Answer: The coefficient of nitrogen in the given equation is 2.
Explanation: The reaction for the oxidation of methamphentamine with oxygen gas in the body is given by:

By Stoichiometry,
4 moles of methamphentamine reacts with 55 moles of oxygen gas to produce 40 moles of carbon dioxide gas, 30 moles of water and 2 moles of nitrogen gas.
Coefficient of 
Coefficient of 
Coefficient of 
Coefficient of 
Coefficient of 
Hence, the coefficient of nitrogen in the given equation is 2.
Answer:
2Na=Ca(OH)000.1 AgBr=2KF 2KBr=LiNO
Cu+ p=29 e=28 n=34
S2- p=16 e=18 n=16
Pb4+ p=82 e=78 p=125
I hope i did it right :))
a) 56g
<h3>Calculation:</h3>
At STP,
22.4 L of N₂ = 1 mol
We have given 44.8 L of N₂, therefore,
44.8 L of N₂ = 
=
mol
We know that,
1 mol of N₂ = 28 g
Hence,
2 mol of N₂ = 28 × 2
= 56g
Hence, there are 56 g of N₂ in 44.8 L of nitrogen gas.
Learn more about calculation at STP here:
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