Three of the primary components of air are carbon dioxide, nitrogen, and oxygen. In a sample containing a mixutre of only these gases at exactly one atmosphere pressure, the partial pressure of carbon dixoide and nitrogen are given as Pco2 = 0.285torr and Pn2=593.525torr. What is the partial pressure of oxygen?

Okay, I tried working this one, and all I can get out of this problem is

0.285+593.525= 593.810torr
I need help

Determine the partial pressure of oxygen collected by water displacement if the water temperature is 20.0C and the total pressure of the gas in the collection bottle is 730 torr.

Po2=712.5torr <-- is that right?

A sample of gas is collected over water at a temperature of 35.0C when the barometric pressure reading is 742.0 torr. What is the partial pressure of the dry gas?

I got 699.8torr for the dry gas

A sample of oxygen is collected in a 175mL container over water at 15C, and the barometer reads 752.0 torr. What volume would the dry gas occupy at
770.0 torr and 15C

I'm thinking this has to do something with Boyles law? I need help.

Suppose that 120.mL of argon is collected over water at 25C and 780.0 torr. Compute the volume of the dry argon at STP.

Would this be involved with Charles law?

Three of the primary components of air are carbon dioxide, nitrogen, and oxygen. In a sample containing a mixutre of only these gases at exactly one atmosphere pressure, the partial pressure of carbon dixoide and nitrogen are given as Pco2 = 0.285torr and Pn2=593.525torr. What is the partial pressure of oxygen?

Okay, I tried working this one, and all I can get out of this problem is

0.285+593.525= 593.810torr
I need help
Dalton's law says that the total pressure is the sum of the partial pressures.

Determine the partial pressure of oxygen collected by water displacement if the water temperature is 20.0C and the total pressure of the gas in the collection bottle is 730 torr.

Po2=712.5torr <-- is that right?

yes

A sample of gas is collected over water at a temperature of 35.0C when the barometric pressure reading is 742.0 torr. What is the partial pressure of the dry gas?

I got 699.8torr for the dry gas
yes. Watch the number of significant figures if your prof is picky about that, although I think you are allowed 4 in this problem.

A sample of oxygen is collected in a 175mL container over water at 15C, and the barometer reads 752.0 torr. What volume would the dry gas occupy at
770.0 torr and 15C

I'm thinking this has to do something with Boyles law? I need help.

Use PV=nRT or (P1V1/T1) = (P2V2/T2)

Suppose that 120.mL of argon is collected over water at 25C and 780.0 torr. Compute the volume of the dry argon at STP.

Would this be involved with Charles law?
(P1V1/T1) = (P2V2/T2)
Don't forget to use T in Kelvin. Subtract the vapor pressure of water from the total pressure for this problem as well as the previous one.

To solve the problems step by step:

1. To find the partial pressure of oxygen, first, we need to know the total pressure of the sample. In this case, it is given that the total pressure is "exactly one atmosphere pressure." One atmosphere is equal to 760 torr.

The sum of the partial pressures of the three components should equal the total pressure.
So, Pco2 + Pn2 + Po2 = Total Pressure

Given:
Pco2 = 0.285 torr
Pn2 = 593.525 torr
Total Pressure = 760 torr

To find Po2, rearrange the equation and solve:
Po2 = Total Pressure - (Pco2 + Pn2)
Po2 = 760 torr - (0.285 torr + 593.525 torr)
Po2 = 166.19 torr

Therefore, the partial pressure of oxygen is 166.19 torr.

2. To determine the partial pressure of oxygen collected by water displacement, we need to consider the vapor pressure of water at the given temperature.

At 20.0°C, the vapor pressure of water is 17.5 torr (approximately).

Given:
Total pressure of the gas in the collection bottle = 730 torr

The partial pressures of carbon dioxide, nitrogen, and oxygen sum up to the total pressure. So,
Pco2 + Pn2 + Po2 = Total Pressure

Given:
Pco2 and Pn2 are not provided.

Therefore, it's not possible to determine the partial pressure of oxygen using the information given in the question.

3. To find the partial pressure of the dry gas, use Dalton's law of partial pressures.

Given:
Barometric pressure = 742.0 torr
Water vapor pressure at 35.0°C = 25.2 torr (approximately)

The partial pressure of the dry gas is equal to the difference between the barometric pressure and the water vapor pressure:
Partial pressure of the dry gas = Barometric pressure - Water vapor pressure
Partial pressure of the dry gas = 742.0 torr - 25.2 torr
Partial pressure of the dry gas = 716.8 torr

Therefore, the partial pressure of the dry gas is 716.8 torr.

