Can someone check these homework answers please?

32. The barometric pressure measured outside an airplane at 9 km was 259 mmHg. Calculate the pressure in kPa.

34.5 kPa

36. A diving bell is a container open at the bottom. As the bell descends, the water level inside changes so that the pressure inside equals the pressure outside. Initially, the volume of the air is 8.58 m^3 at 1.020 atm and 20 degrees C. What is the volume at 1.584 atm and 20 degrees C.

5.52 m^3

40. A mole of gas at 0 degrees C and 760 mmHg occupies 22.41 L. What is the volume at 20 degrees C and 760 mmHg?

24.1 L

44. A sample of 62.3 cm^3 of argon gas at 18 degrees C was contained at a pressure of 155 kPa in a J-shaped tube with mercury. Later the temperature changed. When the mercury level was adjusted to give the same pressure of argon, the gas volume changed to 50.9 cm^3. What was the final temperature of the argon?

243. K

And I need help with this problem.

Methanol can be produced in industrial plants by reacting carbon dioxide with hydrogen in the presence of a catalyst. Water is the other product. how many volums of hydrogen are required for each volume of carbon dioxide when each gas is at the same temperature and pressure?

Is this just a mol mol comparison within the formula? What is is the unbalanced formula?

For #40 I get 24.05 which I would round to 24.0.

For #44 I used (V1/T1) = (V2/T2)
62.3/291 = 50.9/T2 and for T2 I obtained 237.9 which rounds to 238 K.
For the last problem,
CO2 + 3H2 ==>CH3OH + H2O
When everything is present as a gas (I think this is at elevated temperature so H2O will be a gas), you need not go through the mole procedure but you may take a short cut and use volumes directly. Therefore, 3 volumes of H2 are required for each volume of CO2; i.e., 1 liter of CO2 will take 3 liters of H2 (and will produce 1 liter CH3OH and 1 liter of H2O

To check the homework answers:

32. The barometric pressure measured outside the airplane at 9 km is given as 259 mmHg. We need to calculate the pressure in kPa.
To convert mmHg to kPa, we use the conversion factor: 1 mmHg = 0.133322 kPa.

Calculation:
Pressure (kPa) = Pressure (mmHg) x (0.133322 kPa / 1 mmHg)
= 259 mmHg x (0.133322 kPa / 1 mmHg)
≈ 34.5 kPa

So, the answer is 34.5 kPa.

36. The volume of air inside a diving bell changes as it descends to maintain equal pressure inside and outside.
Given:
Initial volume (V₁) = 8.58 m^3
Initial pressure (P₁) = 1.020 atm
Final pressure (P₂) = 1.584 atm

Since the temperature remains constant at 20 degrees C, we can use Boyle's Law.

Boyle's Law states that, at constant temperature, the product of initial volume and initial pressure is equal to the product of final volume and final pressure.

Mathematically,
V₁ * P₁ = V₂ * P₂

Substituting the given values:
8.58 m^3 * 1.020 atm = V₂ * 1.584 atm

Solve for V₂:
V₂ = (8.58 m^3 * 1.020 atm) / 1.584 atm
≈ 5.52 m^3

So, the volume at 1.584 atm is approximately 5.52 m^3.

40. The question states that a mole of gas at 0 degrees C and 760 mmHg occupies 22.41 L. We need to calculate the volume at 20 degrees C and 760 mmHg.
Since temperature and pressure are constant, we can use Charles's Law.

Charles's Law states that, at constant pressure, the volume of a gas is directly proportional to its temperature in kelvin.

Mathematically,
(V₁ / T₁) = (V₂ / T₂)

Substituting the given values:
(22.41 L / 273 K) = (V₂ / 293 K)

Solve for V₂:
V₂ = (22.41 L * 293 K) / 273 K
≈ 24.1 L

So, the volume at 20 degrees C and 760 mmHg is approximately 24.1 L.

