kang chuen tat


{ City } penang
< Country > malaysia
* Profession * biochemical engineer
User No # 116977
Total Questions Posted # 219
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Answers / { kang chuen tat }

Question { 1671 }

MICROBIOLOGICAL ENGINEERING - QUESTION 28.3 : In the calculation of the growth of bacteria, absorbance, A in spectrophotometry is used. According to Beer-Lambert Law, A = e x l x c where A is the absorbance of the solution (no unit), l is the distance of light travels through the solution (in cm), e is the molar absorptivity or the molar extinction coefficient [ in L / (mol.cm) ]. For a particular solute and fixed path length : As / Ao = Cs / Co where Ao is the observed signal for a known concentration Co, and As is the observed signal for a sample concentration Cs. (a) For a cell concentration of 560 cells / mL, a spectrophotometre gives an absorbance reading of 1.0. A mixture of concentration 3600000 cells / mL can be diluted in several operations, with each operation having a dilution of 1:20. How many dilutions should be made so that the concentration of this mixture can be calculated within a range of A = 0.0 to 1.0. (b) In another experiment, a sample tube of 1 cm in width is used. Let A = 0.06 and e = 0.0012 ml / (cell.cm). Find the cell concentration of the sample.


Answer

MICROBIOLOGICAL ENGINEERING - ANSWER 28.3 : (a) Let Ao = 1.0, Co = 560 cells / mL, Cs = 3600000 cells / mL. By using the equation As / Ao = Cs / Co, then As = Ao x Cs / Co = 1.0 x 3600000 / 560 = 6428.57. For a dilution of 1:20, first stage dilution gives A = 6428.57 / 20 = 321.43, second stage dilution gives A = 321.43 / 20 = 16.07, third stage dilution gives A = 16.07 / 20 = 0.8, which is within a range of A = 0.0 to 1.0. Total stages of dilution needed = 3. (b) By using the equation A = e x l x c, then c = A / (e x l) = 0.06 / { [ 0.0012 ml / (cell.cm) ] x 1 cm } = 50 cells / mL. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1409 }

BIOCHEMICAL ENGINEERING INSTRUMENTATION - EXAMPLE 29.1 : In the application of Theory of Spectrometry in spectrophotometer, let n = N x C x V, V = A x t, e = a x N where n = number of molecules, N = Avogadro's number, V = volume of cuvette, A = area of cuvette, t = thickness of cuvette, C = concentration of fluid in the cuvette, e = extinction coefficient, a = effective area of molecule. (a) By using calculus in dI = -I x a x N x C x dt, prove that ln (I / Io) = -a x N x C x t, where dI is the small difference in I and dt is the small difference in t. I = intensity of light. Io = initial intensity of light. (b) Show by calculations that ln (Io / I) = e x C x t based on the answer in the previous question (a). (c) Find the equation of log (Io / I) as a function of e, C and t based on the answer in the previous question (b).


Answer

BIOCHEMICAL ENGINEERING INSTRUMENTATION - ANSWER 29.1 : (a) Let dI = -I x a x N x C x dt, then dI / I = -a x N x C x dt. Integrate both sides of the equation gives ln (I / Io) = -a x N x C x t. (b) When ln (I / Io) = -a x N x C x t and e = a x N, then ln (I / Io) = -e x C x t. If ln (I / Io) = - ln (Io / I), then ln (Io / I) = e x C x t. (c) Let ln (Io / I) = log (Io / I) / log E, where E = 2.718 approximately. Then log (Io / I) = log E x ln (Io / I) = log E x e x C x t. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1459 }

Question 110 - When an algebraic product is defined on the space, the Lie bracket is the commutator [x,y] = xy - yx according to Lie algebra in mathematics. If [p,x] f(x) = px f(x) - xp f(x), p = -ih d / dx, find the value of [p,x] in term of i and h.


Answer

Answer 110 - Let [p,x] f(x) = px f(x) - xp f(x) = (-ih d / dx) x f(x) - x (-ih d / dx) f(x) = -ih { d / dx [ x f(x) ] - x d f(x) / dx } = -ih [ f(x) ] + x d f(x) / dx - x d f(x) / dx = -ih [ f(x) ]. Then [p,x] = -ih. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1274 }

BIOCHEMICAL ENGINEERING INSTRUMENTATION - EXAMPLE 29.2 : (a) In order to determine the slurry volume (L) in chromatography, arrange the formula using the following parameters : slurry concentration (%), slurry volume (L), settled chromatography medium volume (L), 100. (b) In order to determine the chromatography medium volume (L) in gel filtration chromatography, arrange the formula using the following parameters : sample volume to be processed (L), chromatography medium volume (L), sample volume in percent of total column volume (%), 100.


