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What is the pressure of 4 moles of helium in a 50 liter tank at 308k?
Using the ideal gas law (PV = nRT), where P is pressure, V is volume, n is number of moles, R is the ideal gas constant, and T is temperature in Kelvin, we can solve for pressure. Plugging in the values, the pressure of the 4 moles of helium in a 50 liter tank at 308 K is approximately 81.6 atm.
How many moles of gas are present in a 10L container when 3.84atm of pressure is exerted at 35 Celsius?
Assuming that gas being used is an ideal gas, the gas law equation PV=nRT can be used. P=Pressure (atm) V=Volume (Liters) n=Moles R=0.08206 (a gas constant) T=Temperature (Kelvin) In this problem we know that there is a 10L container with 3.84atm of pressure at 35 C. So P=3.84atm, V=10L, and R=0.08206. To calculate T, you need to convert from Celsius (C) to Kelvin (K). Using the formula K=273+C, you should get T=308K. Now you know all of the variables you can work the problem out. Solve the equation for moles(n) by dividing each side by R and T: (PV/RT)=n Plug in your numbers and solve with a calculator: n=(3.84)(10)/(0.08206)(308) =(38.4)/(25.27448) =1.5193190918 Leaving us a rounded answer of 1.52n in the container
Why co2 cannot be liquified at 308k?
At 308K, carbon dioxide is above its critical temperature of 304.25K. This means that it cannot exist as a liquid under those conditions, as it would be above its critical point and would behave as a supercritical fluid instead of distinct liquid and gas phases.
What is the pressure of 4 moles of helium in a 50 liter tank at 308k?
Using the ideal gas law (PV = nRT), where P is pressure, V is volume, n is number of moles, R is the ideal gas constant, and T is temperature in Kelvin, we can solve for pressure. Plugging in the values, the pressure of the 4 moles of helium in a 50 liter tank at 308 K is approximately 81.6 atm.
How many moles of ammonia gas are found in a 202 mL container at 35C and 750 mmHg?
To find the number of moles of ammonia gas, you can use the ideal gas law equation: PV = nRT. Convert the volume to liters (202mL = 0.202L) and the temperature to Kelvin (35°C + 273 = 308K). Plug in the values: (0.750 atm) * (0.202 L) = n * (0.0821 Latm/molK) * (308K), solve for n to find the number of moles of ammonia gas.
How many moles of gas are present in a 10L container when 3.84atm of pressure is exerted at 35 Celsius?
Assuming that gas being used is an ideal gas, the gas law equation PV=nRT can be used. P=Pressure (atm) V=Volume (Liters) n=Moles R=0.08206 (a gas constant) T=Temperature (Kelvin) In this problem we know that there is a 10L container with 3.84atm of pressure at 35 C. So P=3.84atm, V=10L, and R=0.08206. To calculate T, you need to convert from Celsius (C) to Kelvin (K). Using the formula K=273+C, you should get T=308K. Now you know all of the variables you can work the problem out. Solve the equation for moles(n) by dividing each side by R and T: (PV/RT)=n Plug in your numbers and solve with a calculator: n=(3.84)(10)/(0.08206)(308) =(38.4)/(25.27448) =1.5193190918 Leaving us a rounded answer of 1.52n in the container
Why co2 cannot be liquified at 308k?
At 308K, carbon dioxide is above its critical temperature of 304.25K. This means that it cannot exist as a liquid under those conditions, as it would be above its critical point and would behave as a supercritical fluid instead of distinct liquid and gas phases.
What is the volume in liters occupied by 1.21g of Freon-12 gas at 0.980 ATM and 308k?
To calculate the volume of a gas using the ideal gas law, we use the formula V = (nRT)/P, where V is the volume, n is the number of moles of gas, R is the ideal gas constant, T is the temperature in Kelvin, and P is the pressure. First, we need to convert the mass of Freon-12 to moles using its molar mass. Then, we can plug in the values for n (moles), R (0.0821 L·atm/mol·K), T (308 K), and P (0.980 atm) to find the volume in liters.
10.0 grams of a gas occupies 12.5 liters at a pressure of 42.0 mm Hg What is the volume when the pressure has increased to 75.0 mm Hg?
Using Boyle's Law, p1*V1= p2*V2. This means that the pressure multiplied by the volume remains constant whilst the temperature is the same. Therefore; p1=42.0mm Hg, V1= 12.5L and so the product of the two is 525. If the pressure is now 75 mm Hg the volume must be 525/75= 7 liters. The 10.0 grams of gas information is not needed.
