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Answer:
Explanation:
This type of experiment was carried out in 1960s on rodents, it was partially successful but was perceived impractical and dangerous for humans,it is possible theoretically.
Oxygen is broken down or dissolves in a thin film of fluid in the alveoli, surprisingly in normal breathing liquid composed of dissolved oxygen is involved. Evidently respiratory gas must be able to dissolve in this liquid and in concentration required to keep the partial pressure necessary to power diffusion.
Answer:
The mass in grams of glucose produced when 132.0 g of CO2 reacts with an excess of water is 90.1 grams
Explanation:
The chemical equation for the reaction is
6H₂O + 6CO₂ → C₆H₁₂O₆ + 6O₂
From the reaction, it is seen that 6 moles of H₂O reacts ith 6 moles of CO₂ to produce 1 mole of glucose C₆H₁₂O₆ and 6 moles oxygen gas
The molar mass of CO₂ = 44.01 g/mol
There fpre 132.0 g contains 132.0/44.01 moles or ≅ 3 moles
However since 6 moles of CO₂ produces 1 mole of O₂, then 3 moles of CO₂ will prduce 1/6×3 or 0.5 moles of C₆H₁₂O₆
and since the molar mass (or the mass of one mole) of C₆H₁₂O₆ is 180.2 grams/mole then 0.5 mole of C₆H₁₂O₆ will have a mass of
mass of 1 mole C₆H₁₂O₆ = 180.2 g
mass of 0.5 mole C₆H₁₂O₆ = 180.2 g × 0.5 = 90.1 grams
Mass of glucose produced = 90.1 grams
