Answer:
Most substituted alkene is produced as a major product
Explanation:
- Dehydration of 3-methyl-2-butanol proceeds through E1 mechanism to form alkenes.
- Most substituted alkene is produced as major product because of presence of highest number of hyperconjugative hydrogen atoms corresponding to the produced double bond (Saytzeff product).
- Here, a H-shift also occurs in one of the intermediate step during dehydration to produce more stable tertiary carbocation.
- Reaction mechanism has been shown below.
Answer:
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Explanation:
Answer is: mass <span>of 4,30 moles of sodium</span> is 98800 mg.
n(Na) = 4,30 mol.
m(Na) = ?
m(Na) = n(Na) · M(Na).
m(Na) = 4,30 mol · 23 g/mol.
m(Na) = 98,90 g.
m(Na) = 98,90 g · 1000 mg/1g.
m(Na) = 98900 mg.
n - amount of substance.
m - mass of substance.
M - molar mass of substance.
Answer:
Explanation:
Hello,
In this case, we write the reaction again:
In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:
Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:
But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:
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