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stealth61 [152]
3 years ago
11

L avanzar hacia la derecha por el período 5, el tamaño atómico, la energía de ionización y los electrones de valencia: A. dismin

uye, aumenta y aumentan, respectivamente. B. aumenta, disminuye y disminuye, respectivamente. C. disminuye, disminuye y aumentan, respectivamente. D. disminuye, aumenta y disminuyen, respectivamente.
Chemistry
1 answer:
rusak2 [61]3 years ago
4 0

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>

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sweet-ann [11.9K]

Answer:

135 kPa

Explanation:

4 0
3 years ago
What is the hybridization of the central atom in each of the following? 1. Beryllium chloride 2. Nitrogen dioxide 3. Carbon tetr
Lina20 [59]

Answer :

(1) The hybridization of central atom beryllium in BeCl_2  is, sp

(2) The hybridization of central atom nitrogen in NO_2  is, sp^2

(3) The hybridization of central atom carbon in CCl_4  is, sp^3

(4) The hybridization of central atom xenon in XeF_4  is, sp^3d^2

Explanation :

Formula used  :

\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

Now we have to determine the hybridization of the following molecules.

(1) The given molecule is, BeCl_2

\text{Number of electrons}=\frac{1}{2}\times [2+2]=2

The number of electron pair are 2 that means the hybridization will be sp and the electronic geometry of the molecule will be linear.

(2) The given molecule is, NO_2

\text{Number of electrons}=\frac{1}{2}\times [4]=2

If the sum of the number of sigma bonds, lone pair of electrons and odd electrons present is equal to three then the hybridization will be, sp^2.

In nitrogen dioxide, there are two sigma bonds and one lone electron pair. So, the hybridization will be, sp^2.

(3) The given molecule is, CCl_4

\text{Number of electrons}=\frac{1}{2}\times [4+4]=4

The number of electron pair are 4 that means the hybridization will be sp^3 and the electronic geometry of the molecule will be tetrahedral.

(4) The given molecule is, XeF_4

\text{Number of electrons}=\frac{1}{2}\times [8+4]=6

Bond pair electrons = 4

Lone pair electrons = 6 - 4 = 2

The number of electrons are 6 that means the hybridization will be sp^3d^2 and the electronic geometry of the molecule will be octahedral.

But as there are four atoms around the central xenon atom, the fifth and sixth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be square planar.

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3 years ago
What are the factors that affect the rate of dissolution?
Liula [17]

Option (C) is the correct answer.

When solute is added to the solvent, results in the formation of solution and the process is said to be dissolution.

Dissolution process depends upon the three factors: temperature, stirring and surface area.

Stirring increases the rate of dissolving as when stirring a solute in the solvent, stirring helps in the distribution of solute throughout the solvent which results in speed up the dissolving rate.

Dissolution also depends upon the temperature, in warm solvent, solute dissolves readily whereas in cool solvent, solute dissolves slowly as particles have more movement energy.

Greater the surface area, there will be more contact between the solute particles and solvent particles which increases the rate of dissolving. Smaller the surface are, there will be less contact between the particles which decreases the rate of dissolving. For example: granulated sugar dissolves more readily due to larger surface area and one ice cube dissolves slowly due to smaller surface area.

Hence, rate of dissolution depends upon temperature, stirring, surface area.

4 0
3 years ago
Which statement about the relationship between density and particle attraction is true?
maw [93]

A. As the density of a substance increases, the attraction among its particles increases.

Explanation:

The relationship between density and particle attraction is that as the density of a substance increases, the attraction among its particles increases.

  • Density is the function of the mass and volume of body.
  • Mass is the amount of particles a substance contains. It is directly related to the density.
  • Volume is the amount of space a substance occupies. It is inversely related to the density.
  • A substance will have a large volume if it is poorly packed and the attraction between particles is very weak or low.
  • Particles that are well packed have high attraction between their particles. This reduces their volume and the density will increase.

Learn more:

Density brainly.com/question/3764212

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4 0
3 years ago
Enthalpy of formation (kJ/mol) C6H12O6(s)-1260 O2 (g)0 CO2 (g)-393.5 H2O (l)-285.8 Calculate the enthalpy of combustion per mole
trapecia [35]

Answer : The enthalpy of combustion per mole of C_6H_{12}O_6 is -2815.8 kJ/mol

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]

The equilibrium reaction follows:

C_6H_{12}O_6(s)+6O_2(g)\rightleftharpoons 6CO_2(g)+6H_2O(l)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(C_6H_{12}O_6)}\times \Delta H^o_f_{(C_6H_{12}O_6)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]

We are given:

\Delta H^o_f_{(C_6H_{12}O_6(s))}=-1260kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(6\times -393.5)+(6\times -285.8)]-[(1\times -1260)+(6\times 0)]=-2815.8kJ/mol

Therefore, the enthalpy of combustion per mole of C_6H_{12}O_6 is -2815.8 kJ/mol

6 0
4 years ago
Read 2 more answers
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