O.8 moles of 122.4 L of neon at STP.
1 mole of gas occupies 22.4 L at STP. So 56/22.4 = 2.5 moles. Neon has atomic mass of 20 and so 2.5 moles x 20 atomic weight = 50 so the mass is 50g.
16,8 L of Xe gas at STP is equivalent to 0,754 moles.
8,4 liters of nitrous oxide at STP contain 2,65 moles.
Use the ideal gas equation. PV = nRT (1 atmosphere)(X volume) = (0.350 moles Ne)(0.08206 L*atm/mol*K)(298.15 K) Volume = 8.56 Liters of neon gas ========================
22.4 liters will have 1 mole of Helium at STP. So, 6 liters will have 0.23 moles
PV=nRT P=1 ATM at STP n is # of moles R is ideal gas constant (here use, .0821 L*ATM/(mol * K) T = 273 K at STP V = volume Plug and chug
Assuming ideal behaviour, 1 mole of any gas occupies 22.4L at STP. So, moles of 10L = 10/22.4 moles = 0.4464 moles
At STP/NTP, 10dm3 of oxygen contains 0.45 moles
1,75/21,4 = 0,0818 moles
The answer is 0,2675 moles.
The answer is 2,68 moles.
The Density of Neon at STP is: a 0.89994 mg/cm-3.
The volume of 5.0 moles of 02 at STP is 100 litres.
The answer is 0,305 moles at 1 at and 25 0C.
If it occupies 3.16 liters at STP, then that's 0.14 moles. 2.82g/0.14mol=20.14g/1mol, and that's the atomic mass of neon, so the gas is neon, and its atomic mass is 20.14 grams/mole.
As far as I understood and found in my chemistry book, the answer to this question is 0.638: 14.3 L X 1 mole / 22.4 L= 0.638
3 moles at STP are present. 1 mole of a gas occupies 22.4 liters at STP. So 67.2/22.4 = 3.
The amount of oxygen is 0,067 moles.
Using 6.02x10^23 for Avogadros number, and 22.4 for the conversion factor at STP, there are 3.84x10^23 atoms of neon present.
1 mole of gas occupies 22.4 liters at STP. 564/22.4 = 25.18 moles (2 decimal places)
STP stands for standard temperature and pressure. Neon is a chemical element that is a gas at STP. The symbol for neon is Ne and it has the atomic number 10.
1 mole = 22.414 liters So, 5 moles = 112.07 liters
67.2 (L) / 22.4 (L/mole) = 3.00 mole of ANY gas at STP
0.43 moles, since 1 mole of gas occupies 22.4 L at STP.
At STP, one moles will occupy 22.4 liters.
The number of argon moles is 0,25.
The answer is 2,5 moles.
1 MOLE of gas occupies 22.4 Liters at STP so a 4 liter flask conatins 4/22.4 = 0.1786 moles
The answer is 4,1 CO2 moles.
At standard temperature and pressure (STP), 1 mole of any gas occupies about 22.4 liters of space. If you have 3.75L of a gas, then 3.75/22.4=0.17 moles.
1 mole occupies 22.414 litres at STP
At STP it is a gas. Neon is a gas at room temperature.
Neon gas is a noble gas / inert gas at STP
1 mole of an ideal gas occupies 22.4 liters at STP - so 2 moles occupies 44.8 liters.
Ideal gas equation. PV = nRT ===============
1 mole of any gas at STP is 22.4L, so 22.4L of nitrogen gas is 1 mole of nitrogen gas.
That will depend on the temperature and pressure. At STP, you will have 5x22.4 = 112 liters
density= mass/ volume. number of moles= volume x RTP or STP
1 mole occupies 22.4 liters. So 2/22.4 = 0.0893 moles
You find the answer by mulitplying the 8 moles by 22.4, L/mol, which is the volume of any one mole of gas at STP. So the answer is 179.2 L
PV = nRT ⟹ n = PV/RT = 1 * 18.65 / (0.082 * 273.15) = 0.8321 moles.
5.6 dm3 is divided by 22.4 dm3 to give number of moles: =5.6dm3/22.4moles per dm3 =0.25 moles So 0.25 moles of C3H4 react.
In STP a mole has a volume of 22.4dm3.so 1.5 mole has a volume of 33.6dm3
1 mole of an ideal gas occupies 22.4 liters at STP. So 112 liters would be occupied by 5 moles of an ideal gas.
1 mole of gas at STP occupies 22.4Lt. of space. So 25 mole will occupie 22.4*25=560lt.
PV = nRT (1atm)(300 L) = n(moles)(0.08206 L*atm/mol*K)(298.15 K) = 12.3 moles of CO2
22.4L=1 mole of gas at STP
For gases, at STP, 1 mole = 22.4L.
There is a mole of H2 in 24dm3 (24 liters) of H2 in STP. So there will be 1/24 (0.0417, accurate to 3 sig fig) mole of H2 gas in 1 liter of H2 in STP.
A cubic decimeter is a liter, so we're talking 50L of oxygen gas at STP. 1mol of any gas at STP occupies 22.4L of space, so 50/22.4 = about 2.2mol of oxygen.