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Delicious77 [7]
3 years ago
8

What effect does pneumonia have on the diffusion of O2 and CO2 across the membranes in the lungs? a. slows diffusion b. complete

ly blocks diffusion c. accelerates diffusion d. has no effect e. reverses diffusion direction
Chemistry
1 answer:
GarryVolchara [31]3 years ago
5 0

Answer:

a. slows diffusion

Explanation:

Gas exchange on respiratory surfaces in the body (the lungs) occurs through a process known as diffusion. Blood which is low in oxygen and high in carbondioxide (carried from cells) goes through an exchange in the lung's alveoli (where oxygen concentration is high and carbondioxide is low). The oxygen in the alveoli diffuses into the blood, while the carbondioxide in the blood diffuses into the alveoli. This diffusion is possible because of the concentration gradient across the membranes.

Pneumonia is the inflammation of the lungs due to injury or infection. Liquid (pus) accumulates in the alveoli (a natural immune response to the infection or injury), a condition known as pulmonary edema which makes it harder for gases to be exchanged between the blood and the alveoli, thereby making breathing difficult. This slows down diffusion and if the condition is severe enough, can cause a respiratory failure where oxygen levels in the blood are critically low and carbondioxide levels are very high.

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Dissolving 5.28 g of an impure sample of calcium carbonate in hydrochloric acid produced 1.14 L of carbon dioxide at 20.0 °C and
swat32

Answer:

\%\ mass\ of\ CaCO_3=93.37\ \%

Explanation:

Given that:

Pressure = 791 mmHg

Temperature = 20.0°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (20 + 273.15) K = 293.15 K  

T = 293.15 K  

Volume = 100 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 62.3637 L.mmHg/K.mol  

Applying the equation as:

791 mmHg × 1.14 L = n × 62.3637 L.mmHg/K.mol  × 293.15 K  

⇒n of CO_2 produced =  0.0493 moles

According to the reaction:-

CaCO_3 + 2 HCl\rightarrow CaCl_2 + H_2O + CO_2

1 mole of carbon dioxide is produced 1 mole of calcium carbonate reacts

0.0493 mole of carbon dioxide is produced 0.0493 mole of calcium carbonate reacts

Moles of calcium carbonate reacted = 0.0493 moles

Molar mass of CaCO_3 = 100.0869 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.0493\ mol= \frac{Mass}{100.0869\ g/mol}

Mass_{CaCO_3}=4.93\ g

Impure sample mass = 5.28 g

Percent mass is percentage by the mass of the compound present in the sample.

\%\ mass\ of\ CaCO_3=\frac{Mass_{CaCO_3}}{Total\ mass}\times 100

\%\ mass\ of\ CaCO_3=\frac{4.93}{5.28}\times 100

\%\ mass\ of\ CaCO_3=93.37\ \%

3 0
3 years ago
Plz help ASAP. thx for you answers​ ALGEBRA 2 sorry
iren [92.7K]

So, the answer to 27.) would be <em>2x.</em> Both 6x and 2x can be divided by 2x, but they can't go any higher without the end-answer becoming a fraction. As such, 2x is the greatest common factor.

For 28.), x and x^2 can't be like terms, since like terms have the same variable and exponent :)

Hope I could help!

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