- molar mass: 34,08 g
- density: 1,363 g/cm3
Mass of H2S: 1,363 x 9,36 = 12,758 g
34,08--------------------------1 mol
12,758-------------------------x
x = 0,374 moles
Wiki User
∙ 7y agoWiki User
∙ 12y agoAt STP? I must assume STP since you mention no temperature or pressure. The formula would be the same anyway, so.......
PV = nRT
(1 atm)(9.63 L) = n(0.08206 L*atm/mol*K)(298.15 K)
9.63/24.466 = n(moles H2)
0.394 moles of hydrogen gas
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At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 liters. Therefore, in 10.3 L of helium gas at STP, the number of moles would be 10.3 L / 22.4 L/mol = 0.460 moles.
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∙ 10y ago0.43 moles, since 1 mole of gas occupies 22.4 L at STP.
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∙ 12y agoIn 11.2 L gas (of any kind, even mixtures such as air) at STP contains always 0.50 mole gaseous particles.
Wiki User
∙ 12y agoThere 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.
Maria Ramirez
0.460
Mable Zboncak
Wiki User
∙ 11y ago.430 mol
Wiki User
∙ 13y ago.459 mol
152
0.25 moles
4 g of Helium = 1 mole.So, 56 g of helium = 14 moles
8,4 liters of nitrous oxide at STP contain 2,65 moles.
16,8 L of Xe gas at STP is equivalent to 0,754 moles.
152
The volume is 0.887 L.
1 mole occupies 22.414 liters So, 3.30 moles will occupy 73.966 liters.
0.25 moles
4 g of Helium = 1 mole.So, 56 g of helium = 14 moles
8,4 liters of nitrous oxide at STP contain 2,65 moles.
16,8 L of Xe gas at STP is equivalent to 0,754 moles.
At STP, 1 mole of any gas occupies 22.4 L. Therefore, in a 5L sample of argon at STP, there would be 5/22.4 moles of argon, which is approximately 0.223 moles.
The answer is 2,68 moles.
The answer is 0,2675 moles.
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, in 1.00 L, there would be 1/22.4 = 0.0446 moles of helium.
The volume of 10.9 mol of helium at STP is 50 litres.