boiling point - condensation point
is the answer i would choose because it makes more scene
Answer:
A
Explanation:
molarity=moles of solute/liter of solution
molarity=0.26/0.3
molarity=0.87molar
Answer: Option (d) is the correct answer.
Explanation:
It is given that molecular formula is
. Now, we will calculate the degree of unsaturation as follows.
Degree of unsaturation = 
= 
= 9 - 8 + 1
= 2
As the degree of unsaturation comes out to be 2. It means that this compound will contain one ring and one double bond.
Yes, this compound could be an alkyne as for alkyne D.B.E = 2.
But this compound cannot be a cycloalkane because for a cycloalkane D.B.E = 1 which is due to the ring only.
Thus, we can conclude that it is a cycloalkane is not a structural possibility for this hydrocarbon.
<u>Answer:</u> The mass of second isotope of indium is 114.904 amu
<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 mass of isotope 2 of indium be 'x'
Mass of isotope 1 = 112.904 amu
Percentage abundance of isotope 1 = 4.28 %
Fractional abundance of isotope 1 = 0.0428
Mass of isotope 2 = x amu
Percentage abundance of isotope 2 = [100 - 4.28] = 95.72 %
Fractional abundance of isotope 2 = 0.9572
Average atomic mass of indium = 114.818 amu
Putting values in equation 1, we get:
![114.818=[(112.904\times 0.0428)+(x\times 0.9572)]\\\\x=114.904amu](https://tex.z-dn.net/?f=114.818%3D%5B%28112.904%5Ctimes%200.0428%29%2B%28x%5Ctimes%200.9572%29%5D%5C%5C%5C%5Cx%3D114.904amu)
Hence, the mass of second isotope of indium is 114.904 amu