Answer / kangchuentat

QUANTUM BIOLOGY - ANSWER 33.5 : (a) Mode = 120 (2 readings, greater than others with 1 reading). Median = 70 (since 70 is the middle, when the values are rearranged orderly as 10, 60, 70, 120, 120). Min = (sum of readings) / (number of readings) = (10 + 70 + 60 + 120 + 120) / 5 = 380 / 5 = 76. (b) Assumption : Microtubule is a cylinder. Let V = 3.142 x h x d x d / 4 where V = volume, d = diameter, h = height. Then V = 3.142 x h x d x d / 4 = 3.142 x n x 25 n x 25 n / 4 = 490.9375 n x n x n = 4.909375 x 10^2 x 10^(-9) x 10^(-9) x 10^(-9) = 4.909375 x 10^(-25) meters, where ^ is the symbol of power, n is nano or 10^(-9). 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.

What is the pattern of interview questions and which field is of high importance viz., Mass Transfer, Heat Transfer, Petroleum Refining etc ?

2 Answers   Cairn, IOCL, Reliance,

---

Is there a difference in mtd (mean temperature difference) between 'e' and 'j' (divided flow) type shell and tube heat exchangers?

0 Answers  

---

QUANTUM COMPUTING - EXAMPLE 32.4 : A system of linear congruences consists of 3 equations : X ≡ 1 (mod 2), X ≡ 3 (mod 3), X ≡ 4 (mod 5). X has positive values. (a)(i) List the values of these equations from 1 to approximately 40. (ii) Find the first smallest value and second smallest value of X. (iii) Guess the third smallest value of X. (b) Let X ≡ Aa (mod Ma), X ≡ Ab (mod Mb), X ≡ Ac (mod Mc). According to Chinese remainder theorem, X ≡ (Aa x Ya x Md + Ab x Yb x Me + Ac x Yc x Mf) [ mod (Ma x Mb x Mc) ]. (i) Show that Ma, Mb and Mc have the greatest common divisor of Ma x Mb x Mc. (ii) Find the values of Md, Me and Mf if Md = Mb x Mc, Me = Ma x Mc and Mf = Ma x Mb. (iii) Find the values of Ya, Yb and Yc if Ya = Remainder of (Md / Ma), Yb = Remainder of (Me / Mb) and Yc = Remainder of (Mf / Mc). (iv) Use Chinese remainder theorem to find X.

1 Answers  

---

CHEMICAL MATERIAL BALANCE – EXAMPLE 2.2 : Three hundred gallons of a mixture containing 75.0 wt % ethanol and 25 wt % water (mixture specific gravity = 0.877) and a quantity of a 40.0 wt % ethanol - 60 wt % water mixture (specific gravity = 0.952) are blended to produce a mixture containing 60.0 wt % ethanol. The specific gravity of a substance is the ratio of density of a substance compared to the density of water. The symbol of weight percent is wt %. (a) Estimate the specific gravity of the 60 % mixture by assuming that y = mx c where y is wt % ethanol, x is mixture specific gravity. Values for m and c are constants. (b) Determine the required volume of the 40 % mixture.

1 Answers  

---

X is a solid having a white colour at room temperature. It has a density about 2g/cc. Although it has melting point near 325 degree Celsius, its properties start degrading above 260 degree Celsius. The coefficient of friction is very low about 0.1. It has very good dielectric properties especially at higher radio frequencies. It has a very high bulk resistivity. It is chemically inert. It is also resistant to van der Waals force. It is hydrophobic as well as lipophobic. Creep or ‘Cold Flow’ has been observed in X.

0 Answers   Ford,

---

What are the steps required to design a vapor-liquid separator or flash drum?

0 Answers  

---

Are there any methods of preventing cracking of carbon steel welds in refining environments?

0 Answers  

---

hello everyone do anyone have interview questions for any chemical industry please email me at

0 Answers   ONGC,

---

Explain the procedure to estimate the friction factor involved in heat exchanger tubes?

0 Answers  

---

QUANTUM COMPUTING - EXAMPLE 32.5 : In quantum teleportation, let (C0 + D1) (00 + 11) = C000 + C011 + D100 + D111. Ba = 00 + 11, Bc = 10 + 01, Be = 00 - 11, Bm = 10 - 01. (a) Find the values of 00, 01, 10 and 11 in term of Ba, Bc, Be and Bm. (b) Prove by calculation that (C0 + D1) (00 + 11) = 0.5 [ Ba (C0 + D1) + Bc (C1 + D0) + Be (C0 - D1) + Bm (-C1 + D0) ].

1 Answers  

---

Hi, Please give me chemical engineering paper for IOCL exam for entire written exam, GD and personel interview model questions to my email id. (chemistnathan@rediffmail.com) Rgds, Ragu

0 Answers   HAL, IOCL,

---

Question 90 - 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.

1 Answers  

---