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professor190 [17]
3 years ago
8

A mixture of gases at 2.99 atm can You have two gases, and , at the same temperature. Determine the ratio of effusion rates of a

nd .ists of 13.2 moles of hydrogen gas and 19.1 moles of helium gas. Determine the partial pressure of the helium gas.
Chemistry
1 answer:
san4es73 [151]3 years ago
3 0

Answer:

Given total pressure of the gas mixture(hydrogen and helium) is 2.99 atm

Number of moles of hydrogen is ---- 13.2 mol

Number of moles of helium is ---- 19.1 mol

Determine the partial pressure of the helium gas.

Ratio of effusion rates of the two gases.

Explanation:

According to Dalton's law of partial pressures,

partial pressure of a component gas in a mixture is:

partial pressure of a gas = total pressure x mole fraction

mole fraction of helium gas is:

mole fraction of helium gas = \frac{number of moles of helium gas}{total number of mioles} \\=>mole fraction of He= \frac{19.1mol}{(19.1+13.2)mol} \\=>mole fraction of He = 0.591\\

Partial pressure of He gas is:

Partial pressure of He =mole fraction of He * total pressure\\                                      =0.591 x 2.99atm\\                                      =1.77atm

Effusion rate of a gas is inversely proportional to its square root of its molecular mass.

\frac{rate of effusion of H2 gas}{rate of effusion of He gas} =\sqrt{\frac{molar mass of He gas}{molar mass of H2 gas} } \\=> \frac{rate of effusion of H2 gas}{rate of effusion of He gas}=\sqrt{\frac{4g.}{2g} } \\=>\frac{rate of effusion of H2 gas}{rate of effusion of He gas}=1.414:1

Hence, rates of effusion of H2:He is 1.414:1.

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Answer:

Number of molecules = 1.8267×10^20

Explanation:

From the question, we can deuced that the gases behave ideally, the we can make use of the ideal gas equation, which is expressed below;

PV = nRT

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Given:

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T = 290.0 k,

; V = 1.07 cm³ = 0.001 L

( 6.75 atm)(0.00107 L) = n(0.0821 L·atm/mol·K)(290K)

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<h3>Answer:</h3>

72.19 g HCl

<h3>Explanation:</h3>

We are given the reaction between hydrogen gas and chlorine gas as;

H₂(g) + Cl₂(g) → 2HCl(g)

Mass of hydrogen gas that reacts is 2 g

We are required to determine the amount of HCl produced

<h3>Step 1: Determine the number of moles of H₂ used </h3>

Moles = Mass ÷ molar mass

Molar mass of Hydrogen gas = 2.02 g/mol

Thus;

Moles = 2 g ÷ 2.02 g/mol

          = 0.99 moles

<h3>Step 2: Moles of HCl produced </h3>

From the reaction equation, 1 mole of hydrogen gas reacts to produce 2 moles of HCl

Therefore, the mole ratio of H₂ : HCl = 1 : 2

Thus, moles of HCl = 0.99 moles × 2

                                = 1.98 moles

<h3>Step 3: Mass of HCl produced </h3>

To calculate mass we multiply the number of moles by molar mass

Molar mass of HCl =36.46 g/mol

Therefore;

Mass of HCl = 36.46 g/mol × 1.98 moles

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Thus, the amount of HCl produced during the reaction is 72.19 g

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