Question

A person accidentally swallows a drop of liquid oxygen, O2(l), which has a density of 1.149 g/mL. Assuming the drop has a volume of 0.053 mL, what volume of gas will be produced in the person's stomach at body temperature (37°C) and a pressure of 1.0 atm?

Answer 1

Answer:

First, let's determine how many moles of oxygen we have.

Atomic weight oxygen = 15.999

Molar mass O2 = 2*15.999 = 31.998 g/mol

We have 3 drops at 0.050 ml each for a total volume of 3*0.050ml = 0.150 ml

Since the density is 1.149 g/mol,

we have 1.149 g/ml * 0.150 ml = 0.17235 g of O2

Divide the number of grams by the molar mass to get the number of moles 0.17235 g / 31.998 g/mol = 0.005386274 mol

Now we can use the ideal gas law. The equation PV = nRT where P = pressure (1.0 atm) V = volume n = number of moles (0.005386274 mol) R = ideal gas constant (0.082057338 L*atm/(K*mol) ) T = Absolute temperature ( 30 + 273.15 = 303.15 K)

Now take the formula and solve for V, then substitute the known values and solve.

PV = nRT V = nRT/P V = 0.005386274 mol * 0.082057338 L*atm/(K*mol) * 303.15 K / 1.0 atm V = 0.000441983 L*atm/(K*) * 303.15 K / 1.0 atm V = 0.133987239 L*atm / 1.0 atm V = 0.133987239 L

So the volume (rounded to 3 significant figures) will be 134 ml.

Answer 2

You need to divid 02(1) in to the first number that’s there