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3 years ago

What must the condition of the alveoli be in order for gas exchange to occur

Physics

1 answer:

aliya0001 [1]3 years ago

8 0

Answer:

There are 5 main conditions of the alveoli for proper gaseous exchange to in the lungs.

Explanation

Alveoli is found in the lungs of mammals, birds and reptiles. It is the part of the lungs where gaseous exchange really occurs.

Alveoli has a shape of that of balloon but indeed smaller than the real balloon. It allows the passage of oxygen and carbon dioxide

The five main conditions includes:

The Alveoli must have large surface to volume ratio which increase the gases that could be exchanged
The wall must be thin. This can shorten diffusion distance.
Alveoli must be very moist so that oxygen and carbon dioxide can pass through the solution
Alveoli must be well supplied with blood
Alveoli must be permeable

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16. The energy absorbed in 10minutes by an electrical heater is 1.5 MJ. The supply voltage is 240 V. calculate: a) The current d

dsp73

Answer:

If by 1.5 MJ you mean 1.5E6 Joules then

W = P t = power X time

W / t = P power

P = 1.5E6 J / 600 sec = 2500 J / s

P = I V

a) I = 2500 J/s / (240 J/c) = 10.4 C / sec = 10.4 amps

b) Q = I t = 10.4 C / sec * 300 sec = 3120 Coulombs

c) E = P * t = 2500 J / sec * 100 hr * 3600 sec / hr = 9.0E8 Joules

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3 years ago

Consider a container of oxygen gas at a temperature of 23°C that is 1.00 m tall. Compare the gravitational potential energy of a

Sergio039 [100]

Answer:

Yes, it is reasonable to neglect it.

Explanation:

Hello,

In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):

$m=1molec*\frac{1mol}{6.022x10^{23}molec} *\frac{32g}{1mol}*\frac{1kg}{1000g}=5.31x10^{-26}kg$

After that, we compute the potential energy 1.00 m above the reference point:

$U=mhg=5.31x10^{-26}kg*1.00m*9.8\frac{m}{s^2}=5.2x10^{-25}J$

Then, we compute the average kinetic energy at the specified temperature:

$K=\frac{3}{2}\frac{R}{Na}T$

Whereas $N_A$ stands for the Avogadro's number for which we have:

$K=\frac{3}{2} \frac{8.314\frac{J}{mol*K}}{6.022x10^{23}/mol}*(23+273)K\\ \\K=6.13x10^{-21}J$

In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.

Regards.

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3 years ago

A sample of chlorine has two naturally occurring isotopes. The isotope Cl-35 (mass 35.0 amu) makes up 75.8% of the sample, and t

irinina [24]

Answer:

$M_{av}=35.521amu$

Explanation:

As in any sample you will have 75.8% of Cl-35 iosotopes and 24.3% of Cl-37 iosotopes you can get the average atomic mass as:

$M_{av}=(35amu*75.8+37amu*24.3)/100=35.521amu$

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How could a slow moving train had the same momentum as a high-speed bullet

Flura [38]

Answer:

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2 years ago

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topjm [15]

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3 years ago