This molecule can be synthesized from an alkyne anion and an alkyl bromide. However, there are two ways in which this molecule c
an be formed. One way uses a higher molecular weight alkyne anion (Part 1) and the other uses a lower molecular weight anion (Part 2). Draw the two versions in the boxes below. Omit spectator ions.
1 answer:
Answer:
See figures
Explanation:
In this case, we will have different options around the<u> triple bond</u>. On part 1 we have to use the <u>bigger anion</u>, so we have to use the anion in which the <u>benzene ring is included </u>therefore the <u>ethyl</u> would be the <u>halide</u> (see figure 1).
On part 2 we have to use the <u>smaller anion</u>, so the <u>ethyl group</u> would be the anion and the other part of the molecule would be the <u>alkyl halide</u> (see figure 2).
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Answer:
Explanation: For the given reaction:
Concentrations of the ions are not given so we need to think about another way to calculate Kc.
We can calculate the free energy change using the standard cell potential as:
can be calculated using standard reduction potentials.
Standard reduction potential for zinc is -0.76 V and for silver, it is +0.78 V.
=
Reduction takes place at anode and oxidation at cathode. As silver is reduced, it is cathode. Zinc is oxidized and so it is anode.
= 0.78 V - (-0.76 V)
= 0.78 V + 0.76 V
= 1.54 V
Value of n is two as two moles of electrons are transferred in the cell reaction F is Faraday constant and its value is 96485 C/mol of electron .
= -297173.8 J
Now we can calculate Kc using the formula:
T = 25+273 = 298 K
R =
--297173.8 = -(8.314*298)lnKc
297173.8 = 2477.572*lnKc
lnKc = 119.946