Answer:
b. hydride shift from C-3 to C-2.
Explanation:
Markovnikov's rule states that *in the addition of a protic acid HX or other polar reagent to an asymmetric alkene, the acid hydrogen (H) or electropositive part gets attached to the carbon with more hydrogen substituents, and the halide (X) group or electronegative part gets attached to the carbon with more alkyl substituents* (wikipedia).
This rule implies that the hydrogen of HBr will be attached to C-1 and the carbocation will be on C-2. Remember that the order of stability of carbocations is tertiary > secondary > primary > methyl. A hydride shift can yield a tertiary carbocation.
C-3 is a tertiary carbon atom. If the hydride on carbon 3 shifts to carbon 2, a tertiary and more stable carbocation is formed. This accounts for the major product in the reaction.
Answer:
The boiling point of the fluoromethane (CH3F) is higher than that of fluorine (F2).
Answer:
![V_{NO}=2.8L](https://tex.z-dn.net/?f=V_%7BNO%7D%3D2.8L)
Explanation:
Hello!
In this case, we consider the given reaction to realize there is a 5:4 mole ratio between oxygen and nitrogen monoxide; thus, we infer that, at STP conditions, such mole ratio is eligible as a volume ratio too; therefore, the produced liters of nitrogen monoxide gas is:
![V_{NO}=3.5LO_2*\frac{4LNO}{5LO_2} \\\\V_{NO}=2.8L](https://tex.z-dn.net/?f=V_%7BNO%7D%3D3.5LO_2%2A%5Cfrac%7B4LNO%7D%7B5LO_2%7D%20%5C%5C%5C%5CV_%7BNO%7D%3D2.8L)
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Answer:
L=0.85M
Explanation:
ca3 (po4)
66gX1mol/310.18g=2.2178mol/2.5
Hello,
Here is your answer:
The proper answer to this question is option C "stigma".
Here is how:
The stigma is responsible for producing pollen in a plant.
Your answer is C.
If you need anymore help feel free to ask me!
Hope this helps!