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.
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The atmosphere is considered homogeneous. It isn’t exactly on the smallest scales but that doesn’t matter. Homogenous means the composition will be the same in any sample taken from the substance. And clearly, the atmosphere is mostly gas. So the last answer is right
The correct answer is measurement in Ci/Bq, the amount of radioactive materials released into the environment, and <span>number of disintegrations of radioactive atoms in a radioactive material over a period of time. I hope this helps.</span>
