Answer: Benzene is less reactive than methylbenzoate and more reactive than Nitrobenzene
Explanation:
This is because the methyl group on the benzene ring is an electron donating group leading to the activation of the ring and subsequently leading to more canonical resonance structure at the intermediate stage of the reaction enhancing the faster reactivity
However for the Nitrobenzene the nitro group is an electron withdrawing group leading to a slower activation and less resonance canonical structure at the reaction intermediate leading to a slower reaction than the reaction of benzene without the nitro group
Here we have to get the atom, in the molecule fentanyl, which will behave as a hydrogen (H) bond acceptor from the donor water molecule.
The -O atom in the molecule will behave as the hydrogen bond acceptor from water.
The H-bond always forms between the proton (here water i.e. H₂O) and the proton acceptor atom in a molecule which is essentially is an electronegative element.
In the fentanyl molecule there are two nitrogen (N) atom and one oxygen (O) atom. We know the electronegativity of oxygen is more than the nitrogen. Thus it is expected that the O atom will be H-bond acceptor in the molecule.
The hydrogen bond with water molecule is shown in the figure.
Answer:
0.0159m
Explanation:
9 M
Explanation:
Lead(II) chloride,
PbCl
2
, is an insoluble ionic compound, which means that it does not dissociate completely in lead(II) cations and chloride anions when placed in aqueous solution.
Instead of dissociating completely, an equilibrium rection governed by the solubility product constant,
K
sp
, will be established between the solid lead(II) chloride and the dissolved ions.
PbCl
2(s]
⇌
Pb
2
+
(aq]
+
2
Cl
−
(aq]
Now, the molar solubility of the compound,
s
, represents the number of moles of lead(II) chloride that will dissolve in aqueous solution at a particular temperature.
Notice that every mole of lead(II) chloride will produce
1
mole of lead(II) cations and
2
moles of chloride anions. Use an ICE table to find the molar solubility of the solid
PbCl
2(s]
⇌
Pb
2
+
(aq]
+
2
Cl
−
(aq]
I
−
0
0
C
x
−
(+s)
(
+
2
s
)
E
x
−
s
2
s
By definition, the solubility product constant will be equal to
K
sp
=
[
Pb
2
+
]
⋅
[
Cl
−
]
2
K
sp
=
s
⋅
(
2
s
)
2
=
s
3
This means that the molar solubility of lead(II) chloride will be
4
s
3
=
1.6
⋅
10
−
5
⇒
s
= √
1.6
4
⋅
10
−
5 =
0.0159 M