Since we are told that 1L of air contains 0.21L of oxygen, you can use the conversion (0.21L O₂)/(1L air). That means that you can just multiply 6.0L by 0.21L to get 1.26L of O₂.
that means that the lungs can hold about 1.26L of oxygen.
I hope this helps. Let me know if anything is unclear.
Answer:
28.43 min
Explanation:
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given that:
The rate constant, k = 
min⁻¹
Initial concentration = 0.1 M
Final concentration = 
M
Time = ?
Applying in the above equation, we get that:-
The answers to this problem is C
<span>46.06844 g/mol
yup
yup
yup</span>
