Explanation:
Formula to calculate osmotic pressure is as follows.
Osmotic pressure = concentration × gas constant × temperature( in K)
Temperature = 
= (25 + 273) K
= 298.15 K
Osmotic pressure = 531 mm Hg or 0.698 atm (as 1 mm Hg = 0.00131)
Putting the given values into the above formula as follows.
0.698 = 
C = 0.0285
This also means that,
= 0.0285
So, moles = 0.0285 × volume (in L)
= 0.0285 × 0.100
= 
Now, let us assume that mass of 
= x grams
And, mass of 
= (1.00 - x)
So, moles of 
= 
Now, moles of 
= 
= 
= x = 0.346
Therefore, we can conclude that amount of present is 0.346 g and amount of 
present is (1 - 0.346) g = 0.654 g.
Molarity = moles of solute/volume of solution in liters.
The solute here is NaCl, of which we have 46.5 g. To calculate the molarity of an NaCl solution, we need to know the number of moles of NaCl. To convert from grams to moles, we divide the mass by the molar mass of NaCl. The molar mass of NaCl is the sum of the atomic masses of Na and Cl: 23 amu + 35 amu = 58 amu. For our purposes, we can regard amu as equivalent to grams/mole.
(46.5 g)/(58 g/mol) = 0.8017 moles NaCl.
Now that we know both the number of moles of our NaCl solute and the volume of the solution, we can calculate the molarity:
(0.8017 moles NaCl)/(2.2 L) = 0.364 M.
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.
Learn more about Respiratory acidosis here;
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Answer: A 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.
Explanation:
Given: 
= 0.20 M, 
= 15.0 mL
= 0.10 M, 
= ?
Formula used is as follows.
Substitute the values into above formula s follows.
![M_{1}V_{1} = M_{2}V_{2}\\0.20 M ]times 15.0 mL = 0.10 M ]times V_{2}\\V_{2} = 30 mL](https://tex.z-dn.net/?f=M_%7B1%7DV_%7B1%7D%20%3D%20M_%7B2%7DV_%7B2%7D%5C%5C0.20%20M%20%5Dtimes%2015.0%20mL%20%3D%200.10%20M%20%5Dtimes%20V_%7B2%7D%5C%5CV_%7B2%7D%20%3D%2030%20mL)
Thus, we can conclude that a 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.
Answer:
Calculate the pH of a buffer prepared by mixing 30.0 mL of 0.10 M acetic acid and 40.0 mL of 0.10 M sodium acetate.
