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solong [7]
1 year ago
10

The ideal gas model assumes that ideal gas particles: (select all that apply) take up a lot of volume take up negligible volume

are attracted to other gas particles are not attracted to other gas particles
Chemistry
2 answers:
GarryVolchara [31]1 year ago
7 0

Answer is: take up negligible volume and are not attracted to other gas particles.

An ideal gas is:

1) made up of molecules which are in constant random motion in straight lines.  

2) all collisions are perfectly elastic, there is no loss of kinetic energy during the collision.

3) follows ideal gas law: p·V = n·R·T.

4) the gas particles have negligible volume.

Ede4ka [16]1 year ago
4 0
Negligible volume of gas molecules and are not attracted to other gas particles.
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A gas sample occupies 3.50 liters of volume at 20.°c. what volume will this gas occupy at 100.°c (reported to three significant
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Explanation:

According to Charle's law, at constant pressure the volume of an ideal gas is directly proportional to the temperature.

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Hence, it is given that V_{1} is 3.50 liters, T_{1} is 20 degree celsius, and T_{2} is 100 degree celsius.

Therefore, calculate V_{2} as follows.

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1 year ago
Read 2 more answers
Electrons in __________ bonds remain localized between two atoms. electrons in __________ bonds can become delocalized between m
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Electrons in sigma <span>bonds remain localized between two atoms. Sigma </span><span>bond results from the formation of </span><span>a molecular orbital </span><span>by the end to </span><span>end overlap of atomic </span>orbitals. Electrons<span> in pi</span> bonds can become delocalized between more than two atoms. Pi bonds result from the formation of molecular orbital by side to side overlap of atomic orbitals. 

 

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3. Scott and James work at a grocery store. After the grocery store closed, they were playing a game with a shopping cart and Sc
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A 1.0 g sample of a cashew was burned in a calorimeter containing 1000. g of water, and the temperature of the water changed fro
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The correct answer is option C.

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Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water :  Q

Q=1000 g\times 4.184 J/g^oC\times 5^oC=20,920 J

Heat released on  combustion of 1.0 gram of cashew is -20,920 J.

3.0 g sample of a marshmallows  :

Heat released on  combustion of 3.0 g sample of a marshmallows = -Q'

We have mass of water = m = 2000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water :  Q'

Q'=2000 g\times 4.184 J/g^oC\times 5^oC=41,840 J

Heat released on 3.0 g sample of a marshmallows= -Q' = -41,840 J

Heat released on 1.0 g sample of a marshmallows : q

q =\frac{-Q'}{3} = \frac{-41,840 J}{3}=-13,946.67 J

Heat released on  combustion of 1.0 gram of marshmallows -13,946.67 J.

-20,920 J. > -13,946.67 J

The combustion of 1.0 g of cashew releases more energy than the combustion of 1.0 g of marshmallow.

5 0
1 year ago
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