Answer

BIOCHEMICAL ENGINEERING INSTRUMENTATION - ANSWER 29.2 : (a) By fraction of concentration, [ settled chromatography medium volume (L) ] / [ slurry volume (L) ] = slurry concentration (%) / 100. Then [ slurry volume (L) ] = [ settled chromatography medium volume (L) ] / [ slurry concentration (%) / 100 ]. (b) By fraction of volume, [ sample volume to be processed (L) ] / [ chromatography medium volume (L) ] = sample volume in percent total volume (%) / 100. Then chromatography medium volume (L) = [ sample volume to be processed (L) ] / [ sample volume in percent of total column volume (%) / 100 ]. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1422 }

BIOCHEMICAL ENGINEERING INSTRUMENTATION - EXAMPLE 29.3 : In thin layer chromatography (TLC), let retention factor Rf = (distance traveled by solute) / (distance traveled by solvent). Silica gel is used as stationary phase which is more polar than the hexane solvent as mobile phase. If the Rf values of 3 solutes A, B and C are 1.5, 0.75 and 1 respectively, compare the polarities of A, B and C with reasons.


Answer

BIOCHEMICAL ENGINEERING INSTRUMENTATION - ANSWER 29.3 : Polarities : A < C < B. A is least polar since it travels farthest with the mobile phase, whereas B is most polar as it is attracted to the stationary phase and travels least. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1475 }

BIOCHEMICAL ENGINEERING INSTRUMENTATION - EXAMPLE 29.4 : The resolution of separation, Rs for chromatography is given by the formula Rs = (difference in retention time) / (average width at the base). In a chromatogram, 3 peaks a, b and c are found. Average widths W at the bases of the solutes are : Wa = 20 s, Wb = 40 s, Wc = 30 s. Resolutions of separation, Rs for solutes b and c in comparison to a are 2 and 4 respectively. The differences in retention times T for b and c in comparison to a are (Tb - Ta) and (Tc - Ta), Ta = Tc - Tb : (a) Form 2 equations involving Rs as a function of Wa, Wb, Wc, Ta, Tb and Tc. (b) Find the values of Ta, Tb and Tc.


Answer

BIOCHEMICAL ENGINEERING INSTRUMENTATION - ANSWER 29.4 : Reference formula Rs = (difference in retention time) / (average width at the base) is used. (a) First equation : Rs = (Tb - Ta) / [ (Wa + Wb) / 2 ] = 2 (Tb - Ta) / (Wa + Wb). Second equation : Rs = (Tc - Ta) / [ (Wa + Wc) / 2 ] = 2 (Tc - Ta) / (Wa + Wc). (b) Substitute Ta = Tc - Tb, Wa = 20 s, Wb = 40 s and Wc = 30 s into first equation and second equation. First equation : Rs = 2 = 2 [ Tb - (Tc - Tb) ] / (20 + 40) = (2 Tb - Tc) / 30, 2 Tb - Tc = 60. Second equation : Rs = 4 = 2 [ Tc - (Tc - Tb) ] / (20 + 30) = Tb / 25, Tb = 100 s. Substitute Tb = 100 s into first equation gives 2 Tb - Tc = 2 x 100 - Tc = 200 - Tc = 60, then Tc = 200 - 60 = 140 s. Then Ta = Tc - Tb = 140 - 100 = 40 s. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1345 }

BIOCHEMICAL ENGINEERING INSTRUMENTATION - EXAMPLE 29.5 : The following formula is used as a reference : (analyte signal) / (internal standard signal) = (f-factor) x (concentration of analyte) / (concentration of internal standard). A solution containing 3 mM of analyte and 4 mM of internal standard gave peak signals of 2 and 3 mamps respectively. Another similar solution containing 2 mM of analyte and 1 mM of internal standard gave peak signals of 1 and 4 mamps respectively. Find the average f-factor.


