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djverab [1.8K]
3 years ago
12

Observe yourself breathing and count the number of times you inhale per second. During each breath you probably inhale 0.66 L of

air. Assume the pressure is 1 atm and the temperature is 20∘C. Only about 21% of air is O2. 1. How many oxygen molecules do you inhale if you are at sea level?
Physics
1 answer:
Pavel [41]3 years ago
7 0

To solve this exercise it is necessary to apply the concepts related to Robert Boyle's law where:

PV=nRT

Where,

P = Pressure

V = Volume

T = Temperature

n = amount of substance

R = Ideal gas constant

We start by calculating the volume of inhaled O_2 for it:

V = 21\% * 0.66L

V = 0.1386L

Our values are given as

P = 1atm

T=293K R = 0.083145kJ*mol^{-1}K^{-1}

Using the equation to find n, we have:

PV=nRT

n = \frac{PV}{RT}

n = \frac{(1)(0.1386)}{(0.0821)(293)}

n = 5.761*10^{-3}mol

Number of molecules would be found through Avogadro number, then

\#Molecules = 5.761*10^{-3}*6.022*10^{23}

\#Molecules = 3.469*10^{21} molecules

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3 years ago
Convert the volume 8.06 in.3 to m3, recalling that1in. =2.54cmand100cm=1m. Answer in units of m3.
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1 in=2.54 cm=(2.54 cm)(1 m/100 cm)=0.0254 m
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Therefore:

8.06 in³=(8.06 in³)(1.6387064 x 10⁻⁵ m³ / 1 in³)≈1.321 x 10⁻⁴ m³.

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3 years ago
The velocity time graph of an object is shown below. How far does the object travel in the time interval t =4 s to t = 6 s?
Trava [24]

The distance covered by the object between t =4 s and t = 6 s is 4 m

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Learn more about distance:

brainly.com/question/3969582

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3 0
3 years ago
A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik
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Answer:

The options are approximations of the exact answers:

A) 1\times10^6N/C

B) 2\times10^{-13}N

C) 1\times10^{-2}V

D) Toward the inner wall

E) 3\times10^{-17}J

Explanation:

A) The electric field in a parallel plate capacitor is given by the formula E=\frac{\sigma}{k\epsilon_0}, where \epsilon_0=8.85\times10^{-12}C/Vm and in our case \sigma=10^5C/m^2 and, for air,k=1.00059, so we have:

E=\frac{10^5C/m^2}{(1.00059)(8.85\times10^{-12}C/Vm)}=1.13\times10^6N/C

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F=(1.6\times10^{-19}C)(1.13\times10^6N/C)=1.81\times10^{-13}N

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K=(3\times10^{-15}C)(1.13\times10^{-2}V)=3.39\times10^{-17}J

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