Answer:
0.92 kg
Explanation:
The volume occupied by the air is:

The moles of air are:

The heat required to heat the air by 10.0 °C (or 10.0 K) is:

Methane's heat of combustion is 55.5 MJ/kg. The mass of methane required to heat the air is:

The empirical formula is N₂O₅.
The empirical formula is the <em>simplest whole-number ratio of atoms</em> in a compound.
The ratio of atoms is the same as the ratio of moles, so our job is to calculate the <em>molar ratio of N:O</em>.
I like to summarize the calculations in a table.
<u>Element</u> <u>Moles</u> <u>Ratio¹ </u> <u> ×2² </u> <u>Integers</u>³
N 1.85 1 2 2
O 4.63 2.503 5.005 5
¹To get the molar ratio, you divide each number of moles by the smallest number (1.85).
²Multiply these values by a number (2) that makes the numbers in the ratio close to integers.
³Round off the number in the ratio to integers (2 and 5).
The empirical formula is N₂O₅.
Answer:
I believe it's the lowest portion of the atmosphere