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.
The amount of the 240 g sample of the radioisotope that will remain after 525 billion years is 7.5 g
<h3>How to the number of half-lives that has elapsed</h3>
- Half-life (t½) = 105 billion years
- Time (t) = 525 billion years
- Number of half-lives (n) = ?
n = t / t½
n = 525 / 105
n = 5
<h3>How to determine the amount remaining</h3>
- Original amount (N₀) = 240 g
- Number of half-lives (n) = 5
- Amount remaining (N) = ?
N = N₀ / 2ⁿ
N = 240 / 2⁵
N = 240 / 32
N = 7.5 g
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4.0
i think it has something to do with molar ratios and finding the limiting reactant
4.0 mol NO * 2 mol NO2/2 mol NO = 4.0 moles of NO2
4.0 mol O2 * 2 mol NO2/1 mol O2 = 8.0 moles of NO2
so the limiting reactant (the reactant that runs out the quickest leaving an excess) is NO
once the limiting reactant is found, we can use that data for that substance to calculate the amount of product
4.0 mol NO * 2 mol NO2/2 mole NO = 4.0 moles of NO2
