Answer / kangchuentat

(a) There are 3 types of seeds produced : RR, Rr, rR, rr where RR, Rr, rR (3) are dominantly round and rr (1) is recessively wrinkled. P (wrinkled) = 1 / (3 + 1) = 1 / 4. (b) There are 2 types of seeds produced : Yy, Yy, yy, yy where Yy (2) are dominantly yellow and yy (2) are recessively green. P (green) = 2 / (2 + 2) = 1 / 2. (c) Let RR x Rr produces RR, Rr, RR, Rr, then P (Rr) = 2 / 4 = 1 / 2. Let Yy x Yy produces YY, Yy, yY, yy, then P (Yy) = 2 / 4 = 1 / 2. P (RrYy) = P (Rr) x P (Yy) = 1 / 2 x 1 / 2 = 1 / 4. 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.

plz send the interview questions in the written test asked by hindustan petroleum corporation ltd.Iam a b>tech in chemical engineering.My written test is on feb 2008.urgent plz.send the questions to avazithraja@rediffmail.com

9 Answers   AZTEC, Bapex, HPCL,

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In a triple effect evaporator, the heat transfer for an evaporator is calculated as q = UA (TI – TF) where TI is the initial temperature, TF is the final temperature; U and A are constants. Given that heat transfer for the first evaporator : q(1) = UA (TI – TB); second evaporator : q(2) = UA (TB – TC); third evaporator : q(3) = UA (TC – TF) where q(x) is the heat transfer function, TB is the temperature of second inlet and TC is the temperature of third inlet, prove that the overall heat transfer Q = q(1) q(2) q(3) = UA (TI – TF).

1 Answers  

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repected sir i have given gate exam today.......... if i ll get the better rank in that ....what is the next process .....shall i give another one test......or wat....i dint get properlly..... plz help me or mail me on Nish_gasti@yahoo.co.in

0 Answers   IOCL,

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CHEMICAL ENERGY BALANCE - EXAMPLE 11.4 : Calculate the bubble temperature T at P = 85-kPa for a binary liquid with x(1) = 0.4. The liquid solution is ideal. The saturation pressures are Psat(1) = exp [ 14.3 - 2945 / (T + 224) ], Psat(2) = exp [ 14.2 - 2943 / (T + 209) ] where T is in degree Celsius. Please take note that x(1) + x(2) = 1. Please take note that y(1) + y(2) = 1, y(1) = [ x(1) * Psat(1) ] / P, y(2) = [ x(2) * Psat(2) ] / P, * is multiplication. P is in kPa.

1 Answers  

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Hi, I want to validate an analytical procedure of a specified impurity, but this impurity is unavailable for me. can I use the API instead of impurity in the linearity test? in Which reference can I find such information?

0 Answers   Pharma,

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what is first order of reaction?

0 Answers   Deepak Fertilisers,

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What factors should be compared when evaluating cooling tower bids?

2 Answers  

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A distillation column separates 10000 kg / hr of a mixture containing equal mass of benzene and toluene. The product D recovered from the condenser at the top of the column contains 95 % benzene, and the bottom W from the column contains 96 % toluene. The vapor V entering the condenser from the top of the column is 8000 kg / hr. A portion of the product from the condenser is returned to the column as reflux R, and the rest is withdrawn as the final product D. Assume that V, R, and D are identical in composition since V is condensed completely. Find the ratio of the amount refluxed R to the product withdrawn D. Hint : Solve the simultaneous equations as follow in order to find the answer (R / D) : 10000 = D + W; 10000 (0.5) = D (0.95) + W (0.04); 8000 = R + D.

2 Answers  

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Are there flow velocity restrictions to avoid static charge build up in pipelines?

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Question 80 - Liquid octane has a density of 703 kilograms per cubic metre and molar mass of 114.23 grams per mole. Its specific heat capacity is 255.68 J / (mol K). (a) Find the energy in J needed to increase the temperature of 1 cubic metre of octane for 1 Kelvin. (b) At 20 degree Celsius, the solubility of liquid octane in water is 0.007 mg / L as stated in a handbook. For a mixture of 1 L of liquid octane and 1 L of water, prove by calculations that liquid octane is almost insoluble in water.

1 Answers  

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Explain various protein purification techniques?

2 Answers  

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QUANTUM COMPUTING - EXAMPLE 32.7 : If | ± > = 0.707 ( | 0 > ± | 1 > ), prove that | Ψ (t = 0) > = | 0 > = 0.707 ( | + > + | - > ).

1 Answers  

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