It's only a small difference (103 degrees versus 104 degrees in water),
and I believe the usual rationalization is that since F is more
electronegative than H, the electrons in the O-F bond spend more time
away from the O (and close to the F) than the electrons in the O-H bond.
That shifts the effective center of the repulsive force between the
bonding pairs away from the O, and hence away from each other. So the
repulsion between the bonding pairs is slightly less, while the
repulsion between the lone pairs on the O is the same -- the result is
the angle between the bonds is a little less.
Hope this helps!
Answer:
Approximately 
.
Explanation:
Convert both volumes to standard units (that is: liters.)
.
.
Number of moles of 
initially present (in the 
solution at 
.)
.
Number of moles of 
from the titration:
.
neutralizes at a 
ratio:
.
Hence, 
.
.
Answer:
6 orbitals
Explanation:
you take the total number of orbitals there would normally be and add that to three and you get a total of 6 orbitals in axial.
We need to know the value of van't hoff factor.
The van't hoff factor is: 2.66 or 2.7 (approximately)
(NH₄)₂SO₄ is an ionic compound, so it dissociates in solution and produces 3 ionic species. Therefore van't hoff factor is more than one.
From the equation: Δ
=i 
.m, where Δ= elevation of boiling point=102.5 - 100=2.5°C.
m=molality of solute=1.83 m (Given)
= Ebullioscopic constant or Boiling point elevation constant= 0.512°C/m (Given)
i= Van't Hoff factor
So, 2.5= i X 0.512 X 1.83
i=
i=2.66= 2.7 (approx.)
Answer:
Here's what I get
Explanation:
CH₃CH₂CH₂CH₂CH₂CH₃ — hexane
CH₂=CHCH₂CH₂CH₂CH₃ — hex-1-ene is the preferred IUPAC name (PIN). 1-Hexene is accepted
CH₃C≡CCH₃ — but-2-yne (PIN); 2-butyne is accepted
CH₃CH(CH₃)CH₂CH₂CH₃ — 2-methylpentane
CH₃CH₂CHCICH₂CH₃ — 3-chloropentane
