Answer:
D because i did this last week and got it right.
A) B). That’s it I hope it’s help.
Answer:
<u>During Metaphase</u>
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Explanation:
During Metaphase the cell chromosome align themselves in the middle of the cell.This occur due to a cellular process called "Tug of War".
The chromosome which have been replicated and joined at the central point called centromere are called sister chromatids
Prior to metaphase , Kinetochore type of protein are formed around the centromere. Long protein filament called kinetochore are extended from poles to other end of the cell attached to kinetochore.
Therse is important checkpoint in the middle of mitosis called<u> metaphase.</u>
At this point cell ensure that , it is ready to divide or not.
Once the cell ensure that everything is ready to divide. Only after then , yhe cell enters the fourth phase called <u>anaphase.</u>
Answer:
13.5 g
Explanation:
This question is solved easily if we remember that the number of moles is obtained by dividing the mass into the atomic weight or molar mass depending if we are referring to elements or molecules.
Therefore, the mass of aluminum in the reaction will the 0.050 mol Al times the atomic weight of aluminum.
number of moles = n = mass of Al / Atomic Weight Al
⇒ mass Al = n x Atomic Weight Al = 0.050 mol x 27 g mol⁻¹
= 13.5 g
We have three significant figures in 0.050 and therefore we should have three significant figures in our answer.
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.
