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.
The volume of the gas that will be produced in the person's stomach at body temperature (37°C) and a pressure of 1.0 atm is 48.36 mL
We'll begin by calculating the mass of the liquid oxygen. This can be obtained as follow:
Density = 1.149 g/mL
Volume = 0.053 mL
Mass = Density × Volume
Mass of O₂ = 1.149 × 0.053
Next, we shall determine the number of mole in 0.060897 g of O₂.
Mass of O₂ = 0.060897 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mole of O₂ =?
Mole = mass / molar mass
Mole of O₂ = 0.060897 / 32
Finally, we shall determine the volume of the gas produced in the stomach. This can be obtained as follow:
Mole of O₂ (n) = 0.0019 mole
Temperature (T) = 37 °C = 37 + 273 = 310 K
Pressure (P) = 1 atm
Gas constant (R) = 0.0821 atm.L /Kmol
1 × V = 0.0019 × 0.0821 × 310
V = 0.04836 L
Multiply by 1000 to express in mL
V = 0.04836 × 1000
Therefore, the volume of the gas produced in the person's stomach is 48.36 mL
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