.
17. If L = 0, Ml Should Be = 0.
18. Outermost Shell Is At N = 4; Taking The Electrons At The Outermost Shell Of 4s2 And 4p3, 2 +
3=5.
31. T = 1600 °C; Br2(G) → 2br(G)
Nbr2 = 1.05 Moles ; Tv = 2 L; Α = 0.025 Br2(g) 2Br(g)
[Br2] = 1.05/2 = 0.525 M
0.525 0
Kc = {([Br]2)/Br
2}Eq = 1.3462 X M 10-3 0.525(1-α) 0.525x2α
Kp = Kc(Rt)Σv
Kp = 1.3462 X 10-3(0.08206 X 1873.15)2-1 = 0.2069 Atm 0.511875 M 0.02625 M
36. Vph2o° = 23.756 Torr (25°C); Msolute = 18.913 G; Msolvent = 36 G H2o; Vpsoln = 20.234
Torr Mw Solute = ?
, Vpsoln = Vph2o°Xsolvent
20.234 = 23.756(X); X = 0.8517 Therefore, The Mole Fraction Of The Solute Is
0.1483 36 G/ 18.02g/Mole = 1.9978 Mole Solvent; 0.8517 = 1.9978/(1.9978 +
Nsolute)
Nsolute = 0.3479 Moles
Mw = 18.913 G/ 0.3479 Moles = 54.3693 G/Mol ~ 56 G/Mol
37. 30% Wt Urea In Water.
∆𝑇 = 𝐾𝑓𝑚𝑖
Since Urea Is Non-Electrolyte, I = 1;
M = (30 G Urea/60.07 G/Mol)/70 G
H2o/1000g/Kg M = 7.1345 Mol/Kg
𝐶
∆𝑇 = (7.3145 𝑚) (1.86° 𝑚 ) (1) = 13.2702 𝐶°; Tsoln = 0-13.2702 = -13.2702°C
38. M = 0.205 M With Respect To Urea
200 G Of Solution Is Diluted With 250 G Of Water
First Find The Number Of Moles Of Urea Present In A 200 G
Solution. 0.205/1000 G = (X/60.07)/(200 – X); X = 2.4329 G Of
Urea
M’ = (2.4329 G/60.07 G/Mole)/(200 G – 2.4329 G + 250
G)/1000g/Kg M’ = 0.09049 M
𝐶
𝑇𝑓 = 0℃ − (0.09049 𝑚) (1.86° 𝑚 ) (1) = -0.1683 °C
39. Π = 38 Mm Hg (At 273.15 K)
Δtb =?
Π = Mrt
M = (38mm Hg/760 Mm Hg/1 Atm)/(0.08206 Atm-L/Mol-K)(273.15 K)
M = 2.2307 X 10-3 M
∆𝑇𝑏 = 𝐾𝑏𝑚𝑖 𝑎𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝑡ℎ𝑎𝑡 𝑚 ~ 𝑀; = (0.512 𝐶° ) (2.2307 X 10−3𝑚)(1) = 0.001142 𝐶°
𝑚
40. Ra-226 ----- → He + 222 86rn (Α Emission Atomic Mass Decreased By 4, Atomic Number By 2)
41. Half-Life Is 29 Years. Fraction (X) Remained After 100 Years
= ? Since Half-Life Exhibits First Order Reactions,
K = Ln2/T1/2 = Ln2/29 = 0.0239 Years-1
Ln|Ao| - Ln|A| = Kt = Ln|1| - Ln|X| =
0.0239(100) X = 0.0916 Remains
42. Another First Order
Reaction: Ln|1| -
Ln|0.01| = 50(K) K =
0.092103 Years-1
T1/2 = Ln2/K = 7.5257 Years
43. Ra-226 Half-Life = 1600 Years
Α Emission; T= 10 Mins; 10 Mg Sample Of Ra-226. 1 Year = 365 Days.
Since It Still Follows First Order Reaction, K = Ln2/1600 = 4.3322 X 10-4
Years-1 For T = 10 Mins, 10/(60)(24)(365) = 1.9026 X 10-5 Years Thus,
Kt = 8.24233 X 10-9 Disintegration
For 10 Mg, 10 Mg/(226)(1000 Mg/G) = 4.4248 X 10-5 Moles = 2.6646 X 1019 Particles
(2.6646 X 1019 Particles)(8.24233 X 10-9 Disintegration) = 2.1963 X 1011 Disintegrations
44. Activity = Number Of Disintegrations Per
Second 1ci = 3.7x1010
Disintegration/Atom-Sec
K = Ln2/(1600 X 365 X 24 X 3600) = 1.3737 X 10-11 Sec-1
Kn = (1.3737x10-11)(2.6646 X 1019) = 366041979.3 Particles/S
Activity = 366041979.3 Dps/3.7 X 1010 Dps/Ci = 9.8930 Ci
, 46. First Order Reaction; K-40 Decayed To Ar-40 T1/2 = 1.27x109 Years. Age Of Rock = ?
Mar/Mk = 4.
K = Ln2/1.27x109 Years = 5.4579 X 10-10/Year
Ln|K+Ar| – Ln|K| = Kt = Ln|4k + K| - Ln|K| = T(5.4579 X 10-10/Year)
Cancel K As Ln|5| = T(5.4579 X 10-10/Year); T = 2.9488 X109 Years.
49. For C-13 (13.003355), What Is The Binding Energy In Mev?
Using Ebind = Δmco2, For C-13 With Number Of Protons = 12 And Neutron = 1;
12(1.007276466 Amu) = 12.08731759 (From (Mp)(Na)(1000))
1(1.008664915 Amu) = 1.008664915 (From (Mn)(Na)(1000))
M’ = Proton + Neutron = 13.09598251
Ebind = Δmco2 = (13.003355 - 13.09598251)(299792458)2 = -8.3249 X 1015 Amu M2/S2
= (-8.3249 X 1015 Amu M2/S2)/(6.022x1023 Amu/G)(1000g/Kg) = -1.3824 X10-11 J
Ebind/Nucleon = 1.3824 X10-11 J/13 = (1.0634 X 10-12 J/Nucleon)
= (1.0634 X 10-12 J/Nucleon)/(1.602176 X10-19 J/Ev) = 6.6372 Mev ~ 7 Mev
50. Since It Also Follows The Fo Kinetics, 1g Of Sr = 0.0111 Moles And 0.953 G = 0.01058888 Mol
Activity = ?
Ln|0.0111111| - Ln|0.01058888888| = K(2x365x24x3600); K = 7.632604 X 10-10 Dps
Initial Particle Concentration = (0.01111 X 6.02214179 X 1023) = 6.6913 X 1021 Particles
Kn = (7.632604 X 10-10 Dps)( 6.6913 X 1021 Particles) = 5.107182022 X1012 P-Dps
Activity = 5.107182022 X1012 P-Dps/3.7x1010 Dps/Ci = 138.0319 Ci ~ 140 Ci
