To determine the number of atoms in 560 cm3 of ammonia at STP (Standard Temperature and Pressure), we first need to calculate the number of moles of ammonia present. The molar volume of a gas at STP is 22.4 L/mol, which is equivalent to 22,400 cm3/mol. Therefore, 560 cm3 is equal to 0.025 moles of ammonia. Next, we use Avogadro's number, 6.022 x 10^23 atoms/mol, to find that there are approximately 1.51 x 10^22 ammonia molecules in 560 cm3 at STP.
At standard temperature and pressure (STP), the volume occupied by 1 mole of any ideal gas is 22.4 liters. Therefore, the volume of 1.42 moles of ammonia at STP would be 1.42 * 22.4 liters = 31.808 liters.
At STP (standard temperature and pressure), 1 mole of gas occupies 22.4 L. Xenon's molar mass is approximately 131.29 g/mol. Using this, we find that 10.0 L of Xe gas at STP would contain approximately 0.446 moles of Xe atoms, which is about 2.69 x 10^23 xenon atoms.
At STP (Standard Temperature and Pressure), the volume of 1 mole of any gas is 22.4 liters. Since hydrogen gas exists as H2 molecules, 67.2 liters of hydrogen gas at STP contains 3 moles of H2 molecules. Since each H2 molecule contains 2 hydrogen atoms, there are 6 moles of hydrogen atoms, which is equivalent to 6 x 6.022 x 10^23 atoms of hydrogen.
No, fluorine is a diatomic gas at standard temperature and pressure (STP), meaning it exists as F2 molecules, not as single F atoms.
At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 liters. 10.0 liters of Xe gas at STP would therefore contain 10.0/22.4 = 0.4464 moles of Xe. 1 mole of Xe contains 6.022 x 10^23 atoms. Therefore, 10.0 liters of Xe gas at STP would contain 0.4464 x 6.022 x 10^23 = 2.69 x 10^23 xenon atoms.
At standard temperature and pressure (STP), the volume occupied by 1 mole of any ideal gas is 22.4 liters. Therefore, the volume of 1.42 moles of ammonia at STP would be 1.42 * 22.4 liters = 31.808 liters.
The volume of ammonia is 19,5 L.
At STP (standard temperature and pressure), 1 mole of gas occupies 22.4 L. Xenon's molar mass is approximately 131.29 g/mol. Using this, we find that 10.0 L of Xe gas at STP would contain approximately 0.446 moles of Xe atoms, which is about 2.69 x 10^23 xenon atoms.
Assuming that you mean the reaction of nitrogen and hydrogen to form ammonia N2 + 3H2 -> 2NH3 1 mole of nitrogen forms 2 moles of ammonia- so 4.08 l of nitrogen will be consumed to form 8.16 moles of ammonia. This assumes both are pretty ideal gases which is a reasonabale approximation.
At STP (Standard Temperature and Pressure), the volume of 1 mole of any gas is 22.4 liters. Since hydrogen gas exists as H2 molecules, 67.2 liters of hydrogen gas at STP contains 3 moles of H2 molecules. Since each H2 molecule contains 2 hydrogen atoms, there are 6 moles of hydrogen atoms, which is equivalent to 6 x 6.022 x 10^23 atoms of hydrogen.
No, fluorine is a diatomic gas at standard temperature and pressure (STP), meaning it exists as F2 molecules, not as single F atoms.
At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 liters. 10.0 liters of Xe gas at STP would therefore contain 10.0/22.4 = 0.4464 moles of Xe. 1 mole of Xe contains 6.022 x 10^23 atoms. Therefore, 10.0 liters of Xe gas at STP would contain 0.4464 x 6.022 x 10^23 = 2.69 x 10^23 xenon atoms.
If these atoms are loose unities, not bonded in a molecule, it would be about 22 dm3 at 0oC and standard pressure (STP). If the atoms would have formed gas molecules, e.g. CO2 or SF6, the answer would be different, because 6.02 x 1023 molecules always take in a volume of about 22 dm3 at STP. But with CO2 you would have then 18 x 1023 atoms because there are 3 atoms in one molecule.
Using the combined gas law, we can relate the initial and final conditions of the gas: P1V1/T1 = P2V2/T2. At STP (Standard Temperature and Pressure), the conditions are 1 atm and 0 degrees Celsius. Convert 1250 mm Hg to atm and 75 degrees Celsius to Kelvin. With this information, you can then calculate the final volume of the ammonia gas at STP.
Using stoichiometry, 16.2 L of H2 gas at STP equals about .7228 moles (1 L of gas at STP has a volume of 22.41 L), and there are 6.02 x 1023molecules of hydrogen in a mole, so we have (6.02 x 1023molecules/mol)(.7228 mol) = 4.35 x 1023 hydrogen molecules. There are two hydrogen atoms in each molecule, so the answer is (4.35 x 1023molecules H2)(2 H atoms/molecule) = 8.70 x 1023 H atoms in 16.2 L.
Since the reaction produces 2 moles of ammonia for every 3 moles of hydrogen used, the mole ratio is 3:2. Thus, for every 2 liters of ammonia, 3 liters of hydrogen are required. This means that (2/3) x L of hydrogen were used to produce L of ammonia in the Haber process at STP.
The molar mass of ammonia (NH3) is 17 g/mol. Therefore, one mole of ammonia weighs 17 grams. Since one liter of gas at STP (standard temperature and pressure) contains 22.4 liters per mole, this means that a liter of ammonia gas would weigh 17 grams.