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A 200 ml sample of hydrogen is collected over water at 745.5 torr and 290K If the vapor pressure of water at 290k is 14.5 torr What is the volume of dry hydrogen at STP The answer is 181m?
To find the volume of dry hydrogen at STP, we need to correct for the presence of water vapor. First, calculate the pressure of dry hydrogen by subtracting the vapor pressure of water from the total pressure: 745.5 torr - 14.5 torr = 731 torr. Then, apply the ideal gas law to solve for the volume of dry hydrogen at STP: V = (200 ml * 731 torr * 273 K) / (290 K * 760 torr) ≈ 181 ml.
What volume of air contains 12.7g of oxygen gas at 273 K and 1.00 ATM?
Using the ideal gas law, PV = nRT, we can calculate the moles of oxygen gas present. Then, using the molar mass of oxygen, we can find the volume of air required to contain 12.7g of oxygen gas at 273 K and 1.00 atm.
Calculate the number of CO molecules in 1.1 L of this air at a pressure of 760 torr and a temperature of 18 Celsius?
To calculate the number of CO molecules, first convert the temperature to Kelvin (18°C + 273 = 291K). Then, use the ideal gas law equation PV = nRT to find the number of moles of CO present. Finally, since 1 mole of a gas contains 6.022 x 10^23 molecules, you can convert the moles of CO to molecules.
What amount of oxygen is needed to produce 12000 mols of nitrogen monoxide gas?
As oxygen is a diatomic gas it would take 6000 moles. As oxygen gas is 32 g/mole this would be 192000 grams or 192kg. At STP this would be a volume of (1)V =6000R(273) P=1atm R=8.314 v=13,618,332m3
What volume will 40.0L of He at 50.00 C and 1201 torr occupy at STP?
Use the combined gas law and solve. (PV/T)1 = (PV/T)2 P1 = 1201 torr, V1 = 40L, T1= 50C = 323K must be kelvin for gas laws STP implies P2 = 760 torr, T2 = 273K, V2 is the unknown Substitute and solve.... (1201x50 / 373) = (760 V2 / 273) rearrange with algebra and solve V2 = (1201x50x273)/(373x760) answer will be in liters
Which changes in pressure and temperature occur as a given mass of gas at 380 torr and 546 K is changed to STP known?
To change from 380 torr and 546 K to standard temperature and pressure (STP: 1 atm and 273 K), the pressure decreases from 380 torr to 1 atm, and the temperature decreases from 546 K to 273 K. This change can be calculated using the ideal gas law and the principles of gas behavior.
What is the volume of a gas if 8.24 moles of the gas exerts a pressure of 1720 torr at 43C?
There are a couple of ways to do this, so i'll just pick one. 1 mole of gas occupies 22.4 L at 760 torr and 0C (standard temp and press) (PV/nT)a = (PV/nT)b <-- temp must be kelvin (760 x 22.4L)/(1 x 273) = (1720 x V)/(8.24 x 316) rearrange to solve for volume V = (760 x 22.4L)/(1 x 273) x (8.24 x 316)/1720 V = 94.4 Liters
A 200 ml sample of hydrogen is collected over water at 745.5 torr and 290K If the vapor pressure of water at 290k is 14.5 torr What is the volume of dry hydrogen at STP The answer is 181m?
To find the volume of dry hydrogen at STP, we need to correct for the presence of water vapor. First, calculate the pressure of dry hydrogen by subtracting the vapor pressure of water from the total pressure: 745.5 torr - 14.5 torr = 731 torr. Then, apply the ideal gas law to solve for the volume of dry hydrogen at STP: V = (200 ml * 731 torr * 273 K) / (290 K * 760 torr) ≈ 181 ml.
The temperature at which the vapor pressure of a liquid has an external pressure of 1 atm is?
0 degrees Celsius, 273 degrees Kelvin, 32 degrees Fahrenheit.
What does STP stand for in chemistry?
It means Standard Temperature and Pressure.-Standard temperature is equal to 0 °C, which is 273 K.-Standard pressure is equal to 1 atm 760 mmhg or torr and 101.3 KPa.
What volume of air contains g of oxygen gas at 273 K and 1.00 ATM?
Use temperature and pressure to define the density, air density is 1.293 kg/m3 at 273 K and 1 ATM. You know the mass of air is 14.3 g, convert that to kg first. Density = mass/volume, just solve volume from known density and mass.
What volume of air contains 12.7g of oxygen gas at 273 K and 1.00 ATM?
Using the ideal gas law, PV = nRT, we can calculate the moles of oxygen gas present. Then, using the molar mass of oxygen, we can find the volume of air required to contain 12.7g of oxygen gas at 273 K and 1.00 atm.
Calculate the number of CO molecules in 1.1 L of this air at a pressure of 760 torr and a temperature of 18 Celsius?
To calculate the number of CO molecules, first convert the temperature to Kelvin (18°C + 273 = 291K). Then, use the ideal gas law equation PV = nRT to find the number of moles of CO present. Finally, since 1 mole of a gas contains 6.022 x 10^23 molecules, you can convert the moles of CO to molecules.
What is the volume at 273 K and 1 Bar?
At 273 K (0°C) and 1 bar pressure, the molar volume of an ideal gas is approximately 24.79 L/mol. This value represents the volume occupied by one mole of the gas under these conditions.
What is 273 as a fraction?
It is 273/1It is 273/1It is 273/1It is 273/1
What amount of oxygen is needed to produce 12000 mols of nitrogen monoxide gas?
As oxygen is a diatomic gas it would take 6000 moles. As oxygen gas is 32 g/mole this would be 192000 grams or 192kg. At STP this would be a volume of (1)V =6000R(273) P=1atm R=8.314 v=13,618,332m3
What volume will 40.0L of He at 50.00 C and 1201 torr occupy at STP?
Use the combined gas law and solve. (PV/T)1 = (PV/T)2 P1 = 1201 torr, V1 = 40L, T1= 50C = 323K must be kelvin for gas laws STP implies P2 = 760 torr, T2 = 273K, V2 is the unknown Substitute and solve.... (1201x50 / 373) = (760 V2 / 273) rearrange with algebra and solve V2 = (1201x50x273)/(373x760) answer will be in liters
