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myrzilka [38]
2 years ago
6

Define an ideal gas and explain under which conditions you may reasonably approximate a real gas as an ideal gas. Also mention c

onditions under which it would be inappropriate to approximate a real gas as an ideal gas. (you may need to use a search engineto help you out here)
Chemistry
1 answer:
TEA [102]2 years ago
5 0

Answer:

High temperature and low pressure

Explanation:

According to the kinetic molecular theory, gases are composed of small particles called molecules which are in constant motion.

At high temperature and low pressure, gas molecules possess high kinetic energy and move at high velocities hence intermolecular interaction is almost none existent and real gases approach the behavior of ideal gases.

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Calculate q (in units of joules) when 1.850 g of water is heated from 22 °C to 33 °C. Report only the numerical portion of your
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When q is the heat energy in joules (J)

so, according to this formula, we can get q (in joule unit):

q = M*C*ΔT

when M is the mass of the water sample = 1.85 g

C is the specific heat capacity of water = 4.18 J/g.°C

and Δ T is the difference in temperature (Tf-Ti) = 33 - 22 = 11°C

So, by substitution, we will get the value of q ( in Joule):

∴ q = 1.85 g * 4.18 J/g.°C * 11 °C

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5 0
3 years ago
How is humility related to air pressure
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Humid air has higher pressure because of the heaviness of water humid air is lighter so it has lower pressure.
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2 years ago
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How many liters of carbon dioxide will 0.5 moles of lithium hydroxide (LiOH) absorb?
Elden [556K]

Answer : The volume of carbon dioxide will be, 5.6122 L

Solution : Given,

Moles of LiOH = 0.5 moles

Molar mass of carbon dioxide = 44 g/mole

Density of carbon dioxide = 0.00196 g/ml

First we have to calculate the mass of carbon dioxide.

The balanced reaction will be,

2LiOH+CO_2\rightarrow Li_2CO_3+H_2O

From the reaction we conclude that

As, 2 moles of LiOH absorbs 44 grams of carbon dioxide

So, 0.5 moles of LiOH absorbs \frac{0.5moles}{2moles}\times 44g=11 grams of carbon dioxide

Mass of carbon dioxide = 11 g

Density of carbon dioxide = 0.00196 g/ml

Now we have to calculate the volume of carbon dioxide.

Density=\frac{Mass}{volume}

0.00196g/ml=\frac{11g}{volume}

Volume of carbon dioxide = 5612.24 ml = 5.6122 L      (1 L = 1000 ml)

Therefore, the volume of carbon dioxide will be, 5.6122 L

5 0
2 years ago
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