Answer

BIOCHEMICAL ENGINEERING INSTRUMENTATION - ANSWER 29.5 : Let (f-factor) = (analyte signal) (concentration of internal standard) / [ (internal standard signal) (concentration of analyte) ]. For first solution, (f-factor) = (2 mamps) (4 mM) / [ (3 mamps) (3 mM) ] = 8 / 9. For second solution, (f-factor) = (1 mamps) (1 mM) / [ (4 mamps) (2 mM) ] = 1 / 8. Average : Average f-factor = (8 / 9 + 1 / 8) / 2 = (64 + 9) / (72 x 2) = 73 / 144. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1797 }

BIOCHEMICAL ENGINEERING INSTRUMENTATION - EXAMPLE 29.6 : In infrared spectrum, one of the factors affecting peak location is the mass of the atoms. The stretching frequency of a bond connected to a lighter atom will be greater than the same bond connected to a heavier atom. (a) For halogens like florine (F), chlorine (Cl), bromine (Br), iodine (I) and astatine (At), what is their IUPAC group number? Hint : The proton numbers for F, Cl, Br, I and At are 9, 17, 35, 53 and 85 respectively. (b) For the compounds of H-F, H-Cl, H-Br, H-I and H-At, which one has the lowest stretching frequency and which one has the highest stretching frequency? State the reasons.


Answer

BIOCHEMICAL ENGINEERING INSTRUMENTATION - ANSWER 29.6 : (a) 17. (b) Lowest stretching frequency : H-At. Highest stretching frequency : H-F. Reasons : F has the lowest proton number 9 and At has the highest proton number 85. For the compound of H-X, the stretching frequency of the bond decreases when the mass of X increases and vice versa. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1574 }

ENGINEERING PHYSICS - EXAMPLE 30.1 : Two viruses a and b with masses of Ma and Mb are moving at velocities of Va and Vb respectively, facing towards each other and collide. After collision both masses of Ma and Mb disappear. (a) Find the total momentum available for both a and b. Hint : momentum = mass x velocity = M x V. (b) Guess the total energy E generated from the disappearance of a and b. Let c to be the velocity of light. Hint : E is equal to M c square.


Answer

ENGINEERING PHYSICS - ANSWER 30.1 : (a) Momentum of a = Ma x Va. Momentum of b = Mb x Vb. Total momentum of a and b = Ma x Va + Mb x Vb. (b) After mass disappearance : Energy of a = Ma x c x c. Energy of b = Mb x c x c. Then E = Ma x c x c + Mb x c x c = (Ma + Mb)(c x c). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1356 }

ENGINEERING PHYSICS - EXAMPLE 30.2 : The Planck-Einstein relation connects the particulate photon energy E with its associated wave frequency f to produce E = hf. Let h to be the Planck constant. The frequency f, wavelength L and speed of light c are related by E = hc / L. With p denoting the linear momentum of a particle, the de Broglie wavelength L of the particle is given by L = h / p. (a) Find the equation of E as a function of p and c. (b) If E has a unit of electron-volt and f has a unit of 1 / second, then what is the unit of h?


Answer

ENGINEERING PHYSICS - ANSWER 30.2 : (a) Let first equation : E = hc / L, second equation : L = h / p. Substitute the second equation into first equation will produce E = hc / L = hc / (h / p) = pc. (b) E = hf then h = E / f. Dimensional analysis of h will produce E / f of electron-volt / (1 / second) or electron-volt-second. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1331 }

ENGINEERING PHYSICS - EXAMPLE 30.3 : (a) The quantum number m is given by m = -s, -s + 1. If s = 0.5, find the values of m. (b) | T > = (cos T) | V > + (sin T) | H >. The V and H states form a basis for all polarizations. Let cos T = 0.8. (i) If (sin T)(sin T) + (cos T)(cos T) = 1, find the value of sin T. (ii) For | T > = a | V > + b | H >, where a x a represents the probability of | V > and b x b represents the probability of | H >. Which one is more abundant, | V > or | H >? (iii) Find the value of T without using any mathematical tools.


