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blondinia [14]
3 years ago
11

Any girls out there wear leggings

Chemistry
1 answer:
erastovalidia [21]3 years ago
7 0

Answer:

yes because I seen one yesterday

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An individual is hospitalized and the initial blood work indicates high levels of hco3- in the blood and a ph of 7. 47. This wou
expeople1 [14]

An individual is hospitalized and the initial blood work indicates high levels of HCO_{3} ^{-} 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 PCO_{2} 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 HCO_{3} that tubular cells reabsorb from the tubular fluid, the amount of H^{+} that collecting duct cells secrete while also producing HCO_{3} , and the amount of NH_{3} 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.

Learn more about Respiratory acidosis here;

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5 0
1 year ago
Why was Rutherford's Gold Foil Experiment so important?
Oksanka [162]

Answer:

Answer is 'B' I think

Explanation:

Rutherford's Gold Foil Experiment proved the existence of a small massive center to atoms, which would later be known as the nucleus of an atom. Through previous experiments of shooting alpha particles, Rutherford knew they had considerable mass and speed.

4 0
3 years ago
in order of the human to maintain homeostasis, the breakdown of glucose to release energy must be followed by the?
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2 years ago
A mixture consists of 28% oxygen, 14% hydrogen, and 58% nitrogen by volume. A sample of this mixture has a pressure of 4.0 atm i
stepladder [879]

Answer:

C) 1.3 mol

Explanation:

Using gas law we can find the initial moles of the sample of the mixture, as follows:

PV = nRT

PV / RT = n

<em>Where P is pressure: 4.0atm</em>

<em>V is volume: 9.6L</em>

<em>R is gas constant: 0.082atmL/molK</em>

<em>T is absolute temperature: 300K</em>

<em>And n are moles of the gas</em>

<em />

PV / RT = n

4.0atm*9.6L / 0.082atmL/molK300K = n

n = 1.56moles of the mixture of the gas are present into the 9.6L container

Now, 14% of this gas is hydrogen that was removed of the system, that is:

1.56mol*14% = 0.22 moles of hydrogen are removed.

Thus, moles of gas that remains in the container are:

1.56mol - 0.22mol = 1.34mol.

Right answer is:

<h3>C) 1.3 mol</h3>

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3 years ago
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