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openvellum. college m course.html? ourseld 14300489&OpenVellum; HMAC-292663 737d1ccc3ee1a6 1 89937 bd61 # 1 0001 Secure https://session.masteringchemistry.com/myt itemView?assignmentproblemID-90389305&offset;=next Experiment 8. Behavior ot Gasses Mol... Problem 10.57- Enhanced - with Feedback e previous | 4 of 5 ne Problem 10.57 - Enhanced- with Feedback Part A The metabolic oxidation of glucose, C6H1206. in our bodies produces C02, which is expelled from our lungs as a gas: Calculate the volume of dry CO2 produced at body temperature (37 °C) and 0 970 atm when 25.0 g of glucose is consumed in this reaction Express your answer using three significant figures. GH1206 (ag) + 602 (g) 6CO2 (g) + 6H20(1) You may want to reference (pages 408-410) Section 10 5 while completing this problem Submit My Answers Give Up Part B Calculate the volume of oxygen you would need, at 1.00 atm and 298 K, to completely oxidize 53 g of glucose Express your answer using three significant figures. Submit

Explanation / Answer

part A

no of mol of glucose = 25/180 = 0.139 mol

no of mol of CO2 produced = 6*0.139 = 0.834 mol

volume of CO2 produced = nRT/P

                    = 0.834*0.0821*(37+273.15)/0.97

       = 21.9 L

Part B

no of mol of glucose = 53/180 = 0.294 mol

no of mol of CO2 produced = 6*0.294 = 1.764 mol

volume of CO2 produced = nRT/P

                    = 1.764*0.0821*298/1

       = 43.2 L

PART A

density(d) = PM/RT = 0.99*28/(0.0821*(34+273.15)) = 1.01 g/L

part B

molarmass(M) = wRT/PV

         = 2.3*0.0821*(34+273.15)/((680/760)*0.87)

        = 74.5 g/mol

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