Well in this
case, silver
nitrate is reduced:
Ag<span>+ </span><span>+ </span>e<span>− </span>→ Ag(s) ↓
Meanwhile, the aluminum
is oxidized forming a positive ion:
Al(s<span>) → </span>Al<span>3+ </span><span>+ 3</span>e−
To get the
overall reaction, we add the half
equations so that the electrons are eliminated:
Al(s<span>) + 3</span>Ag<span>+ </span><span>→ </span>Al<span>3+ </span><span>+ 3</span>Ag(s)
And similarly:
Al(s<span>) + 3</span>AgNO3(aq<span>) → </span>Al(NO3)3(aq<span>) + 3</span>Ag(s<span>)</span>
Its 470mg k dude or dudette
Taking into account the definition of dilution:
- you have to use 8.23 mL of a stock solution of 7.00 M HNO₃ to prepare 0.120 L of 0.480 M HNO₃.
- If you dilute 20.0 mL of the stock solution to a final volume of 0.270 L , the concentration of the diluted solution is 0.518 M.
<h3>Dilution</h3>
Dilution is the process of reducing the concentration of solute in solution, which is accomplished by simply adding more solvent to the solution at the same amount of solute.
In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.
A dilution is mathematically expressed as:
Ci×Vi = Cf×Vf
where
- Ci: initial concentration
- Vi: initial volume
- Cf: final concentration
- Vf: final volume
<h3>Volume of stock solution</h3>
In this case, you know:
- Ci= 7 M
- Vi= ?
- Cf= 0.480 M
- Vf= 0.120 L
Replacing in the definition of dilution:
7 M× Vi= 0.480 M× 0.120 L
Solving:
Vi= (0.480 M× 0.120 L)÷ 7 M
<u><em>Vi= 0.00823 L= 8.23 mL</em></u> (being 1 L= 1000 mL)
Finally, you will need 8.23 mL of the stock solution.
<h3>Concentration of the diluted solution</h3>
In this case, you know:
- Ci= 7 M assuming the stock solution is 7.00 M HNO₃
- Vi= 20 mL= 0.02 L
- Cf= ?
- Vf= 0.270 L
Replacing in the definition of dilution:
7 M× 0.02 L= Cf× 0.270 L
Solving:
(7 M× 0.02 L)÷ 0.270 L= Cf
<u><em>0.518 M= Cf</em></u>
Finally, the concentration is 0.518 M.
Learn more about dilution:
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#SPJ1
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.
