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.
I'm not sure if this is what you're looking for, but I suppose you would have to have the solute and the solution to balance the reaction.
I assume this is a true or false question. I would say true
<u>Answer:</u> The percentage abundance of
and
isotopes are 75.77% and 24.23% respectively.
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
Let the fractional abundance of
isotope be 'x'. So, fractional abundance of
isotope will be '1 - x'
- <u>For
isotope:</u>
Mass of
isotope = 34.9689 amu
Fractional abundance of
isotope = x
- <u>For
isotope:</u>
Mass of
isotope = 36.9659 amu
Fractional abundance of
isotope = 1 - x
- Average atomic mass of chlorine = 35.4527 amu
Putting values in equation 1, we get:
![35.4527=[(34.9689\times x)+(36.9659\times (1-x))]\\\\x=0.7577](https://tex.z-dn.net/?f=35.4527%3D%5B%2834.9689%5Ctimes%20x%29%2B%2836.9659%5Ctimes%20%281-x%29%29%5D%5C%5C%5C%5Cx%3D0.7577)
Percentage abundance of
isotope = ![0.7577\times 100=75.77\%](https://tex.z-dn.net/?f=0.7577%5Ctimes%20100%3D75.77%5C%25)
Percentage abundance of
isotope = ![(1-0.7577)=0.2423\times 100=24.23\%](https://tex.z-dn.net/?f=%281-0.7577%29%3D0.2423%5Ctimes%20100%3D24.23%5C%25)
Hence, the percentage abundance of
and
isotopes are 75.77% and 24.23% respectively.