Use the general Gas Law, but be careful to use correct units!
p.V=n.R.T
so
n = [p.V] / [R.T]
in which:
or = 8.314 [Pa.m3.K-1.mol-1] when SI-units are used.
According to the ideal gas law, all gases occupy about 22.4 liters per moleof space at standard temperature and pressure, so 22.4x2.56=57.34 liters.
At STP (Standard Temperature and Pressure: 0°C and 1 atm pressure), 1 mole of any gas occupies 22.4 L. Therefore, to find the number of moles of O gas in 10.0 L at STP, we divide 10.0 by 22.4 to get 0.4464 moles. The molar mass of oxygen (O₂) is 32 g/mol, so 0.4464 moles of O₂ gas would be 0.4464 moles * 32 g/mol ≈ 14.3 grams.
The pressure is 2,02 atmospheres.
To calculate the volume of 85.5 grams of carbon monoxide (CO) at a specific temperature and pressure, we first need to determine the number of moles of CO using its molar mass (approximately 28.01 g/mol). Dividing 85.5 grams by the molar mass gives about 3.05 moles of CO. Using the ideal gas law (PV=nRT), if we assume standard temperature and pressure (STP: 0°C and 1 atm), one mole occupies 22.4 liters, so 3.05 moles would occupy approximately 68.4 liters. However, for an accurate calculation, temperature and pressure conditions must be specified.
To measure the pressure of a gas in atmospheres at different temperatures in Celsius, you can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin. To predict the temperature at which the pressure would equal zero, you would need to decrease the temperature below the gas's critical temperature, at which point the gas would liquefy and the pressure would drop to zero.
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.
At standard temperature and pressure, 1 mole of any gas will occupy 22.4 liters. Set up a direct proportion of 22.4 liters/1 mole = 1 liter/x moles and solve for x. You get 0.045 moles.
According to the ideal gas law, all gases occupy about 22.4 liters per moleof space at standard temperature and pressure, so 22.4x2.56=57.34 liters.
I'm pretty sure NaOH is a solid under those conditions! You just need to know the density of NaOH. The temperature and pressure are irrelevant if it was a gas, you could use the Ideal Gas Law. See the Related Questions link to the left of this answer about that.
The amount of oxygen is 0,067 moles.
Using the ideal gas law PV=nRT P= pressure 1atm at stp V=volume our unknown n=number of moles 2.88 R= the gas law constant .08026 L*atm*/(k*mol) T=temperature 273.15K at stp using algebra to solve for V the formula is V=(nRT)/P = (2.88*.08026*273.15)/1 = 64.55L
At STP (Standard Temperature and Pressure: 0°C and 1 atm pressure), 1 mole of any gas occupies 22.4 L. Therefore, to find the number of moles of O gas in 10.0 L at STP, we divide 10.0 by 22.4 to get 0.4464 moles. The molar mass of oxygen (O₂) is 32 g/mol, so 0.4464 moles of O₂ gas would be 0.4464 moles * 32 g/mol ≈ 14.3 grams.
The pressure would decrease to zero because a gas at 0 degrees Celsius is frozen, and frozen gas does not exert any pressure due to lack of molecular motion.
The answer would depend on the pressure attained by the cooker.
the reaction would shift to favor the side with the most moles of gas. (apex).
The pressure is 2,02 atmospheres.
At STP (standard temperature and pressure: 0°C and 1 atm pressure), 1 mole of any gas occupies 22.4 liters. The molar mass of hydrogen gas (H2) is 2 g/mol. With 0.00919 g of H2, you have 0.004595 moles. So, at STP, it would occupy approximately 0.1029 liters, which is the same as 102.9 milliliters.