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Sloan [31]
3 years ago
14

How do you solve this problem

Chemistry
1 answer:
ad-work [718]3 years ago
3 0

Answer: Volume of gas in the stomach, V = 0.0318L or 31.8mL

Explanation:

The number of moles of oxygen will remain constant even though the liquid oxygen will undergo a change of state to gaseous inside the person's stomach due to an increase in temperature.

<em>Number of moles of oxygen gas = mass/molar mass</em>

molar mass of oxygen gas = 32 g/mol

mass of oxygen gas = density * volume

mass of oxygen gas = 1.149 g/ml * 0.035 ml

mass of oxygen gas = 0.040215 g

Number of moles of oxygen gas = 0.0402 g/(32 g/mol)

Number of moles of oxygen gas = 0.00125 moles

<em>Using the ideal gas equation, PV=nRT</em>

where P = 1.0 atm, V = ?, n = 0.00125 moles, R = 0.082 L*atm/K*mol, T = (37 + 273)K = 310 K

<em>V = nRT/P</em>

V = (0.00125moles) * (0.082 L*atm/K*mol) * (310 K) / 1 atm

V = 0.0318L or 31.8mL

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