Answer:
They are already matched for you. It goes
1.
2.
3.
in the order of the questions.
H2SO3 or sulfurous acid is actually a strong acid. We know
for a fact that strong acids completely dissociate into its component ions in a
solution, that is:
<span>H2SO3 --> 2H+ + SO3-</span>
<span>So from the equation above, there are 2 moles of H+</span>
its me again this is how you find the answer. note: i dont have the periodic table to see the exact atomic mass to find the molar mass however this is the answer:
so the actual formula is Mg(OH) with a 2 as a subscript because there are 2 Mg. so with Mg(OH)2 the Molar mass is 58.32g/mol. 58.32 g/mol x 7.1x1024 = 4.1x1026g
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.