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Debora [2.8K]
2 years ago
6

Ideal gases are often studied at standard ambient temperature and pressure (satp). The international union of pure and applied c

hemistry (iupac) defines satp to be t = 25° c and p = 100 kpa.
Physics
1 answer:
11Alexandr11 [23.1K]2 years ago
5 0

This problem provides information about the pressure and temperature ideal gases are studied at. The answer to the questions are that all molecules have the same density, 2.43x10²⁵ mol/m³ and 2.43x10¹⁹ mol/cm³.

<h3>Idela gases</h3>

In science, we can start studying gases with the concept of ideal gas, as they do not collide one to another and are assumed to be perfect spheres with no relevant interactions.

In such a way, one can conclude that the <u>number density of all ideal gasses at SATP is the same</u>, as they are assumed to be perfect spheres with equal volumes per molecule.

Moreover, when calculating the number of molecules per cubic meter, one must use the ideal gas equation as:

PV=nRT\\\\\frac{N}{V}= \frac{P*N_A}{RT}

And plug in the numbers we are given:

\frac{N}{V}= \frac{100kPa*\frac{1000Pa}{1kPa}*6.022x10^{23}molec/mol}{8.314\frac{Pa*m^3}{mol*K}*298K}=2.43x10^{25}molec/m^3

Lastly, we can calculate the molecules per cubic centimeter by performing the following conversion:

2.43x10^{25}\frac{molec}{m^3}*(\frac{1m}{100cm} )^3\\ \\=2.43x10^{19}\frac{molec}{cm^3}

Learn more about ideal gases: brainly.com/question/26450101

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Answer:

44,000 Nm^2/C

Explanation:

The electric flux through a certain surface is given by (for a uniform field):

\Phi = EA cos \theta

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In this problem, we have:

E=1.1\cdot 10^4 N/C is the electric field

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\theta=0^{\circ}, since the electric field is perpendicular to the surface

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A substance did not change it's chemical nature in reaction which mostly likely describe the reaction
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If the substance doesn't change chemically, it is a physical reaction.
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What is an object’s acceleration if it is moving at 30 m/s and comes to a stop in 5 s? –30 m/s2 –6 m/s2 30 m/s2 6 m/s2
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Radar uses radio waves of a wavelength of 2.9 m . The time interval for one radiation pulse is 100 times larger than the time of
Mandarinka [93]

Answer:

145 m

Explanation:

Given:

Wavelength (λ) = 2.9 m  

we know,

c = f × λ  

where,

c = speed of light ; 3.0 x 10⁸ m/s

f = frequency  

thus,

f=\frac{c}{\lambda}

substituting the values in the equation we get,

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f = 1.03 x 10⁸Hz  

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