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Vsevolod [243]
3 years ago
6

The frequency of stretching vibrations is correlated to the strength and stiffness of the bond between two atoms. This can be th

ought of as a ball-and-spring model. Using this knowledge, rank the following bonds in each part of the question by increasing frequency.
(a) Alkyne
(b) Alkane
(c) alkene
Chemistry
2 answers:
larisa86 [58]3 years ago
8 0

Answer: (A) < (C) < (B)

Ranking in order of increasing frequency.

Explanation:

Leona [35]3 years ago
5 0

Answer:

a > c > b

Explanation:

As higher is the strength and stiffness of the bond between two atoms, more stable it is, and more difficult is to these bonds vibrate. So, the stretching vibration decreases when the strength and stiffness increases.

As more bonds are done between the atoms, more strength, and stiffness they have. So, the order of increase is:

simple bond > double bond > triple bond

And the increased frequency of vibration is:

triple bond > double bond > simple bond

An alkane is a hydrocarbon that has only simple bonds between carbons, an alkene is a hydrocarbon with one double bond between carbon, and an alkyne is a hydrocarbon with one triple bond. So, the increase in vibration of them is:

alkyne (a) > alkene (c) > alkane (b)

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Answer:

M=0.15

Explanation:

138 g AgNO -> 1 mol AgNO

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What amount of energy is required to change a spherical drop of water with a diameter of 1.80 mm to three smaller spherical drop
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This is a straightforward question related to the surface energy of the droplet. 

<span>You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. </span>

<span>With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. </span>

<span>The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ </span>

<span>The five smaller droplets need to have the same volume as the original. Therefore </span>

<span>5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. </span>

<span>Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. </span>

<span>Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². </span>

<span>The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ </span>
<span>From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. </span>

<span>Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets. </span>
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3 years ago
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