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Part time jobs for engineer students working or studying

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Part time jobs for engineer students working or studying..

Answer / dhanashree

Part time job for engg. student is studying experiance . It
gives lot of knwoledge both theroticaly & practically.

Is This Answer Correct ?    4 Yes 2 No

Part time jobs for engineer students working or studying..

Answer / shashikant reddy

To get a actual practical knowledge student should have working experiance.Neverthelees they have tens of subject for studing.

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More Chemical Engineering Interview Questions

Question 77 - The molecular weights M in kg / mol of 3 different monomers a, b and c in a polymer are Ma = 14, Mb = 16 and Mc = 18, with their respective quantities in N units having the ratio of Na : Nb : Nc = 2 : 3 : 5. (a) Find the numerical average molecular weight of the polymer by using the formula (Ma Na + Mb Nb + Mc Nc) / (Na + Nb + Nc). (b) Find the weighted average molecular weight of the polymer by using the formula (Ma Na Ma + Mb Nb Mb + Mc Nc Mc) / (Ma Na + Mb Nb + Mc Nc). (c) Calculate the polydispersity Q by using the answer in (b) divided by answer in (a). (d) Find the volumetric average molecular weight of the polymer by using the formula (Ma Na Ma Ma + Mb Nb Mb Mb + Mc Nc Mc Mc) / (Ma Na Ma + Mb Nb Mb + Mc Nc Mc). (e) Estimate the polydispersity Q by using the answer in (d) divided by answer in (b).

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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.

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Question 48 - Let a ^ 2 = a x a and a ^ 3 = a x a x a where ^ is power function. Niobium is a metal with a body-centered cubic structure. The length of the unit cell structure is b = 0.3349 nm. (a) Find the volume for a unit cell structure for niobium. (b) There are 2 atoms per unit cell structure of niobium. The metal has a molar mass of 92.9 g / mol. One mole of the metal consists of 6.02 x 10 ^ 23 atoms. Find the mass of niobium per unit cell and the density of niobium.

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DIFFERENTIAL EQUATIONS - EXAMPLE 20.1 : By using Excel program either on laptop or desktop PC, solve the differential equation dy / dx = -2y + x + 4 with h = 0.005, initial values : x = 0, y = 1. The 4th order Runge-Kutta method provides : y(N + 1) = y(N) + (1/6) (k1 + 2k2 +2k3 + k4), k1 = h [ -2y(N) + x(N) + 4 ], k2 = h { -2 [ y(N) + k1 / 2 ] + x(N) + h / 2 + 4 }, k3 = h { -2 [ y(N) + k2 / 2 ] + x(N) + h / 2 + 4 }, k4 = h { -2 [ y(N) + k3 ] + x(N) + h + 4 }. What is the value of y at x = 0.5?

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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.

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