Answer:
a)CH₄, BH₃, and CCl₄
Explanation:
<u>London dispersion forces:-
</u>
The bond for example, in the molecule is F-F, which is non-polar in nature because the two fluorine atoms have same electronegativity values.
The intermolecular force acting in the molecule are induced dipole-dipole forces or London Dispersion forces / van der Waals forces which are the weakest intermolecular force.
Out of the given options, H₂O , NH₃ exhibits hydrogen bonding which is:-
<u>Hydrogen bonding:-
</u>
Hydrogen bonding is a special type of the dipole-dipole interaction and it occurs between hydrogen atom that is bonded to highly electronegative atom which is either fluorine, oxygen or nitrogen atom.
Thus option B and C rules out.
<u>Hence, the correct option which represents the molecules which would exhibit only London forces is:- a)CH₄, BH₃, and CCl₄</u>
H > 6,000
height greater than 6,000
Answer: 1.3% many crosslinks as isoprene units,
Explanation:
Given:
mass pf natural rubber= 200.0g
mass of sulphur = 4.8g
molar mass of sulphur =32g/mol
molar mass of isoprene = C5H8=( 12x5) +(1x8)= 68g/mol
Solution: we first find no of moles present in each using
no of moles =
Isoprene: 200.0g x [1mole / 68g] = 2.94moles.
Sulfur: 4.8g x [1mole / 32g] x [1 mole crosslinks / 4 moles S] = 0.0375 moles crosslinks.
to find % crosslinked units, we have
0.0375 / 2.94 = 1.3% as many crosslinks as isoprene units,
The final destination to where some of the electrons go to at the end of cellular respiration would be D. Oxygen. Assuming that this aerobic cellular respiration, the final electron acceptor is that of oxygen.