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levacccp [35]
2 years ago
9

4. The atmosphere is composed of about 78% nitrogen, 21% oxygen, and 1% argon. Typical atmospheric pressure in Boulder, Colorado

is about 0.83 atm. What is the pressure contributed by each gas?​
Physics
1 answer:
photoshop1234 [79]2 years ago
6 0

<u>ANS</u>

<u>Step</u> <u>1</u> :Explanation of required formula.

According to Dalton's Law of Partial Pressures, the partial pressure of a component of a gaseous mixture depends on the mole ratio of said component and the total pressure of the gaseous mixture

i.e Pi=Xi × Ptotal,

Here we don't know exactly how many moles of the mixture we have,

but we know that 78.0% of all the molecules present in the mixture are nitrogen molecules, 21.0% are oxygen molecules, and 1% are molecules of Ar gas.

As we know, a mole is simply a very large collection of molecules. In order to have one mole of a substance, we need to have 6.022 × 1023 molecules of that substance.

This means that the actual number of moles is not important here, because the ratio that exists between the number of molecules is equivalent to the ratio that exists between the number of moles.

Hence,

<u>Step</u> <u>2</u> : Calculate of mole fraction of the mixture.

mole fraction of nitrogen = 78 /100 = 0.78

mole fraction of O2 =

21 /100 = 0.21

mole fraction of Argon =

1 /100 = 0.01.

<u>Step</u> <u>3</u> : Calculate the pressure contributed by each of the mixture.

The pressure contributed by N2 = mole fraction of N2 × Total pressure = 0.78 × 0.83 atm = 0.6474 atm

The pressure contributed by O2 = 0.21 × 0.83 atm = 0.1743 atm

The pressure contributed by N2 = 0.01 × 0.83 atm = 0.0083 atm.

<u>Tha</u><u>nk</u> <u>You</u> !!!!!!

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<h3>How to calculate the photon energy?</h3>

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Mathematically, the Planck-Einstein equation can be calculated by using this formula:

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Two identical spheres are released from a device at time t = 0 from the same ... Sphere A has no initial velocity and falls straight down. ... (b) On the axes below, sketch and label a graph of the horizontal component of the velocity of sphere A and of sphere B as a function of time. ... Which ball has the greater vertical velocity

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A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police
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It will take 15.55s for the police car to pass the SUV

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We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

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2. x=x_{0}+v_{0}t+\frac{at^{2}}{2}

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

vt = v_{0}t+\frac{at^2}{2}\\\\   16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}

We simplify the fraction present and rearrange for our formula so that it equals 0:

0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0

In the very last step we factored a common factor t. There is two possible solutions to the equation at t=0 and:

0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\  0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at t=0 s (when the SUV passed the police car) and t=15.56s(when the police car catches up to the SUV)

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