3.3. The specific gravity of gasoline is approximately 0.70 (a) Estimate the mass (kg) of 50.0 liters of gasoline. (b) The mass flow rate of gasoline exiting a refinery tank is 1150 kg/min. Estimate the volumetric flow rate in liters/s. (e) Estimate the average mass flow rate (lbm/min) delivered by a gasoline pump. (d) Gasoline and kerosene (specific gravity 0.82) are blended to obtain a mixture with a specific ravity of 0.78. Calculate the volumetric ratio (volume of gasoline/volume of kerosene) of the two compounds in the mixture, assuming Vblend Vgasoline + Vkerosene

3.3. The specific gravity of gasoline is approximately 0.70 (a) Estimate the mass (kg) of 50.0 liters of gasoline. (b) The mass flow rate of gasoline exiting a refinery tank is 1150 kg/min. Estimate the volumetric flow rate in liters/s. (e) Estimate the average mass flow rate (lbm/min) delivered by a gasoline pump. (d) Gasoline and kerosene (specific gravity 0.82) are blended to obtain a mixture with a specific ravity of 0.78. Calculate the volumetric ratio (volume of gasoline/volume of kerosene) of the two compounds in the mixture, assuming Vblend Vgasoline + Vkerosene

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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1. [15] Atmospheric pressure (P) decreases with increasing altitude (h) based on the following exponential function (with e as the base): P = Pº 1 exp Mgh RT where Pº=1 atm at h=0 (sea level) and T'is the average temperature of the atmosphere, M the molar mass of air (29 g mol-¹), and g the standard gravity. Please estimate the boiling point of water at atmospheric temperature T=293.2 K, and height h=3000 m. The heat of vaporization of water at 293.2 K is 44.22 kJ mol-
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Calculations. 1. Standard solutions and Beer's law calibration curve (1) Mass of SA Molar mass of salicylic acid (SA), HOC6H4COOH (2) Moles SA = mass/molar mass 0,165 138.1 0,500 = (0.500/10.0)x(stock solution concentration) /o. 6 0.100 (3) Concentration of SA stock solution = mol/0.100L = (2)/0.100 = SA solution "4" (0.400 mL stock to 10.0 mL) = 0.1658 SA solution "3" (0.300 mL stock to 10.0 mL) = The SA is hydrolyzed and quantitatively transferred to a 100 mL volumetric flask, so: 11997 / 0 SA solution "2" (0.200 mL stock to 10.0 mL) (4) Concentration of SA solution "5" (0.500 mL stock to 10.0 mL) 33 x 2762 = SA solution "1" (0.100 mL stock to 10.0 mL) 138.1 -1.1947 mol SA x 10 11 g/mol 33 2762 -3 mol/L mol/L As=0-888 mol/L A+ = +695 0.422 A2 = 0.351 A₁ = 0,1 mol/L mol/L mol/L A₁ = 0. ter these concentrations and absorbance in an Excel sheet and create a calibration curve. Attach s calibration curve with this report.
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