Chemistry, 21.03.2020 02:57 sherlock19

Gas mixture consists of N2, O2, and Ne, where the mole fraction of N2 is 0.55 and the mole fraction of Ne is 0.25. If the mixture is at STP in a 5.0 L container, how many molecules of O2 are present?

Answers: 1

Show answers

Answers

Answer from: precioussmith76

The number of molecules of O₂ present in the container which consists of N₂, O₂ and Ne is 2.68×10²² molecules

We'll begin by calculating the the mole fraction of O₂. This can be obtained as follow:

Mole fraction of N₂ = 0.55

Mole fraction of Ne = 0.25

Total mole fraction = 1

Mole fraction of O₂ =?

Mole fraction of N₂ + Mole fraction of Ne + Mole fraction of O₂ = 1

0.55 + 0.25 + Mole fraction of O₂ = 1

0.8 + Mole fraction of O₂ = 1

Collect like terms

Mole fraction of O₂ = 1 – 0.8

Mole fraction of O₂ = 0.2Next, we shall determine the total mole of the gases in the container. This can be obtained as follow:

Pressure (P) = 1 atm (at stp)

Temperature (T) = 273 (at stp)

Volume (V) = 5 L

Gas constant (R) = 0.0821 atm.L /Kmol

Number of mole (n) =? PV = nRT

1 × 5 = n × 0.0821 × 273

5 = n × 22.4133

Divide both side by 22.4133

$n = \frac{5}{22.4133}\\\\$

n = 0.223 mole

Thus, the total mole of the gases in the container is 0.223 mole

Next, we shall determine the number of mole of O₂. This can be obtained as follow:

Mole fraction of O₂ = 0.2

Total mole = 0.223 mole

Mole of O₂ =.?

Mole = Total mole × mole fraction

Mole of O₂ = 0.2 × 0.223

Mole of O₂ = 0.0446 mole Finally, we shall determine the number of molecules of O₂ present in the container.

From Avogadro's hypothesis

1 mole of O₂ = 6.02×10²³ molecules

Therefore,

0.0446 mole of O₂ = 0.0446 × 6.02×10²³

0.0446 mole of O₂ = 2.68×10²² molecules

Thus, the number of molecules of O₂ present in the container is 2.68×10²² molecules

Learn more: link

Answer from: cupcake3103670

2.69x10²² molecules of O₂

Explanation:

We start with the knowldge that says: "the sum of mole fractions in a mixture of gases = 1 ". When we start with this, we propose

Mole fraction N₂ + Mole fraction Ne + Mole fraction O₂ = 1

0.55 + 0.25 + Mole fraction O₂ = 1

Mole fraction O₂ = 1 - 0.25 - 0.55 → 0.20

We know that the mixture is at STP in a 5L of volume, so let's calculate the volume. We use the Ideal Gases Law

1 atm . 5L = n . 0.082 . 273 K

n = 1 atm . 5L / 0.082 . 273K → 0.223 moles

These number of moles are the total moles in the mixture. We apply the mole fraction to determine the moles of O₂ that are present in the container.

Mole fraction O₂ = Moles O₂ / Total moles → Mole fraction O₂ . Total moles = moles O₂

0.223 moles . 0.2 = 0.0446 moles of O₂

Let's count the number of molecules

0.0446 moles . 6.02x10²³molecules / 1 mol = 2.69x10²² molecules of O₂

Answer from: Quest

the awnser is group 12

Answer from: Quest

it's 3.2 on edge, i just took the test.

Another question on Chemistry

Chemistry, 22.06.2019 03:30

The atomic radius of sodium is 186 pm and of chlorine is 100 pm. the ionic radius for na+ is 102 pm and for cl– is 181 pm. in going from na to cl in period 3, why does the atomic radius decrease while the ionic radius increases? a. the inner electrons in the sodium cation shield its valence electrons more effectively than the inner electrons in the chloride anion do. b. the inner electrons shield the valence electrons more effectively in the chlorine atom than in the chloride anion. c. the outermost electrons in chloride experience a smaller effective nuclear charge than those in the sodium cation do. d. the outermost electrons in chloride experience a larger effective nuclear charge than those in the sodium cation do. e. monatomic ions are bigger than the atoms from which they are formed.

Answers: 2

Answer

Chemistry, 22.06.2019 09:30

In apex! a liquid heated beyond a certain temperature becomes

Answers: 1

Answer

Chemistry, 22.06.2019 10:40

Ammonia and oxygen react to form nitrogen monoxide and water, like this: 4nh3 (g) + 5o2 (g) → 4no (g) + 6h2o (g) also, a chemist finds that at a certain temperature the equilibrium mixture of ammonia, oxygen, nitrogen monoxide, and water has the following composition: compound pressure at equilibrium nh3 65.1atm o2 31.3atm no 62.7atm h2o 65.8atm compound pressure at equilibrium nh3 65.3 atm o2 7.79 atm no 12.1 atm h2o 65.8 atm calculate the value of the equilibrium constant kp for this reaction. round your answer to 2 significant

Answers: 2

Answer

Chemistry, 22.06.2019 18:00

Answer asap need it by wednesday morning carry out the following calculations on ph and ka of from data. i. calculate the ph of 0.02m hcl ii. calculate the ph of 0.036m naoh iii. calculate the ph of 0.36m ca(oh)2 iv. calculate the ph of 0.16m ch3cooh which has ka = 1.74 x 10-5 mol dm-3 v. calculate ka for weak acid ha which has a ph of 3.65 at 0.30m concentration vi. calculate the ka of a solution made by mixing 15.0 cm3 0.2m ha and 60.0 cm3 0.31m a-. [ph= 3.80] vii. calculate the ph of a solution made by mixing 15.0 cm3 0.1m naoh and 35.0 cm3 0.2m hcooh. [ka = 1.82 x 10-4 m]

Answers: 1

Answer

You know the right answer?

Gas mixture consists of N2, O2, and Ne, where the mole fraction of N2 is 0.55 and the mole fraction...

Questions