16.A mixture of carbon dioxide and hydrogen gases contains carbon dioxide at a partial pressure of 217 mm Hg and hydrogen at a partial pressure of 703 mm Hg. What is the mole fraction of each gas in the mixture?

XCO2 =

XH2 =

Answers

Answer 1

x CO₂  = 0.236

x H₂  = 0.764

Further explanation

Given

P CO₂ = 217 mmHg

P H₂ = 703 mmHg

Required

The mole fraction

Solution

Dalton's law :

The total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases  

Can be formulated:  

P tot = P1 + P2 + P3 ....  

P tot for the mixture :

= 217 mmHg + 703 mmHg

= 920 mmHg

Mole fraction of each gas :

x CO₂ = 217 mmHg/920 mmHg = 0.236

x H₂ = 703 mmHg/920 mmHg = 0.764


Related Questions

Base your answer on the information below. The hydrocarbon 2-methylpropane reacts with iodine as represented by the balanced equation below. At standard pressure, the boiling point of 2-methylpropane is lower than the boiling point of 2-iodo-2-methylpropane. Explain the difference in the boiling points of 2-methylpropane and 2-iodo-2-methylpropane in terms of both molecular polarity and intermolecular forces.

Answers

Answer:

See explanation

Explanation:

The boiling point of a substance is affected by the nature of bonding in the molecule as well as the nature of intermolecular forces between molecules of the substance.

2-methylpropane has only pure covalent and nonpolar C-C and C-H bonds. As a result of this, the molecule is nonpolar and the only intermolecular forces present are weak dispersion forces. Therefore, 2-methylpropane has a very low boiling point.

As for 2-iodo-2-methylpropane, there is a polar C-I bond. This now implies that the intermolecular forces present are both dispersion forces and dipole interaction. As a result of the presence of stronger dipole interaction between 2-iodo-2-methylpropane molecules, the compound has a higher boiling point than  2-methylpropane.

17.Oxygen gas can be prepared by heating potassium chlorate according to the following equation:

2KClO3(s)2KCl(s) + 3O2(g)

The product gas, O2, is collected over water at a temperature of 25 °C and a pressure of 755 mm Hg. If the wet O2 gas formed occupies a volume of 6.22 L, the number of grams of O2 formed is ________
g. The vapor pressure of water is 23.8 mm Hg at 25 °C.

Answers

Mass of O₂ formed = 7.84 g

Further explanation

Given

Reaction

2KClO₃(s) ⇒2KCl(s) + 3O₂(g)

P water = 23.8 mmHg

P tot = 755 mmHg

V = 6.22 L

T = 25 + 273 = 298 K

Required

mass of O₂

Solution

P tot = P O₂ + P water

P O₂ = P tot - P water

P O₂ = 755 - 23.8

P O₂ = 731.2mmHg = 0.962 atm

Moles O₂ :

Ideal gas law :

n = PV/RT

n = 0.962 x 6.22 / 0.082 x 298

n = 0.245

Mass O₂ :

= mol x MW

= 0.245 x 32

= 7.84 g

what type of reaction is Au2S+H2---> 2Au+H2S

Answers

Answer:

single replacement reaction is type of reaction is Au2S+H2---> 2Au+H2S.

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