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frosja888 [35]
3 years ago
13

A client comes to the emergency department with status asthmaticus. His respiratory rate is 48 breaths/minute, and he is wheezin

g. An arterial blood gas analysis reveals a pH of 7.52, a partial pressure of arterial carbon dioxide (PaCO2) of 30 mm Hg, PaO2 of 70 mm Hg, and bicarbonate (HCO3??') of 26 mEq/L. What disorder is indicated by these findings?
Chemistry
1 answer:
Tema [17]3 years ago
5 0

Answer:

The disorder is called Respiratory alkalosis

Explanation:

A client with asthma can hyperventilate (excess of O2). This oxygen excess can lead also to a lack of carbon dioxide (CO2) (acid substance) which can causes alkalosis (increasing the pH values in blood and body).

In this medical condition (respiratory alkalosis), the bicarbonate (HCO3-) hosted in the kidneys has a late response, so the  HCO3- level  in the client remains normal.

The PaCO2 (partial pressure of arterial carbon dioxide) is low in comparison with normal values so that indicates loss of CO2 in the respiratory system.

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Kemmi pipets 25.00 ml of pure 1-propanol (c3h7oh, a liquid organic alcohol) into a 100.0 ml volumetric flask. she dilutes it wit
algol13
Q1)
As Kemmi pipettes a volume of 25.00 ml of the solution
density of pure propanol is 0.803 g/ml
This means that in 1000 ml of solution - 0.803 g of pure propanol
Therefore in 25.00 ml of solution - 0.803 g x 25.00 ml / 1000 ml
                                                     = 0.0201 g
Using molar mass, number of moles can be calculated= 0.0201 g / 60.09 g/mol
                                                                                       = 3.35 x 10⁻⁴ mol
therefore the number of pure propanol moles in exactly 25.00 ml is
3.35 x 10⁻⁴ mol

Q2)
molarity is the concentration of the solution. It can be defined as the number of moles of solute per liter of solution
we know the number of moles in 25.00 ml of solution. When its diluted in a 100.00 ml volumetric flask, number of moles remain constant but now the volume over which the moles of solute are dissolved is increased.
therefore number of moles = 3.35 x 10^(-4) mol
volume over which its dissolved - 100.00 / 10³ dm³
                                                    = 1.0000 x10⁻¹ dm³
the molarity = 3.35 x 10⁻⁴ mol / 1.0000 x10⁻¹ dm³
                    = 3.35 x 10⁻³ mol/dm³
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How many moles of SO3 are produced when 1.5 mol of O2 react with SO2?
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Answer:

3 mole O2

Explanation:

Need balanced equation first:  O2 + 2SO2 --> 2SO3

assuming SO2 is in excess,

1.5 mol O2 (2moles SO3/1mole O2) = 3 mole O2

5 0
2 years ago
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