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.
Answer 80 - (a) 1 cubic metre of liquid octane has 703 kilograms or 703 kilograms x mole / (0.11423 kilograms) = 6154.25 mole. Energy needed = [ 255.68 J / (mol K) ] x (6154.25 mol) K = 1573518.64 J where K is the unit symbol for Kelvin and mol is the unit symbol for mole. (b) Density of liquid octane = 703 kilograms per cubic metre = 0.703 grams per cubic centimetre = 703 g / L. There is 0.007 mg = 0.000007 g or 0.000007 g x (1 L / 703 g) = 0.0000000099573257 L of liquid octane in 1 L of water. With negligible amount of liquid octane in 1 L of water, octane is considered insoluble in water. 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.
| Is This Answer Correct ? | 0 Yes | 0 No |
What are the three classes of organic solvents?
how to convert Nm3/hr to Kg/hr for air, @ 10.4 kg/cm2.g dischare pressure, does density doesn't affect on calculation?
Question 76 - Let C% be the fractional crystallinity, Rs = density of sample, Ra = density of amorphous form and Rc = density of crystalline form. In a polymer, these unknowns could be related by the equation C% = (Rc / Rs) (Rs - Ra) / (Rc - Ra). (a) Find the equation of Rc as a function of C%, Rs and Ra. (b) Two samples of a polymer, C and D exist. For sample C, C% = 0.513 when Rs = 2.215 unit. For sample D, C% = 0.742 when Rs = 2.144 unit. Both samples C and D have the same values of Ra and Rc. Find the values of Ra and Rc in 6 decimal places.
ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.22 : An engineering company intends to produce a small piece of biochemical instrument for sales. Let A = overall fixed cost of production, B = variable cost of production per unit, C = selling price per unit, D = quantity of unit produced. Breakeven Analysis is used where revenue = cost. (a) Explain the role of Breakeven Analysis by using A, B, C and D. (b) Find the value of Contribution Margin in term of A, B, C and / or D. (c) At the breakeven point where revenue = cost, derive an equation of D as a function of A, B and C.
hi, i have completed B.tech Chemical enggneering, plz can anyone provide me GVK BIO company's interview questions. iam attending the interview for the first time..
QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - EXAMPLE 31.4 : In a rigid rotor model in quantum chemistry, the moment of inertia I is given by an Equation E as I = Ma x La x La + Mc x Lc x Lc = m x L x L, where m = (Ma x Mc) / (Ma + Mc) and L = La + Lc, m is the reduced mass, Ma is the mass of a, Mc is the mass of c, La is the radius of a from point O, Lc is the radius of c from point O. Prove by simplest method that Equation E is wrong.
How much water is lost through a commercial cooling tower system with a throughput of about 600 gpm?
What is some good advice for specifying allowable pressure drops in shell and tube exchangers for heavy hydrocarbons?
A concentric, counter-current heat exchanger is used to cool lubricating oil. Water is used as the coolant. The mass flow rate of oil into the heat exchanger is 0.1 kg / s = FO. For oil, the inlet temperature TIO = 100 degree Celsius and the outlet temperature TOO = 55 degree Celsius. For water, the inlet temperature TIW = 35 degree Celsius and the outlet temperature TOW = 42 degree Celsius. What is the mass flow rate of water in kg / s, FW needed to maintain these operating conditions? Constant for heat capacity of oil is CO = 2131 J /(kg K) and for water is CW = 4178 J /(kg K). Use the equation (FO)(CO)(TIO – TOO) = (FW)(CW)(TOW – TIW).
Question 36 - For a mixture of benzene (B), toluene (T) and xylene (X), the equation applies where x for B, T and X will sum up to 1. The equation of x for each component is x = (L / V + 1) (F) / (L / V + K). The data of F for each component are : 0.5 for B, 0.35 for T, 0.15 for X. The data of K for each component are : 1.98 for B, 0.76 for T, 0.24 for X. When x for B + x for T + x for X = 1, find the values of (a) L / V; (b) x for each component of B, T, X respectively. You may use Excel program - Data : What-If-Analysis for Goal Seek to perform the iterative calculations.
pls send me HPCL & reliance test paper for chemical engineering.....
4 Answers BCPL, HPCL, NOCL, Reliance,
What can cause bulk solids to stop flowing from a bin?
Civil Engineering (5086)
Mechanical Engineering (4456)
Electrical Engineering (16639)
Electronics Communications (3918)
Chemical Engineering (1095)
Aeronautical Engineering (239)
Bio Engineering (96)
Metallurgy (361)
Industrial Engineering (259)
Instrumentation (3014)
Automobile Engineering (332)
Mechatronics Engineering (97)
Marine Engineering (124)
Power Plant Engineering (172)
Textile Engineering (575)
Production Engineering (25)
Satellite Systems Engineering (106)
Engineering AllOther (1379)