The answer is 6,197 399 5 liters at 25 0C.
The molar volume of a gas at STP (standard temperature and pressure) is 22.4 L/mol. Therefore, the volume occupied by 2 moles of oxygen would be 44.8 L.
I would assume chlorine gas and standard temperature an atmospheric pressure. Using the ideal gas equation. PV = nRT (1 atm)(X volume) = (2.4 moles Cl2)(0.08206 Mol*K/L*atm)(298.15 K) Volume = 59 Liters of chlorine gas --------------------------------------------
The molar volume of hydrogen is approx. 22,7 L at 100 kPa and 0C.
Since chlorine gas is a diatomic molecule (Cl2), one mole of chlorine gas contains two moles of chlorine atoms. Therefore, 6.00 moles of chlorine atoms would be equivalent to 3.00 moles of chlorine gas.
It is necessary first to determine how many moles of chlorine there are in the volume specified. Using the ideal gas law, 40 cm3 of chlorine is equivalent to 0.129 grams of Cl2, or 3.63 millimoles. Next, the stoichiometric equation is set up. X H2 + Y Cl2 -- reacts --> Z HCl. It can be clearly seen that X = 1, Y = 1, and Z = 2. This means that for every mole of chlorine that reacts, 2 moles of hydrogen chloride is produced. Then, we have 7.26 millimoles of HCl. Reversing the ideal gas law, this means the resulting volume of HCl gas is .224 cm3. The reason for the smaller volume is due to the differences in the molar weight of chlorine and hydrogen chloride (70.906 g/mol and 36.461 g/mol respectively).
This volume is 0,449 L at 0 0C.
This volume is 6,197 399 5 at 25 0C.
The volume of one mole of gas at a standard temperature and pressure is 22.4 liters. Multiply 22.4 liters by 0.25 moles to get a volume of 5.6 liters.
The volume that 2.4 moles of chlorine gas would occupy depends on the temperature and pressure of the gas, according to the ideal gas law (PV = nRT). At standard temperature and pressure (STP), which is 0°C and 1 atm pressure, 2.4 moles of chlorine gas would occupy approximately 53.75 liters.
The volume occupied by gas molecules is negligible when compared to volume occupied by the gas.The collisions between gas molecules-gas molecules and gas molecules-walls of the container are perfectly elastic.
The molar volume of a gas at STP (standard temperature and pressure) is 22.4 L/mol. Therefore, the volume occupied by 2 moles of oxygen would be 44.8 L.
No, the volume occupied by one mole of a gas at a given temperature and pressure is the same for all gases, according to Avogadro's hypothesis and the ideal gas law. This is known as the molar volume of a gas, which is approximately 22.4 liters at standard temperature and pressure (STP).
The ideal gas law does not account for the volume occupied by gas particles and the interactions between gas molecules.
I would assume chlorine gas and standard temperature an atmospheric pressure. Using the ideal gas equation. PV = nRT (1 atm)(X volume) = (2.4 moles Cl2)(0.08206 Mol*K/L*atm)(298.15 K) Volume = 59 Liters of chlorine gas --------------------------------------------
The volume of a gas depends on its pressure, temperature, and volume according to the ideal gas law PV = nRT. Without knowing the pressure, temperature, or container size, it's not possible to determine the volume occupied by the 0.48 moles of hydrogen.
Volume is the amount of 3-dimensional space occupied by an object. It could be a liquid, but it can also be a gas. For example, a glass of air.
To calculate the volume of chlorine gas produced, you need to know the molar mass of chlorine and use the ideal gas law equation. First, convert the mass of chlorine gas to moles using its molar mass. Then use the ideal gas law equation PV = nRT, where P is pressure, V is volume, n is moles, R is the ideal gas constant, and T is temperature. Finally, you can solve for V to find the volume in liters.