44. A sample of argon gas has an initial volume of 62.3 cm^3 at 18 degrees C and 155 kPa pressure. The gas volume changes to 50.9 cm^3 when the pressure remains the same. We need to calculate the final temperature of the argon gas.
Since pressure is constant and the volume and temperature change, we can use the combined gas law.

The combined gas law expresses the relationship between pressure, volume, and temperature for a given amount of gas.

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

Substituting the given values:
(155 kPa * 62.3 cm^3) / (18 + 273 K) = (155 kPa * 50.9 cm^3) / T₂

Solve for T₂:
T₂ = (155 kPa * 50.9 cm^3 * (18 + 273 K)) / (155 kPa * 62.3 cm^3)
≈ 243 K

So, the final temperature of the argon gas is approximately 243 K.

Regarding the additional question:
To determine the volumes of hydrogen and carbon dioxide required for their reaction, we need their balanced chemical equation for the reaction. Unfortunately, the unbalanced chemical equation for the reaction is not provided. Without it, we cannot determine the volumes of the gases required for the reaction.

To check the homework answers, let's go through each question individually:

32. The barometric pressure measured outside the airplane at 9 km was given as 259 mmHg. To calculate the pressure in kPa, we need to convert mmHg to kPa. The conversion factor is 1 mmHg = 0.133322 kPa. So, we can multiply 259 mmHg by 0.133322 kPa/mmHg to get the pressure in kPa.

32 * 0.133322 = 4.264464 kPa

So, the pressure in kPa is approximately 4.26 kPa.

34.5 kPa - Answer seems correct.

36. To solve this problem, we can use the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.

We are given the initial volume (V1 = 8.58 m^3), initial pressure (P1 = 1.020 atm), final pressure (P2 = 1.584 atm), and temperature (T).

Since the temperature remains constant, we can cancel it out in the equation. We can set up a proportion using the initial and final pressures:

(P1 * V1) / (P2) = V2

(1.020 atm * 8.58 m^3) / (1.584 atm) = V2

Solving this equation gives:

V2 ≈ 5.52 m^3

So, the volume at the final pressure is approximately 5.52 m^3.

40. This question involves the ideal gas law as well. We are given the initial volume (V1 = 22.41 L), initial temperature (T1 = 0 degrees C), final temperature (T2 = 20 degrees C), and the pressure remains constant (P = 760 mmHg).

First, we need to convert the temperatures to Kelvin. The conversion formula is K = C + 273.15.

T1 = 0 + 273.15 = 273.15 K
T2 = 20 + 273.15 = 293.15 K

Now, we can use the ideal gas law equation:

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

Plugging in the values:

(760 mmHg * 22.41 L) / 273.15 K = (760 mmHg * V2) / 293.15 K

Solving this equation for V2 gives:

V2 ≈ 24.1 L

So, the volume at 20 degrees C and 760 mmHg pressure is approximately 24.1 L.

44. To solve this problem, we need to use the combined gas law, which is derived from the ideal gas law for comparing different sets of conditions.

The combined gas law formula is: (P1 * V1) / T1 = (P2 * V2) / T2

We are given the initial volume (V1 = 62.3 cm^3), initial temperature (T1 = 18 degrees C), final volume (V2 = 50.9 cm^3), and the pressure remains constant (P).

First, we need to convert the temperatures to Kelvin using the formula K = C + 273.15.

T1 = 18 + 273.15 = 291.15 K

Now we can use the combined gas law formula:

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

Plugging in the values:

(155 kPa * 62.3 cm^3) / 291.15 K = (P * 50.9 cm^3) / T2

Solving for T2 gives:

T2 ≈ 243 K

So, the final temperature of the argon is approximately 243 K.

Now, let's address your additional question:

To determine how many volumes of hydrogen are required for each volume of carbon dioxide, we need to balance the equation for the reaction between carbon dioxide and hydrogen. Without knowing the specific reaction, it's difficult to provide an exact answer. However, if we assume that the reaction is:

CO2 + 2H2 → CH3OH + H2O

Then, based on the coefficients in this balanced equation, we can say that for every 1 volume of carbon dioxide, 2 volumes of hydrogen are required.