Which bonds form in the reaction shown in the diagram?
2 answers:
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Answer:
0.382g
Explanation:
Step 1: Write the reduction half-reaction
Al³⁺(aq) + 3 e⁻ ⇒ Al(s)
Step 2: Calculate the mass of Al produced when a current of 100. A passes through the cell for 41.0 s
We will use the following relationships.
- 1 A = 1 C/s
- 1 mole of electrons has a charge of 96486 C (Faraday's constant)
- 1 mole of Al is produced when 3 moles of electrons pass through the cell.
- The molar mass of Al is 26.98 g/mol.
The mass of Al produced is:
Answer: Option (d) is the correct answer.
Explanation:
It is given that molecular formula is . Now, we will calculate the degree of unsaturation as follows.
Degree of unsaturation =
=
= 9 - 8 + 1
= 2
As the degree of unsaturation comes out to be 2. It means that this compound will contain one ring and one double bond.
Yes, this compound could be an alkyne as for alkyne D.B.E = 2.
But this compound cannot be a cycloalkane because for a cycloalkane D.B.E = 1 which is due to the ring only.
Thus, we can conclude that it is a cycloalkane is not a structural possibility for this hydrocarbon.