D Homepage state X N Netflix x C chegg study l Guided solutio x Chapter 1-Gas La
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D Homepage state X N Netflix x C chegg study l Guided solutio x Chapter 1-Gas Law C O wwwwebassign.net/web/Student/Assignment-Responsesllast?dep 15806012 Question 1 2 3 4 5 6 7 8 9 10 11 i 15 16 Total -100 (00%) gnment Submission remaining changes if you submit or change the answer this assignment, you submit answers by question parts. The number of submissions for each question part only gnment Scoring ur last submission is used for your score. O -7 points ZumChem85 E-052. of 0.064 mlr what volume of gas wil be A person accidentally swallows a drop of liquid oxygen, O20, which has a density of 1.149 g/mL Assuming the drop has a volume produced in the person's stomach at body temperature (37eC) and a pressure of 1.0 atm? Need Help? Reed ISutemi Answer Save Progress Practioe Another Version View Previous Question Question 3 of 16 View Ned Question Home My AssignmentsExplanation / Answer
Mass of oxygen liquid drop swallowed is m = Volume of drop * Density of liquid drop.
===> m = 0.064 mL * 1.149 g/mL
===> m = 0.073536 g
Oxygen gas can be assumed as an ideal gas.
Molecular weight of oxygen gas is M = 32 g.
Characteristic gas constant of oxygen gas is Roxygen = (R / M)
where R = Universal Gas Constant = 8.314472 J/(mole - K)
===> Roxygen = 8.314472 / (32 * 10-3) J/(kg - K)
===> Roxygen = 259.82725 J/(kg - K)
Ideal gas equation of state is PV = mRoxygenT
m = Mass of oxygen gas = 0.073536 g
P = Pressure of oxygen gas = 1 atm = 1.01325 * 105 Pascals
V = Volume of oxygen gas
T = Temperature of oxygen gas = 37°C = 310.15 K
===> V = mRoxygenT / P
===> V = 0.073536 * 10-3 kg * 259.82725 J/(kg - K) * 310 K / (1.01325 * 105 Pascals)
===> V = 5.8456 * 10-5 m3 = 58.456 mL
The volume of oxygen produced in the stomach of person is
V = 58.456 mL
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