Given:
A compound with:
Number of carbon atoms = 9
Number of double bonds = 1
A double bond between 5th and 6th carbon
A propyl group (CH2CH2CH3) branching off the 3rd carbon from the left
Try to illustrate the given and observe the formation of the atoms. Now, follow the correct IUPAC naming system. The name of the compound is
4-propyl-1-hexene
Count from the right to the left, the double bond is between the 1st and 2nd carbon, thus, 1-hexene. The propyl branches out the 4th carbon from the right, thus 4-propyl.
Answer: 116g/mole
Explanation:
She didn't get the answer because she didn't add the them well , due to the bracket present.
A) sodium fluoride
B) rubidium oxide
C) boron trichloride
D) dihydrogen selenide
E) tetraphosphate hexoxide
F) iodine trichloride
The following quantities will effect the reaction rate as follows:
1. On increasing Concentration of the reactant: Rate of the reaction will increases.
2. On increasing pressure : Increases the rate of reaction to the side where there are fewer number of molecules.
3.On increasing temperature of an endothermic reaction: Increases the rate of reaction
4. On decreasing temperature of an endothermic reaction: Increases the rate of reaction.
So the answer is increase pressure, decrease temperature, increase concentration will increases the rate of the reaction.
