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aliina [53]
2 years ago
7

What would be the greatest effect of the finite size of molecules on the ideal gas law?

Physics
1 answer:
Musya8 [376]2 years ago
8 0

At high density of the molecules tends to be greater pressure that of molecules and at low density of the molecules tends to be less pressure that of molecules.

It is given that the molecules are in the ideal gas.

It is required to find the greatest effect of the finite size of molecules on the ideal gas law.

<h2>What would be the greatest effect of the finite size of molecules on the ideal gas law?</h2>

According to ideal gas law, at high density of the molecules tends to be greater pressure that of molecules and at low density of the molecules tends to be less pressure that of molecules.

From ideal gas equation,

PV = nRT

Where,

P - Pressure of the molecules

V - Volume of the molecules

n - Total amount of substance

R - Gas constant

T - Temperature of the molecules

As shown from above equation, both the quantities are directly proportional to each other and also ideal gas law follows the kinetic molecular theory  where it suggests that the volume of the molecules have negligible point.

Thus, at high density of the molecules tends to be greater pressure that of molecules and at low density of the molecules tends to be less pressure that of molecules.

Learn more about the ideal gas here:

brainly.com/question/8711877

#SPJ4

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How old is a bone if it still has 50% of its carbon-14 content?
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Answer:

5730 years

Explanation:

The half life of carbon-14 is 5730 years.  If 50% of the carbon-14 remains, then exactly 1 half life has passed.

The half-life equation is:

A = A₀ (½)^(t / T)

where A is the remaining amount,

A₀ is the initial amount,

t is time,

and T is the half life.

In this case, A = ½ A₀ and T = 5730.

½ A₀ = A₀ (½)^(t / 5730)

½ = (½)^(t / 5730)

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t = 5730

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An object exerts a reaction force when it
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all the forces occur in pairs that if one object exerts a force on another object, then the second object exerts an equal and opposite reaction force on the first.

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What dictates the minimum and maximum pressure allowed for plumbing fixtures?
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During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to c
IRINA_888 [86]

During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to come near, the clown turns due east and runs 19.8 m to exit the arena. The magnitude of the clown’s displacement is 27 m.

<u>Explanation: </u>

As the clown is running in the north direction for about 7.7 m and then he turns 49.9 degrees east of north. In the east of north, he covers a distance of 6.4 m and then turns east to exit the arena after covering a distance of 19.8 m. Let’s have a simple diagram to easily understand the problem.

In first step, the clown runs 7.7 m in north direction, so the image will be  as in fig 1. Then he takes a direction of north east and covers a distance of 6.4 m, so the image will be modified as in fig 2. Then after the bull comes, he turns east and runs 19.8 m to exit the arena, so the image will be as in figure 3.

So, the extension of North line and the East line at a point shown as the dotted line in the above image, forms the total displacement as the hypotenuse of a right angled triangle. The extended dotted lines is nothing but the horizontal and vertical components of the angle 49.9 degree.

By using Pythagoras theorem, the total displacement can be found as

\text { Total displacement }=\sqrt{(o p p)^{2}+(a d j)^{2}}

\text { Distance covered by the clown in east direction }=(6.4 \times \cos 49.9)+19.8=23.9 \mathrm{m}

Similarly, the adjacent side of this imaginary triangle is the distance covered by the clown in the North direction.

\text { Distance covered by the clown in north direction }=6.4 \sin 49.9+7.7=12.6 \mathrm{m}

Thus, the total displacement covered by the clown is

\text { Total displacement }=\sqrt{(23.9)^{2}+(12.6)^{2}}=\sqrt{571.21+158.76}=\sqrt{729.97}=27 \mathrm{m}

Thus, the total displacement by the clown is 27 m.

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What is the correct equation for calculating the average atomic mass for 3 isotopes? (pls be 100%of your answer pls no guessing)
mariarad [96]

<u>Answer:</u>

<em>The correct equation for measuring the average microscopic weight  for 3 isotopes is multiply the rate of abundance by each weight and add them.</em>

<u>Explanation:</u>

To calculate the average microscopic mass of element using weights and relative abundance we have to follow the following steps.

  • Take the correct weight of each isotope (that will be in decimal form)
  • Multiply the weight of each isotope by its abundance
  • Add each of the results together.

<em>This gives the required  average microscopic weight of the three isotopes.</em>

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