Answer:
The mouse consumed 0,04064 g of oxygen.
Explanation:
To answer this question you can use gas law (PV=nRT) solving for moles of air that are proportional to moles of oxygen. These moles most be converted to grams with molecular weight:
n = PV/RT
Where, at initial conditions:
P is pressure (761,0torr×) = <em>1,0013 atm</em>
V is volume <em>2,40 L</em>
R is gas constant <em>0,082 atmL/molK</em>
T is temperature <em>306 K</em>
Solving, moles of aire are: <em>0,09577 moles</em>
21% of air is oxygen, thus:
0,09577×0,21 = <em>0,02011 initial moles of oxygen </em>
The final moles of oxygen are:
n = PV/RT
With the same initial conditions just changing pressure:
P is 712,9torr× = <em>0,9380 atm</em>
Final moles are: <em>0,08972 moles of air</em>
The moles of oxygen are:
0,08972×0,21 = <em>0,01884 final moles of oxygen </em>
Thus, moles of oxygen consumed by the mouse are:
0,02011 - 0,01884 = 1,27x10⁻³ moles
In grams:
1,27x10⁻³ moles × = <em>0,04064 g of O₂</em>