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Juli2301 [7.4K]
2 years ago
14

Substitution of an amino group on the para position of acetophenone shifts the cjo frequency from about 1685 to 1652 cm−1 , wher

eas a nitro group attached to the para position yields a cjo frequency of 1693 cm−1 . explain the shift for each substituent from the 1685 cm−1 base value for acetophenone

Chemistry
1 answer:
cluponka [151]2 years ago
6 0

Answer:

Here's what I get.

Explanation:

The frequency of a vibration depends on the strength of the bond (the force constant).

The stronger the bond, the more energy is needed for the vibration, so the frequency (f) and the wavenumber increase.

Acetophenone

Resonance interactions with the aromatic ring give the C=O bond in acetophenone a mix of single- and double-bond character, and the bond frequency = 1685 cm⁻¹.

p-Aminoacetophenone

The +R effect of the amino group increases the single-bond character of the C=O bond. The bond lengthens, so it becomes weaker.

The vibrational energy decreases, so wavenumber decreases to 1652 cm⁻¹.

p-Nitroacetophenone

The nitro group puts a partial positive charge on C-1. The -I effect withdraws electrons from the acetyl group.

As electron density moves toward C-1, the double bond character of the C=O group increases.

The bond length decreases, so the bond becomes stronger, and wavenumber  increases to 1693 cm¹.

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Write an equation that represents the action in water of rubidium hydroxide as an Arrhenius base.
Anika [276]

Answer:

RbOH  → Rb⁺ +  OH⁻

As the hydroxide can gives the OH⁻ in water, it is considered as an Arrhenius's base

Explanation:

Arrhenius theory states that a compound is considered a base, if the compound can generate OH⁻ ions in aqueous solution.

Our compound is the RbOH.

When it is put in water, i can dissociate like this:

RbOH  → Rb⁺ +  OH⁻

As the hydroxide can gives the OH⁻ in water, it is considered as an Arrhenius's base

3 0
2 years ago
Which of the following chemical equations does not correspond to a standard molar enthalpy of formation? a. Mg(s) + C(s) + 3/2 O
goblinko [34]

Answer:

N₂_{s}  + O₂_{g}   →  2NO_{g}

Explanation:

  The given equations are:

      Mg_{s} + C_{s} + \frac{3}{2}O_{g}   →     MgCO₃_{s}

   

      C_{s} +  \frac{1}{2}O₂_{g}    →   CO_{g}

      N₂_{s}  + O₂_{g}   →  2NO_{g}

     

      N₂_{s}  + 2O₂_{g}    →  N₂O_{g}

      H₂_{g} + \frac{1}{2}O_{g}    →    HO_{l}

From the given equations only N₂_{s}  + O₂_{g}   →  2NO_{g} does not correspond to an expression of the standard molar enthalpy of formation.

  • The heat of formation is the heat liberated or absorbed when one mole of a compound is formed from its constituent elements under standard conditions.

 The equation must result in the formation of only one mole of the compound.

It could have been properly written as

   \frac{1}{2}N₂_{s}  +  \frac{1}{2}O₂_{g}   →  NO_{g}

5 0
2 years ago
How can the existence of spectra help to prove that energy levels in atoms exist?
Liula [17]
The existence of spectra can help the existence of atoms by expanding and multiplying in the law of London Dispersion
6 0
2 years ago
In the nuclear transmutation, 168O (?, α)137N, what is the bombarding particle? In the nuclear transmutation, O (?, )N, what is
xeze [42]

Answer:

The bombarding particle is a Proton

Explanation:

A Nuclear transmutation reaction occurs when radioactive element decay, usually converting them from one element/isotope into another element. Transmutation is the process which causes decay, generally, alpha or beta.

¹⁶₈O(P,alpha) ¹³₇N, can be written as

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Where x can be anything, balancing the equation in order to give us the correct amount of proton number and nucleus number

16 + x = 13 + 4

x = 17 – 16 = 1, Hence we can say that x = ¹₁P

<u>¹⁶₈O + ¹₁P goes to ¹³₇N + ⁴₂He</u>

Here we can clearly see the bombarding particle is ¹₁P (proton). The ejected particle being ⁴₂He which is also known as an alpha particle

5 0
2 years ago
Which represents the balanced equation for the beta minus emission of phosphorus 32
vlabodo [156]
³²₁₅P → ³²₁₆S + e⁻ + ν

or

³²₁₅P → ³²₁₆S + β⁻ + ν
5 0
2 years ago
Read 2 more answers
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