Aluminum atomic number 13 becomes Neon with an atomic number of 10 when it loses three valence electrons.
Colligative
properties calculations are used for this type of problem. Calculations are as
follows:<span>
</span>
<span>ΔT(freezing point)
= (Kf)m
ΔT(freezing point)
= 1.86 °C kg / mol (0.705)
ΔT(freezing point) = 1.3113 °C
</span>
<span>
</span>
<span>Hope this answers the question. Have a nice day.</span>
Those reactions in which Alkyl Halide reacts with the solvent without the involvement of any acid or base is called as
Solvolysis. In given problem <em>tert</em>-Butyl Bromide is a tertiary Alkyl Halide and we know well that tertiary alkyl halides undergo
SN¹ and
E¹ elimination reaction due to the formation of
stable tertiary carbocation. In given example after the formation of carbocation when Isopropyl act as
nucleophile it will produce
ether and when it acts as a
base it will produce
unsaturated compound. The reaction along with both products is shown below,
Hey there!:
K = Ka * Kb / Kw
Ka = 1.8*10⁻⁴
Kb = 10⁻¹⁴ / 6.8*10⁻⁴
K = 1.8*10⁻⁴ * ( 10⁻¹⁴/ 6.8*10⁻⁴ ) * ( 1 / 10⁻¹⁴ )
K = = 1.8 / 6.8
K = 0.265
Answer A
Therefore:
K is less than on the forward reaction is not favorable .
Hope That helps!
Lithium dihydrogen phosphate