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ipn [44]
3 years ago
10

A gas mixture contains 0.150 mol of O2 gas, 0.116 mol of N2 gas, and 0.211 mol of Ar gas in a 0.500 L flask at 298 K. What is th

e partial pressure of N2 the mixture?

Chemistry
1 answer:
miskamm [114]3 years ago
4 0

Answer:

The partial pressure of N₂ in the mixture is 5.67 atm

Explanation:

According to Dalton's law, the partial pressure of a gas in a mixture of gases is given by the formula;

Partial pressure of gas A = Mole fraction of gas A × Total pressure of the mixture

We are required to calculate the partial pressure of N₂ in the mixture;

We are going to use the following simple steps;

Step 1. Calculating the total pressure of the mixture

We are given;

Moles of O₂ = 0.150 mol

Moles of N₂ = 0.116 mol

Moles of Ar = 0.211 mol

Thus , total moles of the mixture =  0.150 mol +0.116 mol + 0.211 mol

                                                      = 0.477 mol

Volume of the flask = 0.5 L

Temperature = 298 K

But; PV = nRT , where n is the number of moles and R is the ideal gas constant,0.082057 L.atm/mol.K

Thus; P = nRT ÷ V

            = (0.477 mol × 0.082057 × 298 )÷ 0.5

            = 23.328 atm

Step 2: Mole fraction of N₂

Mole fraction of a gas = Moles of a gas ÷ Total moles of the mixture

Mole fraction of N₂ = 0.116 mol ÷ 0.477 mol

                               = 0.243

Step 3: Partial pressure of N₂

P(N₂) = Mole fraction of N₂ × P(total)

       = 0.243 × 23.328 atm

       = 5.669 atm

       = 5.67 atm ( 2 d.p.)

Therefore, the partial pressure of N₂ in the mixture is 5.67 atm

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How many grams of water will be produced when 1.6 moles of ethanol (CH3CH2OH) are burned completely? Enter a number only (no uni
AfilCa [17]

Answer:

86

Explanation:

The reaction that takes place is:

  • C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

First we <u>convert moles of ethanol to moles of water</u>:

  • 1.6 mol ethanol * \frac{3molH_2O}{1molEthanol} = 4.8 mol H₂O

Then we <u>convert moles of water to grams of water</u>, using its molar mass:

  • 4.8 mol H₂O * 18 g/mol = 86.4 g

So 84.6 grams of water will be produced.

4 0
3 years ago
What process takes place when hydrogen-3 and hydrogen-2 combine to form
vazorg [7]

Answer:

A alpha decay

hope this helps you :D

4 0
3 years ago
A mixture containing nitrogen and hydrogen weighs 3.49 g and occupies a volume of 7.45 L at 305 K and 1.03 atm. Calculate the ma
-BARSIC- [3]

Answer:

Mass percent N₂ = 89%

Mass percent H₂ = 11%

Explanation:

First we <u>use PV=nRT to calculate n</u>, which is the total number of moles of nitrogen and hydrogen:

  • 1.03 atm * 7.45 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305 K
  • n = 0.307 mol

So now we know that

  • MolH₂ + MolN₂ = 0.307 mol

and

  • MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49 g

So we have a <u>system of two equations and two unknowns</u>. We use algebra to solve it:

Express MolH₂ in terms of MolN₂:

  • MolH₂ + MolN₂ = 0.307 mol
  • MolH₂ = 0.307 - MolN₂

Replace that value in the second equation:

  • MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49
  • (0.307-MolN₂) * 2 + MolN₂ * 28 = 3.49
  • 0.614 - 2MolN₂ + 28molN₂ = 3.49
  • 0.614 + 26MolN₂ = 3.49
  • MolN₂ = 0.111 mol

Now we calculate MolH₂:

  • MolH₂ + MolN₂ = 0.307 mol
  • MolH₂ + 0.111 = 0.307
  • MolH₂ = 0.196 mol

Finally, we convert each of those mol numbers to mass, to <u>calculate the mass percent</u>:

  • N₂ ⇒ 0.111 mol * 28 g/mol = 3.108 g N₂
  • H₂ ⇒ 0.196 mol * 2 g/mol = 0.392 g H₂

Mass % N₂ = 3.108/3.49 * 100% = 89.05% ≅ 89%

Mass % H₂ = 0.392/3.49 * 100% = 11.15% ≅ 11%

5 0
3 years ago
What is the molarity of a solution prepared by dissolving 10.0g of kno3 in 250 ml of solution
maw [93]

Answer:

Mass of KNO3= 10g

Molar mass of KNO3 = 101.1032g/mol

Volume = 250ml = 0.25L

No of mole on of KNO3 = mass of KNO3/Molar mass of KNO3

no of mole of KNO3 = 10/101.1032

No of mole of KNO3 = 0.09891

molarity of KNO3 = no of mole of KNO3/Vol (L)

Molarity = 0.09891/0.25 = 0.3956M

Molarity of KNO3 = 0.3956M

8 0
3 years ago
How does a balanced chemical equation demonstrate conservation of mass?
Mariulka [41]

Answer:

A balanced chemical equation will demonstrate the law of conservation of mass because the amount of elements on one side (reactants) will equal the amount of elements on the other side (products).

Explanation:

CH₄ + 4Cl₂ -> CCl₄ + 4HCl

Reactants:

1 C

4 H

8 Cl

Products:

1 C

4 H

8 Cl

6 0
2 years ago
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