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dezoksy [38]
2 years ago
14

Consider two flasks at 25 degree Celsius, one contains an ideal gas and the other contains the real gas SO3. Which statement reg

arding these gases is true?
A. As the temperature approaches 0 K, the volume of the ideal gas will be larger than the volume of SO3 because ideal gases lack inter-molecular forces.
B. As the temperature is decreased, the ideal gas will liquefy first because ideal gases have larger values of the van der Waals coefficient b.
C. As the temperature approaches 0 K, the volume of the ideal gas will be smaller than the volume of SO3 because ideal gases have larger values of the van der Waals coefficient a.
D. After the temperature has been lowered about 100 K, the pressure of the ideal gas will be smaller than the pressure of SO3 because the van der Waals coefficient a for SO3 is large.
E. After the temperature is increased about 100 K, the pressure of the ideal gas will be smaller than the pressure of SO3 because the van der Waals coefficient a for SO3 is lar
Chemistry
1 answer:
Strike441 [17]2 years ago
3 0

Answer: Option (A) is the correct answer.

Explanation:

In real gases, there exists force of attraction between the molecules at low temperature and high pressure. This is because at low temperature there occurs a decrease in kinetic energy of gas molecules and high pressure causes the molecules to come closer to each other.

As a result, forces of attraction increases as molecules come closer to each other and therefore, gases deviate from an ideal gas behavior.

And, at low pressure and high temperature there exists no force of attraction or repulsion between the molecules of a gas because they have high kinetic energy. Hence, gases behave ideally at these conditions.

Thus, we can conclude that the statement as the temperature approaches 0 K, the volume of the ideal gas will be larger than the volume of SO_{3} because ideal gases lack inter-molecular forces, is true.

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Octane has a density of 0.703 g/ml. Calculate the mass of CO2(g) produced by burning one
STatiana [176]

The mass of CO₂ gas produced during the combustion of one gallon of octane is 8.21 kg.

The given parameters:

  • <em>Density of the octane, ρ = 0.703 g/ml</em>
  • <em>Volume of octane, v = 3.79 liters</em>

<em />

The mass of the octane burnt is calculated as follows;

m = \rho V\\\\m = 0.703 \ \frac{g}{ml} \times 3.79 \ L \ \frac{1000 \ ml}{L} \\\\m = 2,664.37 \ g

The combustion reaction of octane is given as;

2C_8H_{18} +  \ 25O_2 \ --> \ 16CO_2 \ + \ 18H_2O

From the reaction above:

228.46 g of octane -------------------> 704 g of  CO₂ gas

2,664.37 of octane --------------------> ? of CO₂ gas

= \frac{2,664.37 \times 704}{228.46} \\\\= 8,210.3 \ g\\\\= 8.21 \ kg

Thus, the mass of CO₂ gas produced during the combustion of one gallon of octane is 8.21 kg.

Learn more about combustion of organic compounds here: brainly.com/question/13272422

8 0
2 years ago
How many moles of nitrogen trifluoride (NF3) can be produced from 9.65 mole of Fluorine gas (F2)
user100 [1]

Answer:

6.43 moles of NF₃.

Explanation:

The balanced equation for the reaction is given below:

N₂ + 3F₂ —> 2NF₃

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Finally, we shall determine the number of mole of nitrogen trifluoride (NF₃) produced by the reaction of 9.65 moles of Fluorine gas (F₂). This can be obtained as follow:

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Therefore, 9.65 moles of F₂ will react to to produce = (9.65 × 2)/3 = 6.43 moles of NF₃.

Thus, 6.43 moles of NF₃ were obtained from the reaction.

4 0
3 years ago
Why a neutral atom has the same number of protons and electrons
Alex777 [14]

Explanation:

When an atom has an equal number of electrons and protons, it has an equal number of negative electric charges (the electrons) and positive electric charges (the protons). The total electric charge of the atom is therefore zero and the atom is said to be neutral. ... Chemically, we say that the atoms have formed bonds.

5 0
3 years ago
Due to the small and highly electronegative nature of fluorine, the oxyacids of the this element are much less common and less s
steposvetlana [31]
We are told we have an oxyacid of the formula HOFO. We will assume the atoms are in this order and will draw a proper lewis structure for this compound by first drawing bonds between each of the 4 atoms and then place the remaining electron pairs on each atom:
      ..    ..    ..
H - O - F - O:
      ··   ··    ··
We can calculate the formal charge of an atom using the following formula:

Formal charge = [# of valence electrons] - [# of non-bonded electrons + # of bonds]

H: Formal charge = [1]-[0+1] = 0

O: Formal charge = [6]-[4+2] = 0

F: Formal charge = [7]-[4+2] = +1

O: Formal charge = [6]-[6+1] = -1

As we can see the overall charge of the molecule is neutral since the fluorine as a +1 charge and the oxygen a -1 charge.
4 0
3 years ago
B) All atoms contain<br>charged electrons.<br>​
andrey2020 [161]

Answer:

b) All atoms contain

charged electrons.

Explanation:

8 0
2 years ago
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