Answer:
Both reactions share a common intermediate and differ only in the leaving group
Explanation:
The elimination reaction of tertiary alkyl halides usually occur by E1 mechanism. In E1 mechanism, the substrate undergoes ionization leading to the loss of a leaving group and formation of a carbocation.
Loss of a proton from the carbocation completes the reaction mechanism yielding the desired alkene.
In the cases of t-butanol and t-butyl bromide, the mechanism is the same. The both reactions proceed by E1 mechanism. The leaving groups in each case are water and chloride ion respectively.
<span>KCl<span>O3</span><span>(s)</span>+Δ→KCl<span>(s)</span>+<span>32</span><span>O2</span><span>(g)</span></span>
Approx. <span>3L</span> of dioxygen gas will be evolved.
Explanation:
We assume that the reaction as written proceeds quantitatively.
Moles of <span>KCl<span>O3</span><span>(s)</span></span> = <span><span>10.0⋅g</span><span>122.55⋅g⋅mo<span>l<span>−1</span></span></span></span> = <span>0.0816⋅mol</span>
And thus <span><span>32</span>×0.0816⋅mol</span> dioxygen are produced, i.e. <span>0.122⋅mol</span>.
At STP, an Ideal Gas occupies a volume of <span>22.4⋅L⋅mo<span>l<span>−1</span></span></span>.
And thus, volume of gas produced = <span>22.4⋅L⋅mo<span>l<span>−1</span></span>×0.0816⋅mol≅3L</span>
Note that this reaction would not work well without catalysis, typically <span>Mn<span>O2</span></span>.
Answer:
it's urine option ( C ) .
Percentage recovery gives us an idea of the amount of pure substance recovered after the chemical reaction. Percentage recovery can be more than 100 % or less than 100 %. Usually, in any experiment performed the weight percentage recovery will be less than 100. Percent recovery values greater than 100 show that the recovered compound is contaminated.
Amount of acetaminophen initially taken = 350 mg
Amount of acetaminophen obtained after recovery =185 mg
= 
= 52.9%
Answer:
<h3>A-5N B-6N C-7N D-8N</h3>
Explanation:
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