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:
Where,
is the concentration at time t
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>