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What is the pressure in mm Hg of a gas at 2atm?
1 atm of pressure equals 760 mm Hg, so 2 atm = 1520 mm Hg.
Sixty Liters of gas were collected over water when the barometer read 670 60 L of a gas collected over water when the barometer read 670 mm Hg and the temperature was 20 C What volume would the dr?
since we know that the total pressure is 670 mmhg and the pressure of water at 20 c is 17.5 mmhg, we use dalton's law. 670=17.5+ gas pressure 652.5=gas pressure by definition then we use the law P1V1T2=P2V2T1 tp find V2=47.9969 L
If 18.0 g O2 has a temperature of 350 K and a pressure of 550 mm Hg what is its volume?
To find the volume of the O2 gas, you can use the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. First, convert the pressure to atm (550 mm Hg = 0.724 atm) and the temperature to Kelvin (350 K). Then use the equation to solve for volume.
A container filled with gas has a volume of 185 ml and a pressure of 310 mm Hg The desired new volume is 74.0 ml What is the required new pressure?
Using the combined gas law (P1V1 = P2V2), we can solve for the new pressure: P2 = (P1*V1) / V2. Plugging in the values: P2 = (310 mm Hg * 185 ml) / 74.0 ml = 775 mm Hg. The required new pressure is 775 mm Hg.
40 liters of gas were collected over water when the barometer read 622.0 mm Hg and temp was 20C what volume wuould dry gas occupy at standard conditions?
To find the volume of the dry gas at standard conditions (0°C and 1 atm), we need to correct for the water vapor using the vapor pressure of water at 20°C. The vapor pressure of water at 20°C is 17.5 mm Hg. Therefore, the pressure of the dry gas is 622.0 mm Hg (total pressure) - 17.5 mm Hg (water vapor pressure) = 604.5 mm Hg. Using the ideal gas law, we can calculate the volume of the dry gas at standard conditions.
A balloon contains a gas with a pressure of 580 mm Hg and a temperature of 227 degrees C What is the new pressure mm Hg of the gas when the temperature drops to 27 degrees C?
375mmhg
What would be the new pressure if a 400 mL gas sample at 380 mm Hg is expanded to 800 mL with no change of temperature?
According to Boyle's Law, if you double the volume of a gas at constant temperature, the pressure is halved. So, the pressure would decrease to 190 mm Hg when the gas sample is expanded to 800 mL.
What is the partial pressure of hydrogen gas in a mixture of hydrogen and helium if the total pressure is 600 mm Hg and the partial pressure of helium is 439 mm Hg?
The partial pressure of hydrogen gas can be calculated by subtracting the partial pressure of helium from the total pressure. Therefore, the partial pressure of hydrogen gas would be 161 mm Hg (600 mm Hg - 439 mm Hg = 161 mm Hg).
The volume of a gas collected when the temperature is 11.0 C and the pressure is 710 millimeter Hg measures 14.8 mL What is the calculated volume of the gas at 20.0 C and 740 millimeter Hg?
13.7, using P1V1T2=P2V2T1.
What is the difference in pressure between a gas at 720mm and a gas at standard pressure?
Standard pressure is 760mm Hg, so the difference in pressure is 40mm Hg (760mm - 720mm).
What is the pressure in mm Hg of a gas at 2atm?
1 atm of pressure equals 760 mm Hg, so 2 atm = 1520 mm Hg.
What is the partial pressure of hydrogen gas in a mixture of hydrogen and helium if the total pressure is 600 mm Hg and the partial pressure of helium is 439 Torr?
To convert Torr to mm Hg, divide by 1.33. So, the partial pressure of helium in mm Hg is 439 Torr / 1.33 = 330 mm Hg. To find the partial pressure of hydrogen, subtract the partial pressure of helium from the total pressure: 600 mm Hg - 330 mm Hg = 270 mm Hg. Hence, the partial pressure of hydrogen gas is 270 mm Hg.
Sixty Liters of gas were collected over water when the barometer read 670 60 L of a gas collected over water when the barometer read 670 mm Hg and the temperature was 20 C What volume would the dr?
since we know that the total pressure is 670 mmhg and the pressure of water at 20 c is 17.5 mmhg, we use dalton's law. 670=17.5+ gas pressure 652.5=gas pressure by definition then we use the law P1V1T2=P2V2T1 tp find V2=47.9969 L
If 18.0 g O2 has a temperature of 350 K and a pressure of 550 mm Hg what is its volume?
To find the volume of the O2 gas, you can use the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. First, convert the pressure to atm (550 mm Hg = 0.724 atm) and the temperature to Kelvin (350 K). Then use the equation to solve for volume.
If 7.56 L of a gas are collected at a pressure of 745 mm Hg then what volume will this gas occupy at standard atmospheric pressure if the temperature remains the same?
To find the volume at standard pressure, we can use the combined gas law equation, which states that (P1 x V1) / T1 = (P2 x V2) / T2. Assuming standard pressure is 760 mm Hg, we have: (745 mm Hg x 7.56 L) / T = (760 mm Hg x V2) / T. Solving for V2, we get V2 = (745 x 7.56 x T) / 760. Since the temperature is constant, the volume at standard pressure will be 745 x 7.56 = 5644.2 L.
A gas occupies 30 L at 760 mm Hg what is its volume at 350 mm Hg?
The phrase "760 mm Hg" is physicists' shorthand for "an atmospheric pressure equal to that needed to support a column of mercury [chemical symbol Hg] of length 760 mm". This is approximately average atmospheric pressure at sea level. As the pressure decreases from "760 mm Hg" to "350 mm Hg", the volume of the gas will increase (assuming a constant temperature). The new volume can be determined using Boyle's Law: New Volume = 30 x 760 / 350 = 65.143 Litres
Why would the pressure of a sample of gas at a constant volume fall mm Hg?
The pressure of a gas at constant volume can fall due to a decrease in temperature, as described by Gay-Lussac's law. When the temperature of the gas decreases, the kinetic energy of the gas molecules reduces, leading to fewer collisions with the container walls and consequently lower pressure. Additionally, if the gas is allowed to expand or if there are any leaks in the system, the pressure would also decrease.
