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Svetach [21]
4 years ago
12

Why do real gases not behave exactly like ideal gases?

Chemistry
1 answer:
mihalych1998 [28]4 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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p  and d orbitals

Explanation:

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Pls help me <br><br> What is the mass in grams of 2.64 mol of sulfur dioxide, SO2?
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Answer: The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles SO2, or 64.0638 grams.

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3 years ago
Using the following equation, 2C2H6 +7O2 --&gt;4CO2 +6H2O, if 2.5g C2H6 react with 170g of O2, how many grams of water will be p
kirill [66]

The mass of water (H₂O) that would be produced is 4.5 g

<h3>Stoichiometry </h3>

From the question, we are to determine the mass of water that would be produced.

From the given balanced chemical equation

2C₂H₆ +7O₂ → 4CO₂ +6H₂O

This means

2 moles of C₂H₆ reacts with 7 moles of O₂ to produce 4 moles of CO₂ and 6 moles of H₂O

Now, we will determine the number of moles of each reactant present

  • For Ethane (C₂H₆)

Mass = 2.5 g

Molar mass = 30.07 g

Using the formula,

Number\ of\ moles = \frac{Mass}{Molar\ mass}

Number of moles of C₂H₆ present = \frac{2.5}{30.07}

Number of moles of C₂H₆ present = 0.08314 mole

  • For Oxygen (O₂)

Mass = 170g

Molar mass = 31.999 g/mol

Number of moles of O₂ present = \frac{170}{31.999}

Number of moles of O₂ present = 5.3127 moles

Since

2 moles of C₂H₆ reacts with 7 moles of O₂

Then,

0.08314 mole of C₂H₆ will react with \frac{7 \times 0.08314 }{2}

 \frac{7 \times 0.08314 }{2} = 0.58198 mole

Therefore,

0.08314 mole of C₂H₆ reacts with 0.58198 mole of O₂ to produce 3 × 0.08314 moles of H₂O

3 × 0.08314 = 0.24942 mole

Thus, the number of moles of water (H₂O) produced is 0.24942 mole

Now, for the mass of water that would be produced,

Using the formula,

Mass = Number of moles × Molar mass

Molar mass of water = 18.015 g/mol

Then,

Mass of water that would be produced = 0.24942 × 18.015

Mass of water that would be produced = 4.4933 g

Mass of water that would be produced ≅ 4.5 g

Hence, the mass of water (H₂O) that would be produced is 4.5 g

Learn more on Stoichiometry here: brainly.com/question/14271082

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2 years ago
Consider two liquids, labeled A and B, that are both pure substances. Liquid A has
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E is the correct answer
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