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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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Explanation:

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Which stament about energy is mostly correct
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A solution is made by dissolving 10.20 grams of glucose (C6H12O6) in 355 grams of water. What is the freezing point depression o
Ludmilka [50]

Answer:

0.297 °C

Step-by-step explanation:

The formula for the <em>freezing point depression </em>ΔT_f is

ΔT_f = iK_f·b

i is the van’t Hoff factor: the number of moles of particles you get from a solute.

For glucose,

       glucose(s) ⟶ glucose(aq)

1 mole glucose ⟶ 1 mol particles     i = 1

Data:

Mass of glucose = 10.20 g

  Mass of water = 355 g

                 ΔT_f = 1.86 °C·kg·mol⁻¹

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n = 10.20 g × (1 mol/180.16 g)

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(b) <em>Kilograms of water </em>

m = 355 g × (1 kg/1000 g)

   = 0.355 kg

(c) <em>Molal concentration </em>

b = moles of solute/kilograms of solvent

  = 0.056 62 mol/0.355 kg

  = 0.1595 mol·kg⁻¹

(d) <em>Freezing point depression </em>

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3 years ago
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Consider an element Z that has two naturally occuring isotopes with the following percent abundances: the isotope with a mass nu
rodikova [14]

Answer:

Z=22.70

Explanation:

It is given that,

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So, the average atomic mass for element Z is 22.70.

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