Answer:
The reaction would take place with racemization.
Explanation:
The compound (R)-3-bromo 3-methylhexane contains has the leaving group attached to a tertiary carbon atom. When the leaving group is attached to a tertiary carbon atom, the compound undergoes nucleophilic substitution via SN1 mechanism. This mechanism involves the formation of a planar carbocation intermediate. The nucleophile may attack this intermediate from either faces, leading to racemization of the product.
E
θ
Cell
=
+
2.115
l
V
Cathode
Mg
2
+
/
Mg
Anode
Ni
2
+
/
Ni
Explanation:
Look up the reduction potential for each cell in question on a table of standard electrode potential like this one from Chemistry LibreTexts. [1]
Mg
2
+
(
a
q
)
+
2
l
e
−
→
Mg
(
s
)
−
E
θ
=
−
2.372
l
V
Ni
2
+
(
a
q
)
+
2
l
e
−
→
Ni
(
s
)
−
E
θ
=
−
0.257
l
V
The standard reduction potential
E
θ
resembles the electrode's strength as an oxidizing agent and equivalently its tendency to get reduced. The reduction potential of a Platinum-Hydrogen Electrode under standard conditions (
298
l
K
,
1.00
l
kPa
) is defined as
0
l
V
for reference. [2]
A cell with a high reduction potential indicates a strong oxidizing agent- vice versa for a cell with low reduction potentials.
Two half cells connected with an external circuit and a salt bridge make a galvanic cell; the half-cell with the higher
E
θ
and thus higher likelihood to be reduced will experience reduction and act as the cathode, whereas the half-cell with a lower
E
θ
will experience oxidation and act the anode.
E
θ
(
Ni
2
+
/
Ni
)
>
E
θ
(
Mg
2
+
/
Mg
)
Therefore in this galvanic cell, the
Ni
2
+
/
Ni
half-cell will experience reduction and act as the cathode and the
Mg
2
+
/
Mg
the anode.
The standard cell potential of a galvanic cell equals the standard reduction potential of the cathode minus that of the anode. That is:
E
θ
cell
=
E
θ
(
Cathode
)
−
E
θ
(
Anode
)
E
θ
cell
=
−
0.257
−
(
−
2.372
)
E
θ
cell
=
+
2.115
Indicating that connecting the two cells will generate a potential difference of
+
2.115
l
V
across the two cells.
To determine the mass of the sample, first find the volume difference after and before the aluminum was placed, the volume change is equal to the volume of the submerged object, in this case aluminum.
Then knowing volume of aluminum and the density of it, we can solve for the mass.
D = m/v
Dv = m
2.7 g/ml • 8 ml = 21.6 grams.
Answer:
A). 92.02g
Explanation:
Equation of the reaction;
N2 (g)+ 2O2(g)------> 2NO2(g)
Note that the balanced reaction equation is the first step in solving any problem on stoichiometry. Once the reaction equation is correct, the question can be easily solved.
Reaction of one mole of nitrogen gas with two moles of oxygen gas yields two moles of nitrogen dioxide.
Mass of two moles of nitrogen dioxide= 2[14 + 2(16)] = 2[14+32]= 2[46]= 92 gmol-1
Therefore; Mass of two moles of nitrogen dioxide is 92
