An individual is hospitalized and the initial blood work indicates high levels of 
in the blood and a pH of 7. 47. This would indicate the individual probably has compensated respiratory acidosis.
A chronic illness usually leads to compensated respiratory acidosis because the kidneys have time to adjust to the delayed onset. Even if the 
is elevated in a compensated respiratory acidosis, the pH is within the usual range.
The kidneys counteract a respiratory acidosis by increasing the amount of 
that tubular cells reabsorb from the tubular fluid, the amount of 
that collecting duct cells secrete while also producing , and the amount of 
buffer that is formed through ammoniagenesis.
Respiratory acidosis is frequently brought on by hypoventilation as a result of: breathing depression , paralysis of the respiratory muscles, diseases of the chest wall , abnormalities of the lung parenchyma and abdominal squeezing.
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Answer:
The correct answer is: Ka= 5.0 x 10⁻⁶
Explanation:
The ionization of a weak monoprotic acid HA is given by the following equilibrium: HA ⇄ H⁺ + A⁻. At the beginning (t= 0) we have 0.200 M of HA. Then, a certain amount (x) is dissociated into H⁺ and A⁻, as is detailed in the following table:
HA ⇄ H⁺ + A⁻
t= 0 0.200 M 0 0
t -x x x
t= eq 0.200M -x x x
At equilibrium, we have the following ionization constant expression (Ka):
Ka= ![\frac{ [H^{+}] [A^{-} ]}{ [HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%20%5BH%5E%7B%2B%7D%5D%20%20%5BA%5E%7B-%7D%20%5D%7D%7B%20%5BHA%5D%7D)
Ka= 
Ka= 
From the definition of pH, we know that:
pH= - log [H⁺]
In this case, [H⁺]= x, so:
pH= -log x
3.0= -log x
⇒x = 10⁻³
We introduce the value of x (10⁻³) in the previous expression and then we can calculate the ionization constant Ka as follows:
Ka= 
= 
= 5.025 x 10⁻⁶= 5.0 x 10⁻⁶
If it is located at the second to last row of the periodic table (the halogen family), has seven electrons on it's outer shell, and has an oxidation number of -1, it is a halogen.
Hope this helps : D
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