Answer:
The average velocity of the airplane for this trip is 1684.21 km/h
Explanation:
Average velocity is the rate of change of displacement with time. That is,
Average velocity = 
= Δx / Δt = 
Now we will calculate the time taken by the airplane for the first motion before it encounters a wind.
From,
Velocity = 
Time = 
Therefore, Time = 
Time = 2.1h
This is the time taken before the airplane encounters a wind.
Hence, t1 = 2.1h
Now, For the time taken by the airplane when it encounters a wind
Also from,
Velocity =
Time =
Therefore, Time = 
Time = 1.625h
Hence, t2 = 1.625h
Now, to calculate the average velocity
Average velocity =
x1= 2100, x2= 1300, t1= 2.1h and t2= 1.625h
Hence, Average velocity = 
Average velocity = 1684.21 km/h
I found this on google
“The periodic table is important because its is organized to provide a great deal of information about elements and how they relate to one another in one-easy-to-use reference. The table can be used to predict the properties of elements, even those that have not been discovered.”
I hope this helps
Answer:
?
Explanation:
can you describe the question a little more please?
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.
