Assuming you mean oxygen gas, the number of molecules can be found by first finding the number of moles = mass of oxygen (4g) / Molecular mass of oxygen gas (32 g mol-1) This tells us there is 0.125 mol of oxygen gas present. The number of molecules present is given by the number of moles x the avogadro constant (6.022x10^23) So the number of oxygen gas molecules present is equal to 0.125 x 6.022x10^23 = 7.5275x10^22 molecules
To find the number of liters of oxygen gas in 11.3 grams, you need to convert the grams to moles using the molar mass of oxygen (16 g/mol). Then, you can use the ideal gas law to find the volume of gas at standard temperature and pressure (STP), which is 1 mole of gas occupying 22.4 liters at 0 degrees Celsius and 1 atm pressure.
Standard temperature and pressure (STP) for oxygen is defined as a temperature of 0 degrees Celsius (273.15 Kelvin) and a pressure of 1 atmosphere (101.325 kilopascals). At STP, one mole of oxygen gas occupies a volume of 22.4 liters.
The colorless gas is likely nitrous oxide (N2O) because it decomposed into nitrogen gas and oxygen gas, which are the components of nitrous oxide. The balanced chemical equation for the reaction is 2N2O -> 2N2 + O2.
There are 2 moles of oxygen gas and approximately 6.023 x 10^23 molecules in 64 g of oxygen gas.
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, in 50 dm^3 of O2 gas, there would be 50/22.4 = 2.23 moles of O2 molecules.
1 mole of any gas occupies 22.4 liters at standard temperature and pressure (STP). Therefore, 68.5 liters of oxygen gas at STP would be 68.5/22.4 = 3.06 moles of oxygen gas.
At STP (standard temperature and pressure), the molar volume of a gas is 22.4 liters/mol. So, 113.97 liters of oxygen gas would be equivalent to 113.97/22.4 = 5.08 moles of oxygen. Since each mole of oxygen contains 2 oxygen atoms, the number of oxygen molecules present would be 5.08 moles * 6.02 x 10^23 molecules/mole = approximately 3.06 x 10^24 oxygen molecules.
The gas sample that has the greatest number of molecules is the one with the largest amount of substance, which is measured in moles. At STP (standard temperature and pressure), one mole of any gas occupies a volume of 22.4 liters. Therefore, the gas sample with the largest volume at STP will have the greatest number of molecules.
There are 6.02 x 10^23 molecules in one mole of a substance (Avogadro's number). At STP, 22.4 liters of any ideal gas contains 1 mole of gas. Therefore, there are 6.02 x 10^23 butane molecules in 22.4 liters of C4H10 gas at STP.
A gas at STP.
part of the oxygen group....gas, nonmetal...
Like most diatomic gas molecules oxygen will not dissolve very well in water (less than 10 mg/L at STP).
One liter of oxygen (O2) gas at standard temperature and pressure (STP) \ill contain 1/22.4 of a mole of molecules. STP is defined as 0 degrees Celsius at 1 atm of pressure.
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 ideal gas occupies 22.4 liters. Since 11.2 liters represent half the volume of a mole, we have 0.5 moles of O2 gas. Using the molar mass of oxygen (O2), which is around 32 g/mol, we find that 0.5 moles of O2 gas would be equivalent to approximately 16 grams.
Using the ideal gas law (PV = nRT), we can calculate the volume of gas at STP. First, we need to convert the number of molecules to moles by dividing by Avogadro's number. Then, we can use the volume of 1 mole of gas at STP, which is 22.4 liters. Calculate V = (5.4x10^24 / 6.022x10^23) * 22.4 to find the volume in liters.