They form molecules which can be in solution form if diluted in water, but some do form solutions on exposure to the atmosphere i.e they are deliquescent like pellets of sodium hydroxide
To determine a planet's mass, astronomers typically measure the minuscule movement of the star caused by the gravitational tug of an orbiting planet. For planets the massof Earth detecting such a tiny tug is extraordinarily challenging with current technology
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:
A
-1440J
Explanation:
Hello,
This question requires us to calculate the work done on a object to move it from point A to point B
Data
Mass = 60kg
Initial velocity (V1) = 8.0m/s
Final velocity (V2) = 4.0m/s
Workdone on an object is equal to force applied on the object to move it through a particular distance.
Work done = force × distance
Force (F) = mass × acceleration
Distance = s
F = Ma
Work done = M× a × s
But a = velocity (v) / time (t)
Work done = mvs / t
But velocity = distance/ time
Work done = mv × v/
Work done = mv²
Work done = ½mv²
Workdone = ½M(V2² - V1²)
Workdone = ½ × 60 (4² - 8²)
Work done = 30 × (16 - 64)
Workdone = 30 × (-48)
Work done = -1440J
Work done = -1.44kJ
The workdone on the object is equal to -1.44kJ
