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:
the animal
Explanation:
a rodent thas in the ground
The new volume of the oxygen would be 17.4 L. I think it's 17.4 L but I don't know. I hope it was helpful.
First, recognize that this is an elimination reaction in which hydroxide must leave and a double bond must form in its place. It is likely an E2 reaction. Here is an efficient mechanism:
1) Pre-reaction: Protonate the -OH to make it a good leaving group, water. H2SO4 or any strong H+ donor works. The water is positively charged but still connected to the compound.
2) E2: Use a sterically hindered base, such as tert-butoxide (tButO-) to abstract the hydrogen from the secondary carbon. [You want a sterically hindered base because a strong, non-sterically hindered base could also abstract a hydrogen from one of the two methyl groups on the tertiary carbon, and that leads to unwanted products, which is not efficient]. As the proton of hydrogen is abstracted, water leaves at the same time, creating an intermediate tertiary carbocation, and the 2 electrons in the C-H bond immediately are used to make a double bond towards the partial positive charge.
In the products we see the major product and water, as expected. Even though you have an intermediate, remember that an E2 mechanism technically happens in one step after -OH protonation.
Answer:
See the answers below
Explanation:
1) 100. mL of solution containing 19.5 g of NaCl (3.3M)
2) 100. mL of 3.00 M NaCl solution (3 M)
3) 150. mL of solution containing 19.5 g of NaCl (2.2 M)
4) Number 1 and 5 have the same concentration (1.5M)
MW of NaCl = 23 + 36 = 59 g
For number 3
59 g ------------------- 1 mol
19,5 g ----------------- x
x = 19.5 x 1/59 = 0.33 mol
Molarity (M) = 0.33 mol/0.150 l = 2.2 M
For number 4,
Molarity (M) = 0.33mol/0.10 l = 3.3 M
For number 5
Molarity (M) = 0.450/0.3 = 1.5 M
