Hey there!
Here is your answer:
<u><em>The proper answer to this question is option D "sec-butyl bromide".</em></u>Reason:
<u><em>S</em></u><span><u><em>ec-butyl bromide looks often like this:</em></u>
<em>Therefore the answer is option D!</em>If you need anymore help feel free to ask me!
Hope this helps!
~Nonportrit</span>
Substances are chemically combined and cannot be separated by normal physical means. Mixtures, however, are simply physically combined and can be separated into their component parts.
Answer:
See image attached and explanation
Explanation:
I have attached a detailed mechanism of the reaction to this answer. This reaction occurs by SN1 mechanism. It implies that the transition state involves a carbocation.
However, the initial carbocation formed is a primary carbacation. Remember that the order of stability of carbocations is methyl< primary < secondary< tertiary. This means that tertiary carbocations are the most stable carbocations. Tertiary carbocations are those in which the carbon atom bearing the carbon atom is attached to three other carbon atoms.
In the mechanism below, the substrate converts from a primary to a tertiary cabocation (most stable) by a 1,2-alkyl shift as shown giving the 3-ethoxy-3-methylpentane product.
1 mol is (always) the same number of units: 6.02 * 10^23 units.
So 0.5 moles of any gas has the same number of molecules.
Also, we know by the ideal gas laws that a give number of molecules of any gas will occupy the same volume.
Given that the three gases have the same number of atoms in the molecular fomula (2), 0.5 moles will also have the same number of atoms.
g.f.w. stands for grams formula weight and that is different for all of them, because gfw is calcualted from the atomics masses of each atom in the molecule.
Then at STP (standard temperature and pressure) conditions 0.50 moles of any of the gas:
- will contain the same number of molecules,
- will contain the same number of atoms
- will occupy the same volume.