Hemoglobin has a much greater affinity for carbon monoxide than oxygen. which principle explains why a hyperbaric chamber (conta
Hemoglobin has a much greater affinity for carbon monoxide than oxygen. In a hyperbaric chamber (containing high levels of oxygen) can treat carbon monoxide poisoning, by displacing carbon monoxide from Hemoglobin competitively.
Hemoglobin has a much greater affinity for carbon monoxide than oxygen. This is because, a coordinate bond is formed with Carbon monoxide and Haem structure of the hemoglobin.
Carbon monoxide with Hemoglobin is called as Carboxy haemoglobin.
Presence of oxygen displaces the Carbon monoxide with Hemoglobin that is formed due to poisoning.
Hyperbaric chamber is a chamber which contains pure oxygen in a chamber. The atmospheric pressure is kept about three to four times than the normal, such that the replacement of Carbon monoxide from Haem can occur as fast as possible since this reduces the half life of the Carboxy haemoglobin.
It is advisable not to treat Carbon monoxide poisoning yourself.
Hyperbaric oxygen is used to treat the following conditions as well:
- Infections
- Wounds
- Air bubble is blood
Learn more about Carbon Monoxide here, brainly.com/question/11313918
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Answer:
485.76 g of CO₂ can be made by this combustion
Explanation:
Combustion reaction:
2 C₄H₁₀(g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g)
If we only have the amount of butane, we assume the oxygen is the excess reagent.
Ratio is 2:8. Let's make a rule of three:
2 moles of butane can produce 8 moles of dioxide
Therefore, 2.76 moles of butane must produce (2.76 . 8)/ 2 = 11.04 moles of CO₂
We convert the moles to mass → 11.04 mol . 44g / 1 mol = 485.76 g
Explanation:
Mass = volume × density
Mass = 652 cm³ × 21.45 g/cm³
= 13985.4 g
Explanation:
Answer:
alpha decay
Explanation:
Answer:
108.6 g
Explanation:
- 2NaN₃(s) → 2Na(s) + 3N₂(g)
First we use the <em>PV=nRT formula</em> to <u>calculate the number of nitrogen moles</u>:
- P = 1.00 atm
- V = 56.0 L
- n = ?
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 0 °C ⇒ 0 + 273.2 = 273.2 K
<u>Inputting the data</u>:
- 1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 K
- n = 2.5 mol
Then we <u>convert 2.5 moles of N₂ into moles of NaN₃</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 2.5 mol N₂ *
= 1.67 mol NaN₃
Finally we <u>convert 1.67 moles of NaN₃ into grams</u>, using its <em>molar mass</em>:
- 1.67 mol * 65 g/mol = 108.6 g