1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Anit [1.1K]
3 years ago
11

A mixture of gases containing 0.20 mol of SO2 and 0.20 mol of O2 in a 4.0 L flask reacts to form SO3. If the temperature is 25ºC

, what is the pressure in the flask after reaction is complete?
Chemistry
1 answer:
diamong [38]3 years ago
5 0

Answer : The pressure in the flask after reaction complete is, 2.4 atm

Explanation :

To calculate the pressure in the flask after reaction is complete we are using ideal gas equation.

PV=n_TRT\\\\P=(n_1+n_2)\times \frac{RT}{V}

where,

P = final pressure in the flask = ?

R = gas constant = 0.0821 L.atm/mol.K

T = temperature = 25^oC=273+25=298K

V = volume = 4.0 L

n_1 = moles of SO_2 = 0.20 mol

n_2 = moles of O_2 = 0.20 mol

Now put all the given values in the above expression, we get:

P=(0.20+0.20)mol\times \frac{(0.0821L.atm/mol.K)\times (298K)}{4.0L}

P=2.4atm

Thus, the pressure in the flask after reaction complete is, 2.4 atm

You might be interested in
Consider two flasks at 25 degree Celsius, one contains an ideal gas and the other contains the real gas SO3. Which statement reg
Strike441 [17]

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.

3 0
3 years ago
HELPPP PLEASE!!
Rzqust [24]

Answer:

Herbivores would increase causing producers to decrease.

Explanation:

If the carnivores are gone, herbivores wouldn't get eaten and increase, and more herbivores mean less producers available.

6 0
2 years ago
What is the mass measurement shown on the scales of this balance?
otez555 [7]
The answer is definitely not 70.923
6 0
3 years ago
Read 2 more answers
4.41 g of propane gas (C3H8) is injected into a bomb calorimeter and ignited with excess oxygen, according to the reaction below
gayaneshka [121]

Answer :  The heat of the reaction is -221.6 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter

q_{rxn}=-q_{cal}

q_{cal}=c_{cal}\times \Delta T

where,

q_{rxn} = heat released by the reaction = ?

q_{cal} = heat absorbed by the calorimeter

c_{cal} = specific heat of calorimeter = 97.1kJ/^oC=97100J/^oC

\Delta T = change in temperature = (T_{final}-T_{initial})=(27.282-25.000)=2.282^oC

Now put all the given values in the above formula, we get:

q_{cal}=(97100J/^oC)\times (2.282^oC)

q_{cal}=221582.2J=221.6kJ

As, q_{rxn}=-q_{cal}

So, q_{rxn}=-221.6kJ

Thus, the heat of the reaction is -221.6 kJ

6 0
3 years ago
A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h
frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]

Where;

\frac{f_2}{f_1}  is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1}  = 10^{0.7289}\\\\\frac{f_2}{f_1}  = 5.356

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

5 0
3 years ago
Other questions:
  • A student has a rectangular block. It is 2 cm wide. 3 cm tall and 25cm long. It has a mass of 600 g. First, calculate the volume
    14·1 answer
  • Which method would increase the solubility of a gas?
    5·2 answers
  • What exponent should you use to represent one hundred thousand?
    6·1 answer
  • A household variety of coal is
    10·1 answer
  • The small 2 next to Nitrogen is called the ____.
    7·2 answers
  • AlCl3 is lewis acid. Explain how?
    5·1 answer
  • please explain why boiling point melting are not considered chemical properties but are considered physical properties of matter
    8·1 answer
  • Mass and energy are conserved:
    14·1 answer
  • What happens when co2 dissolves in water​
    8·2 answers
  • What is the density of an object with a mass of 53.7 g and a volume of 27.19 ml?
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!