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Svetach [21]
3 years ago
12

Why do real gases not behave exactly like ideal gases?

Chemistry
1 answer:
mihalych1998 [28]3 years ago
8 0

Answer:

Real gas particles have significant volume

Real gas particles have more complex interactions than ideal gas particles.

Explanation:

An ideal gas is an imaginary concept and a gas behaves almost ideally at certain pressure and temperature conditions.

The gas in real deviates from the ideal behavior as some of the assumptions made for ideal gases are not true in case of real gases.

Real gas particles have significant volume as compared to vessel unlike ideal gases.

There are interactions present in between real gas molecules at high pressure conditions.

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A sample of gas has an initial volume of 20 L and an initial pressure of 2.5 atm. If the pressure changes to 3.1 atm, what is th
Studentka2010 [4]

The new volume is mathematically given as

V2= 16.12 L

<h3>What is the new volume?</h3>

Question Parameters:

A sample of gas has an initial volume of 20 L

and an initial pressure of 2.5 atm.

If the pressure changes to 3.1 atm

Generally, the equation for Pressure   is mathematically given as

P1V1 = P2 V2

2.5 * 20 = 3.1 * V2

V2 = 50/3.1

V2= 16.12 L

For more information on volume

brainly.com/question/1578538

6 0
2 years ago
the value of kc for the following reaction is 0.630 at 409 K N2O4(g) --&gt; 2NO2(g) if a reaction vessel at that temperature int
stealth61 [152]

<u>Answer:</u> The concentration of nitrogen dioxide at equilibrium is 0.063 M

<u>Explanation:</u>

We are given:

Initial concentration of nitrogen dioxide = 0.0250 M

Initial concentration of dinitrogen tetraoxide = 0.0250 M

For the given chemical equation:

                        N_2O_4(g)\rightleftharpoons 2NO_2(g)

<u>Initial:</u>                0.025        0.025

<u>At eqllm:</u>         0.025-x     0.025+2x

The expression of K_c for above equation follows:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

We are given:

K_c=0.630

Putting values in above expression, we get:

0.630=\frac{(0.025+2x)^2}{(0.025-x)}\\\\x=-0.2013,0.019

Neglecting the negative value of 'x', because concentration cannot be negative

So, equilibrium concentration of nitrogen dioxide = (0.025 + 2x) = [0.025 + 2(0.019)] = 0.063 M

Hence, the concentration of nitrogen dioxide at equilibrium is 0.063 M

6 0
3 years ago
Is anyone good at chemistry if so can someone help me please ?<br><br> (NO LINKS)
Daniel [21]

The reactions are a bit poorly written. While it's true that aqueous H₂CO₃ is produced in this neutralization reaction, the H₂CO₃ rapidly decomposes to yield CO₂(g) and H₂O(l). Writing the product as H₂CO₃(aq) in the net ionic equation is unnecessarily confusing since it portrays the substance as nonionizing yet water-soluble.

In any case, the Na⁺ and the Cl⁻ are the spectator ions here.  

7 0
3 years ago
Which of the following is always a reactant in a combustion reaction? (5 points)
aleksandr82 [10.1K]

Answer:

A

Explanation:

Combustion reaction always has something hydrocarbon plus oxygen gas which equals to carbon dioxide and water. Always being a reactant in a combustion reaction is oxygen, since reactants are what we start off the reaction with.

4 0
2 years ago
9. What is the process of nuclear fusion? *
Pavlova-9 [17]

Answer:

the process of carrying light

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