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2 years ago

Control of Blood pH by respiratory rate.

Chemistry

1 answer:

nata0808 [166]2 years ago

6 0

Answer:

A) During this procedure ( hypoventilation ) The CO2 in the arterial blood vessels and the lungs increases and this drives the PH level in the system lower, and the equilibrium will shift to the right. this is because the Blood-PH level is controlled by CO2 - bicarbonate buffer system

B) The blood PH may rise to 7.60 during Hyperventilation because the removal of CO2 from the lungs causes the increase in $H^+$ which is directly proportional to the increase in Blood PH levels

C) Hyper ventilation before a dash would be useful because it will remove excessive Hydrogen ions and and raise the Blood PH levels in preparedness of the production of acids like Lactic acid

Explanation:

$CO_{2} +H_{2} O$ ⇄ $H^+ + HCO^-_{3}$

B) The blood PH may rise to 7.60 during Hyperventilation because the removal of CO2 from the lungs causes the increase in $H^+$ which is directly proportional to the increase in Blood PH levels

C) Hyper ventilation before a dash would be useful because it will remove excessive Hydrogen ions and and raise the Blood PH levels in preparedness of the production of acids like Lactic acid

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2 years ago

Balance the following equation. Then, given the moles of reactant or product below, determine the corresponding amount in moles

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Answer:

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Explanation:

This the unbalanced reaction

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3 years ago

For the following species, draw a Lewis structure for one important resonance form. Include all lone pair electrons and assign f

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Answer:

The Lewis structures are shown in the figure

Explanation:

The lewis structure will be drawn using following steps

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ii) will assign one bond (two electrons in between two atoms)

iii) then distribute the rest of the valence electrons as lone pair or shared pair based on completion of octet.

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2 years ago

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2 years ago

(b) Use the first law of thermodynamic to calculate AU for the following situations: (i) A coiled spring unwinds producing 153 J

il63 [147K]

Answer:

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(v) ΔU = 3.25 KJ

Explanation:

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ΔU = Q + W

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⇒ ΔU = Q = 1 KJ

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⇒ ΔU = Q + W

∴ Q = 15.6 KJ

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⇒ ΔU = 15.6 KJ + ( - 12.35 KJ )

⇒ ΔU = 3.25 KJ

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