Answer

ENGINEERING PHYSICS - ANSWER 30.3 : (a) When s = 0.5, m = -s = -0.5 and m = -s + 1 = -0.5 + 1 = 0.5. (b)(i)(sin T)(sin T) = 1 - (cos T)(cos T) = 1 - 0.8 x 0.8 = 0.36. Then sin T = 0.6. (ii) Let a = cos T = 0.8, probability of | V > = a x a = 0.8 x 0.8 = 0.64 (more abundant) and b = sin T = 0.6, probability of | H > = b x b = 0.6 x 0.6 = 0.36 (less abundant). (iii) cos T = 0.8, then T = arc cos 0.8. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1442 }

ENGINEERING PHYSICS - EXAMPLE 30.5 : (a) Let | A > = (Aa Ab Ac), | B > = (Ba Bb Bc), | C > = (Ca Cb Cc). Find | A > + | C > - | B > in term of Aa, Ab, Ac, Ba, Bb, Bc, Ca, Cb and Cc. (b) Let d | E > = d (Ea Eb Ec) = (d Ea d Eb d Ec). If | E > = (6 7 8), find the value of 10 | E >.


Answer

ENGINEERING PHYSICS - ANSWER 30.5 : (a) | A > + | C > - | B > = (Aa + Ca - Ba Ab + Cb - Bb Ac + Cc - Bc). (b) Let d = 10, Ea = 6, Eb = 7, Ec = 8. Then 10 | E > = (10 x Ea 10 x Eb 10 x Ec) = (10 x 6 10 x 7 10 x 8) = (60 70 80). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1364 }

ENGINEERING PHYSICS - EXAMPLE 30.4 : (a) Time evolution in Heisenberg picture, according to Ehrenfest theorem is m (d / dt) < r > = < p >, where m = mass, r = position, p = momentum of a particle. If v = velocity, prove that m < v > = < p >. (b) Lande g-factor is given by Gj = Gl [ J (J + 1) - S (S + 1) + L (L + 1) ] / [ 2J (J + 1) ] + Gs [ J (J + 1) + S (S + 1) - L (L + 1) ] / [ 2J (J + 1) ]. If Gl = 1 and under approximation of Gs = 2, prove by calculation that Gj = (3/2) + [ S (S + 1) - L (L + 1) ] / [ 2J (J + 1) ].


Answer

ENGINEERING PHYSICS - ANSWER 30.4 : (a) Let (d / dt) < r > = < v >. Substitute it into m (d / dt) < r > = < p > so that m < v > = < p >, proven by momentum = mass x velocity. (b) Gj = Gl [ J (J + 1) - S (S + 1) + L (L + 1) ] / [ 2J (J + 1) ] + Gs [ J (J + 1) + S (S + 1) - L (L + 1) ] / [ 2J (J + 1) ]. When Gl = 1 and Gs = 2, Gj = [ J (J + 1) - S (S + 1) + L (L + 1) ] / [ 2J (J + 1) ] + 2 [ J (J + 1) + S (S + 1) - L (L + 1) ] / [ 2J (J + 1) ] = [ J (J + 1) + 2J (J + 1) ] / [ 2J (J + 1) ] + [ 2S (S + 1) - 2L (L + 1) - S (S + 1) + L (L + 1) ] / [ 2J (J + 1) ] = [ 3J (J + 1) ] / [ 2J (J + 1) ] + [ S (S + 1) - L (L + 1) ] / [ 2J (J + 1) ] = (3/2) + [ S (S + 1) - L (L + 1) ] / [ 2J (J + 1) ] (proven). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1437 }

Question 111 - There are 6 spin orbitals in p subshell in a ground state carbon atom. Only 2 electrons fill the p subshell. Number of different ways for n electrons to occupy the k spin orbitals are k! / [ (n!) (k-n)! ]. Find the number of different configurations of electrons to occupy the p subshell in a carbon atom.


Answer

Answer 111 - Let k = 6, n = 2. Number of ways = k! / [ (n!) (k-n)! ] = 6! / [ (2!) (6-2)! ] = 6! / (2! 4!) = 15. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1506 }

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - EXAMPLE 31.1 : As an approximation, let v = Zc / 137 where v is the radial velocity for 1 s electron of an element, c is the speed of light, Z is the atomic number. For gold with Z = 79, find the radial velocity of its 1 s electron, in term of c and percentage of the speed of light. (b) As an approximation, let A x A = 1 - Z x Z / 18769 where A is the ratio of the relativistic and non-relativistic Bohr radius. Find the value of A.


Answer

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - ANSWER 31.1 : (a) Approximately v = Zc / 137 = (79 / 137) c = 0.5766 c or 57.66 % of the speed of light. (b) Approximately A x A = 1 - Z x Z / 18769 = 1 - 79 x 79 / 18769 = 0.667 then A = 0.817. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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