1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
elena55 [62]
3 years ago
7

An athlete at high performance inhales 4.0L of air at 1 atm and 298 K. The inhaled and exhaled air contain 0.5% and 6.2% by volu

me of water,respectively. For a respiration rate of 40 breaths per minute, how many moles of water per minute are expelled from the body through the lungs?
Physics
1 answer:
LenaWriter [7]3 years ago
8 0

To solve this problem we will calculate the total volume of inhaled and exhaled water. From the ideal gas equation we will find the total number of moles of water.

An athlete at high performance inhales 4.0L of air at 1atm and 298K.

The inhaled and exhaled air contain 0.5% and 6.2% by volume of water, respectively.

During inhalation, volume of water taken is

V_i = (4L)(0.5\%)

V_i = 0.02L

During exhalation, volume of water expelled is

V_e = (4L)(6.2\%)

V_e = 0.248L

During 40 breathes, total volume of water taken is

V_{it} = (40L)(0.02L) = 0.8L

During 40 breathes, total volume of water expelled out is

V_{et} = (40L)(0.248L) = 9.92L

Therefore resultant volume of water expelled out from the lung is

\Delta V = 9.92L-0.8L = 9.12

From the body through the lung we have that

n = \frac{PV}{RT}

Here,

P = Pressure

R= Gas ideal constant

T= Temperature

V = Volume

Replacing,

n = \frac{(1atm)(9.12L)}{(8.314J/mol \cdot K)(298K)}

n = 0.373mol/min

Therefore the moles of water per minute are expelled from the body through the lungs is 0.373mol/min

You might be interested in
Constants A capacitor is connected across an ac source that has voltage amplitude 59 0 V and frequency 77 0 Hz Part C What is th
Vladimir79 [104]

Answer:

C = 1.77 \times 10^{-4} F

Explanation:

As we know that in AC circuit we have

V = i x_c

here we have

V = 59 V

i = 5.05 A

so we will have

x_c = \frac{59}{5.05}

x_c = 11.68 ohm

also we know that

x_c = \frac{1}{\omega C}

here we will have

11.68 = \frac{1}{(2\pi 77)C}

C = 1.77 \times 10^{-4} F

7 0
2 years ago
A drag racing vehicle travels from zero to a hundred miles per hour and 5 seconds north what is acceleration
avanturin [10]
Acceleration(a) is the change in velocity(Δv) over the change in time(Δt). so just divide your velocity and time.
6 0
2 years ago
Please help, I do not understand
Anettt [7]
I think the key here is to be exquisitely careful at all times, and
any time we make any move, keep our units with it.

We're given two angular speeds, and we need to solve for a time.

Outer (slower) planet:
Angular speed =  ω  rad/sec
Time per unit angle =  (1/ω)  sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/ω sec/rad) · (2π rad) = 2π/ω seconds .

Inner (faster) planet:
Angular speed =  2ω  rad/sec
Time per unit angle =  (1/2ω)  sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/2ω sec/rad) · (2π rad) = 2π/2ω sec = π/ω seconds.

So far so good.  We have the outer planet taking 2π/ω seconds for one
complete revolution, and the inner planet doing it in only π/ω seconds ...
half the time for double the angular speed.  Perfect !

At this point, I know what I'm thinking, but it's hard to explain.
I'm pretty sure that the planets are in line on the same side whenever the
total elapsed time is something like a common multiple of their periods.
What I mean is:

They're in line, SOMEwhere on the circles, when

     (a fraction of one orbit) = (the same fraction of the other orbit)    
AND
     the total elapsed time is a common multiple of their periods.

Wait !  Ignore all of that.  I'm doing a good job of confusing myself, and
probably you too.  It may be simpler than that.  (I hope so.)  Throw away
those last few paragraphs.

The planets are in line again as soon as the faster one has 'lapped'
the slower one ... gone around one more time.  
So, however many of the longer period have passed, ONE MORE
of the shorter period have passed.  We're just looking for the Least
Common Multiple of the two periods.

      K (2π/ω seconds)  =  (K+1) (π/ω seconds)

                     2Kπ/ω   =    Kπ/ω + π/ω

Subtract  Kπ/ω :    Kπ/ω = π/ω

Multiply by  ω/π :      K  =  1

(Now I have a feeling that I have just finished re-inventing the wheel.)

And there we have it:

     In the time it takes the slower planet to revolve once,
     the faster planet revolves twice, and catches up with it.
    
     It will be  2π/ω  seconds before the planets line up again.
    
     When they do, they are again in the same position as shown
     in the drawing.

To describe it another way . . . 

     When Kanye has completed its first revolution ...

     Bieber has made it halfway around.

     Bieber is crawling the rest of the way to the starting point while ...

     Kanye is doing another complete revolution.

     Kanye laps Bieber just as they both reach the starting point ...

     Bieber for the first time, Kanye for the second time.


You're welcome.  The generous bounty of 5 points is very gracious,
and is appreciated.  The warm cloudy water and green breadcrust
are also delicious.
5 0
2 years ago
The mass of the skier including his equipment is 75 kg in the ski race, the total vertical change in height is 880m
Bogdan [553]
75 percent off of water and please water the light water and water water and then go back and please water pollution please 880m
8 0
3 years ago
Which statement accurately represents the arrangement of electrons in Bohr’s atomic model?
noname [10]

Answer: Electrons move around the nucleus in fixed orbits of equal levels of energy

Explanation:

The statement that accurately represents the arrangement of electrons in Bohr’s atomic model is that the electrons move around the nucleus in fixed orbits of equal levels of energy.

It should be noted that the electrons have a fixed energy level when they travel around the nucleus in with energies which varies for different levels.

Higher energy levels are depicted by the orbits that are far from the nucleus. There's emission of light when the electrons then return back to a lower energy level.

8 0
3 years ago
Other questions:
  • What is a synonym for inertia?
    14·2 answers
  • A horizontal force of 90.0 N is required to push a 75kg object along a horizontal surface at a constant speed. What is the magni
    13·1 answer
  • Which is the largest gas giant?
    7·1 answer
  • A net force of 100 n is moving a mass with an acceleration of 5 m/s2. what is the mass of the object? 0.05 kg 20 kg 95 kg 500 kg
    9·2 answers
  • Your grandfather clock's pendulum has a length of 0.9930 m. if the clock loses 60 seconds per day, should you increase or decrea
    7·2 answers
  • Turning up the Heat
    12·1 answer
  • Environmental factors that can affect activity selection include
    6·2 answers
  • Which symbol and unit of measurement are used for resistance?
    14·2 answers
  • What is the velocity of an 80 kg person skating across the ice with a kinetic energy of 16, 000 joules?
    14·1 answer
  • You tie a string to the ceiling and attach a weight to the end. You hold the weight next to your face but not touching it and th
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!