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Nataliya [291]
3 years ago
12

How is the attraction between molecules of a solid and a gas different?m

Chemistry
2 answers:
alexdok [17]3 years ago
5 0

Answer:

Q1) solids are strong, gasses are weak

Explanation:

  • Solids are Tightly packed molecule with no gap or space in between.
  • Gasses  are loosely packed and move freely in our atmosphere.
  1. Examples of solids are Wood, steel, plastic etc
  2. Example of gasses are oxygen, Carbon dioxide, nitrogen

Please mark me as brainliest

Thank you :-)

stepan [7]3 years ago
3 0

Answer:

Solids are strong, gases are weak

Explanation:

The attraction between molecules of a solid and gas are different in that those of solids are strong and gases are very weak.

The intermolecular attraction in gases are the weakest of the three states of matter. For the solid, they present the strongest attraction.

  • The intermolecular attraction is very important for characterizing the nature of the physical changes matter undergoes.
  • Solids requires a great deal of energy to cause physical changes to them.
  • Gases do not require too much energy.
  • The distances between gas molecules is very great compared to the closeness of the solid molecules.
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How many liters of gas will be in the closed reaction flask when 36.0L of ethane (C2H6) is allowed to react with 105.0L of oxyge
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Answer:- Volume of the gas in the flask after the reaction is 156.0 L.

Solution:-  The balanced equation for the combustion of ethane is:

2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.

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36.0LC_2H_6(\frac{7LO_2}{2LC_2H_6})

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let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:

105.0LO_2(\frac{4LCO_2}{7L O_2})

= 60.0 L CO_2

Similarly, let's calculate the volume of water vapors formed:

105.0L O_2(\frac{6L H_2O}{7L O_2})

= 90.0 L H_2O

Since ethane is present in excess, the remaining volume of it would also be present in the flask.

Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:

105.0LO_2(\frac{2LC_2H_6}{7LO_2})

= 30.0 L C_2H_6

Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L

Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L

Hence. the answer is 156.0 L.

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