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DIA [1.3K]
3 years ago
5

The strength of an acid is affected by the polarity of the bond connected to the acidic hydrogen. The more highly polarized this

bond, the more easily the hydrogen is ionized. Electronegative atoms or groups of atoms present in the structure of an acid can act to withdraw electrons and produce additional polarization. Two common groups of acids to which this principle can be applied are oxoacids and carboxylic acids. In the latter group, the length of the hydrocarbon chain in a carboxylic acid has very little effect on acid strength Longer chains may slightly diminish acidity. Bases act as hydrogen ion acceptors because of the unshared electron pass in their structure. Any group present in a base that withdraws electrons makes these electron pairs less available to accept a hydrogen ion. In contrast, any group that can act as an electron donating group such as hydrocarbon groups (usually represented as II) can increase the base strength. Thus, the addition of electronegative atoms or groups of atoms to the structure of a base decreases the base strength and electron donating groups increase base strength. Many common weak bases are derivatives of ammonia, in which H atom(s) of NH_2 are replaced with other groups.
Arrange the following oxoacids in order of decreasing acid strength. Rank from strongest to weakest acid.
1. HBrO
2. HClO
3. HClO2
4. HClO3
Chemistry
1 answer:
Shtirlitz [24]3 years ago
5 0

Answer:

The answer is "\bold{HClO_3 > HClO_2 >HClO > HBrO}"

Explanation:

We arrange oxoacids to decrease the intensity of acids in this question. Or we may conclude all this from strongest to weakest acids they order oxoacids, that's why above given order is correct.

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expeople1 [14]

your answer is c hope this helps

6 0
3 years ago
In IR spectroscopy, we normally talk about "frequencies" when in reality we are referring to wavenumbers. What is the mathematic
Svetach [21]

Answer:

Here's what I get.

Explanation:

(b) Wavenumber and wavelength

The wavenumber is the distance over which a cycle repeats, that is, it is the number of waves in a unit distance.

\bar \nu = \dfrac{1}{\lambda}

Thus, if λ = 3 µm,

\bar \nu = \dfrac{1}{3 \times 10^{-6} \text{ m}}= 3.3 \times 10^{5}\text{ m}^{-1} = \textbf{3300 cm}^{-1}

(a) Wavenumber and frequency

Since

λ = c/f and 1/λ = f/c

the relation between wavenumber and frequency is

\bar \nu = \mathbf{\dfrac{f}{c}}

Thus, if f = 90 THz

\bar \nu = \dfrac{90 \times 10^{12} \text{ s}^{-1}}{3 \times 10^{8} \text{ m$\cdot$ s}^{-1}}= 3 \times 10^{5} \text{ m}^{-1} = \textbf{3000 cm}^{-1}

(c) Units

(i) Frequency

The units are s⁻¹ or Hz.

(ii) Wavelength

The SI base unit is metres, but infrared wavelengths are usually measured in micrometres (roughly 2.5 µm to 20 µm).

(iii) Wavenumber

The SI base unit is m⁻¹, but infrared wavenumbers are usually measured in cm⁻¹ (roughly 4000 cm⁻¹ to 500 cm⁻¹).

8 0
3 years ago
L avanzar hacia la derecha por el período 5, el tamaño atómico, la energía de ionización y los electrones de valencia: A. dismin
rusak2 [61]

Answer:

Al avanzar hacia la derecha por el período 5, el tamaño atómico, la energía de ionización y los electrones de valencia: A. disminuye, aumenta y aumentan, respectivamente.

Explanation:

El radio atómico representa la distancia que existe entre el núcleo y la capa de valencia, es decir la más externa. Por medio del radio atómico es posible determinar el tamaño del átomo.  En un período el tamaño atómico disminuye de izquierda a derecha pues en este sentido aumenta el  número atómico aumentando la carga nuclear mientras que el efecto pantalla y el número de  niveles permanecen constantes. En otras palabras, disminuye de izquierda a derecha debido a la atracción que ejerce el núcleo sobre los electrones de los orbitales más externos, disminuyendo así la distancia núcleo-electrón.

<u><em>Al avanzar hacia la derecha por el período 5, el tamaño atómico disminuye.</em></u>

La energía de ionización es la necesaria para remover un electrón a un átomo en estado  gaseoso. Mientras más lejos del núcleo esté el electrón, es más fácil removerlo porque se necesita  menos energía. Al aumentar el número atómico de los elementos de un  mismo período, se incrementa la atracción nuclear sobre el  electrón más externo, ya que disminuye el radio atómico y  aumenta la carga nuclear efectiva sobre él. Entonces en un período, al aumentar el número atómico, la energía de ionización aumenta.

<u><em>Entonces, al avanzar hacia la derecha por el período 5, la energía de ionización y los electrones de valencia aumenta.</em></u>

Los electrones de valencia  son los electrones que están en la última capa electrónica (llamados orbitales de valencia) y tienen una alta posibilidad de participar en una reacción química.

En cada período aparecen los elementos cuyo último nivel de su configuración electrónica coincide con el número del período, ordenados por orden creciente de número atómico. Por ejemplo, el período 3 incluye los elementos cuyos electrones más externos están en el nivel 3.

Los electrones de valencia aumentan en número a medida que se avanza en un período.

<u><em> Entonces, al avanzar hacia la derecha por el período 5, los electrones de valencia aumentan.</em></u>

4 0
3 years ago
Identify the item that does not have kinetic energy.
d1i1m1o1n [39]

Answer:

D

Explanation:

there is no other answer choice that does not move. This is a fool-proof question because all the other answer choices contain movement except the butterfly resting

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What is the driving force for the movement of earths plates?
ElenaW [278]
Heat and gravity are fundamental to the process
The energy source for plate tectonics is Earth's internal heat while the forces moving the plates are the “ridge push” and “slab pull” gravity forces.

Hope I helped! :)
4 0
3 years ago
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