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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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(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g
Anarel [89]

Answer:

Part A

 The volume of the gaseous product  is  V = 787L

Part B

The volume of the the engine’s gaseous exhaust is  V_e = 2178 \ L

Explanation:

Part A

From the question we are told that

    The temperature is  T = 350^oC = 350 +273 =623K

     The pressure is  P = 735 \ torr = \frac{735}{760} =  0.967\ atm

     The of  C_8 H_{18} = 100.0g

The chemical equation for this combustion is

               2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}

 The number of moles of  C_8 H_{18} that reacted is mathematically represented as

               n = \frac{mass \ of \  C_8H_{18}  }{Molar \  mass \ of  C_8H_{18} }

The molar mass of  C_8 H_{18} is constant value which is

                  M = 114.23 \ g/mole  

So          n = \frac{100  }{114.23} }

             n = 0.8754 \ moles

The gaseous product in the reaction is CO_2_{(g)} and water vapour

Now from the reaction

    2 moles of C_8 H_{18}  will react with 25 moles of O_2 to give (16 + 18) moles of CO_2_{(g)} and  H_2 O_{(g)}

So

    1 mole of C_8 H_{18} will  react with 12.5 moles of  O_2 to give 17 moles of CO_2_{(g)} and  H_2 O_{(g)}

This implies that

    0.8754 moles of C_8 H_{18} will react with (12.5 * 0.8754 ) moles of O_2 to give  (17 * 0.8754) of CO_2_{(g)} and  H_2 O_{(g)}

So the no of moles of gaseous product is

         N_g = 17 * 0.8754

         N_g = 14.88 \ moles

From the ideal gas law

       PV = N_gRT

making V the subject

        V = \frac{N_gRT}{P}

Where R is the gas constant with a value R = 0.08206 \  L\cdot atm /K \cdot mole

Substituting values

          V = \frac{14.88* 0.08206 *623}{0.967}

          V = 787L

Part B

From the reaction the number of moles of oxygen that reacted is

         N_o = 0.8754 * 12.5

         N_o = 10.94 \ moles

The volume is

      V_o  = \frac{10.94 * 0.08206 *623}{0.967}

      V_o  = 579 \ L

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

         V_e = V_o * \frac{0.79}{0.21}

Substituting values

       V_e = 579 * \frac{0.79}{0.21}

       V_e = 2178 \ L

3 0
3 years ago
When HClO2 is dissolved in water, it partially dissociates according to the equation
Natasha_Volkova [10]

Answer:

x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates

Explanation:

Recall that , depression present in freezing point is calculated with the formulae = solute particles Molarity x KF

0.3473 = m * 1.86

Solving, m = 0.187 m

Moles of HClO2 = mass / molar mass = 5.85 / 68.5 = 0.0854 mol

Molality = moles / mass of water in kg = 0.0854 / 1 = 0.0854 m

Initial molality

Assuming that a % x of the solute dissociates, we have the ICE table:

                 HClO2         H+    +   ClO2-

initial concentration:       0.0854                    0             0

final concentration:      0.0854(1-x/100)   0.0854x/100   0.0854x / 100

We see that sum of molality of equilibrium mixture = freezing point molality

0.0854( 1 - x/100 + x/100 + x/100) = 0.187

2.1897 = 1 + x / 100

x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates

3 0
4 years ago
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