4. To find the volume of the dry gas at a different pressure, we can use Boyle's law.

Given:
Initial volume of oxygen gas = 175 mL
Initial barometric pressure = 752.0 torr
Target pressure = 770.0 torr
Temperature = 15°C

Using Boyle's law:
P1 * V1 = P2 * V2

Convert the temperatures to Kelvin:
Initial temperature = 15°C + 273.15 = 288.15 K

Now let's calculate the volume of the dry gas at the target pressure:
(Volume of initial dry gas * Initial pressure) / Target pressure = Volume of dry gas at target pressure

(Volume of initial dry gas * 752.0 torr) / 770.0 torr = Volume of dry gas at 770.0 torr

(Volume of initial dry gas * 752.0 torr) = (Volume of dry gas at 770.0 torr * 770.0 torr)

Volume of dry gas at 770.0 torr = (Volume of initial dry gas * 752.0 torr) / 770.0 torr

Plugging in the given values:
Volume of dry gas at 770.0 torr = (175 mL * 752.0 torr) / 770.0 torr

Simplifying the equation gives:
Volume of dry gas at 770.0 torr = 170.9 mL

Therefore, the dry gas would occupy a volume of 170.9 mL at 770.0 torr and 15°C.

5. To find the volume of dry argon at STP, we can use the combined gas law.

Given:
Volume of argon collected over water at 25°C = 120.0 mL
Barometric pressure = 780.0 torr

We need to find the volume at STP, which is 0°C and 1 atm (760 torr).

Using the combined gas law equation:
P1 * V1 / T1 = P2 * V2 / T2

Rearranging the equation:
(Volume of argon at STP) = (Volume of argon collected) * (Pressure at STP) * (Temperature at STP) / (Pressure at the time of collection) * (Temperature at the time of collection)

Now, plug in the given values:
(Volume of argon at STP) = (120.0 mL) * (760 torr) * (273.15 K) / (780.0 torr) * (298.15 K)

Simplifying the equation gives:
Volume of dry argon at STP = 112.2 mL

Therefore, the volume of the dry argon at STP is 112.2 mL.

To find the partial pressure of a specific gas in a mixture, you need to use the concept of Dalton's law of partial pressures. According to Dalton's law, the total pressure of a gas mixture is equal to the sum of the partial pressures of each individual gas.

In the first problem, you are given the partial pressures of carbon dioxide (Pco2 = 0.285 torr) and nitrogen (Pn2 = 593.525 torr) and you need to find the partial pressure of oxygen. To find the partial pressure of oxygen (Po2), you can use the equation:

Total Pressure = Pco2 + Pn2 + Po2

Since the total pressure is given as exactly one atmosphere, which is approximately 760 torr, you can rewrite the equation as:

760 torr = 0.285 torr + 593.525 torr + Po2

Now, you can solve for Po2 by subtracting the sum of Pco2 and Pn2 from the total pressure:

Po2 = 760 torr - (0.285 torr + 593.525 torr)

By calculating this expression, you should find the partial pressure of oxygen.

Similarly, in the second problem, you need to find the partial pressure of oxygen (Po2) given the total pressure of the gas in the collection bottle (730 torr). To find Po2, you can use the equation:

Total Pressure = Po2 + Pwater

The partial pressure of water vapor (Pwater) can be determined using the water vapor pressure table at the given temperature of 20.0C. Subtracting the water vapor pressure from the total pressure will give you the partial pressure of oxygen (Po2).

In the third problem, you are given the barometric pressure reading (742.0 torr) but need to find the partial pressure of the dry gas. To do this, you need to consider the vapor pressure of water at the given temperature of 35.0C and subtract it from the barometric pressure to obtain the partial pressure of the dry gas.

In the fourth problem, you are given the volume of the gas collected over water and need to find the volume of the dry gas at a different pressure (770.0 torr) and temperature (15C). To solve this, you can use the combined gas law, which incorporates Boyle's and Charles' laws, to relate the initial and final conditions. By using the formula and inputting the given values, you can find the volume of the dry gas.

In the last problem, you need to find the volume of the dry argon at STP (standard temperature and pressure). STP is defined as 0 degrees Celsius (273.15K) and 1 atmosphere (760 torr) of pressure. To solve this, you can make use of Avogadro's law, which states that equal volumes of gases, at the same temperature and pressure, contain equal numbers of molecules. Given the volume of argon at the given conditions and the ratio of pressures between the given conditions and STP, you can calculate the volume of the dry argon at STP.