Answer:
See images below.
Explanation:
The sodium ethoxide is a strong nucleophile and a strong base. When it acts as a strong nucleophile (a substance that has unpaired electrons and needs to share or donate it to be stable), an elimination reaction occurs. When it acts as a strong base, a substitution reaction occurs.
For a halide, the elimination takes place when the halogen is no steric hindrance, so the halogen must be in position 1. Then the compound A must be the must steric hindrance compound with formula C₇H₁₅Br, which is the compound showed below (3-Bromo-3-ethyl-pentane), to promote only elimination product. It happens because it's a tertiary halide.
When it reacts with sodium ethoxide (H₃C-H₂C-O-Na), the Br⁻ will leave the compound with close hydrogen, and a double bond will be formed between the carbons, so compound B must be the one showed below (3-Ethyl-2-pentene).
When the compound B is treated with dilute sulfuric acid, an addition reaction will occur, so the double bond will be broken and a hydrogen and a hydroxyl must be added to the compound. For the Markov-Nikov rule, the hydrogen will be in the most hydrated carbon. So, compound C will be the 3-Ethyl-3-pentanol showed below.
Answer:
The answer is
<h2>0.07 g/mL</h2>Explanation:
The density of a substance can be found by using the formula
From the question
mass of metal = 5.11 g
volume = 71.46 mL
The density of the metal is
We have the final answer as
<h3>0.07 g/mL</h3>Hope this helps you