It is 0.720 meters cause if the manufacturers of liters contain 2.27 inches it would make a deeply filled of 0.660
One way of knowing that oxygen was the gas removed from the volume of air and not another is to know what the volume of air is made of first. When the composition of the volume of air is already identified, then next would be the process of separating these elements from each other and as to which is to be separated first. This would usually lead to knowing their masses, their boiling and freezing points, the temperatures at which they condense, and so on. This is to identify their differences to each other and use those differences to successfully separate those elements to each other.
Answer:

Explanation:
First of all we need to calculate the heat that the water in the cooler is able to release:

Where:
- Cp is the mass heat capacity of water
- V is the volume
is the density


To calculate the mass of CO2 that sublimes:

Knowing that the enthalpy of sublimation for the CO2 is: 


Remember this:
1) n is principal quantum number and represents the energy level.
2) l is the second quantum number and represent the type of orbital.
3) l can take values from 0 to n - 1
4) each number of l is associated with a type of orbital. This table shows the equivalence:
l number type of orbital
0 s
1 p
2 d
3 f
With that, you can tell that n = 2 permits l = 0 and 1, which is orbitals s and p.
Therefore, the answer is the option D) s, p.
It takes 33.4 s for the concentration of A to fall to one-fourth of its original value.
A <em>half-life</em> is the time it takes for the concentration to fall to half its original value.
Assume the initial concentration is 1.00 mol/L. Then,

The concentration drops to one-fourth of its initial value in two half-lives.
∴ Time = 2 × 16.7 s = 33.4 s