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A 49.0-L cylinder contains a gas mixture at 13700 kPa and 21 °C. Given the percent by weight is 5.00% HCl, 1.00% H₂, and 94% Ne, the partial pressures of the gases are:
A gas mixture is 5.00% by weight HCl, 1.00% H₂, and 94% Ne.
Let's suppose we have 100 grams of sample. Thus, we would have 5.00 g of HCl, 1.00 g of H₂, and 94 g of Ne.
We can calculate the moles of each gas by dividing its mass by its molar mass.
HCl: 5.00 g / (36.46 g/mol) = 0.137 mol
H₂: 1.00 g / (1.01 g/mol) = 0.990 mol
Ne: 94 g / (20.18 g/mol) = 4.66 mol
n = nHCl + nH₂ + nNe = 0.137 mol + 0.990 mol + 4.66 mol = 5.79 mol
The mole fraction (χ) is the ratio of the number of moles of one component of a solution or other mixture to the total number of moles.
χ(HCl) = 0.137 mol/5.79 mol = 0.0237
χ(H₂) = 0.990 mol/5.79 mol = 0.171
χ(Ne) = 4.66 mol/5.79 mol = 0.805
The partial pressure of a gas (p) is the pressure of a gas in a mixture of gases.
The partial pressure of a gas is equal to the product of the total pressure and the mole fraction of the gas.
pHCl = 13700 kPa × 0.0237 = 325 kPa
pH₂ = 13700 kPa × 0.171 = 2.34 × 10³ kPa
pNe = 13700 kPa × 0.805 = 1.10 × 10⁴ kPa
A 49.0-L cylinder contains a gas mixture at 13700 kPa and 21 °C. Given the percent by weight is 5.00% HCl, 1.00% H₂, and 94% Ne, the partial pressures of the gases are:
The question is incomplete. The most likely complete question is:
A gas mixture for use in some lasers contains 5.00 % by weight HCl, 1.00 % H₂, and 94 % Ne. The mixture is sold in cylinders that have a volume of 49.0 L and a pressure of 13700 kPa at 21.0 ∘C.
What is the partial pressure (in kilopascals) of each gas in the mixture?
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