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Burka [1]
2 years ago
7

Freediving is an activity in which a person dives, sometimes to great depth, without the use of scuba gear. The diver must hold

his or her breath for the duration of the dive. The record depth, with no equipment such as diving fins, is 101 m; the official record time is over 11 minutes. Historically, freediving has been used by pearl divers and sponge divers. Some freedivers hyperventilate (breathe rapidly and deeply) before diving. Hyperventilation can change the concentration of CO2 in the blood and may increase the length of time that a person feels like he or she can hold his or her breath. How does hyperventilation affect blood pH?
Chemistry
1 answer:
Luden [163]2 years ago
6 0

Explanation:

Carbon dioxide is eliminated from the body as a gas through exhaled air. When the volumes of air breathed into and out of the lungs increases above what is normal (hyperventilation), more carbon dioxide than normal is eliminated.

When carbon dioxide is dissolved in water forms carbonic acid which is an acid, having less CO2 in the system affects the balance of the normal pH tips towards the blood pH to become alkaline (higher pH), this condition is called respiratory alkalosis.

We can observe the equilibrium in this equation:

CO2 + H2O ← H2CO3 ← H+ + HCO3

<em>Here we observe how losing CO2 during hyperventilation leads to a decrease of H+ and therefore an increase in pH</em>.

I hope you find this information useful and interesting! Good luck!

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Decomposition reaction results in the formation of the products by splitting the reactants. Moles of mercury(ii) oxide needed are 15.63 moles.

<h3>What are moles?</h3>

Moles are the ratio of mass and the molar mass of the compound or the molecule.

Moles of oxygen are calculated as:

\begin{aligned}\rm n &= \rm \dfrac {mass}{molar\; mass}\\\\&= \dfrac{250}{32}\\\\&= 7.8125 \;\rm moles\end{aligned}

The balanced chemical reaction can be shown as,

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From the reaction, it can be said that 1 mole of oxygen requires 2 moles of mercury oxide.

Moles of mercury oxide are calculated as:

\begin{aligned}  \rm n HgO& = (7.8125 \;\rm moles \; O_{2}) \times (\dfrac{2 \;\rm moles\;  HgO}{1 \;\rm mole \; O_{2}})\\\\\rm n HgO &= 15.625 \;\rm moles \end{aligned}

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