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love history [14]
2 years ago
7

Under what conditions does the Ideal Gas Law not apply and gases are considered real? Check all that apply. High pressure low pr

essure high temperature low temperature.
Chemistry
2 answers:
Ulleksa [173]2 years ago
8 0

Answer:

its A and D

Explanation:

Assoli18 [71]2 years ago
7 0

Considering the definition of ideal gas and gas laws, The Ideal Gas Law not apply and gases are considering real under conditions of high pressure low pressure.

<h3>Definition of ideal gas</h3>

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other.

In other words, an ideal gas is defined as one in which all collisions between atoms or molecules are perfectly elastic and in which there are no intermolecular attractive forces, so they do not interact with each other.

Gases in general are ideal when they are at high temperatures and low pressures.

<h3>Definition of gas laws</h3>

On the other hand, the gas laws are a set of chemical and physical laws that allow us to determine the behavior of gases in a closed system. The parameters evaluated in these laws are pressure, volume, temperature, and moles.

Summary

The Ideal Gas Law not apply and gases are considering real under conditions of high pressure low pressure.

Learn more about  Ideal Gas Law:

brainly.com/question/4147359?referrer=searchResults

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What is the mass, in grams, of a 12.0cm³ sample of aluminum? The density of aluminum is 2.70g/cm³
Levart [38]

Answer:

The answer is

<h2>32.4 g</h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume of aluminum = 12 cm³

Density = 2.70 g/cm³

The mass of aluminum is

mass = 2.7 × 12

We have the final answer as

<h3>32.4 g</h3>

Hope this helps you

4 0
3 years ago
Calculate the solubility of mn(oh)2 in grams per liter when buffered at ph=7.0. assume that buffer capacity is not exhausted
Vanyuwa [196]
When PH + POH = 14 
∴ POH = 14 -7 = 7

when POH = -㏒[OH-]

          7    = -㏒ [OH-]
∴[OH-] = 10^-7

by using ICE table:

           Mn(OH)2(s) ⇄  Mn2+ (aq)  + 2OH-(aq)
initial                              0                     10^-7
change                           +X                      +2X
Equ                                 X                  (10^-7 + 2X)

when Ksp = [Mn2+][OH-]^2

when Ksp of Mn(OH)2 = 4.6 x 10^-14

by substitution:

4.6 x 10^-14 = X*(10^-7+2X)^2  by solving this equation for X

∴ X =2.3 x 10-5 M

∴ The solubility of Mn(OH)2 in grams per liter (when the molar mass of Mn(OH)2 = 88.953 g/mol
= 2.3 x10^-5 moles/L * 88.953 g/mol

= 0.002 g/ L
3 0
3 years ago
Read 2 more answers
Cosmic rays are
IrinaK [193]

A. High energy radiation produced in the ozone layer. (:

6 0
3 years ago
The partial pressure of N2 in the air is 593 mm Hg at 1 atm. What is the partial pressure of N2 in a bubble of air a scuba diver
Evgen [1.6K]

Answer: Partial pressure of N_{2} at a depth of 132 ft below sea level is 2964 mm Hg.

Explanation:

It is known that 1 atm = 760 mm Hg.

Also,   P_{N_{2}} = x_{N_{2}}P

where,    P_{N_{2}} = partial pressure of N_{2}

                 P = atmospheric pressure

            x_{N_{2}} = mole fraction of N_{2}

Putting the given values into the above formula as follows.

      P_{N_{2}} = x_{N_{2}}P

    593 mm Hg = x_{N_{2}} \times 760 mm Hg

       x_{N_{2}} = 0.780

Now, at a depth of 132 ft below the surface of the water where pressure is 5.0 atm. So, partial pressure of N_{2} is as follows.

         P_{N_{2}} = x_{N_{2}}P

                  = 0.78 \times 5 atm \times \frac{760 mm Hg}{1 atm}

                  = 2964 mm Hg

Therefore, we can conclude that partial pressure of N_{2} at a depth of 132 ft below sea level is 2964 mm Hg.

8 0
3 years ago
How many mols of chlorine are in 120g of chlorine gas
xz_007 [3.2K]

Answer:

Number of moles of chlorine = 3.38 mol

Explanation:

Given data:

Mass of chlorine = 120 g

Moles of chlorine = ?

Solution:

Formula:

Number of moles = mass/molar mass

Molar mass of chlorine = 35.5 g/mol

Now we will put the values in formula.

Number of moles = 120 g/ 35.5 g/mol

Number of moles = 3.38 mol

6 0
3 years ago
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