The answer is 40.
We can solve this by finding out the number of protons, and neutrons. Atomic number of an element means the number of protons in that element. So, the atom has 30 protons if the atomic number is 30.
On the other hand, mass number is the total number of protons and neutrons, but not electrons, because they're too light comparing to the other 2. Therefore, we can simply solve the number of neutrons in the atom by subtracting the number of protons from the mass number. 70 - 30 = 40.
Therfore, the number of neutrons is 40.
Answer:
The Correct increasing order of solubility is O2 < Br2 < LiCl < Methanol (CH3OH)
Explanation:
Solubility of compounds or molecules are solely dependent on its inter molecular forces or bonding present in them.
Molecules with Hydrogen bonding usually very soluble in water. Ionic compounds are also very soluble in water because they form ions in solutions. Molecules that possess van der waal forces are usually insoluble in water because they are non-polar.
- O2 (oxygen gas) and Br2 (bromine gas) have van der waal forces in them. Van der waal forces are stronger in Br2 (bromine gas) than O2 (oxygen gas) because Br2 has more number of electrons.
- LiCl is ionic in nature which makes it dissolve in water readily. it easily forms its ions (Li+ and Cl- ) in solutions.
- Methanol (CH3OH) has the highest solubility in water compared to LiCl, Br2 and O2 because it contains Hydrogen bonding which is strongest of all inter molecular forces.
Answer:
The boiling point of the fluoromethane (CH3F) is higher than that of fluorine (F2).
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.
