10 seconds = 8grams
then just divide by 2 another 4 times...
= 0.5grams after 50 seconds
The question is incomplete, the complete question is;
Why is a terminal alkyne favored when sodium amide (NaNH2) is used in an elimination reaction with 2,3-dichlorohexane? product. A) The terminal alkyne is more stable than the internal alkyne and is naturally the favored B) The terminal alkyne is not favored in this reaction. C) The resonance favors the formation of the terminal rather than internal alkyne. D) The strong base deprotonates the terminal alkyne and removes it from the equilibrium.
E) The positions of the Cl atoms induce the net formation of the terminal alkyne.
Answer:
E) The positions of the Cl atoms induce the net formation of the terminal alkyne.
Explanation:
In this reaction, sterric hindrance plays a very important role. We know that sodamide is a strong base, it tends to attack at the most accessible position.
The first deprotonation yields an alkene. The strong base attacks at the terminal position again and yields the terminal alkyne. Thus the structure of the dihalide makes the terminal hydrogen atoms most accessible to the base. Hence the answer.
Since we are already given the balanced equation:
→ 
We can derive the molar ratios as: 1:2:1:1
That being said, we are given 0.172 moles of bromine (
), so it has a ratio of 1:1 with sodium bromide (
).
So we can take from that ratio, that when 0.172 moles of bromine are used, we are, in turn, going to get 0.172 moles of sodium bromide produced.
Jamal’s hair is behaving that way because he is in a cold area.
