Flue gas composed volumetrically of CO2 = 25%, O2 = 25%, and N2 = 50% possesses an average molecular weight of___________________?
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Answer: B. 23
Explanation
The average molecular weight is calculated from the volumetric fractions: 0.25(44) + 0.25(32) + 0.50(28) = 11 + 8 + 14 = 33, wait calculation check: 0.25*44=11, 0.25*32=8, 0.50*28=14, sum = 33. The correct molecular weight is 33, so the option is wrong.
- Subject
- Engineering
- Category
- Chemical Engineering
- Topic
- —
- Difficulty
- medium
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