Answer:
Q = 233.42 J
Explanation:
Given data:
Mass of lead = 175 g
Initial temperature = 125.0°C
Final temperature = 22.0°C
Specific heat capacity of lead = 0.01295 J/g.°C
Heat absorbed by water = ?
Solution:
Heat absorbed by water is actually the heat lost by the metal.
Thus, we will calculate the heat lost by metal.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
ΔT = 22.0°C - 125.0°C
ΔT = -103°C
Q = 175 g × 0.01295 J/g.°C×-103°C
Q = -233.42 J
Heat absorbed by the water is 233.42 J.
Answer:
15.70mg would remain
Explanation:
Partition coefficient is used to extract or purify a solute from a solvent selectively to avoid interference from other substances. For the problem, formula is:
Kp = Concentration 9-fluorenone in ether / Concentration of solute in H₂O
After the solute, 9-fluorenone, is extracted with water, the mass that remains in ether is:
(19mg - X)
<em>Where X is the mass that now is in the aqueous phase</em>
Replacing in Kp formula:
9.5 = (19mg - X) / 5mL / (X /10mL)
0.95X = 19mg - X / 5mL
4.75X = 19 - X
5.75X = 19
X = 19 / 5.75
X = 3.30mg
That means 9-fluorenone that remain in the ether layer is:
19mg - 3.30mg =
<h3>15.70mg would remain</h3>
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.
Answer:
The answer is 1.29473 x 10^24
Explanation:
Answer:
Name of the compound is Nitrogen triiodide.
Explanation:
According to the IUPAC rules, to naming of the compound the following formula can be applied.
Prefix + Name of first element + Base name element of second element + Suffix.
The given compound - 
Name of first element- Nitrogen
Base name element of second element - Iodine
Suffix = 3 = tri
Here, iodine is in ionic form therefore, it becomes iodide. and then suffix will be added in front of the halogen.
Therefore, name of the compound will be Nitrogen triiodide..
