Answer:
gde
Explanation:
We are attempting to synthesize 1-butyne from 1-chlorobutane. Since 1-chlorobutane is a primary alkyl halide, 1-butene is formed when 1-chlorobutane is reacted with a bulky base such as t -BuOK or t -BuOH in presence of strong heat. This is an E2 reaction.
Secondly, the 1-butene is reacted with bromine in carbon tetrachloride. The vicinal dihalide (1,2-dibromobutane) is formed. This can now undergo further elimination reactions in the presence of sodamide and strong heat to yield 1-butyne which is the desired product. These reactions involve the elimination of the first HBr molecule to give an alkenyl bromide. A second elimination now gives the terminal alkyne.
Answer:
Demo Mole Quantities
58.5g NaCl(mol/58.5g)(6.02 x 1023/mol) = 6.02 x 1023 Na
+
Cl21 pre-1982 pennies (after 1982 pennies are mostly zinc with copper coating)
63.5g Cu( mol/ 63.5g)(6.02 x 1023/mol) = 6.02 x 1023 Cu
19.0g Al (mol/27.0g)(6.02 x 1023/mol) = 4.24 x 1023 Al
Explanation:
Moles = n/v where n is the moles of solute and v being the liters of solution.
We can put in the information provided to find the molarity.
Moles = .45/3.0 = .15
So we now know that the molarity of that solution is .15!
I hope I helped you :). Make sure to memorize that formula because it's not that hard as long as you know what to plug in.
Well, all of this we owe it to Bohr who analyzed the atomic emission spectrum of hydrogen and he could probe matematically that it was a result of movement of e- from an especific energy level to a lower one. The understanding of levels of energy took to the development of the atomic theory
