Chemistry College

A sample of gas is put into a rigid (fixed volume) container at 3 oC and a pressure of 38.5 kPa. The container is then placed in an oven at 267 oC.

What pressure would you expect to measure for the gas in the container at this higher temperature?

Answers

Answer 1

We would expect to measure a pressure of approximately 75.25 kPa for the gas in the container at the higher temperature of 267 oC.

To determine the expected pressure of the gas in the container at the higher temperature, we can use the combined gas law, which relates the initial and final conditions of temperature and pressure in a fixed volume system. The combined gas law equation is given as:

(P1 * V1) / T1 = (P2 * V2) / T2

Where:

P1 = Initial pressure

V1 = Initial volume (which is fixed in this case)

T1 = Initial temperature

P2 = Final pressure (to be determined)

V2 = Final volume (which is fixed in this case)

T2 = Final temperature

In this scenario, the initial conditions are given as 3 oC (which is equivalent to 276 K) and 38.5 kPa. The final temperature is 267 oC (which is equivalent to 540 K). Since the volume is fixed, we can substitute the given values into the equation:

(38.5 kPa * V1) / 276 K = (P2 * V1) / 540 K

Simplifying the equation, we can cancel out V1:

38.5 / 276 = P2 / 540

Solving for P2:

P2 = (38.5 / 276) * 540 ≈ 75.25 kPa

Therefore, we would expect to measure a pressure of approximately 75.25 kPa for the gas in the container at the higher temperature of 267 oC.

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Related Questions

-Convert 6.02 x 1020 formula units of MgCl₂ to mol of MgCl₂:​

Answers

6.02 x [tex]10^{20[/tex] formula units of MgCl₂ is equal to 0.1 moles of MgCl₂.

To convert formula units of MgCl₂ to moles of MgCl₂, we need to use Avogadro's number, which relates the number of formula units to the number of moles.

Avogadro's number (NA) is approximately 6.022 x 10^23 formula units per mole.

Given that we have 6.02 x 10^20 formula units of MgCl₂, we can set up a conversion factor to convert to moles:

(6.02 x 10^20 formula units MgCl₂) * (1 mol MgCl₂ / (6.022 x 10^23 formula units MgCl₂))

The formula units of MgCl₂ cancel out, and we are left with moles of MgCl₂:

(6.02 x 10^20) * (1 mol / 6.022 x 10^23) = 0.1 mol

Therefore, 6.02 x 10^20 formula units of MgCl₂ is equal to 0.1 moles of MgCl₂.

It's important to note that this conversion assumes that each formula unit of MgCl₂ represents one mole of MgCl₂. This is based on the stoichiometry of the compound, where there is one mole of MgCl₂ for every one formula unit.

Additionally, this conversion is valid for any substance, not just MgCl₂, as long as you know the value of Avogadro's number and the number of formula units or particles you have.

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A bottle of rubbing alcohol contains both 2-propanol and water.
These liquids can be separated by the process of distillation because the 2-propanol and water
A) have combined chemically and retain their different boiling points
B) have combined physically and have the same boiling point
C) have combined physically and retain their different boiling
DOmIS
D) have combined chemically and have the same boiling point

Answers

These liquids can be separated by the process of distillation because the 2-propanol and water option B) have combined physically and have the same boiling point.

A bottle of rubbing alcohol contains both 2-propanol and water.

These liquids can be separated by the process of distillation because the 2-propanol and water have combined physically and retain their different boiling points.

The process of separating two miscible liquids with different boiling points is called distillation.

The procedure is based on the idea that the liquids have various boiling points.

Thus, during the boiling process, one of the fluids vaporizes faster than the other, and that vapor is then collected and allowed to condense, producing pure liquid.

Alcohol and water are examples of two miscible liquids that can be separated by distillation.

The process of distillation operates on the fact that the liquid with the lowest boiling point is vaporized first.

Alcohol has a lower boiling point than water, so alcohol is vaporized first in distillation, leaving the water behind.

A bottle of rubbing alcohol contains both 2-propanol and water.

These liquids can be separated by the process of distillation because the 2-propanol and water have combined physically and retain their different boiling points.

This means that 2-propanol and water are physically combined; that is, they do not form any new chemical bonds with each other.

Consequently, the molecules of the substances remain separate from each other.

Furthermore, both water and 2-propanol have various boiling points.

The boiling point of 2-propanol is 82.6 °C, while that of water is 100 °C. Thus, they can be separated using distillation.

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Which of the following set of quantum numbers (ordered n, ℓ, mℓ ) are possible for an electron in an atom? Check all that apply
a. 2, 1, 3
b. 5, 3, -3
c. 4, 3, -2
d. -4, 3, 1
e. 2, 1, -2
f. 3, 2, 2
g. 3, 3, 1

Answers

the possible quantum numbers (ordered n, ℓ, mℓ ) are:Option B.5, 3, -3 and Option C. 4, 3, -2

The quantum numbers n, ℓ, mℓ represent respectively the principal quantum number, the orbital angular momentum quantum number and the magnetic quantum number.

These are the three most important quantum numbers. T

here is another quantum number called the spin quantum number, denoted by ms.

Let's see which of the given quantum number sets is possible.2, 1, 3 is not possible because for ℓ = 1, mℓ can only be -1, 0, or 1. 5, 3, -3 is possible.4, 3, -2 is possible. -4, 3, 1 is not possible.

For any value of ℓ, mℓ must be between -ℓ and +ℓ. e. 2, 1, -2 is not possible because for ℓ = 1, mℓ can only be -1, 0, or 1. f. 3, 2, 2 is not possible because for ℓ = 2, mℓ can only be -2, -1, 0, +1, or +2. g. 3, 3, 1 is not possible because for any value of ℓ, mℓ must be between -ℓ and +ℓ.

Therefore, the possible quantum numbers (ordered n, ℓ, mℓ ) are:5, 3, -34, 3, -2

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59.6 ml of air is at 20.5 °C. What is its volume at 73.9 °C?

Answers

The volume of air at 73.9 °C would be approximately 70.91 ml.

To solve this problem, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature when pressure and amount of gas are kept constant.

Let's assume that the pressure and amount of air remain constant.

Given:

Initial volume (V1) = 59.6 ml

Initial temperature (T1) = 20.5 °C = 20.5 + 273.15 K = 293.65 K

Final temperature (T2) = 73.9 °C = 73.9 + 273.15 K = 347.05 K

Using Charles's Law, we can set up the following proportion:

V1 / T1 = V2 / T2

Plugging in the values, we have:

59.6 ml / 293.65 K = V2 / 347.05 K

To solve for V2 (the final volume), we can rearrange the equation:

V2 = (59.6 ml * 347.05 K) / 293.65 K

V2 = 70.91 ml

Therefore, the volume of air at 73.9 °C would be approximately 70.91 ml.

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