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Xelga [282]
3 years ago
11

A rigid stainless steel chamber contains 180 Torr of methane, CH 4 , and excess oxygen, O 2 , at 160.0 °C. A spark is ignited in

side the chamber, completely combusting the methane. What is the change in total pressure within the chamber following the reaction? Assume a constant temperature throughout the process.
Chemistry
1 answer:
monitta3 years ago
4 0

Answer:

No pressure will change in the chamber at reactions end

Explanation:

CH4 + 2O2 ---> CO2 + 2H2O

According the above reaction

There are 3 moles total of gaseous reactants, and 3 moles total of gaseous products, so this reaction results in no change of volume. Therefore, since this happens at constant volume and constant temperature, the pressures are directly related to the moles used and produced. This way, no pressure will change in the chamber at reactions end

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The concentration of hydroxide ions is less than the concentration of hydronium ions for acidic solutions. True False
wolverine [178]

Answer:

True

Explanation:

It's true because the pH is a measure of how basic or acid a solution is. In an acidic medium, the pH scales goes from 0 to 7. While in a basic medium goes from 7 to 14. The lower the pH value of the most acid the solution is.

1. The expression pH = -log(molar concentration of hydronium) allow to calculate the pH of a solution.

2. On the other hand, the expression pOH = -log(molar concentration of hydroxide) allow to determine the pOH of a solution.

The values of pH and pOH always obey the following expression:

pH + pOH = 14

Thus if for instance the pH becomes smaller the pOH must become bigger in order to fulfill the equation. Which means that the concentration of hydronium ions is greater than the hydroxide concentration.

For example, in an acidic medium:

if pH= 3, pOH= 11

In this case the molar concentration of hydronium is 0,001M. And the molar concentration of hydroxide ions is just 0,00000000001M.

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3 years ago
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All animals make body heat. True or false?
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Answer:

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Explanation:

Some animals are cold blooded

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3 years ago
For the following reaction, KP = 0.455 at 945°C: At equilibrium, is 1.78 atm. What is the equilibrium partial pressure of CH4 in
Alborosie

Answer:

See explanation below

Explanation:

The question is incomplete. However, here's the missing part of the question:

<em>"For the following reaction, Kp = 0.455 at 945 °C: </em>

<em>C(s) + 2H2(g) <--> CH4(g). </em>

<em>At equilibrium the partial pressure of H2 is 1.78 atm. What is the equilibrium partial pressure of CH4(g)?"</em>

With these question, and knowing the value of equilibrium of this reaction we can calculate the partial pressure of CH4.

The expression of Kp for this reaction is:

Kp = PpCH4 / (PpH2)²

We know the value of Kp and pressure of hydrogen, so, let's solve for CH4:

PpCH4 = Kp * PpH2²

*: You should note that we don't use Carbon here, because it's solid, and solids and liquids do not contribute in the expression of equilibrium, mainly because their concentration is constant and near to 1.

Now solving for PpCH4:

PpCH4 = 0.455 * (1.78)²

<u><em>PpCH4 = 1.44 atm</em></u>

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