if i am correct the volume of nitrogen gas has to equal to molecules density, making the substance 1.27 liters :)
Considering the definition of mole fraction, the mole fraction of O₂ in the mixture is 0.434.
<h3>Definition of mole fraction</h3>
The molar fraction is a way of measuring the concentration that expresses the proportion in which a substance is found with respect to the total moles of the solution.
In other words, the mole fraction expresses the concentration of solute in a solution as the ratio of moles of substance to total moles of solution:

<h3>Mole fraction of O₂ in this mixture</h3>
In this case, you know a gas mixture used for anesthesia contains 4.60 mol oxygen (O₂) and 6.00 mol nitrous oxide (N₂O).
So, the total moles of the solution can be calculated as:
Total moles = moles of oxygen (O₂) + moles of nitrous oxide (N₂O)
Then:
Total moles= 4.60 moles + 6 moles
Total moles= 10.60 moles
Finally, the more fraction of O₂ can be calculated as follow:


Solving:
<u><em>Mole fraction O₂ = 0.434</em></u>
Finally, the mole fraction of O₂ in the mixture is 0.434.
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Answer:
2n² means 2 × n × n . the power on the n represent the no. of time it is multiplied.
for ex 3x² = 3 × x × x
4x⁴ = 4 × X× X × X× X
Answer:
Answers with detail are given below
Explanation:
1) Given data:
Mass of Rb₃Rn = 76.19 g
Number of moles = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 478.43 g/mol
Number of moles = 76.19 g/ 478.43 g/mol
Number of moles = 0.16 mol
2) Given data:
Mass of FrBi₂ = 120.02 g
Number of moles = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 640.96 g/mol
Number of moles = 120.02 g/640.96 g/mol
Number of moles = 0.19 mol
3) Given data:
Mass of Zn₂F₃ = 88.24 g
Number of moles = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 187.73 g/mol
Number of moles = 88.24 g/ 187.73 g/mol
Number of moles = 0.47 mol
4) Given data:
Number of moles of Sb₄Cl = 1.20 mol
Mass of Sb₄Cl = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 522.49 g/mol
Mass = Number of moles × molar mass
Mass = 1.20 mol × 522.49 g/mol
Mass = 626.99 g
Answer:
Basically, paramagnetic and diamagnetic refer to the way a chemical species interacts with a magnetic field. More specifically, it refers to whether or not a chemical species has any unpaired electrons or not.
A diamagnetic species has no unpaired electrons, while a paramagnetic species has one or more unpaired electrons.
Now, I won't go into too much detail about crystal field theory in general, since I assume that you're familiar with it.
So, you're dealing with the hexafluorocobaltate(III) ion, [CoF6]3â’, and the hexacyanocobaltate(III) ion, [Co(CN)6]3â’.
You know that [CoF6]3â’ is paramagnetic and that [Co(CN)6]3â’ is diamagnetic, which means that you're going to have to determine why the former ion has unpaired electrons and the latter does not.
Both complex ions contain the cobalt(III) cation, Co3+, which has the following electron configuration
Co3+:1s22s22p63s23p63d6
For an isolated cobalt(III) cation, all these five 3d-orbitals are degenerate. The thing to remember now is that the position of the ligand on the spectrochemical series will determine how these d-orbtals will split.
More specifically, you can say that
a strong field ligand will produce a more significant splitting energy, Δ a weak field ligand will produce a less significant splitting energy, Δ
Now, the spectrochemical series looks like this
http://chemedu.pu.edu.tw/genchem/delement/9.htmhttp://chemedu.pu.edu.tw/genchem/delement/9.htm
Notice that the cyanide ion, CNâ’, is higher on the spectrochemical series than the fluoride ion, Fâ’. This means that the cyanide ion ligands will cause a more significant energy gap between the eg and t2g orbitals when compared with the fluoride ion ligands.
http://wps.prenhall.com/wps/media/objects/3313/3393071/blb2405.htmlhttp://wps.prenhall.com/wps/media...
In the case of the hexafluorocobaltate(III) ion, the splitting energy is smaller than the electron pairing energy, and so it is energetically favorable to promote two electrons from the t2g orbitals to the eg orbitals → a high spin complex will be formed.
This will ensure that the hexafluorocobaltate(III) ion will have unpaired electrons, and thus be paramagnetic.
On the other hand, in the case of the hexacyanocobaltate(III) ion, the splitting energy is higher than the electron pairing energy, and so it is energetically favorable to pair up those four electrons in the t2g orbitals → a low spin complex is formed.
Since it has no unpaired electrons, the hexacyanocobaltate(III) ion will be diamagnetic.