Answer:
I think it's B
Explanation:
apologies if I get this wrong
Answer:
Iodide> Bromide > chloride > flouride
Explanation:
During a nucleophilic substitution reaction, a nucleophilie replaces another in a molecule.
This process may occur via an ionic mechanism (SN1) or via a concerted mechanism (SN2).
In either case, the ease of departure of the leaving group is determined by the nature of the C-X bond. The stronger the C-X bond, the worse the leaving group will be in nucleophilic substitution. The order of strength of C-X bond is F>Cl>Br>I.
Hence, iodine displays the weakest C-X bond strength and it is thus, a very good leaving group in nucleophillic substitution while fluorine displays a very high C-X bond strength hence it is a bad leaving group in nucleophilic substitution.
Therefore, the ease of the use of halide ions as leaving groups follows the trend; Iodide> Bromide > chloride > flouride
Answer:
By visiting other households with cats.
Explanation:
This will give Brian a variety of other houses and determine if it is truly cats or just alleries from other items. This is the most direct way to get Brian the answer he is looking for.
The reaction involved in this problem is called the combustion reaction where a hydrocarbon reacts with oxygen to product carbon dioxide and water. The reaction of C2H5OH would be as follows:
C2H5OH + 3O2 = 2CO2 + 3H2O
To determine the number of molecules of CO2 that is formed, we need to determine the number of moles produced from the initial amount of C2H5OH and the relation from the reaction. Then we multiply avogadros number which is equal to 6.022x10^23 molecules per mole.
2.00 g C2H5OH ( 1 mol C2H5OH / 46.08 g C2H5OH ) ( 2 mol CO2 / 1 mol C2H5OH ) = 0.0868 mol CO2
0.0868 mol CO2 ( 6.022x10^23 molecules / mol ) = 5.23x10^22 molecules CO2
