Answer:
The warmer, lighter air rises, bringing cooler, heavier air to low altitudes.
Air at higher altitudes doesn't have as much air weighing down on it from above.
Explanation:
In short - air pressure is the result of the cumulative force that air molecules act on objects below them due to Earth's gravity. The higher the altitude, the less air molecules there are to act a force below them, and therefore, there's less air pressure at higher altitudes.
Answer:
17.1195 grams of nitric acid are produced.
Explanation:

Moles of nitrogen dioxide :

According to reaction 3 moles of nitrogen dioxides gives 2 moles of nitric acid.
Then 0.5434 moles of nitrogen dioxides will give:
of nitric acid.
Mass of 0.3623 moles of nitric acid :

Theoretical yield = 22.8260 g
Experimental yield = ?


Experimental yield of nitric acid = 17.1195 g
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.
Answer:
i think keep the same rat i`m not sure
Explanation:
Answer:
Solution Concentration: Molarity Moles of solute in one liter of solution Used because it describes how many molecules of solute are in each liter of solution Tro: Chemistry: A Molecular Approach, 2/e 6 amount of solute (in moles) Molarity, M = amount of solution (in L) moles of solute M = L of solution or simply 4
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